doane solutions

Chapter 1

Overview of Statistics

1.1 Answers will vary.

1.2 a. Answers will vary.b. An hour with an expert at the beginning of a project could be the smartest move a manager can make.

When your team lacks certain critical skills, or when an unbiased or informed view cannot be found inside your organization. Expert consultants can handle domineering or indecisive team members, personality clashes, fears about adverse findings, and local politics. As in any business decision, the costs of paying for statistical assistance must be weighed against the benefits. Costs are: statistician’s time, more time invested in the beginning of a project which may mean results are not immediate. Benefits include: better sampling strategies which can result in more useful data, a better understanding of what information can be extracted from the data, greater confidence in the results.

1.3 a. The average business school graduate should expect to use computers to manipulate the data.

b. Answers will vary.

1.4 a. Answers will vary. Why Study: In fact, most college graduates will use statistics every day. Why Not Study: It is difficult to become a statistical “expert” after taking one introductory college course. A business person should hire statistical experts and have faith that those who are using statistics are doing it correctly.

b. Answers will vary.c. To arrive at an absurd result, and then conclude the original assumption must have been wrong, since it

gave us this absurd result. This is also known as proof by contradiction. It makes use of the law of excluded middle — a statement which cannot be false, must then be true. If you state that you will never use statistics in your business profession then you might conclude that you shouldn’t study statistics. However, the original assumption of never using statistics is wrong, therefore the conclusion of not needing to study statistics is also wrong.

1.5 a. An understanding of statistics helps one determine what data is necessary by requiring us to state our questions up front. We can then determine the proper amount of data needed, sample sizes, to confidently answer our questions.

b. Answers will vary.c. Answers will vary.

1.6 a. Yes, the summary is succinct. No, the purpose was not clear. Why do we want to know the weight of Tootsie Rolls? Yes, the sampling method was explained. Yes, the findings were clearly stated. No implications of the study were noted. Yes, jargon is a problem. Non statisticians will not know what a FPCF is or what a confidence interval is. To improve this report the writer should restate in laymen’s terms.

b. Yes, the summary is succinct. No, the purpose was not clear. Why do we want to know the proportion of pages with advertisements? Yes, the sampling method was explained. Yes, the findings were clearly stated. Jargon is not a problem here as it was previously. To improve, the writer should state the purpose.

1.7 a. The graph is more helpful. The visual illustration of the distribution focuses the individual on the experience of the many (in this case 16 to 20 years and 21 to 25 years). We can quickly see that the typical financial planner has between 16 to 25 years experience.

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1.10 a. It is not obvious that there is a direct cause and effect relationship between an individual choosing to use a radar detector and that individual choosing to vote and wear a seatbelt. There may be unidentified variables that are related to these three individual characteristics. Users of radar detectors may drive faster and thus recognize risks and therefore are more likely to wear seatbelts. Users may also be more prone to vote, since they are concerned more about government policies and who will support their desire to have government influence minimized.

b. Increasing the use of radar detectors may not influence those who obey laws and are less concerned with government limitations.

1.11 a. No, the method did not “work” in the sense that he increased his chances of winning by picking the numbers the way he did. Every combination of six numbers has the same chance of winning. The fact that this winner chose his numbers based on his families’ birthdays and school grade does not increase the chance of his winning.

b. Someone who picks 1-2-3-4-5-6 has just as much chance of winning as anyone else (see (a)).

1.12 a. The phrase “much more” is not quantified. The study report is not mentioned. There is no way to determine the veracity of this statement. Six causes of car accidents could be poor weather, road construction, heavy traffic, inexperienced driver, engine failure, drinking. Smoking is not on this list.

b. Smokers might pay less attention to driving when lighting a cigarette.

1.13 Many people have math “phobia” and because statistics involves math the subject can sound scary. The subject of statistics has a reputation for being difficult and this can cause fear of the unknown. There is usually not the same fear towards an ethics class. This is because there is much more emphasis on unethical behavior in the media, to the extent that ethical behavior and an understanding of how to be ethical is widely accepted as a requirement to graduate and then succeed.

1.14 Random sampling of cans of sauce for a specific manufacturer can be used to assess quality control.

1.15 a. The consultant can analyze the responses from the 80 purchasing managers noting that the linen supplier should not make any conclusions about the managers who did not respond. The consultant should not use the responses to ambiguous questions. She should suggest that the supplier redesign both the survey questions and the survey methods to increase the response rates.

b. An imperfect analysis would be a mistake because the supplier may make changes to their business that upset those customers not responding to the survey or those customers not sent a survey.

c. As a consultant it would be important to point out the problems with the survey instrument and the survey method and suggest alternatives for improvement.

1.16 All of these involve taking samples from the population of interest and estimating the value of the variable of interest.

1.17 a. Class attendance, time spent studying, natural ability of student, interest level in subject, instructor’s ability, performance in course prerequisites. Smoking is not on the list.

b. Most likely students who earn A’s are also making good decisions about their health. Students who smoke might also be making poor choices surrounding their study habits.

c. Giving up smoking alone may not stop a student from using poor study habits nor is it likely to increase their interest in a topic.

1.18 Curiosity, parents’ who smoke, friends who smoke, seeing teenagers smoke in movies and TV, boredom, wanting to look cool. Yes, seeing movie and TV stars smoking was on the list.

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1.19 a. We need to know the total number of philosophy majors to evaluate this.b. We don’t know the number of students in each major from this table.c. This statement suffers from self-selection bias. There are likely many more marketing majors who choose

to take the GMAT and therefore a wider range of abilities than the abilities of physics majors who choose to take the GMAT.

d.The GMAT is just one indicator of managerial skill and ability. It is not the only predictor of success in management.

1.20 a. The graph is much more useful. We can clearly see that as chest measurement increases body fat also increases. It is not a perfectly linear relationship but the relationship is there nevertheless.

b. The last two data points on the far right show a very high chest measurement but the body fat percentage has leveled off.

1.21 a. “Its R2 value is quite close to 1 indicating it is a good fit to the actual data. I feel that G.E. is one of the most respected corporations in the world because of its strong management and name recognition. Its valuable assets make it poised for steady growth over the next decade.”

b. “If a country’s unemployment rate is too high, it could cause a down turn in its economy’s structure.”c. “You cannot have a negative number of people unemployed; therefore, this forecast is very unlikely.”d. “This is not a well designed graph because its title is too long and there are no labels on the axes.”e. “This graph has no clear border to give it a sense of containment. It is dealing with three separate pieces of

information. In this graph, the same data is presented, but in a deceptive manner. The sources do not contain enough detail.”


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Chapter 2

Data Collection

2.1 Observation – single data point. Variable – characteristic about an individual.


2.3 a. attribute b. attribute

c. discrete numericald. continuous numericale. discrete numericalf. discrete numericalg. continuous numerical

2.4 a. continuous numerical b. discrete numerical c. attributed. continuous numericale. attributed. discrete numerical



2.7 a. ratiob. ordinal

c. nominal d. interval e. ratio f. ordinal

2.8 a. ratiob. ratioc. intervald. nominale. nominalf. nominal


2.10 a. ordinal or intervalb. ordinalc. nominald. ratio

2.11 a. cross-sectional b. time series c. time series d. cross-sectional.

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2.12 a. time seriesb. cross-sectionalc. time seriesd. cross-sectional

2.13 a. time series b. cross-sectional. c. time series. d. cross-sectional.


2.15 a. Census b. Sample c. Sample d. Census

2.16 a. parameterb. parameterc. statistic.d. statistic

2.17 a. Sample b. Census c. Sample d. Census

2.18 Use the formula: N= 20*na. N = 20*10 = 200 b. N = 20*10 = 1000c. N = 20*100 = 20000

2.19 a. Convenience b. Systematic c. Judgment or biased

2.20 a. No. In the rush to leave the theater, stop at the restroom, use their cell phone, etc. it would not be possible for everyone to have an equal chance to be included in the sample.

b. Might only get those who didn’t like the movie and couldn’t wait to leave. There might not be a large enough crowd to get every 10th person to be representative and leaving the theatre is not a linearly organized event. Might have underrepresented sample by only selecting those with earrings.

c. Only those who liked the movie or really hated the movie might respond, a bias due to self-selection.


2.22 a. 0.50b. Answers will vary.c. Due to random variation the sample may not be representative.


2.24 a. Response bias.b. Self-selection bias, coverage error. c. coverage error, self-selection bias.

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2.25 a. Telephone or Web b. Direct observation c. Interview, Web, or mail d. Interview or Web

2.26 a. Mailb. Direct observation, through customer invoices/receipts c. Mail d. Interview

2.27 Version 1: Most would say yes. Version 2: More varied responses.

2.28 Does not include all possible responses or allow for the responder to pick something other than those

presented.

2.29 a. Continuous numerical b. Attribute. c. Discrete numerical. d. Discrete numerical. e. Continuous numerical.

2.30 a. ordinal (seeds represent a ranking of the players)b. ratioc. ratiod. ratioe. ratio, zero is meaningful.



2.33 Q1 Attribute, nominal

Q2 Continuous, ratio

Q3 Attribute, nominal


Q5 Discrete, ratio

Q6 Discrete, ratioQ7 Attribute, nominalQ8 Attribute, interval


Q10 Discrete, ratio


Q12 Discrete, ratio


Q14 Discrete, ratio


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Q16 Discrete, ratio

Q17 Attribute, interval


Q19 Attribute, interval


2.34 a. Census. b. Sample: too costly to track each can c. Census,: can count them all quickly and cheaply d. Census: as long as the company can easily generate the value from its human resource center.

2.35 a. Statisticb. Parameterc. Statisticd. Parameter


2.37 a. Number of employees or industryb. There may be differences in profitability based on number of employees or industry type therefore we

should be sure to take a sample that includes both types of industries. c. Under representation of chemical companies.

2.38 a. Cluster sampling. Easier to define geographic areas within a state where gasoline is sold. Gasoline stations are not everywhere, thus simple random sample or stratified sampling doesn’t make sense.

b. Population is finite. It is listable.

2.39 Use mail or telephone. Census not possible.

2.40 a. Could use cluster sampling as grocery stores are in well defined locations. Identify clusters within each state.

b. The sample frame is all stores in the US selling peanut butter. This population is very large, approaching infinity, but could still be listed.

c. A census is not possible given the size and scope of the investigation.

2.41 a. Cluster sampling b. Finite and listable c. Yes.

2.42 a. No. It would have been too costly and taken too much time to observe everyone who used the restroom. b. The population is finite but not listable. c. Judgment d. Direct observatione. Interviewer bias.

2.43 a. Cluster Sampling b. It doesn’t change the results but you cannot use the results to make conclusions about all salmon advertised

as wild.

2.44 a. Answers will vary.b. Convenience.c. No. The population is too large.d. Population can be treated as infinite and unlistable.

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2.45 a. Telephone or mail, b. Finite and listable

2.46 Simple random sample or systematic sampling.

2.47 a. No b. Ordering of the list could influence the make up of the first sample.

2.48 a. Judgment or convenienceb. Non-response bias is always present in surveys. Coverage error may occur since we don’t know who has

radar detectors and who doesn’t before hand so may over represent one group.c. No causation shown so conclusions are not trustworthy.

2.49 a. Cluster sampling, neighborhoods are natural clusters. c. Picking a day near a holiday with heavy trash.

2.50 a. Convenience sampling.b. Based on such a small sample, that may not be representative of the entire population, it would be incorrect

to make such a statement. c.Perhaps, if the block is representative of the city, or area with in the city, or even in his local neighborhood,

then such an inference might be valid, but confined to a specific geographic area. d. Coverage

2.51 a. Systematic b. Simple random sample

c. Simple random sample or systematicd. Simple random sample or systematice. Stratified

2.52 a. Systematic: every 5th person who emerges from the office; or obtain data on n randomly selected patients and visits and analyze.

b. Direct observation for a specific time period, such as all day Wednesday.c. n convenient placesd. Last n flightse. Direct observation of gasoline prices at selected stations over a two week period.

2.53 a. Sales, store type b. Yes c. Simple random sample

2.54 a. No, one has to sample because the population is infinite and unlistable. A census is not possible. b. One could stratify by state or county because geographic regions may differ.

2.55 a. No, the population is too large therefore sampling is required. b. Systematic.

2.56 Judgment sampling or systematic sampling were the most likely sampling methods. A census is not possible because the population is too large.

2.57 Convenience sample because any other method would have been more expensive and time consuming.

2.58 a. Judgment sampling.b. Simple random sample would be impossible because it would be impossible to identify the individuals in

the population.

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2.59 Education and income could affect who uses the no-call list. a. They won’t reach those who purchase such services. Same response for b and c.

2.60 Selection (only those who survived would be in the sample); coverage: may include those who were least exposed to such hazards.

2.61 a. Ordinal b. That the intervals are equal.

2.62 For each question, the difficulty is deciding what the possible responses should be and giving a realistic range of responses.

2.63 a. Rate the effectiveness of this professor. 1 – Excellent to 5 – Poor. b. Rate your satisfaction with the President’s economic policy. 1 – Very Satisfied to 5 – Very dissatisfied. c. How long did you wait to see your doctor? Less than 15 minutes, between 15 and 30 minutes, between 30

minutes and 1 hour, more than 1 hour.

2.64 a. It depends on the questions asked. It is possible that more could agree the law should be upheld, even though on moral grounds they oppose it.

b. Setting aside your moral and personal beliefs, given that abortion is legal, should the laws be upheld? Setting aside the fact that abortion is legal, do you believe that killing an unborn child is moral?

c. Do you believe abortion should stay legal?

2.65 Answers will vary, one consideration would be to ask the questions as a yes or no and then provide a list of “whys” or ask the respondent to list reasons for yes or no answer.

2.66 Ordinal measure. There is no numerical scale and the intervals are not considered equal.

2.67 a. Likert scale.b. Should add a “middle category” that states Neither Agree Nor Disagree and remove “Undecided” category.

2.68 a. A constrained response scale.b. A Likert scale would be better.c. Self-selection bias. People with very bad experiences might respond more often than people with

acceptable experiences.

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Chapter 3

Describing Data Visually

3.1

Approximately symmetric, but can be viewed as skewed to the left.

3.2

Distribution appears symmetric.

3.3Sarah’s Calls:

Bob’s Calls:

Sarah’s makes more calls than Bob and her calls are shorter in duration.

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3.4 a. 7 bins of 20

b. Answers will vary. Too few bins (less than five) or too many bins (more than 15) might hide the skewness in the distribution.

3.5 a. 6 bins of 100


3.6 a. 4 bins of 10

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3.7 Sample default graph given. Answers will vary as to modification.

3.8 Default graph presented, answers will vary with respect to modifications made.

3.9 Default graphs presented, answers will vary with respect to modifications made.

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3.10 Default graphs for a, b, and c.

a.

b. c.

3.11 a. Sample default graph presented.


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b. The relationship is negative, linear and strong.


b. There is a strong, positive relationship between midterm exam scores and final exam scores.


b. There is a weak, positive linear relationship.3.15 a. Sample default graph presented.

b. There is weak, negative linear relationship.

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3.16 Sample default graphs presented for a, b, and c.

3.17 Sample default graphs presented for a, b, and c.

3.18 Sample default graphs presented for a and b.

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3.19 a.

b.

Frequency Distribution - Quantitative

Nurse/Bed cumulative

lower upper midpoint width frequency percent frequenc

y percent

0.8 ≤ 1 0.9 0.2 2 5 2 5

1 ≤ 1.2 1.1 0.2 5 12.5 7 17.5

1.2 ≤ 1.4 1.3 0.2 13 32.5 20 50

1.4 ≤ 1.6 1.5 0.2 10 25 30 75

1.6 ≤ 1.8 1.7 0.2 4 10 34 85

1.8 ≤ 2 1.9 0.2 3 7.5 37 92.5

2 ≤ 2.2 2.1 0.2 1 2.5 38 95

2.2 ≤ 2.4 2.3 0.2 1 2.5 39 97.5

2.4 ≤ 2.6 2.5 0.2 0 0 39 97.5

2.6 ≤ 2.8 2.7 0.2 1 2.5 40 100

c. The distribution is skewed to the right. Almost half the observations are between 1.2 and 1.6. (Note: Dotplot and histogram were generated on Megastat. Frequency distribution was calculated using Excel’s Data Analysis tool.)

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3.20 a

b.


Data cumulative

lower upper midpoint width frequency percent frequency percent

0 < 5 3 5 20 23.0 20 23.0

5 < 10 8 5 23 26.4 43 49.4

10 < 15 13 5 15 17.2 58 66.7

15 < 20 18 5 7 8.0 65 74.7

20 < 25 23 5 7 8.0 72 82.8

25 < 30 28 5 7 8.0 79 90.8

30 < 35 33 5 3 3.4 82 94.3

35 < 40 38 5 3 3.4 85 97.7

40 < 45 43 5 0 0.0 85 97.7

45 < 50 47 5 2 2.3 87 100.0

c. Distribution is skewed to the right. More games seem to be decided by smaller margins of victory than large margins of victory.

3.21 a.

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b.


Data cumulative

lower

upper midpoint width frequency percent frequency percent

1 < 2 2 1 24 36.9 24 36.9

2 < 3 3 1 12 18.5 36 55.4

3 < 4 4 1 9 13.8 45 69.2

4 < 5 5 1 2 3.1 47 72.3

5 < 6 6 1 3 4.6 50 76.9

6 < 7 7 1 4 6.2 54 83.1

7 < 8 8 1 1 1.5 55 84.6

8 < 9 9 1 1 1.5 56 86.2

9 < 10 10 1 0 0.0 56 86.2

10 < 11 11 1 1 1.5 57 87.7

11 < 12 12 1 0 0.0 57 87.7

12 < 13 13 1 1 1.5 58 89.2

13 < 14 14 1 3 4.6 61 93.8

14 < 15 15 1 0 0.0 61 93.8

15 < 16 16 1 0 0.0 61 93.8

16 < 17 17 1 0 0.0 61 93.8

17 < 18 18 1 0 0.0 61 93.8

18 < 19 19 1 1 1.5 62 95.4

19 < 20 20 1 0 0.0 62 95.4

20 < 21 21 1 1 1.5 63 96.9

21 < 22 22 1 0 0.0 63 96.9

22 < 23 23 1 0 0.0 63 96.9

23 < 24 24 1 0 0.0 63 96.9

24 < 25 25 1 0 0.0 63 96.9

25 < 26 26 1 0 0.0 63 96.9

26 < 27 27 1 1 1.5 64 98.5

27 < 28 28 1 0 0.0 64 98.5

28 < 29 29 1 0 0.0 64 98.5

29 < 30 29 1 1 1.5 65 100.0

c. The distribution is skewed to the right. Three-fourths of the calls are 6 minutes or less in duration.

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3.22 a.

b.


Calories Per Gram cumulative

lower

upper

midpoint

width frequency percent frequency percent

1.60 < 1.80 1.70 0.20 1 6.7 1 6.7

1.80 < 2.00 1.90 0.20 1 6.7 2 13.3

2.00 < 2.20 2.10 0.20 3 20.0 5 33.3

2.20 < 2.40 2.30 0.20 1 6.7 6 40.0

2.40 < 2.60 2.50 0.20 4 26.7 10 66.7

2.60 < 2.80 2.70 0.20 4 26.7 14 93.3

2.80 < 3.00 2.90 0.20 1 6.7 15 100.0

c. The distribution appears to be skewed to the left. The sample size is not large enough to draw valid inferences about shape.

3.23 a.

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b.


Data cumulative

lower

upper midpoint width frequency percent frequency percent

80 < 90 85 10 1 2.7 1 2.7

90 < 100 95 10 3 8.1 4 10.8

100 < 110 105 10 8 21.6 12 32.4

110 < 120 115 10 4 10.8 16 43.2

120 < 130 125 10 6 16.2 22 59.5

130 < 140 135 10 4 10.8 26 70.3

140 < 150 145 10 4 10.8 30 81.1

150 < 160 155 10 5 13.5 35 94.6

160 < 170 165 10 1 2.7 36 97.3

170 < 180 175 10 1 2.7 37 100.0

c. The shape is somewhat symmetrical.

3.24 a. Column chart with 3D visual effect.b. Strengths: labels on X and Y axes; Good proportions; no distracting pictures; Weaknesses: Title doesn’t

make sense; Non-zero origin; 3D effect does not add to presentation; tick marks do not stand out, measurement units on Y axis missing; vague source

c. Correct weaknesses as noted, 2Dcolumn chart would be less distracting, zero origin on Y axis would show realistic differences in years.

3.25 a. Bar chart with 3D visual effect.b. Strengths: Good proportions, no distracting pictures. Weaknesses: No labels on X and Y axes, Title

unclear, 3D effect does not add to presentation.c. Correct weaknesses as noted. 2D bar chart with zero origin on X axis would improve chart.

3.26 a. Line chartb. Strengths: labels on X and Y axes, measurement included on Y axis, Good title; No distracting pictures;

good use of gridlines Weaknesses: magnitude difference between net income and sales, no source. c. Correct weaknesses as noted, use of logarithmic scale would correct proportion issue and show growth

rates more clearly.

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3.27 a. Map b. They are appropriate when patterns of variation across space are of interest. Not sure that the question

being addressed requires a map. It is difficult to assess the question from the information on the map. The actual share rather than above or below the national average would have more meaning.

c. A pie chart showing distribution by largest states and an “other” category. Set of pie charts by region, with values on national shares. A histogram of percentages would show distribution.

3.28 a. Standard pie chart.b. Strengths: source identified, answers the question posed. Weaknesses: “Other” category quite large.c. Might change title: Distribution of Advertising Dollars in the United States, 2001. Would add the total

dollars spent on advertising as well.

3.29 a. Exploded pie chart.b. Strengths include details on names and values, differentiating between types of countries, good use of

color.c. Improvement might be percentage shares instead of actual values, and include just the total level of

imports. Might consider a sorted column chart with OPEC and Non-OPEC countries color coded.

3.30 a. Line chart.b. Strengths: labels on X and Y axes, use of gridlines. Weaknesses: distracting pictures, dramatic title, non-

zero origin, no source.c. Correct weaknesses as noted, keep graph as line graph.

3.31 a. Pictogram. b. Weakness: Title unclear. Uses two “pieces of the paw” to illustrate a single number. Strengths: Visually

appealing, values and labels easy to understand. c. Should use a better title and try to use a single piece to represent each value. A pie chart with exploding

piece for Shelter/Rescue might be just as effective. Or perhaps a Pareto chart would work well.

3.32 a. Pictogramb. Strengths: labels on X and Y axes, zero origin, Short descriptive title, detailed, accurate source.

Weaknesses: use of Area Trick to make 2000 appear much larger than 1980 values.c. Correct weaknesses as noted, regular column chart would be better choice.

3.33 a. Figure B is the better choice, even though it has a non-zero origin. The regression line fitted shows the actual change in noise level due to a one unit increase in airspeed (nautical miles per hour). The inclusion of this line and its equation overcomes any “exaggeration” in the trend from the use of a non-zero origin.

b. A one nautical mile per hour increase leads to .0765 increase in decibels. A one hundred mile per hour increase leads to a 7.650 decibel increase.

c. Yes, it does seem logical.d. It appears that the average level of noise is around 95 decibels given an airspeed of about 350. Most jets

cruise at around 400 or more, so the noise level is generally between that of a hair dryer and chain saw.

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3.34 a.

b. A table showing the data might be useful given the small number of countries.

3.35 a.

b. A table, bar, or column chart would also work.

3.36 a.

b. A table or column chart would also be informative.

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3.37 a.

b. A column chart would also work.

3.38 a.

b. A pie chart would also work

3.39 a.

b. A pie chart would also work.

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3.40 a.

b. A bar chart would also be informative.

3.41 a.

b. A bar chart would also be informative.

3.42 a.

b. Side by side pie charts or a bar chart would also work.

3.43 a.

b. A bar chart or column chart would also work.

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3.44 a.

b. A bar chart would also work.

3.45 a

b. A line plot would also work.

3.46 a.

b. A column chart would also work.

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3.47 a.


3.48 a.


3.49 a.


3.50 No, a table is the best way to display the data. Showing four different years would make a graph too cluttered.

3.51 No, a table is the best way to display the data. The rows may not add up to 100 due to rounding.

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Chapter 4

Descriptive Statistics

4.1 b.

4.2 b.

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Descriptive statistics

Data

count 32

mean27.3

4

sample variance61.6

5

sample standard deviation 7.85

minimum 9

maximum 42

range 33


Data

count 28

mean 107.25

sample variance 1,043.75


minimum 52

maximum 176

range 124

4.3 b.

4.4 a.

Quiz 1 Quiz 2 Quiz 3 Quiz 4

Count 10 10 10 10

Mean 72.00 72.00 76.00 76.00

Median 72.00 72.00 72.00 86.50

Mode 60.00 65.00 72.00 none

b. No, they don’t agree for all quizzes. The mean and the median are the same for quiz 1 and quiz 2, and the mode and the median are the same for quiz 3.

c. The mode is an unreliable measure of central tendency for quantitative data. Where the mean and median disagree, one should look at the shape of the distribution to see which measure is more appropriate.

d. Quiz 1 and Quiz 2 have a symmetric distribution. Quiz 3 is skewed right and Quiz 4 is skewed left. e. Students on average did better on quizzes 3 and 4.

4.5 a.

Descriptive Statistic Data

count 32

mean27.3

4

mode26.0

0

median26.0

0

28


Data

count 65

mean 4.48

sample variance 34.47


minimum 1

maximum 29

range 28

b. The mode and median are the same, but the mean is greater than the median.c. The data are skewed to the left. See 4.1.d. The mode is not a reliable measure of central tendency for quantitative variables.

4.6 a.

Descriptive Statistics Data

count 28

mean 107.25

median 106.00

mode 95.00 b. The measures are close in value.c. Symmetric. See the dot plot and histogram in 4.2.d. The mode is not a reliable measure of central tendency.

4.7 b.

Descriptive Statistics Data

count 65

mean 4.48

median 2.00

mode 1.00

c. The mean is more than twice the median and the median is twice the mode. d. The data are skewed to the right based on the histogram in 4.3.e. Because the data is heavily skewed, the median is a better measure of central tendency.

4.8 b.

Descriptive Statistics Mon Tue Wed Thu

mean 4 4.7 3.8 1.9

median 5 5 5 1

trimmed mean 4 4.7 3.8 1.9

geometric mean 2.89 3.76 NA 1.26

midrange 5 5 5 5.5

c. The geometric mean is very different from the other measures. The mean, median, and midrange are close in value.

d. The mean or median are better measures of central tendency for this type of data. More empty seats on Monday and Tuesday than on Wednesday and Thursday. If one stands by on Thursday, there is little chance that they will get on a flight compared to earlier in the week.

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4.9 a.

mean 27.34

midrange 25.50

geometric mean 26.08

trim mean 27.47

b. The mean and the trimmed mean are similar, but they are both greater than the midrange and the geometric range.

4.10 a.

mean 107.25


trimmed mean 106

midrange 114

b. The mean and trimmed mean are similar and fall between the geometric mean and midrange.

4.11 a.

mean 4.48

midrange 15.00

geometric mean 2.60

trimmed mean 3.61

b. The mean is greater than the geometric and trimmed mean. The midrange is almost 4 times the mean. This is not surprising given the distribution of the data and a few calls lasting for a very long duration.

c. The data are skewed to the right. d. The mean or trimmed mean are appropriate measures of central tendency for this data set.

4.12 39.9%. See calculation below.

1

1= −nn

xG

x = ((60.6/15.8)^(1/4))-1 = 0.399

4.13 a.

Sample A: Sample B: Sample C:

Mean 7 62 1001

Sample Standard Deviation 1 1 1

b. The midpoint of each sample is the mean. The other 2 data points are exactly 1 standard deviation from the mean. The idea is to illustrate that the standard deviation is not a function of the value of the mean.

30

4.14 a.

Data Set A: Data Set B: Data Set C:

Mean 7.0000 7.0000 7.0000

Sample Standard Deviation 1.0000 2.1602 3.8944

Population Standard Deviation 0.8165 2.0000 3.7417

b. The sample standard deviation is larger than the population standard deviation for the same data set. Samples can have similar means, but different standard deviations.

4.15

Stock /s x CV

Stock A 5.25/24.50 21.43%

Stock B 12.25/147.25 8.32%

Stock C 2.08/5.75 36.17%

a. Stock C, the one with the smallest standard deviation and smallest mean, has the greatest relative variation.b. The stocks have different values therefore directly comparing the standard deviations is not a good

comparison of risk. The variation relative to the mean value is more appropriate.

4.16

Quiz 1 Quiz 2 Quiz 3 Quiz 4

Count 10 10 10 10

Mean 72.00 72.00 76.00 76.00

sample standard deviation 13.23 6.67 11.41 27.43 coefficient of variation (CV) 18.38% 9.26% 15.02% 36.09%

b. Scores, on average, are higher for Quiz 3 and Quiz 4. Quiz 2 has the least relative variation and Quiz 4 the most. As the quiz scores increase from quiz to quiz, the variation within a quiz increases.

4.17

Sample standard deviation 5.87

Mean absolute deviation 3.92

4.18 From Megastat:

empirical rule

mean - 1s 19.49

mean + 1s 35.20

percent in interval (68.26%) 68.8%

mean - 2s 11.64

mean + 2s 43.05


mean - 3s 3.79

mean + 3s 50.90


low extremes 0

31

low outliers 0

high outliers 0

high extremes 0

b. There are no outliers based on the empirical rule. No unusual data values. c. Assume the data are normally distributed based on the empirical rule results.d. Yes, the sample size is large enough.

4.19 From MegaStat:

empirical rule

mean - 1s -1.39

mean + 1s 10.35


mean - 2s -7.27

mean + 2s 16.22


mean - 3s -13.14

mean + 3s 22.09


low extremes 0

low outliers 0

high outliers 4

high extremes 4

b. Yes, there are 8 outliers.c. There are more observations in the mean +-1 interval than the empirical rule would indicate, 87.7% vs.

68.26%. There are fewer observations in the mean +- 2 interval, 93.88% vs. 95.44%. Data do not seem to be from a normal distribution.

d. Yes, there are enough sample points to assess normality.

4.20 a.

1st quartile 22.50

Median 26.00

3rd quartile 33.00

midhinge 27.5

b.

c. The median number of customers is 26. Days with 22 or fewer customers are in the bottom quartile. Days with 33 or more customers are in the upper quartile. The midhinge is 27.75 and is a measure of central

tendency. It is a measure of central tendency. The box plot displays the first and third quartiles and the median. The box plot indicates that there are no extreme values or outliers.

4.21 a.

32

1st quartile 1.00

3rd quartile 5.00

interquartile range 4.00

mode 1.00

median 2.00

low extremes 0

low outliers 0

high outliers 4

high extremes 4

midhinge 3

b. The median call length was 2 minutes. Calls lasting less than 1 minute were in the bottom 25%, call lasting more than 5 minutes were in the top 25%. The midhinge was a call length of 3 minutes. The midhinge is a measure of central tendency, and like the median is not influenced by extreme values.

c. The box plot confirms the quartile calculations and reveals that there are 6 calls of unusually large duration, 4 of the them extreme in length.

4.22 a. Mean = 725. Median = 720. Mode = 730. SD = 114.3. Q1 = 662.5. Q3 = 755.b. The typical student pays $725 per month.c. The measures do tend to agree.d. From Megastat:

empirical rule

mean - 1s 610.39

mean + 1s 838.95


mean - 2s 496.10

mean + 2s 953.23


mean - 3s 381.82

mean + 3s 1,067.51


low extremes 0

low outliers 1

high outliers 3

high extremes 0

There are four outliers.e. It is possible that the data are normally distributed based on the empirical rule results.

4.23 a. Mean = 66.2. Median = 48. Mode = 48. Midrange = 108.

33

b. The median is the best measure of central tendency for a data set that is skewed.c. The typical number of pages in a mail order catalog is 48.d. SD = 36.36.e. From Megastat:

empirical rule

mean - 1s 29.74

mean + 1s 102.56


mean - 2s -6.67

mean + 2s 138.97


mean - 3s -43.08

mean + 3s 175.38


low extremes 0

low outliers 0

high outliers 1

high extremes 1

There are two high outliers.

4.24 a.

b.

Mean 17.83

median 18.00

mode 17.00


midrange 20.00

c. The mean is the best measure of central tendency.d. The typical ladder weighs 17 or 18 pounds.

34

4.25 a.

Most travelers buy their tickets between 1 and 21 days before their trip. Half of these buy the ticket about 2 weeks before.

b. Mean = 26.71. Median = 14.5. Mode = 11. Midhinge = 124.5.c. Q1 = 7.75, Q3 = 20.25, Midhinge = 14, and CQV = 44.64%. d. The geometric mean is only valid for data greater than zero.e. The median is the best measure of central tendency because the data is quantitative and heavily skewed

right.

4.26 The mode is the measure of central tendency because the data is categorical.

4.27 a. Stock funds: 1.329x = , median = 1.22, mode = 0.99. Bond funds: 0.875x = , median = 0.85, mode = 0.64.

b. The central tendency of stock fund expense ratios is higher than bond funds. c. Stock funds: s = 0.5933, CV = 44.65%. Bond funds: s = 0.4489, CV = 51.32%. The stock funds have less

variability relative to the mean.d. Stock funds: Q1 = 1.035, Q3 = 1.565, Midhinge = 1.3. Bond funds: Q1 = 0.64, Q3 = 0.99, Midhinge = 0.815.

Stock funds have higher expense ratios in general than bond funds.

4.28 a. Mean = 34.54. Median = 33.0. Mode = 23. Midrange = 42.b. The mean or median would be an appropriate measure of central tendency because the data is fairly

symmetric.c. SD = 10.31. CV = 29.85%.d. The number of raisins in a box is based on weight, not quantity. The size of the raisins vary therefore the

number will vary.

4.29 a. Mean = 52.15. Median = 48.5. Mode = 47.0. Midrange = 60.0.b. Geometric mean = 50.92.c. All measures of central tendency are fairly close therefore use the mean.

4.30 a. Most brands contain between 160 and 230 milligrams of sodium. Five brands are very low in sodium.

b. Mean = 179.9. Median = 195.0. Mode = 225. Midrange = 187.5.c. The median and midrange are better measures than the mean and mode because the data appear to have

outliers.d. The geometric mean cannot be used because there are data values equal to zero.f. There are five low outliers and 1 high outlier.

35

4.31 a.

The dot plot shows that most of the data is centered around 6500 yards. The distribution is skewed to the left.

b. 6,335.52x = , median = 6,400.0, mode = 6,500.0, and midrange = 6,361.5.

c. Best: Median because data is quantitative and skewed left. Worst: Mode worst because the data is quantitative and very few values repeat themselves.

d. This data is not highly skewed. The geometric mean works well for skewed data.

4.32 a.

The dot plot shows a fairly uniform distribution. b. Mean = 7.52. Median = 6.8. Mode = 9.0. Midrange = 9.25.c. The mean or the median are the best measures of central tendency because the data is fairly symmetric. The

worst measure would be the mode because there are very few values that repeat.

4.33 a. Male: Midhinge = 177, CQV = 2.82%. Female: Midhinge = 163.5, CQV = 2.75%. These statistics are appropriate because we have specific percentiles but not the entire data set.

c. Yes, height percentiles do change. The population is slowly increasing in height.

4.34 a. Mean = 95.1. Median = 90.0. There is no mode. b. The median is the best measure of central tendency because the data set has two high outliers.

4.35 a. 3012.44x = , median = 2,550.5. There is no value for the mode.

b. The typical cricket club’s income is approximately £2.5 million.

4.36 The coefficient of variation for plumbing supplier’s vinyl washers is: 6053/24212 = 25%. The coefficient of variation for steam boilers is 1.7/6.8 = 25%. The demand patterns exhibit similar relative variation, even though the standard deviations are very different.

4.37 The coefficient of variation for the lab mouse = 0.9/18 = 5%. The coefficient of variation for the lab rat = 20/300 = 6.67%. The rat has more relative variation based on a higher coefficient of variation. The weights of lab mouse vary less around the mean.

4.38 a. See table below for CV values.

Comparative Returns on Four Types of Investments

Investment Mean Return Standard Deviation Coefficient of Variation

Venture funds (adjusted)

19.2 14.0 72.92%

All common stocks 15.6 14.0 89.74%Real estate 11.5 16.8 146.09%Federal short term paper

6.7 1.9 28.36%

36

b. The standard deviations are an “absolute” not relative measure of dispersion. It is best to use the CV when comparing across variables that have different means.

c. The risk and returns are captured by the CV. Federal short term paper has the lowest CV and hence lowest risk, real estate the greatest risk. Venture funds have lower risk and greater return than common stocks based on the CV.

4.39 a. CV Tuition Plans = 100*2.7%/6.3% = 42.86%. CV SP500 = 100*15.8%/12.9% = 122.48%.b. We use the CV to compare the return and risk. The standard deviation tells us the measure of dispersion

relative to the mean of the distribution only. The standard deviation alone can not be used to compare distributions.

c. The tuition plans have lower returns than the SP 500, but less risk as measured by the CV. This is not surprising since the goal of a tuition plan is to ensure that a minimum amount of money is available at the time the plan matures, thus parents and students are willing to take a lower return in exchange for lower risk.

4.40 a. Midrange = (180+60)/2 = 120.b. Assuming normality is important so that we can estimate the mean with the midrange.c. Caffeine levels in brewed coffee are dependent on many factors including brand of coffee, grind of coffee

beans, and brew time. It is likely that the distribution is skewed to the right.

4.41 a. Midrange = (.92+.79)/2 = .82 b. A normal distribution is plausible here because there are likely to be controls on the level of chlorine added

to the water. There will be some variation around the mean but it will be predictable.

4.42 a. The distribution should be skewed to the right because the mean is greater than the median.b. Most ATM transactions will tend to low in value but a few will be of longer duration.

4.43 a. The distribution should be skewed to the right because the mean is greater than the median.b. Most patrons keep books out for a week or so. There will be a few patrons that keep a book out much

longer.

4.44 a. The distribution should be skewed to the left because the mean is less than the median. b. It appears that most students scored a C or higher but there were a few students that may not have studied

for the exam.

4.45 a. The distribution of the number of DVDs owned by a family would likely be skewed to the right. Most families will own a fairly small number of DVDs but a few families will own many.

b. Mean > median > mode.

4.46 a. The histogram should show a symmetrical distribution.b. Answers will vary.

4.47 a. One would expect the mean to be close in value to the median, or slightly higher. b. In general, the life span would have a normal distribution. If skewed, the distribution is more likely skewed

right than left. Life span is bounded below by zero but is unbounded in the positive direction.

4.48 a. The mean would be greater than the median. There are likely to be a few waiting times that are extremely long.

b. If someone dies while waiting for a transplant that value should not be included in the mean or median calculation.

4.49 a. It is the midrange, not the median. b. The midrange is influenced by outliers. Salaries tend to be skewed to the right. The community should use

the median to base charges.

38

4.50 a. The distribution would be skewed right.b. Switching from the mean to the median would trigger a penalty sooner because the median is less than the

mean.c. The union would oppose this change because they would probably have to pay more penalties.

4.51 a. and c.

Week 1 Week 2 Week 3 Week 4

mean 50.00 50.00 50.00 50.00

sample standard deviation 10.61 10.61 10.61 10.61

median 50.00 52.00 56.00 47.00

b. Based on the mean and standard deviation it appears that the distributions are the same.

d.

e. Based on the medians and dotplots, the distributions are quite different.

4.52 Results will vary by student

4.53 a. For 1990:

From To frequency (F) midpoint (M) F*M F*(M-xbar)^2

1 2 39 1.5 58.5 240.1898

2 3 35 2.5 87.5 76.83767

3 4 27 3.5 94.5 6.264302

4 5 26 4.5 117 6.98517

5 6 24 5.5 132 55.32743

6 7 30 6.5 195 190.2588

7 8 9 7.5 67.5 111.4075

8 9 1 8.5 8.5 20.41526

Total 191

Average = 3.981675Standard Deviation = 1.92993846CV = 0.484705

39

For 2000:


1 2 54 1.5 81 210.3826

2 3 44 2.5 110 41.72649

3 4 23 3.5 80.5 0.015762

4 5 22 4.5 99 23.16691

5 6 26 5.5 143 106.7403

6 7 16 6.5 104 146.5241

7 8 5 7.5 37.5 81.05055

8 9 1 8.5 8.5 25.26247

Total 191

Average = 3.473822Standard Deviation = 1.82795415 CV = 0.526208

b. The average fertility rate is approximately 4 children per women with a standard deviation of 1.9. The central tendency and dispersion have decreased slightly from 1990 to 2000.

c. We could more easily have seen which countries are at the high end of the distribution. d. A frequency table makes it easier to see the distribution and to create a histogram.

4.54 a.


119 120 1 119.5 119.5 12.29671111

120 121 5 120.5 602.5 31.41688889

121 122 16 121.5 1944 36.32071111

122 123 22 122.5 2695 5.647644444

123 124 12 123.5 1482 2.920533333

124 125 9 124.5 1120.5 20.0704

125 126 5 125.5 627.5 31.08355556

126 127 3 126.5 379.5 36.61013333

127 128 2 127.5 255 40.38008889

Total 75


b. The average winning time 123.0 seconds and the standard deviation on winning time is 1.71 seconds. The distribution may be slightly skewed right.

c. The raw data would show us the years that the winning times were much longer than the average.d. Because the overall distribution on time is slightly skewed right it is possible that the times within an

interval are also skewed right. We give equal weight to the midpoints of each interval and our estimate of the mean could be too high.

40

4.55 a.


0 250 2 125 250 1396836.735

250 500 5 375 1875 1715306.122

500 1000 14 750 10500 621607.1429

1000 2000 14 1500 21000 4071607.143

Total 35


b. The average number of degree days is 960 and the standard deviation is 480. c. The raw data would have been useful for creating a histogram to see the shape of the distribution. d. Equal intervals of 250 might have spread the data out too far resulting in classes with frequency of zero.

Yes, it does affect the calculations.

4.56 a.


20 30 1 25 25 1522.529796

30 40 9 35 315 7579.238754

40 50 20 45 900 7234.90965

50 60 17 55 935 1383.006536

60 70 36 65 2340 34.60207612

70 80 67 75 5025 8078.123799

80 90 3 85 255 1320.530565

Total 153


b. The average life expectancy is approximately 64 years with a standard deviation of 13.4 years. The distribution appears to be skewed to the left.

c. The raw data would allow us to see those countries that have a very small life expectancy.d. A frequency table makes it easier to see the distribution and to create a histogram.

4.57 a.


40 50 12 45 540 2771.049596

50 60 116 55 6380 3131.910804

60 80 74 70 5180 7112.648981

80 100 2 90 180 1776.547482

Total 204


b. No the unequal class sizes don’t hamper the calculations. Class sizes are unequal to ensure that no class size has a zero value.

41

4.58 a.


0 3 23 1.5 34.5 3697.366199

3 6 46 4.5 207 4309.350046

6 10 37 8 296 1412.625655

10 20 44 15 660 29.66347078

20 30 31 25 775 3629.967891

30 50 23 40 920 15334.7461

Total 204


b. The unequal class sizes don’t hamper the calculations. Unequal class sizes can be used when a distribution is strongly skewed to avoid classes with zero frequencies.

4.59 a. We can find the median class because we know the frequency within each class. b. Unequal class sizes can be used when a distribution is strongly skewed to avoid classes with zero

frequencies.

4.60 a. We can find the median class because we know the frequency within each class.b. Unequal class sizes can be used when a distribution is strongly skewed to avoid classes with zero

frequencies.

42

Chapter 5

Probability

5.1 a. S = {(V,B), (V,E), (V,O), (M,B), (M,E), (M,O), (A,B), (A,E), (A,O)} b. Events are not equally likely. Border’s probably carries more books than other merchandise.

5.2 a. S = {(S,L), (S,T), (S,B), (P,L), (P,T), (P,B), (C,L), (C,T), (C,B)}b. There are different likelihoods of risk levels among the 3 types of business forms; therefore the different

elementary events will have different likelihoods.

5.3 a. S = {(L,B), (L,B’), (R,B), (R,B’)} b. Events are not equally likely. There are more right handed people than left handed people.

5.4 a. S ={(1,H), (2,H), (3,H), (4,H), (5,H), (6,H), (1,T), (2,T), (3,T), (4,T), (5,T), (6,T)}b. Yes, assuming that we have a fair die and fair coin.

5.5 a. Opinion of experienced stock brokers or empirical.b. From historical data of IPOs or based on judgments.

5.6 a. Subjective b. Opinion of a group of telecommunication stock brokers.

5.7 a. Empiricalb. Historical data of past launches.

5.8 a. Classical

b. There are 36 different outcomes from rolling two dice. There are 6 ways to roll a 7. .0046 = 3

6

36

.

5.9 a. P(A ∪ B) = .4 + .5 − .05 = .85. b. P(A | B) = .05/.50 = .10.c. P(B | A) = .05/.4 = .125.d.

5.10 a. P(A∪B) = P(A) + P(B) - P(A∩B) = .7 + .3 – 0 = 1.0 .

b. P(AB) = P(A∩B) / P(B) = .00 / .30 = .00.

c. The intersection is an empty set because the P(A∩B) = 0.

43

5.11 a. P(S) = .217.b. P(S’) = .783.c. Odds in favor of S: .217/.783 = .277.

d. Odds against S: .783/.217 = 3.61

5.12 a. (.017) /(.983) /100 = .0173 to 1b. 98.3 / .017 = 57.83 to 1

5.13 a. X = 1 if the drug is approved, 0 otherwise.b. X = 1 if batter gets a hit, 0 otherwise.c. X = 1 if breast cancer detected, 0 otherwise.

5.14 a. (admitted unconditionally, admitted conditionally, not admitted)b. (completed pass, incomplete pass, intercepted pass)c. (deposit, withdrawal, bill payment, funds transfer)

5.15 a. P(S’) = 1−.246. There is a 75.4% chance that a female aged 18-24 is a nonsmoker.

b. P(S ∪ C) = .246+ .830 − .232 = .844. There is an 84.4% chance that a female aged 18-24 is a smoker or is Caucasian.c. P(S | C) = .232/.830 = .2795. Given that the female aged 18-24 is a Caucasian, there is a 27.95% chance that they are a smoker.

d. P(S ∩ C’) = P(S) – P(S ∩ C) = .246 − .232 = .014. P(S | C’) = .014/.17 = .0824. Given that the female ages 18-24 is not Caucasian, there is an 8.24% chance that she smokes.

5.16 P(A∩B). = P(A) * P(B) = .40*.50 =0.20

5.17 a. P(AB) = P(A ∩ B) / P(B) = .05/.50 = .10

b. No, A and B are not independent because P(AB) ≠ P(A).

5.18 a. P(A) * P(B) = .40*.60 =.24 and P(A ∩ B) = .24, therefore A and B are independent.

b. P(A) * P(B) = .90*.20 =.18 and P(A ∩ B) = .18, therefore A and B are independent.

c. P(A) * P(B) = .50*.70 = .35 and P(A ∩ B) = .25, therefore A and B are not independent.

5.19 a. P(V ∪ M) = .70 + .60 – .50 = .80.

b. P(V ∩ M) ≠ P(V)*P(M) therefore V and M are not independent.

5.20 a. There is 25% chance that a clock will not ring (a failure, F). Both clocks would have to fail in order to

have him oversleep. Assuming independence: P(F1∩F2) = P(F1) * P(F2) = .25*.25 = .0625. b. The probability that at least one of the clocks rings is 1 – (P(F1)*P(F2)*P(F3)) = 1- (.25*.25*.25) = .9844,

which is less than 99%.

5.21 “Five nines” reliability means P(not failing) = .99999. P(power system failure) = 1 − (.05)3 = .999875. The system does not meet the test.

5.22 a. P(A) = 100/200 = .50. There is a 50% chance that a student is an accounting student.b. P(M) =102/200 = .51. There is a 51% chance that a student is male.

c. P(A ∩ M) = 56/200 = .28. There is a 28% chance that a student is a male accounting major.

d. P(F ∩ S) = 24/200 = .12. There is a 12% chance that a student is a female statistics major.e. P(A | M) = 56 /102 = .549. There is 54.9% chance that a male student is an accounting major.

f. P(A | F) = P(F ∩ A) / P(F)= (44/200)/(98/200) = .4489. There is a 44.89% chance that a female student is an accounting major.

g. P(F | S) = P(F ∩ S) / P(S) (24/200) /(40/200) = .60. There is a 60% chance that a statistics student is female.

44

h. P(E ∪ F) = P(E) + P(F) - P(F ∩ E) 60/200 + 98/100 – 30/100 = 128/200 = 64%. There is 64% chance that a student is an economics major or a female.

5.23 Gender and Major are not independent. For example, P(A ∩ F) = .22. P(A)P(F) = .245. Because the values are not equal, the events are not independent.

5.24 a. P(D3) = 15/38 = .3948.b. P (Y3) = 15/38 = .3948.

c. P(Y3 | D1 = P(Y3∩ D1) / (D1) = (2/38)/(11/38) =.1818.

d. P(D1 | Y3) = P(Y3∩ D1) / P(Y3) = (2/38)/(15/38) =.1333.

5.25Joint Probabilities

P(C1∩S1) = .8 *.7 = .56

P(C2∩S1) = .8 *.2 = .16

P(C3∩S1) = .8 *.1 = .08

Sums to 1.00

P(C1∩S2) = .2*.5 = .10

P(C2∩S2 = .2 *.4 = .08

P(C3∩S2) = .2 *.1 = .02

Sums to 1.00 Sum of 6 joint probablities is 1.00

Pays by debit/credit cardP(C3|S2) = .1



Pays by debit/credit cardP(C2|S1) = .2Takes a shopping cart

P(S1) = .8

Does Not Take a

Shopping Cart P(S2) = .2



45

5.26

5.27 Let A = using the drug. P(A) = .04. P(A’) = .96. Let T be a positive result. False positive: P(T | A’) = .05. False negative: P(T’ | A) = .10. P(T | A) = 1 − .10 = .90. P(T) = (.04)(.90) + (.05)(.96) = .084. P(A | T) = (.9)(.04)/.084 = .4286.

5.28 P(A) = .5 P(B) = .5 P(D) = .04 P(ND) = .96 P(D|A) = .06 P(A|D) = P(D ∩ A) / P(D) = ( P(D|A)* P(A) ) / P(D) = (.06*.5) / .04 = .75.

5.29 Let W = suitcase contains a weapon. P(W) = .001. P(W’) = .999. Let A be the alarm trigger. False positive: P(A | W’) = .02. False negative: P(A’ | W) = .02. P(A | W) = 1 − .02 = .98. P(A) = (.001)(.98) + (.02)(.999) = .02096. P(W | A) = (.98)(.001)/.02096 = .04676.

5.30 a. 26*26* 10*10*10*10 = 6,760,000 unique ids. b. No, that only yields 26,000 unique ids. c. As growth occurs over time, you would not ever have to worry about a duplicate id nor have to generate

new ones.

5.31 a. 106 = 1,000,000.b. 105 = 100,000.c. 106 = 1,000,000

5.32 a. !

( )!rn

nP

n r=

− ) (n=4, r=4) = 24 ways.

b. ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC, BCAD, BCDA, BDAC, BDCA, CABD, CADB, CBAD, CBDA, CDBA, CDAB, DABC, DACB, DBAC, DBCA, DCAB, DCBA

5.33 a. 7! = 5,040 ways.b. No, too many!

5.34 a. n = 8 and r = 3 : 336.b. n = 8 and r = 5 : 6720.c. n = 8 and r = 1 : 8.d. n = 8 and r = 8. : 40320.

46

5.35 a. 8C3 = 56.b. 8C5 = 56.c. 8C1 = 8.d. 8C8 = 1.

5.36 a. !

( )!rn

nP

n r=

−, (n =10, r = 4 )= 5040.

b. !

!( )!rn

nC

r n r=

−, (n=10, r =4 ) = 210.

5.40 a. Sample Space is made up of pairs denoted by x. There are 21 combinations.

Brown Yellow Red Blue Orange Green

Brown x

Yellow x x

Red x x x

Blue x x x x

Orange x x x x x

Green x x x x x x

b. P(BR)*P(BR) = .13*.13 = .0169.c. P(BL)*P(BL) = .24*.24 = .0576.d. P(G)*P(G) = .16*.16 = .0256.e. P(BR)*P(G) = .13*.16 = .0208.

5.41 a. An empirical probability using response frequencies from the survey.b. Odds for failure: .44/.56 = .786.

5.42 a. Subjectiveb. P(loss) = 0.01408

5.43 No, the law of large numbers says that the larger the sample, the closer our sample results will be to the true value. If Tom Brookens increases his times “at bat” he’ll get closer and closer to his true batting average, which is probably close to .176.

5.44 a. Subjectiveb. Bob probably based this estimate on his swimming ability and success of others who have completed this

feat.

5.45 a. Empirical or subjectiveb. Most likely estimated by interviewing ER doctors.c. The sample could have been small and may not have been a representative sample of all doctors.

5.46 a. Empirical b. From a sample of births in the U.S. c. It depends on the size of the sample and how closely the sample represented the US as a whole.

5.47 a. Subjectiveb. Simulated experiment using a computer model.c. The estimate is probably not very accurate. Results are highly dependent on the computer simulation and

the data put into the model.

5.48 a. 114/108 = .8814

47

b. Empirical

48

5.49 a. Empirical or subjective b. Observation or surveyc. The estimate is probably not very accurate. Observation is difficult and survey results may be biased.

5.50 a. .227/.773 = .29 to 1 in using a debit cardb. .773/.227 = 3.4 to 1 against using a debit card

5.51 H1 = First child has high cholesterol, N1= First child has normal cholesterol.

H1H2H3: P(H1H2H3) = (.01)(.1) = .001

H1H2P(H1H2) = H1H2N3

H1 (.1)(.1) =.01 P(H1H2H3) = (.01)(.9) = .009

P(H1) = .1 H1N2H3P(H1N2H3) = (.09)(.1) = .009

H1N2P(H1N2) = H1N2N3(.1)(.9) = .09 P(H1N2N3) = (.09)(.9) = .081

N1H2H3P(N1H2H3) = (.09)(.1) = .009

N1H2P(N1H2) = N1H2N3(.9)(.1) = .09 P(N1H2N3) = (.09)(.9) = .081

N1P(N1) = .9 N1N2H3

P(N1N2H3) = (.81)(.1) = .081N1N2P(N1N2) = N1N2N3(.9)(.9) = .81 P(N1N2N3) = (.81)(.9) = .729

49

5.52 a.

5.53 Odds against an Acura Integra being stolen = .987/.013 = 76 to 1.

5.54 a. .33 / (1-.33) = .493 to 1 being killed. b. .67/.33 = 2.03 to 1 against being killed.

5.55 P(Detroit Wins) = 50/51 = .9804. P(New Jersey Wins) = 5/6 = .8333.

5.56 a. 29 = 512 separate codes. b. 210 = 1024 separate codes. c. (1/512)* 1000 = 1.95 times (approximately 2) or (1/1024)* 1000 = 0.977 (approximately 1). We assume

that each combination is selected independently.

5.57 a. 263103 = 17,576,000. b. 366 = = 2,176,782,336. c. 0 and 1 might be disallowed since they are similar in appearance to letters like O and I .d. Yes, 2.1 billion unique plates should be enough. e. 346 = 1,544,804,416.

5.58 Suppose the correct order for the meals is ABC. The possibilities for incorrect orders include: ACB (1 incorrect meal), BAC (1 incorrect meal), BCA (3 incorrect meals), CAB (3 incorrect meals), and CBA (1 incorrect meal).

a. P(No diner gets the correct meal) = 1/3b. P(Exactly one diner gets the correct meal) = 1/2c. P(Exactly two diners get the correct meal) = 0d. P(All three diners get the correct meal) = 1/6

5.59 7P3 = 210.

5.60 a. The first 4 cards are aces: (4/52)(3/51)(2/50)(1/49) = 3.694E-06.b. Any four cards in the hand are aces. The sample space includes: AAAAN, AAANA, AANAA, ANAAA,

NAAAA. Each of these sample spaces has the same probability as given in part a; therefore, the probability that any four of the five cards are aces is 5*(3.694E-06) = 1.847E-05.

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5.61 a. P(Two aces) = (4/52)(3/51) = 0.00452. b. P(Two red cards) = (26/52)(25/51) = 0.245098. c. P(Two red aces) = (2/52)(1/51) = 0.000754.d. P(Two honor cards) = (20/52)(19/51) = 0.143288.

5.62 Let F denote a failure and S denote a non-failure. Use the multiplicative rule of probability for independent events:

a. P(F1 ∩ F2) = P(F1)*P(F2) = .02*.02 = .0004.

b. P(S1 ∩ S2) = P(S1)*P(S2) = .98*.98 = .9604.c. The sample space for all events is : F1F2, S1F2, F1S2, S1S2 There is only one event that does not contain

at least one failure, S1S2. The probability that one or the other will fail is 1- P(S1 ∩ S2) = 1- .9604 = .0396.

5.63 No, P(A)P(B) ≠ .05.

5.64 P(B1 ∩ B2) = P(B1)*P(B2) = .5*.5 = .25. 5.65 a. Having an independent back up power system for the computers might have eliminated the delayed flights.

b. If the cost due to delayed/cancelled flights, weighted by the risk of a power outage, is greater than $100,000 then the airline can justify the expenditure.

5.66 a. Independentb. These are typically considered dependent. Insurance rates are higher for most men because they are

involved in more accidents.c. Dependent, most calls are during regular business hours when the office is open.

5.67 Assuming independence, P(3 cases won out of next 3) = .73 = .343.

5.68 a. If p is the probability of failure, we can set pk = 0.00001, plug in p = 0.01, take the log of both sides, and solve for k. In this case, k = 2.50 and then round up to the next higher integer. So 3 are required.

b. For p = .10, k = 5, so 5 servers are required.

5.69 Assuming independence, P(4 adults say yes) = .564 = 0.0983.

5.70 a. P(fatal accident over a lifetime) = 1 − P(no fatal accident over a lifetime) = 1 − (3,999,999/4,000,000)50,000

= .012422.b. The probability of an accident each time you get behind the wheel is so small that an individual might take

the risk.

5.71 See the Excel Spreadsheet in Learning Stats: 05-13 Birthday Problem.xls. For 2 riders: P(no match) = .9973.For 10 riders: P(no match) = 0.8831.For 20 riders: P(no match) = 0.5886.For 50 riders: P(no match) = 0.0296.

5.72 See the Excel Spreadsheet in Learning Stats: 05-13 Birthday Problem.xls. If there are 23 riders, P(match) = .50730.

If there are 32 riders, P(match) = .75.

5.73 a. i. .4825. The probability of seeing a car in a shopping mall parking lot is .4825.ii. .25. The probability of seeing a vehicle in the Great Lakes shopping mall is .25.iii. .115. The probability of seeing a parked truck in a shopping mall is .115.iv. .19. The probability of seeing a parked SUV at the Somerset mall is .19.v. .64. The probability of seeing a parked car at the Jamestown mall is .64.vi. .3316. The probability that a parked car is at the Jamestown mall is .3316.

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vii. .09. The probability that a parked vehicle is a car and is at the Great Lakes mall is .09.viii. .015. The probability a parked vehicle is a truck and is at the Oakland mall is .015.ix. .0325. The probability that a parked vehicle is a Minivan and is at the Jamestown mall is .0325.b. Yes, the vehicle type and mall location are dependent. For example, P(T)*P(O) = (.115)(.25) = .02875. P(T

and O) = .015. Because .02875 ≠ .015, the events are dependent.

5.74 a. i. P(S) = 320/1000 = .32. The likelihood of a male 18-24 smoking is .32. ii. P(W) =850/1000 = .85. The likelihood of a male 18-24 being white is .85.iii. P(S | W) = P(S and W)/ P(W) = .29/.85 = .3412. The likelihood of a white male 18-24 being a smoker is .

3412.iv. P(S | B) = P(S and B)/ P(B) = (30/1000)/(150/1000) = .200. The likelihood of a black male 18-24 being a

smoker is .20. v. P(S and W) = 290/1000 = .290. The likelihood of a male 18-24 being a smoker and being white is .290. vi. P(N and B) = 120/1000 = .12 The likelihood of a male 18-24 not smoking and being black is .12.b. The P(S and W) = .29 and the P(S)*P(W) = .32*.85 = .272. The P(S and B) = .030 and the P(B)*(S) = .

32*.15 = .048. Yes, the smoking rates suggest that race and smoking are dependent. d. If smoking is dependent on race, then health officials might target or design special programs based on

race.

5.75 a. i. .5588. The probability that the forecasters predicted a decline in interest rates is .5588.ii. .5294. The probability there was a rise in interest rates is .5294.iii. .3684. Given that the forecasters predicted a decline in interest rates, the probability that there was an

actual decline is .3684.iv. .4. Given that the forecasters predicted an increase in interest rates, the probability that there was an actual

increase is .4.v. .1765. The probability that in a given year there was both a forecasted increase and actual increase in

interest rates is .1765.vi. .2059. The probability that in a given year there was both a forecasted decline and actual decline in interest

rates is .2059.b. No, P(A−) = .4705 and P(A− | F−) = .3684. Interest rates moved down 47% of the time and yet the

forecasters’ predictions of a decline showed a 37% accuracy rate.

5.76 a. i. P(B) = 6/61 = .4098. The likelihood of a climb is .4098.ii. P(L) = 14/61= .2295. The likelihood of a low noise level is .2295.iii. P(H) = 18/61 =.2951. The likelihood of a high noise level is .2951. iv. P(H | C) = 3/8 = .3750. The likelihood of a high noise given you are in the cruise phase is .3750. v. P(H | D) = 14/28 = .5. The likelihood of a high noise given you are in the descent phase is .5.vi. P(D | L) = 6/14 = .4286. The likelihood of being in the descent phase given you experience a low noise

is .4286.vii. P(L and B) = .0984. The likelihood of both a low noise and climbing is .0984. viii. P(L and C)= .0328. The likelihood of both a low noise and cruising is .0328. ix. P(H and C) = .0492. The likelihood of both a high noise and cruising is .0492.b. Flight noise is dependent on flight phase. For example: P(H) = .2951 and P(H|C) = .375. If independent the

two probabilities would be the same.

5.77 a. .7403 ii. .1244 iii. .1385 iv. .6205 v. .7098 vi. .9197 vii. .1485 viii. .1274 ix. .0042.b. Yes, the probability of being a nonsmoker increases with level of education.

5.78 a. P(L) =.1321.b. P(C) = .2075.c. P(M) =.3208.d. P(F) = .3019.e. P(C | F) = 15/48 = .3125.f. P(C | M) = 11/51 = .0692 / .3208 = .2157.

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5.79Cancer No Cancer Totals

Positive Test 4 500 504

Negative Test 0 9496 9496

Totals 4 9996 10000

P(Cancer | Positive Test) = 4/504 = 0.00794.

5.80Authorized Not Auth Totals

Denied Access 19,000 999,999 1,018,999

Allowed Access 18,981,000 1 18,981,001

Totals 19,000,000 1,000,000 20,000,000

P(Authorized | Denied Access) = 19,000/1,018,999 = .01865.

5.81

Accident No Accident Totals

Right-Handed 3240 5760 9000

Left-Handed 520 480 1000

Totals 3760 6240 10000

P(Left-Handed | Accident) = 520/3760 = .1383.

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Chapter 6

Discrete Distributions

6.1 A is a probability distribution, since the sum of P(x) is 1 and all probabilities are nonnegative, while B and C are not probability distributions since the sum of P(x) is .95 for B and 1.30 for C.

6.2 E(X) = 70, V(X) = 100, σ = 10. Distribution is skewed to the right. The worksheet is:

x P(x) xP(x) x−E(X) P(x)[x−E(X)]2

60 0.40 24.00 -10.00 40.00

70 0.30 21.00 0.00 0.00

80 0.20 16.00 10.00 20.00

90 0.10 9.00 20.00 40.00

Total 1.00 70.00 100.00

6.3 E(X) = 2.25, V(X) = 1.6875, σ = 1.299. Distribution is skewed to the right.

x P(x) xP(x) x−E(X) P(x)[x−E(X)]2

0 0.05 0.00 -2.25 0.25

1 0.30 0.30 -1.25 0.47

2 0.25 0.50 -0.25 0.02

3 0.20 0.60 0.75 0.11

4 0.15 0.60 1.75 0.46

5 0.05 0.25 2.75 0.38

Total 1.00 2.25 1.6875

6.4 E(X) = (215,000)(.00000884)+(0)(.99999116) = 1.9006.

6.5 Expected payout = E(X) = 1000(.01)+(0)(.999) = $10, so company adds $25 and charges $35.

6.6 The expected winning is E(X) = 28,000,000(.000000023)+0(.99999998) = $0.644. Since the cost of the ticket

is $1.00, its expected value is −$0.356

6.7 Expected Loss = 250(.3) + 950(.3) + 0(.4) = $360 million.

6.8 Using a = 0000 and b = 9999, (0 9999) / 2 4999.5µ = + = and 2[(9999 0 1) 1]/12 2886.75σ = − + − =

6.9 a. With a = 20 and b = 60, (20 60) / 2 40µ = + = and 2[(60 20 1) 1]/12 11.83σ = − + − =

b. P(X. ≥40) = (1/40)(20) = ½ and P(X≥30) = (1/40)(30) = = ¾.

6.10 Using a = 1 and b = 500000, (1 500000) / 2 250000.5µ = + = and 2[(500000 1 1) 1]/12 144337.6σ = − + − =

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6.11 a. With a = 1 and b = 31, (1 31) / 2 16µ = + = and 2[(31 1 1) 1]/12 8.944σ = − + − =b. Yes, if conception is random within each month.

6.12 a. µ = 1.5, σ = .50 =1+INT(2*RAND())

b. µ = 3.0, σ = 1.414 =1+INT(5*RAND())

c. µ = 49.5, σ = 28.87 =0+INT(100*RAND())d. Answers will vary.

6.13 Answers may vary and 0 and 1 are interchangeable.a. 1 = correct, 0 = incorrectb. 1= insured, 0 = uninsuredc. 1 = busy, 0 = not busyd. 1 = lost weight, 0 no weight loss

6.14 a. π = .5 (desirable)

b. π = .5 (desirable)

c. π = .8 (undesirable)

d. π = .5 (desirable)

6.15 a. µ = (8)(.1) = 0.8, (8)(.1)(1 .1) 0.8485σ = − =

b. µ = (10)(.4) = 4, (10)(.4)(1 .4) 1.5492σ = − =

c. µ = (12)(.5) = 6, (12)(.5)(1 .5) 1.7321σ = − =

d. µ = (30)(.9) = 27, (30)(.9)(1 .9) 1.6432σ = − =

e. µ = (80)(.7) = 56, (80)(.7)(1 .7) 4.0988σ = − =

f. µ = (20)(.8) = 16, (20)(.8)(1 .8) 1.7889σ = − =

6.16 a. P(X = 2) = .1488b. P(X = 1) = .0403c. P(X = 3) = .0015d. P(X = 5) = 0.0074

6.17 a. P(X ≤ 3) = .9437b. P(X > 7) = 1 – P(X ≤ 6) = .1719c. P(X < 3) = P(X ≤ 2) = .0705d. P(X ≤ 10) = .00417

6.18 a. P(X < 4 ) = P(X ≤ 3) = .9744b. P(X ≥ 3) = 1 – P(X ≤ 2) = .5801c. P(X ≤ 9) = .7207d. P(X > 10) = 1 – P(X ≤ 10) = .9183

6.19 a. P(X = 0) = .10737b. P(X ≥ 2) = 1 – P(X ≤ 1) = .62419c. P(X < 3) = P(X ≤ 2) = .6778d. µ = nπ =(10)(.2) = 2

e. σ= (10)(.2)(1 .2)− = 1.2649

f. See below.

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g. Skewed to the right.

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6.20 a. P(X = 0) = .54036b. P(X = 1)= .34128c. P(X = 2) = .09879d. P(X ≤ 2) = .98043e. See below, skewed to the right.

6.21 a. P(X =10) = .00098b. P(X ≥ 5) = 1- P(X ≤ 4) = .62305c. P(X < 3) = P(X ≤ 2) = .05469d. P(X ≤ 6) = .82813

6.22 a. P(X = 8) = .016796b. P(X ≥ 5) = 1 – P(X ≤ 4) = .59409c. P(X ≥ 5) = 1 – P(X ≤ 4) = .59409

d. µ = nπ =(8)(.6) = 4.8 and σ= (8)(.6)(1 .6)− = 1.386

e. It is almost symmetric (slightly left-skewed).

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6.23 a. λ = 1, µ = 1.0 and σ =1

b. λ = 2, µ = 2.0 and σ =1.414

c. λ = 4, µ = 4.0 and σ =2.0

d. λ = 9, µ = 9.0 and σ =3

e. λ = 12, µ = 12.0 and σ =3.464

6.24 a. λ = 0.1, P(X = 2) = .24377

b. λ = 2.2, P(X = 1) =.00452

c. λ = 1.6, P(X = 3) =.13783

d. λ = 4.0, P(X = 6) =.10420

e. λ = 12.0, P(X = 10) =.10484

6.25 a. λ = 4.3, P(X ≤ 3) = .37715

b. λ = 5.2, P(X > 7) =.15508

c. λ = 2.7, P(X < 3) = .49362

d. λ = 11.0, P(X ≤ 10) = .45989

6.26 a. λ = 5.8, P(X < 4) = P(X ≤ 3) = .16996

b. λ = 4.8, P(X ≥ 3) = 1 – P(X ≤ 2) = 1 − .14254 = .85746

c. λ = 7.0, P(X ≤ 9) = .83050

d. λ = 8.0, P(X >10) = 1 − P(X ≤ 10) = .81589

6.27 a. P(X ≥ 1) = 1 – P(X≤ 0) = 1 − .09072 = .90928b. P(X = 0) = .09072

c. P(X > 3) = 1− P(X ≤ 3) = 1 − .77872 = .22128d. Skewed right.

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6.28 a. Cancellations are independent and similar to arrivals.b. P(X = 0) = .22313c. P(X = 1) = .33470

d. P(X > 2) =1 − P(X ≤ 2) =1- .80885 = .19115

e. P(X ≥ 5) =1 − P(X ≤ 4) = 1 − .98142 = .01858

6.29 a. Most likely goals arrive independently.

b. P(X ≥ 1) = 1 – P(X ≤ 0) = 1 − .06721 = .93279c. P(X ≥ 4) = 1 – P(X ≤ 3) = 1 – .71409 = .28591d. Skewed right.

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6.30 a. Not independent events, the warm room leads to yawns from all.b. Answers will vary.

6.31 Let λ = nπ = (500)(.003) = 1.5

a. P(X ≥ 2) = 1 – P(X ≤ 1) = 1 − .55783 = .44217b. P(X < 4) = .93436

c. Use the Poisson when n is large and π is small.

d. Yes, based on our rule of thumb n ≥ 20 and π ≤ .05

6.32 Let λ = nπ = (100000)(.000002) = 2

a. P(X ≥ 1) = 1 − P(X=0) = 1 − .13534 = .86466

b. P(X ≥ 2) = 1 – P(X ≤ 1) = 1 − .40601 = .59399c. Excel could be used, otherwise n is too large for practical calculations.

d. Yes, based on our rule of thumb n ≥ 20 and π ≤ .05

6.33 a. µ = (200)(.03) = 6 letters

b. (200)(.03)(1 .03)σ = − = 2.413

c. For λ = nπ = 6, P(X ≥ 10) = 1 – P(X ≤ 9) = 1 − .91608 = .08392

d. For λ = nπ = 6, P(X ≤ 4 ) = .28506e. Excel could be used, otherwise n is too large for practical calculations.

f. Yes, based on our rule of thumb n ≥ 20 and π ≤ .05

6.34 a. Range 0 to 3, P(X = 3) = .03333b. Range 0 to 3, P(X = 2) = .13158c. Range 0 to 4, P(X = 1) = .44691d. Range 0 to 7, P(X = 3) = .10980

6.35 The distribution is symmetric with a small range (2 to 4).

6.36 a. Let X = number of incorrect answers in sample.b. P(X = 0) = .31741

c. P(X ≥ 1) = 1 − P(X = 0) = 1 − .31741 = .68259

d. P(X ≥ 2) = 1 − P(X ≤ 1) = 1 − .74062 = .29538e. Skewed right.

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6.37 a. Let X = number of incorrect vouchers in sample.b. P(X = 0) = .06726c. P(X = 1) = .25869

d. P(X ≥ 3) = 1 − P(X ≤ 2) = 1 − .69003 = .30997e. Fairly symmetric.

6.38 a. Let X = number of HIV specimens in sample.b. P(X = 0) = .30604

c. P(X < 3) = P(X ≤ 2) = .95430

d. P(X ≥ 2) = 1 − P(X ≤ 1) = 1 − .74324 = .25676e. Skewed right.

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6.39* a. 3/100 < .05, okay to use binomial approximationb. 10/200 > .05, don’t use binomial approximationc. 12/160 > .05 , don’t use binomial approximationd. 7/500 < .05, okay to use binomial approximation

6.40* a. P(X = 0) = .59049 (B) or .58717 (H)

b. P(X ≥ 2) = 1 – P(X ≤ 1) = 1-.91854 = .08146 (B) or .0792 (H)c. n/N =5/200 < 0.05 so binomial approximation is OK.

6.41* a. P(X=0)= 0.34868 (B) or .34516 (H)

b. P(X ≥ 2) = 1 – P(X ≤ 1) = 1 − .73610 = .26390 (B) or .26350 (H)c. P(X < 4) = P(X ≤ 3) = .98720 (B) or .98814 (H)

d. n/N = 10/500 = .02 so we can use the binomial with π = s/N= 50/500 = .1

6.42* a. P(X = 6) = .26214 (B) or .25967 (H)

b. P(X ≥ 4) = 1 − P(X ≤ 3) = 1 − .09888 = .90112 (B) or .90267 (H)

c. n/N = 6/400 < 0.05 so we can use the binomial with π = s/N= 320/400 = .8

6.43* a. P(X = 5) = .03125 when π = .5

b. P(X = 3) = .14063 when π = .25

c. P(X = 4) = .03840 when π = .60

6.44* a. Geometric mean is 1/π = 1/(.20) = 5

b. Using the geometric CDF, P(X ≤ 10) = 1− (1−π)x = 1− (1−.20)10 = .89263

6.45* a. Geometric mean is 1/π = 1/(.50) = 2

b. Using the geometric CDF, P(X ≤ 10) = 1− (1−π)x = 1− (1−.50)10 = .99902

6.46* a. µ = 79.6×2.54 = 202.184 cm

b. σ = 3.24×2.54 = 8.2296 cmc. Rule 1 for the mean and Rule 2 for the std dev.

6.47* a. Applying Rule 3, we add the means for each month to get µ = 9500+7400 + 8600 = $25,500. Applying Rule 4, we add the variances for each month and then take the square root of this sum to find the std dev

for the quarter: σ2 = 1250+1425+1610 = 4285 and so s = (4285).5 = 65.4599 is the std dev for the quarter.b. Rule 4 assumes that the sales for each month, in this case, are independent of each other. This may not

be valid, given that a prior month’s sales usually influence the next month’s sales.

6.48 The probability of a payout is 1 − .99842 = .00158. The expected payout is (.00158)(1,000,000) = $1,580 dollars. To break even, the company would charge $1,580.

6.49 E(X) = (100)(1/6) + (−15)(5/6) = 16.67 − 12.50 = $4.17. On average, you would win more than you lose. If you have to pay more than $4.17 to play, a rational person wouldn’t play (unless very risk loving).

6.50 The expected loss is E(X) = (250)(.02) + (0)(.98) = $5 which exceeds the $4 cost of insurance (assuming you would lose the entire cost of the PDA). Statistically, it is worth it to insure to obtain “worry-free” shipping, despite the small likelihood of a loss.

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6.51 a. If uniform, µ = (1 + 44)/2 = 22.5 and 2[(44 1 1) 1]/12 12.698σ = − + − = . Quite a big difference from

what is expected.b. What was the sample size? One might also want to see a histogram.

6.52 a. If uniform, µ = (1 + 5)/2 =3 and 2[(5 1 1) 1]/12 1.414σ = − + − = .

b. Answers will vary.c. Answers will vary.d. =1+INT(5*RAND())

6.53 a. π = .80 (answers will vary).

b. π = .300 (answers will vary).

c. π = .50 (answers will vary).

d. π = .80 (answers will vary).e. Outcomes of one trial might influence the next. For example, if I fail to make a free throw because I shot

the ball “long”, I will adjust my next shot to be a little “shorter,” hence, violating the independence rule.

6.54 a. P(X = 0) =.06634 b. P(X≥2) = 1 – P(X ≤ 1) = 1- .94276 = .05724

c. Binomial µ = nπ = (8)(.05) = 0.4 and (1 ) 8(.05)(.95) 0.616nσ = π − π = =d. Strongly skewed to the right.

6.55 a. Define X to be the number that fail. P(X = 0) = .59049b. P(X = 1) = .32805.c. Strongly skewed to the right.

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6.56 a. P(X = 0) =.10737b. P(X ≥ 2) = 1 – P(X ≤ 1) = 1– .37581 = .67780c. P(X = 10) = .00000d. Slightly skewed to the right.

6.57 a. P(X = 0) =.06250b. P(X ≥ 2) = 1 – P(X ≤ 1) = 1– .31250 = .68750c. P(X ≤ 2) = .68750d. Symmetric.

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6.58 a. P(X = 0) =.01680b. P(X = 1) = .08958c. P(X = 2) = .20902d. P(X ≤ 2) = .31539e. Slightly skewed right.

6.59 a. =BINOMDIST(3,20,0.3,FALSE)b. =BINOMDIST(7,50,0.1,FALSE)c. =BINOMDIST(6,80,0.05,TRUE)

d. =1−BINOMDIST(29,120,0.2,TRUE)

6.60 a. P(X ≥ 14) = 1 – P(X ≤ 13) = 1 – .942341 = .057659b. P(X ≥ 15) = .0207 therefore a score of 15 would be needed.

6.61 a. P(X = 0) =.48398

b. P(X ≥ 3) = 1 – P(X ≤ 2) = 1− .97166 = .02834

c. For this binomial, µ = nπ = (10)(.07) = 0.7 defaults

6.62 Using Excel: =BINOMDIST(0,14,0.08,FALSE) = 0.311193

6.63 Binomial with n = 16, π = .8:a. P(X ≥ 10) = 1 – P(X ≤ 9) = 1 – .02666 = .97334 b. P(X < 8) = P(X ≤ 7) = .00148

6.64 a. =POISSON(7,10,FALSE) = .0901b. =POISSON(3,10,FALSE) = .0076

c. =1 − POISSON(4,10,TRUE) = .0292

d. =1 −POISSON(10,10,TRUE) = .4170

6.65 Let X = the number of no shows. Then:

a. If n = 10 and π = .10, then P(X = 0) = .34868.

b. If n = 11 and π = .10, then P(X ≥ 1) = 1 – P(X=0) = 1 - .31381 = .68619c. If they sell 11 seats, there is no way that more than 1 will be bumped.

d. Let X = the number who do show up and set π = .90. We want P(X ≥ 10) ≥ .95 so we use Excel’s

function = 1−BINOMDIST(9,n,.9,TRUE) for various values of n. It turns out that n = 13 will suffice.

n P(X ≤ 9) P(X ≥ 10)

11 0.30264 0.69736

12 0.11087 0.88913

13 0.03416 0.96584

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6.66 a. Let X be the number that are not working. As long as no more than 2 are not working, he will have enough. Using Excel’s =BINOMDIST(2,10,0.2,1) we calculate P(X ≤ 2) = .67780.

b. Let X be the number that are working and set π = .8. We want P(X ≥ 8) ≥ .95 so we use Excel’s function

=1−BINOMDIST(7,n,0.2,TRUE) for various values of n. It turns out that n = 13 will suffice.

n P(X ≤ 7) P(X ≥ 8)

10 0.32220 0.67780

11 0.16114 0.83886

12 0.07256 0.92744

13 0.03004 0.96996

6.67 a. Because calls to a fire station within a minute are most likely all about the same fire, the calls are not independent.

b. Answers will vary.

6.68 a. Defects happen randomly and are independent events.b. P(X = 5) = .17479c. P(X ≥ 11) = 1 – P(X ≤ 10) = 1 –0.9823 = .0177d. Right-skewed.

6.69. a. Storms happen at different times throughout the year and seem to be independent occurrences.

b. P(X ≥ 5) = 1−P(X ≤ 4) = 1−.00181 = .99819c. P(X > 20) = 1–P(X ≤ 20) = 1–.95209 = .04791

d. Fairly symmetric due to large λ.

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6.70 a. Near collisions are random and independent events.b. P(X ≥ 1) = 1 – P(X = 0) = 1 – .30119 = .69881

c. P(X > 3) = 1 − P(X ≤ 3) = 1 − .96623 = .03377d. See below.

6.71 a. Assume that cancellations are independent of each other and occur randomly.b. P(X = 0) = .22313 c. P(X = 1) = .33470d. P(X > 2) = 1 – P(X ≤ 2) = 1 – .80885 = .19115e. P(X ≥ 5) = 1 – P(X≤ 4) = 1 – .98142 = .01858

6.72 a. The number of fatal crashers occurs randomly and each crash is independent of the other.b. P(X ≥ 4) = 1 – P(X ≤ 3)= 1 – .69194 = .30806c. P(X ≤ 3) = .69164

d. Given the historical mean of λ = 2.8 for that decade, 4 or more crashes in one year was not very unusual, (31% chance of occurring) assuming independent events.

6.73 a. We assume that paint defects are independent events, distributed randomly over the surface. For this

problem, we would use a mean of λ = 2.4 defects per 3 square meter area.b. P(X = 0) = .09072c. P(X = 1) = .21772d. P(X ≤ 1) = .30844

6.74 a. We assume that paint defects are independent events, distributed randomly over the surface.b. P(X ≤ 4) = .02925

c. PX > 15) = 1− P(X ≤ 15) = 1 − .95126 = .04874

d. Fairly symmetric due to large λ.

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6.75 a. λ = 1/30 supernova per year = 0.033333 and 1/ 30σ = λ = =.182574.

b. P(X ≥ 1) = 1 – P(X = 0) = 1 − λxe-λ/x! = 1 − (.033333)0e−.033333/0! = 1 − .96722 = .03278

c. Appendix B does not have λ = .0333333.

6.76 a. Earthquakes are random and independent events. No one can predict when they will occur.b. P(X < 3) = P(X ≤ 2) = .87949

c. P(X > 5) = 1 − P(X ≤ 5) = 1 − . 0.9985 = .0015

6.77 a. Crashes are unrelated events, can’t predict them, so they do happen randomly. A single crash does not necessarily impact any other car crashes. This assumption may be unrealistic.

b. P(X ≥ 1) = 1 – P(X = 0) = 1 − .13534 = .86466c. P(X < 5) = P(X ≤ 4) = .94735d. Skewed to the right.

6.78* Binomial n = 2500, π = .001 or Poisson with λ = 2.5 leaks per 2500 meters. Using the Poisson distribution:a. P(X = 0) = 0.0828

b. P(X ≥ 3) = 1 – P(X ≤ 2) = 1 − .54381 = .45619c. Skewed right.

d. Skewness = 1 1 (2.5) .400λ = =e. n is too large to be convenient.

f. n ≥ 20 and π ≤ .05 so Poisson is accurate

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6.79* a. n = 200, π = .02. Define X to be the number of twin births in 200 deliveries. E(X) = (200)(.02) = 4.b. P(X = 0) = .01759c. P(X = 1) = .07326

d. Using the Poisson approximation to the Binomial with λ = 4: P(X = 0) = .01832 from = POISSON (0, 4, FALSE)

P(X = 1) = .07179 from = POISSON (1, 4, FALSE)

e. Yes, the approximation is justified. Our rule of thumb is n ≥ 20 and π ≤ .05 which is met here and the probabilities from the Poisson are similar to the binomial.

6.80* a. Binomial P(X = 0) = .00226, Poisson P(X = 0) = .00248b. Binomial P(X = 1) = .01399, Poisson P(X = 1) = .01487c. Binomial P(X = 2) = .04304, Poisson P(X = 2) = .04462

d. Set λ = nπ = (200)(.03) = 6.0

e. Yes, n ≥ 20 and π ≤ .05 and probabilities are similar.

6.81* a. For the binomial µ = nπ = (4386)(.00114) = 5 is the expected number killed.

b. For the binomial, (1 ) (4386)(.00114)(.99886) 2.235nσ = π − π = =

c. Using Poisson approximation with λ = 5.0, P(X < 5) = P(X ≤ 4)= .44049

d. P(X > 10) =1 − P(X ≤ 10) = 1 − .98631 = .01369

e. Yes, the approximation is justified. Our rule of thumb is n ≥ 20 and π ≤ .05 which is met.

6.82* a. nπ = (500)(.02) = 10.

b. Using the Poisson approximation with λ = nπ = (500)(.02) = 10 we get P(X ≤ 5) = .06709.

6.83 a. P(X = 5 | N = 52, s = 13, n = 5) = .000495.b. No, since n/N = 5/52 exceeds .05 (our rule of thumb for a binomial approximation.)

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6.84 a. Sampling without replacement, n/N < 0.05b. Range of X is 0 to 2.c. See the table below.

6.85* a. Geometric mean is 1/π = 1/(.08) = 12.5 cars

b. Geometric std. dev. is 2 2(1 ) / (.92) /(.08) 11.99− π π = = cars

c. Using geometric CDF, P(X ≤ 5) = 1−(1−π)x = 1−(1−.08)5 = .3409

6.86* a. Geometric mean is 1/ π = 1/(.07) = 14.29 operations

b. Using geometric CDF, P(X ≥ 20) = 1 − P(X ≤ 19) = 1 − [1−(1−π)x] = (1−π)x = (1−.07)19 = .2519

6.87* a. Geometric mean is 1/ π = 1/(.05) = 20

b. Using geometric CDF, P(X ≤ 29) = 1− (1−π)x = 1 − (1−.05)29 = 1 − .2259 = .7741

6.88 a. 1/ π = 1/(.02) = 50

b. 2 2(1 ) (.98) (.02) 49.5− π π = =

c. Would have to examine a large number to find first check for abnormality. Since most would be OK, it would be easy to lose concentration. Same applies to airport security inspectors.

6.89 The total number of values in the uniform distribution is n = b−a+1. Since P(x) = 1/(b−a+1) is a constant for

all x, the sum is simply that constant multiplied by n or (b−a+1)/ b−a+1) = 1.

6.90 a. ( ) / 2 (0 9999) / 2 4999.5a bµ = + = + =2 2( 1) 1 (9999 0 1) 1

2886.812 12

b a− + − − + −σ = = =

6.91 a.. (233.1)(0.4536)= 105.734 is the mean in kilogramsb. (34.95)(0.4536) = 15.8533 is the std dev in kilogramsc. Rule 1 for the mean and Rule 2 for the std dev.

6.92 a. By Rule 1, expected total cost is µvQ+F = vµQ+F = (8)(25000) + 15000 = $350,000

By Rule 2, std dev. of total cost is σvQ+F = vσQ = (8)(2000) = $16,000

b. To break even, we want TR − TC = 0 where TR = expected total revenue and TC = expected total cost.

Since TR = (Price)(Quantity) = PQ we set PQ − vQ+F = 0 and solve for P to get P(25000) − 350000 = 0

or P = $14. For a profit of $20,000 we have P(25000) − 370000 = 0 or P = $14.80.

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6.93* a. Using Rule 3: µX+Y = µX + µY = 70+80 = 150

b. Using Rule 4: σX+Y = 2 2 64 36 10X Yσ + σ = + =c. Rule 4 assumes independent test scores. Most likely these variables are not independent. The score the

student got on the first exam may influence the score on the second exam (i.e. studied more, attended class more frequently, sought tutoring).

6.94* Using Rule 3: µX+Y = µX + µY = 20+10+14+6+48 = 98 hours

Using Rule 4: σX+Y = 2 2 16 4 9 4 36 8.31X Yσ + σ = + + + + = (assuming independent steps)

2-sigma interval around the mean µ ± 2σ or 98 ± (2)(8.31). The range is 81.4 to 114.6 hours.

6.95 a. By Rule 1, mean of total cost: µvQ+F = vµQ+F = (2225)(7) + 500 = $16,075

By Rule 2, std dev. of total cost: σvQ+F = vσQ = (2225)(2) = $4,450

By Rule 1, expected revenue is E(PQ) = PµQ = (2850)(7) = $19,950Expected profit is TR – TC = 19,950 – 16,075 = $3,875

6.96* Adding 50 will raise the mean by 50 using Rule 1: µaX + b = aµX + b = (1)(25) + 50 = 75. Multiplying by 3 will

also raise the mean by 50 using Rule 1: µaX + b = aµX + b = (3)(25) + 0 = 75. The first transformation will shift

the distribution to the right without affecting the standard deviation by Rule 2: σaX + b = aσX = (1)(6) = 6. The second transformation will spread out the distribution, since the standard deviation will also increase using

Rule 2: σaX + b = aσX = (3)(6) = 18, and some scores will exceed 100.

6.97* a. This is a binomial with µ = nπ= (.25)(250) = 60

b This is a binomial with σ2 = nπ(1−π) = (240)(.25)(.75) = 45 so σ = 6.7082

c. µ ± 1σ is 60 ± (1)(6.7082) or 53.3 days to 66.7 days

µ ± 2σ is 60 ± (2)(6.7082) or 46.6 days to 73.4 days

These intervals contain about 68% and 95% of the X values if the shape of the binomial is approximately normal. In this case, that is true, as you can see by printing the binomial PDF.

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Chapter 7

Continuous Distributions

7.1 a. Db. Cc. C

7.2 a. Cb. Dc. C

7.3 In order to be a valid PDF, total area under f(x) must equal 1. a. Area = .25(1) = .25 therefore this is not a PDF.b. This is a valid PDF.c. Area = ½(2)(2) = 2 therefore it is not a PDF.

7.4 For a continuous PDF, we use the area under the curve to measure the probability. The area above a single point is defined to be zero so if we summed up all the point probabilities we would have a sum equal to zero. In addition, by definition there are an infinite number of points in the interval over which a continuous random variable is defined.

7.5

a. µ= (0+10)/2 =5 σ = 2(10 0)

12

− = 2.886751

b. µ= (200+100)/2 = 150 σ = 2(200 100)

12

−= 28.86751

c. µ= (1+99)/2= 50 σ = 2(99 1)

12

−= 28.29016

7.6 a. P(X < 10) for U(0,50) = (10-0)/(50-0) = 0.2

b. P(X > 500) for U(0,1000) = (1000-500)/(1000-0) = 0.5

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c. P(25 < X < 45) for U(15,65) = (45-25)/(65-15) = .4

.7.7 P(X=25) = 0 for a continuous uniform distribution. Therefore using a < or ≤ yields the same result.

7.8 a. µ= (2500+4500)/2 = 3500

b. σ = 2(4500 3500)

12

− = 577.3503

c. The first quartile is the midpoint between a and the median: (3500+2500)/2 = 3000. The third quartile is the midpoint between the median and b: (4500+3500)/2 = 4000.

d. P(X < 3000) = P(2500 < X < 3000) = for U(2500,4500) =(3000-2500)/(4500-2500) =0.25.e. P(X > 4000) = P(4000 < X <4500) = for U(2500,4500) = (4500-4000)/(4500-2500) = 0.25.f. P(3000< X < 4000) = for U(2500,4500) =(4000-3000)/(4500-2500) =0.50.

7.9 The curves differ by their mean, standard deviation, and height.

7.10 a. The maximum height is 0.0798. (Plug µ = 75 and σ =5 into the PDF.) b. No, f(x) does not touch the X axis at any point. The distribution is asymptotic.

7.11 It says that for data from a normal distribution we expect

about 68.26% will lie within µ ± 1 σabout 95.44% will lie within µ ± 2 σabout 99.73% will lie within µ ± 3 σ

7.12 a. Yesb. No, distribution could be skewed. Direction of skewness depends on how one defines years of education

and which geographic region one is interested in.c. No, distribution could be skewed right. Most bills will be delivered within a week but there may be a few

that take much longer.d. Yes, but there could be outliers.

7.13 a. .1915.

b. .1915. c. .5000.d. 0

7.14 a. P(Z<2.15) – P(Z<1.22) = .9842 − .8888 = .0945

b. P(Z<3.00) – P(Z<2.00) = .99865 − .9772 = .02145

c. P (Z<2.00) – P(Z<−2.00) = .9772 − .0228 = .9544

d. 1 – P(Z<.50) = 1 − .6915 = .3085

7.15 a. P(Z<2.15) – P(Z<−1.22) = .9842 − .1112 = .8730

b. P(Z<2.00) – P(Z<−3.00) = .9772 − .00135 = .97585

c. P(Z < 2.00) = .9772d. P(Z = 0) = 0.

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7.16 a. NORMDIST(232000,232000,7000,TRUE) = 0.50b. NORMDIST(239000,232000,7000,TRUE) − NORMDIST(232000,232000,7000,TRUE) = 0.341345c. NORMDIST(239000,232000,7000,TRUE) = 0.841345d. NORMDIST(245000,232000,7000,TRUE) = 0.968355e. 1 − NORMDIST(225000,232000,7000,TRUE) = 0.84134474

7.17 a. NORMDIST(300,290,14,TRUE) = 0.762475b. 1 − NORMDIST(250,290,14,TRUE) = 0.997863c. NORMDIST(310,290,14,TRUE) − NORMDIST(275,290,14,TRUE) = 0.781448

7.18 Use Excel’s NORMINV function. NORMINV(0.975,3.3,.13) = 3.554795, NORMINV(.025,3.3,.13) = 3.045. The middle 95% is in the interval 3.045 to 3.555.

b. NORMDIST(3.50,3.3,.13,TRUE) = 0.061967919

7.19 Use Excel’s NORMINV function to give the X value associated with the cumulative probability.a. NORMINV(.9,10,3) =13.84465582b. NORMINV(.5,10,3) = 10c. NORMINV(.95,10,3) = 14.93456043d. NORMINV(.2,10,3) = 7.475136874e. NORMINV(.1,10,3) = 6.155344182f. NORMINV(.25,10,3), NORMINV(.75,10,3) = 7.976531423, 12.02346858g. NORMINV(.93,10,3) = 14.42737385h. NORMINV(.025,10,3), NORMINV(.975,10,3) = 4.120111638, 5.87988836i. NORMINV(.07,10,3) = 5.572626152

7.20 Use Excel’s NORMINV function to give the X value associated with the cumulative probability.a. NORMINV(.9,360,9) = 371.5339675b. NORMINV(.5,360,9) = 360c. NORMINV(.95,360,9) = 374.8036813d. NORMINV(.2,360,9) = 352.4254106e. NORMINV(.1,360,9) = 348.4660325f. NORMINV(.25,360,9), NORMINV(.75,360,9) = 353.9295943, 366.0704057g. NORMINV(.9,360,9) = 371.5339675h. NORMINV(.025,360,9), NORMINV(.975,360,9) = 342.3603349, 377.6396651i. NORMINV(.96,360,9) = 375.7561746

7.21 a. P(X ≥ 8) = NORMDIST(8,6.9,1.2,TRUE) = 0.179659. This probability indicates that the event is not common but not unlikely.

b. NORMINV(.9,6.9,1.2) = 8.437862 poundsc. 95% of birth weights would be between 4.5 and 9.3 pounds. NORMINV(0.025,6.9,1.2),

NORMINV(.975,6.9,1.2) = 4.548045, 9.251955

7.22 a. NORMINV(.95,600,100) = 764.4853476b. NORMINV(.25,600,100) = 532.5510474c. NORMINV(0.1,600,100) , NORMINV(0.9,600,100) = 471.8448, 728.1551939

7.23 a. NORMDIST(110,100,15,TRUE) = 0.747507533b. NORMDIST(2,0,1,TRUE) = 0.977249938c. NORMDIST(5000,6000,1000,TRUE) = 0.15865526d. NORMDIST(450,600,100,TRUE) = 0.066807229

7.24 a. NORMDIST(110,100,15,TRUE) − NORMDIST(80,100,15,TRUE) = .6563b. NORMDIST(2,0,1,TRUE) − NORMDIST(1.5,0,1,TRUE) = .0441c. NORMDIST(7000,6000,1000,TRUE) − NORMDIST(4500,6000,1000,TRUE) = .7745d. NORMDIST(450,600,100,TRUE) − NORMDIST(225,600,100,TRUE) = .0667

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7.25 a. NORMINV(.1,360,9) = 348.4660325b. NORMINV(.32,360,9) = 355.7907105c. NORMINV(.75,360,9) = 366.0704057d. NORMINV(.9,360,9) = 371.5339675e. NORMINV(.999,360,9) = 387.8122732f. NORMINV(.9999,360,9) = 393.4718124

7.26 a. 1 − NORMDIST(60,40,28,TRUE) = 0.2375b. NORMDIST(20,40,28,TRUE) = 0.2375c. 1 − NORMDIST(10,40,28,TRUE) = 0.8580

7.27* The Excel formula could be: NORMSINV(RAND()).

7.28* The Excel formula could be: NORMINV(RAND(),4000,200).

Note: The probabilities below were calculated using Appendix C-2.

7.29* nπ ≥ 5 and n(1-π) ≥ 5 so we can use the normal approximation to the binomial.

µ = nπ = 70, σ = (1 )nπ π− = 8.07

a. P(X < 50) ≈ P(X ≤ 49.5) = P(Z < −2.54) = .0055 b. P(X > 100) ≈ P(X ≥ 100.5) = P(Z > 3.78) = .00008

7.30* nπ ≥ 5 and n(1-π) ≥ 5 (800*.03 = 24, 800*.97 = 776) so we can use the normal approximation.

a. µ = 24 , σ = 4.8249b. P(X ≥ 20) ≈ P(X ≥ 19.5) = .8238c. P(X > 30) ≈ P(X ≥ 30.5) = .0885

7.31* nπ ≥ 5 and n(1-π) ≥ 5 (200*.90 = 180, 200*.1 = 20) so we can use the normal approximation.

µ = 180, σ = 4.2426a. P(X ≥ 175) ≈ P(X ≥ 174.5) = .9032b. P(X < 190) ≈ P(X ≤ 189.5) = .9875

7.32* nπ ≥ 5 and n(1-π) ≥ 5 (8465*.048 = 406.32, 8465*.952 = 776) so we can use the normal approximation.

a. nπ = 406.32

b. P(X ≥ 400) ≈ P(X ≥ 399.5) = .6368c. P(X < 450) ≈ P(X ≤ 449.5) = .9861

7.33* Let µ = λ = 28, σ = λ = 5.29.

a. P(X > 35) ≈ P(X ≥ 35.5) =.0778b. P(X < 25) ≈ P(X ≤ 24.5) = .2546

c. λ ≥ 20 therefore the normal approximation is appropriate.d. .0823 and .3272. Not as close as one might wish.

7.34* Let µ = λ = 150, σ = λ = 12.25

a. P(X ≥ 175) ≈ P(X ≥ 174.5) = .0228b. P(X < 125) ≈ P(X ≤ 124.5) = .0188

c. λ ≥ 20 therefore the normal approximation is appropriate.d. From Excel: 1 − POISSON(174, 150,1) = .0248, POISSON(124, 150,1) = .01652. The probabilities are

fairly close.

7.35 a. P(X > 7) = .1225b. P(X < 2) = .4512

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7.36 a. P(X > 30 minutes) = .1225b. P(X < 15 minutes) = .6501c. P(15 < X < 30) = .8775 − .6501 = .2274

7.37 a. P(X < 60 seconds) = .9975b. P(X > 30 seconds) = .0498c. P(X > 45 seconds) = .0111

7.38 There are 26,280 hours in 3 years. A warranty claim can be filed if the hard drive fails within the first three

years. P(X < 26280) = 1−e−λx =1−e−(1/250000)(26280) = 1−e-0.10512 = 1−.9002 = .0998

7.39 a. P(X > t) = .5. To solve for t: −ln(.5)/4.2 = .1650 hours.b. −ln(.25)/4.2 = .3301 hoursc. −ln(.1)/4.2 = .5482 hours

7.40 a. P(X > t) = .5. To solve for t: −ln(.5)/.5 = 1.3863 minutesb. −ln(.75)/.5 = .5754 minutesc. −ln(.7)/.5 = .7134

7.41 a. P(X > t) = .5. Use λ = 1/20 = .05. To solve for t: −ln(.5)/.05 = 13.86 minutes.b. The distribution on time is skewed to the right therefore the median < mean.c. −ln(.25)/.05 = 27.73 minutes.

7.42 a. P(X > t) = .5. To solve for t: −ln(.9)/.125 = .843 years.b. −ln(.8)/.125 = 1.785 years

7.43* a. µ = (0+25+75)/3 =33.33

b. σ = 15.59c. P(X < 25) = .3333d. Shaded area represents the probability.

7.44* a. µ = (50+105+65)/3 = 73.33

b. σ = 11.61c. P(X > 75) = .4091d. Shaded area represents the probability.

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7.45 a. Db. Cc. C

7.46 a. Area = .5(2) = 1 therefore this is a valid PDF.b. Area = ½(2)(2) = 2 therefore this is not a valid PDF.c. Area = ½(.5)(2)(2) = 1 therefore this is a valid PDF.

7.47 a. µ = 45.

b. σ = 11.547c. P(X > 45) = (65-45)/(65-25) = 0.5d. P(X > 55) = (65-55)/(65-25) = 0.25e. P(30< X <65) = (60-30)/(65-25) = 0.75

7.48 a. µ = 27.5.

b. σ = 12.99c. Q1 = 16.25, Q3 = 38.75d. P(re-swiping) = P(10 < X < 40) = 1 − .6667 = .3333. 33.33% must re-swipe.

7.49 Answers will vary.a. Suggested response: Not normal, unknown shape.b. Would expect distribution to be skewed to the right.c. Normald. Normal

7.50 a. µ = 86.

b. σ = 0.0693c. P(X > .80) = (.98-.80)/(.98-.74) = .75d. P(X < 85) =(.85-.74)/(.98-.74) = .4583e. P(.8 < X <.9) = (.9-.8)/(.98-.74) = .4167f. The chlorine is added to kill bacteria.

7.51 a. NORMINV(.5,450,80) = 450

b. NORMINV(.25,450,80) = 396.041c. NORMINV(.9,450,80) = 552.524d. NORMINV(.2,450,80) = 382.670e. NORMINV(.95,450,80) = 581.588f. NORMINV(.25,450,80), NORMINV(.75,450,80) = 396.041, 503.960g. NORMINV(.2,450,80) =382.670h. NORMINV(.025,450,80), NORMINV(.975,450,80) = 293.203, 606.798i. NORMINV(0.99,450,80) = 636.108

7.52 a. The likelihood of a value greater than the mean is .50.b. This corresponds to P(Z > 1) = .1587c. This corresponds to P(Z > 2) = 0.02275d. This corresponds to P(-2 < Z <2) = .9545

7.53 a. P(X>130) = .2266b. P(X<100) = .2266c. P(X<91) = .1151

7.54 a. P(X<579) = .5b. P(X>590) = .2160c. P(X<600) = .9332

77

7.55* a. P(28<X<32) = .8413 − .1587 = .6826b. P(X<22.5) = 8.84E-05

7.56 a. 1-P(X<120) = .7881b. 1- P(X<180) = .0548c. NORMINV(.95, 140,25) =181.121d. NORMINV(.99,140,25) = 198.159

7.57 P(1.975<X<2.095) = .0455

7.58 a. P(X<40) = .0013b. P(X>55) = .1711c. Assumed a normal distribution.d. P(40<X<62.8) = .9973. .9973*73.1 million = 72.9 million.

7.59 a. P(X>50) = .2177b. P(X<29) = .2965c. P(40<X<50) = .2223d. Assumed that email consults had a symmetrical distribution that was bell shaped. Even though the number

of email consults is a discrete random variable it is possible to approximate with the normal distribution.

7.60 a. P(X>30) = .0038b. P(all three men finish in time) = (1-.0038)3 = .9886

7.61 P(X>90) = .2742. Assume that the distribution on time was normal.

7.62 The next procedure will be delayed if the tubal ligation runs 40 minutes or longer. P(X>40) = .1056

7.63 P(X>5200) = .2016

7.64 a. P(X< 135) = .3085b. P(X > 175) = .0668c. P(125 < X < 165) = .8413 − .1587 = .6826d. With variability, physicians run the risk of not treating a patient with dangerous blood pressure or treating a

patient with healthy blood pressure. Understanding variability allows physicians to minimize the chances of making these two types of errors.

7.65 a. John scored better than only 5.26% of the others.b. Mary scored above average, better than approximately 70% of others.c. Zak scored better than 96.33% of others.d. Frieda scored better than 99.34% of others.

7.66 a. False, the normal distribution is asymptotic. Thus, a value outside the given interval is possible.b. False, the standardized values do allow for meaningful comparison. Z scores are unit free.c. False, the normal distribution is a “family” of distributions, each having the same shape, but different

means and standard deviations.

7.67* a. For route A: P(X<54) = .5. For route B: P(X<54) = .0228. He should take route A.b. For route A: P(X<60) = .8413. For route B: P(X<60) = .5. He should take route A.c. For route A: P(X<66) = .9722. For route B: P(X<66) = .9772. He could take either route. Because the

standard deviation is smaller for route B, the chance of getting to the airport in under 66 minutes is the same for each route.

78

7.68 a. Underfilling the bottle means putting less than 500 ml in the bottle. Find the value of µ for which P(X >

500) = .95. This corresponds to a z = −1.645. Using the z score formula to solve for µ we find that the mean should be set at 508.225 ml.

b. To ensure that 99% contain at least 500 ml, set the mean at 511.63.c. To ensure that 99.9% contain at least 500 ml, set the meant at 515.45.

7.69 Find the value for X such that P(X > x) = .80. This corresponds to a z = −.842. Using the z score formula to solve for x we find that the minimum length should be 11.49 inches.

7.70* a. For method A: P(X<28) = .5. For method B: P(X<28) = .0228. Method A is preferred.b. For method A: P(X<38) = .9938. For method B: P(X<38) = .9987. Method B is preferred.c. For method A: P(X<66) = .9722. For method B: P(X<66) = .9772. Either method is acceptable.

7.71* a. P(X > µ) = .5 (property of the normal distribution). Assuming independence, the probability that both exceed the mean is: .5*.5 =.25.

b. P(X < µ) = .5 (property of the normal distribution). Assuming independence, the probability that both are less than the mean is: .5*.5 =.25.

c. P(X<µ) = .5 (property of the normal distribution). Assuming independence, the probability that one is greater than and one is less than the mean is: .5*.5 =.25. There are two combinations that yield this, so the likelihood is: .25+.25 = .50 that one exceeds the mean and one is less than the mean.

d. P(X = µ) = 0, this is a property of a continuous random variable. The probability that both equal the mean is zero.

7.72* Use the normal approximation of the binomial distribution because we clearly meet the requirement that nπ ≥ 5 and n(1-π) ≥ 5.

a. P(X ≥ 50) ≈ P(X ≥ 49.5) = .0646.b. P(X < 35) ≈ P(X ≤ 34.5) = .1894.

7.73* P(X < 20) ≈ P(X ≤ 19.5) = .102

7.74* Use the normal approximation of the binomial distribution because we clearly meet the requirement that nπ ≥ 5 and n(1-π) ≥ 5. Q1 = 16,966, Q3 = 17,034. Use the NORMINV function in Excel with µ = 17000 and

σ = 50.4975.

7.75* Use the normal approximation of the binomial distribution because we clearly meet the requirement that ≥ 5

and n(1-π) ≥ 5. nπ = 100∗.25 = 25, n(1-π) = 100*.75 = 75. a. Find the value of X such that P(X ≥ x) = .05. This corresponds to a z score equal to 1.645. Using the z score

formula and solving for x we find that the minimum score should be 33.12.b. Find the value of X such that P(X ≥ x) = .01. This corresponds to a z score equal to 2.326. Using the z score

formula and solving for x we find that the minimum score should be 35.07.c. Q1 = 22.08, Q3 = 25.

7.76* Use the normal approximation of the binomial distribution because we clearly meet the requirement that nπ ≥ 5 and n(1-π) ≥ 5. nπ = 200∗.8 =160, n(1-π) = 200*.2 = 40.

a. P(X < 150) ≈ P(X ≤ 149.5) = .0317.b. P(X ≥ 150) ≈ P(X ≥ 149.5) = 1 − .0317 = .9683.

7.77* Use µ = 30 and σ = 5.422.a. P(X ≥ 25) ≈ P(X ≥ 24.5) = .8448b. P(X > 40) ≈ P(X ≥ 40.5) = .0264

7.78* Converting the rate from days to years, λ = 73. Let µ = λ = 73 and σ = λ = 8.544. It is appropriate to use

the normal approximation given that λ ≥ 30.P(X < 60) ≈ P(X ≤ 59.5) = 0570.

79

7.79 a. P(X < 10,000) = 1−e−λx =1−e−(1/10000)(10000) = 0.632120559b. The distribution is skewed to the right therefore the mean is greater than the median. The probability

calculated above makes sense.

7.80 a. P(X>6) = e−(.1*6) = .5488

b. P(X>12) = e−(.1*12) = .3012

c. P(X>24) = e−(.1*24) = .0907d. P(6<X<12) = (1 − .3012) − (1 − .5488) = .2476

7.81 a. P(X > 15,000) = e−λx =e−(1/25000)(15000) = 0.1889b. Ten years is 87600 hours. If the airplane is flown 25% of the time then that would be 21,900 hours of use.

Find P(X<21900) = 1−e−λx =1−e−(1/25000)(21900) = .5836.

7.82 a. P(X<50,000) = e−λx =e−(1/16667)(50000) = .0498 P(X<50,000) = e−λx =e−(1/66667)(50000) = .472b. There has been approximately a tenfold increase in the reliability of the engine between 1982 and 1992.

7.83 a. µ = (300 + 350 + 490)/3 = 380.

b. σ = 2 2 2300 350 490 300*350 300*490 350*490

40.2118

+ + − − −=

c. P(X > 400) =2(490 400)

.3045(490 300)(490 350)

−=

− −.

7.84* a. µ = (50 + 95 + 60)/3 = 68.33.

b. σ = 2 2 250 60 95 50*60 50*95 60*95

9.6518

+ + − − −=

c. P(X <75) = 2(95 75)

1 .7460(96 50)(95 60)

−− =

− −.

d.

7.85* a. µ = (500 + 700 + 2100)/3 = 1100.

b. σ = 2 2 2500 700 2100 500*700 500*2100 700*2100

355.9018

+ + − − −=

c. P(X > 750) = 2(2100 750)

.8136(2100 500)(2100 700)

−=

− −.

80

d.

7.86* a. The z scores were, respectively, 5.75, 4.55, 5.55, and 5.45. b. If the exams scores had a historical mean and standard deviation of 80 and 20 with a normal distribution

then the exam scores reported by the four officers were highly unlikely.

81

Chapter 8

Sampling Distributions and Estimation

8.1 a. n

σ =

32

4= 16

b. n

σ =

32

16 = 8

c. n

σ =

32

64 = 4

8.2 a. 1.96n

σµ ± =

12200 1.960

36± or (196.08, 203.92).

b. 1.96n

σµ ± =

151000 1.960

9± or (990.2, 1009.80).

c. 1.96n

σµ ± =

150 1.960

25± or (49.608, 50.392).

8.3 a. 1.96µ σ± = 4.035 1.96 *0.005± or (4.0252, 4.0448).

b. 1.96n

σµ ± =

0.0054.035 1.96

25± or (4.03304, 4.03696).

c. In either case, we would conclude that our sample came from a population that did not have a population mean equal to 4.035.

8.4 a. 1. No, for n = 1 the 100 samples don’t represent a normal distribution 2.The distribution of the sample means becomes more normally distributed as n increases. 3. The standard error becomes closer to that predicted by the CLT the larger the sample becomes. 4. This demonstration reveals that if numerous samples were taken and analyzed we can confirm the CLT. In the real word, based on our notion of the true mean, we can assess this. We can generate the 95% range and determine if our values are within this range or not. Also, recognize that there is a low probability of this single range not being representative.

b. 1. No, for n = 1 the 100 samples don’t represent a normal distribution 2.The distribution of the sample means becomes more normally distributed as n increases. The standard error becomes closer to that predicted by the CLT the larger the sample becomes. 4. This demonstration reveals that if numerous samples taken and analyzed we can confirm the CLT. In the real word, based on our notion of the true mean, we can assess this. We can generate the 95% range and determine if our values are within this range or not. Also, recognize that there is a low probability of this single range not being representative.

8.5 a. 1.645n

σµ ± =

414 1.645

5± or (11.057, 16.943).

b. 2.576n

σµ ± =

537 2.576

15± or (33.675, 40.325).

82

c. 1.96n

σµ ± =

15121 1.96

25± or (115.12, 126.88).

8.6 Exam 1: 1.96n

σµ ± =

775 1.96

10± or (70.661, 79.339).

Exam 2: 1.96n

σµ ± =

779 1.96

10± or (74.661, 83.339).

Exam 3: 1.96n

σµ ± =

765 1.96

10± or (60.661, 69.339)

The confidence intervals overlap. This suggests that all three exams had the same population mean.

8.7 1.96n

σµ ± =

0.0052.475 1.96

15± or (2.4725, 2.4775).

8.8 a. s

tn

µ ± = 3

24 1.94327

± or (21.797, 26.203).

b. s

tn

µ ± =6

42 2.898218

± or (37.901, 46.099).

c. s

tn

µ ± = 14

119 2.051828

± or (113.571, 124.429).

Note: t values are found using the Excel formula =tinv( ((1−cc)),n−1) where cc is the confidence coefficient. For a. this, this would be = tinv(((1-.90)), 6)

8.9 a. Appendix D = 2.262, Excel = tinv(.05, 9) = 2.2622b. Appendix D = 2.602, Excel = tinv(.02, 15) = 2.6025c. Appendix D = 1.678 ,Excel = tinv(.10, 47) =1.6779

8.10 a. Appendix D = 2.021, Excel = tinv(.05, 40) = 2.0211b. Appendix D = 1.990, Excel = tinv(.05, 80) = 1.9901c. Appendix D = 1.984, Excel = tinv(.05, 100) = 1.984All are fairly close to 1.96.

8.11 a. s

tn

µ ± = 27.79

45.66 2.086021

± or (33.01, 58.31).

b. The confidence interval could be narrower increasing the size of the sample or decreasing the confidence level.

8.12 a. s

tn

µ ± = 3.649

19.88 1.75316

± or (18.276, 21.474).

Note: t values are found using the Excel formula =tinv( ((1-cc)/2),n-1) where cc is the confidence coefficient. For a. this, this would be = tinv(((1-.90)/2), 15)

8.13s

tn

µ ± =78.407

812.5 3.105812

± or (742.20, 882.80).

83

8.14 a. s

tn

µ ± =17541.81

24520 2.262210

± or (11971, 37069).

b. Increase the sample size or decrease the confidence level.c. It is unclear whether this distribution is normal or not. There appear to be outliers.

8.15 1. s

tn

µ ± =4.3716

85 2.262210

± or (81.873, 88.127).

2. s

tn

µ ± =8.127

88.6 2.262210

± or (82.787, 94.414).

3. s

tn

µ ± =3.712

76 2.262210

± or (73.345, 78.655).

b. Confidence intervals 1 and 2 overlap. The scores on exam 3 are very different than the first two. There was a decrease in the average exam score on the third exam.

c. Here the standard deviation is not known, so use the t-distribution.

8.16 a. (1 )nπ − π = .5*30*.5 = 2.739

b. (1 )nπ − π = .2 *50 *.8 = 2.828

c. (1 )nπ − π = .1*100 *.9 = 0.030

d. (1 )nπ − π = .005*500 *.995 = .0032

Normality okay except in d.

8.17 a. 1.96 ( (1 )) / nπ − π = 1.96 (.5(1 .5)) / 250− = .0620

b. 1.96 ( (1 )) / nπ − π = 1.96 (.5(1 .5)) /125− = .0877

c. 1.96 ( (1 )) / nπ − π = 1.96 (.5(1 .5)) / 65− = .1216

8.18*(1 )p p

p zn

−± =

.048*(1 .048).048 1.96

500

−± : .0293, .0667

b. Yes, .048*500 = 24 which is larger than 10.c. The Very Quick Rule would not work well here because p is small.

8.19 a. *(1 )p p

p zn

−± =

.3654*(1 .3654).3654 1.96

52

−± or (.2556, 4752)

b. .3654*52 = 19. Normality assumption is met.

8.20 a. *(1 )p p

p zn

−± =

.419*(1 .419).419 1.645

43

−± or (.2948, .5424)

b. Given np = 18 we can assume normality.

8.21 a. *(1 )p p

p zn

−± =

.046(1 .046).046 1.96

250

−± or (.1507, .3199)

b. Given np = 11.5 we can assume normality.

8.22 a. *(1 )p p

p zn

−± =

.48*(1 .48).48 2.326

50

−± : .3157, .6443

b. Yes, np = 24 which is greater than 10.

84

8.23 a. *(1 )p p

p zn

−± =

.2352*(1 .2353).2353 2.326

136

−± : .1507, .3199

b. Yes, np = 32 which is greater than 10.

c.

2

(1 )z

nE

π π = −

=

21.645

.2353(1 .2353).06

−

= 136

2

(1 )z

nE

π π = −

=

21.96

.2353(1 .2353).03

−

= 768

d. When the desired error decreases and the desired confidence increases, the sample size must increase.

8.24 Using MegaStat:a. 55b. 217c. 865

8.25 Using MegaStat: 25

8.26 Assume a normal distribution. Solve for sigma using σ = (28-20)/4 = 2. From Megastat: n = 11.

8.27 Assume a normal distribution. Solve for sigma using σ = (200-100)/4 = 25. From Megastat: n = 97.

8.28 Assume a Poisson distribution. σ = µ = 2.1213. From Megastat: n = 98.

8.29 a.

2z

nE

σ =

=

21.96*86.75

25

= 47

b. We assumed normality and estimated σ = (3450-3103)/2 = 86.75

8.30 a. Using Megastat: 2165. We use π =.5.b. Sampling method: Perhaps a poll via the Internet.

8.31 a. Using Megastat: 1692. We use π =.5. b. Sampling method: Mailed survey.

8.32 a. Using Megastat: 601. We use π =.5. b. Sampling method: Direct Observation.

8.33 a. Using Megastat: 2401. We use π =.5. b. Sampling method: Random sample via telephone or Internet survey.

8.34 a. From Megastat: (−1,243.75, −86.23)b. (−1,268.2, −61.80) Intervals are similar and show that the true mean difference is less than zero.

c. µ1: (740, 1,462) µ2: (1,246, 2,286) Yes, the intervals overlap.d. According to answers from parts a and b, one would conclude that there is a difference in means.

8.35* a. From Megastat: (−1.16302, 0.830302)b. (−1.15081, 0.79081) Intervals are similar. Assumptions about variances did not make a big difference.

c. µ1: (7.95, 9.33) µ2: (8.1, 9.54) Yes, the intervals overlap.d. From all calculations it appears that there is not a significant difference in the two means.

8.36* From Megastat and assuming equal variances, (−181.84, −16.50). Conclude that undergrads tend to pay less.

85

8.37* From Megastat: (−.1584, .1184). Because zero falls in the interval, we cannot conclude there is a difference in proportions.

86

8.38* From Megastat: (.1352, .6148). Because the interval is greater than zero, we can conclude that π1 > π2.

8.39* From Megastat: (.0063, .1937). Because the interval is greater than zero, we can conclude that π1 > π2.

8.40* a. From Megastat: (53.60, 248.72) b. (81.08, 323.63)

8.41 From Megastat: (1.01338, 1.94627) )Take the square root of the lower and upper CI values given to get the CI for the standard deviation of the

population.)

8.42* From Megastat: (0.731438, 2.321751)

8.43 From Megastat: (5.882, 10.914) (Take the square root of the lower and upper CI values given to get the CI for the standard deviation of the

population.)

8.44 Students should observe that raw data histograms will show a uniform distribution whereas the histogram of sample means shows a shape less uniform and closer to a normal distribution. The average of the raw data will be equal to the average of the sample means. The population mean is 49.5 so we would expect our data set to have a value close to this. The population standard deviation is 28.58 so we would expect the raw data to have a sample standard deviation close to this. We would also expect the standard deviation of the means to be close to 14.29 (28.48/sqrt(4)). The point of this exercise is to observe that the average of sample means is close to the population mean but the standard deviation of sample means is smaller.

8.45 a. Because the diameter is continuous there will always be slight variation in values from nickel to nickel.b. From Megastat: (.8332, .8355)c. The t distribution assumes a normal population, but in practice, this assumption can be relaxed, as long as

the population is not badly skewed. We assume that here.

d. Use

2z

nE

σ =

to estimate the sample size. z = 2.577.and so n = 95

8.46 a. From Megastat: (3.2283, 3.3813)

b. Use

2z

nE

σ =

to estimate the sample size. z = 1.645 and so n = 53.

c. The flow of the mixture might be one factor. There are many other possibilities.

8.47 a. From Megastat: (29.4427, 39.6342)b. The cups may be different size. Having different people take the sample and count can cause lack of

consistency in sampling process. Different boxes may have different numbers of raisin. c. Producer most likely regulates raisins by weight, not count.d. A quality control system would increase consistency by monitoring system so producer knows what is

expected from their process and how their process varies. Then producer can work on minimizing variation by eliminating causes of variation.

8.48 a. From Megastat: (266.76, 426.24)

b. Zero values suggest a left skewed distribution.

c. Use

2z

nE

σ =

with z = 2.577 to get n = 1927.

d. Because n > N, increase the desired error. For example, if E = 50 then n = 78.

87

8.49 a. From Megastat: (19.249, 20.689)b. Fuel economy can also vary due to tire pressure and weather. There may be more than sampling variability

contributing to differences in sample means.

8.50 a. From Megastat: (7.292, 9.98)b. There are possible outliers that make the normal distribution questionable.

c. Use

2z

nE

σ =

with z = 2.326 to get n = 38.

8.51 a. From Megastat: (33.013, 58.315) b. With repair costs it is possible the distribution is skewed to the right. Also, the population size is small

relative to the sample size which might cause problems.

c. Use

2z

nE

σ =

with z = 1.96 to get n = 119.

d. From Megastat: (21.26, 40.14)

8.52 a. From Megastat: (3,230, 3,326)

b. Use

2z

nE

σ =

with z = 1.96 to get n = 91.

c. The line chart shows a decrease in the number of steps over time.

8.53 a. From Megastat: (29.078, 29.982) b. Normality is a common distribution for height but at younger ages it is possible to see high outliers.

c. Use

2z

nE

σ =

with z = 1.96 to get n = 116.

8.54 a. From Megastat: (74.02, 86.81)b. The sample is somewhat small and the length of the commercial could be a function of the type of time out

called.

8.55 a. From Megastat: (48.515, 56.965)b. The distribution is more likely to be skewed to the right with a few CDs having very long playing times.

c. Use

2z

nE

σ =

with z = 1.96 to get n = 75.

88

8.56 a. Estimated standard deviation using the uniform approximation using 2( )

12

b a−σ = and normal

approximation using 4

b a−σ = where b is the maximum and a is the minimum of the range.

UniformDistribution

NormalDistribution

Chromium

0.0635 0.055

Selenium

0.0004 0.00035

0.0043 0.00375

Fluoride

0.0289 0.0250

b. An estimate of the standard deviation is necessary to calculate the sample size needed for a desired error and confidence level.

8.57 a. From Megastat: (.125, .255)b. Normality can be assumed. np = 19.

c. Use

2

(1 )z

nE

π π = −

with z = 1.645 to get n = 463.

d. A quality control manager needs to understand that the sample proportion will usually be different from the population proportion but that the way the sample proportion varies is predictable.

8.58 a. From Megastat: (.039, .117)b. Different industries may have different quantities and types of records, especially if they have many

government contracts.

8.59 a. From Megastat: (.092, .134) b. Yes, np = 69.

c. Use

2

(1 )z

nE

π π = −

with z = 1.645 to get n = 677.

d. Yes, the results could be very different today. There is a stronger focus on nutrition today than there was 10-15 years ago.

8.60 a. From Megastat: (.176, .294)

b. Normality assumption holds. np = 47 and n(1−p) = 153.c. No, the Very Quick Rule suggests that p should be close to .5. In this example, p = .235.d. n = 304.e. Frequent sampling of the noodle mix would help the manufacturer identify problems and stay on target.

8.61 a. From Megastat: (.595, .733) b. No, viewers of the late night program are a unique group of television watchers compared to the rest of the

population. Not all TV watchers stay up late to watch the late night programs.

8.62 a. From Megastat: ((.012, .014)b. The sample size is large enough to make the normal assumption valid.

89

8.63 a. standard error = *(1 )p p

n

−= .0127

b. From Megastat: (.121, .171) c. No, np = 112.d. VQR: (.1103, .1821)

8.64 a. From Megastat: (.093, .130)b. Normality assumption holds.c. VQR: (.0753, .1472). This interval is wider than the one found in part a.d. This is one batch of popcorn kernels, not a random sample from the food producer. It is not clear that the

person taking the sample used random sampling techniques. We also do not know the age of the popcorn that was popped.

8.65 a. From Megastat: (.001, .002)b. No, we would like to know the proportion of all mosquitoes killed, not the proportion of mosquitoes killed

out of bugs killed.

8.66 a. From Megastat: (.616, .824) b. Yes, the sample size is large enough to use the normal assumption.c. Contacting the longshoremen’s union might help the sampling process.

8.67* With normality: (.011, .087). With binomial (.0181, .1031).

8.68* With normality (−.002, .402). With binomial (.0433, .4809). Normality is not justified because n is too small.

8.69* a. From Megastat: (.914, 1.006)b. The upper limit on the confidence interval is greater than 1.c. Normality is not justified therefore use the binomial approach.d. From MINITAB: (.8629, .9951)

8.70 a. margin of error = .5(1 .5)

1.96 .02052277

−=

b. From Megastat: (.423, .457)c. Because the interval falls below .5 we would conclude that it is unlikely 50% of the voters opposed the

signing.

8.71 a. Margin of error = .5(1 .5)

1.96 .04600

−= Assume that p = .5 and a 95% confidence level.

8.72* From Megastat: (10.82, 16.96)

8.73* From Megastat: (−2.51, .11). Because zero falls in the interval we cannot conclude that there is a difference in learning methods.

8.74* a. From Megastat: (.976, 2.13)b. Assume equal variances and normal populations.

8.75* From Megastat: (−.1063, .0081). Because zero falls in the interval we cannot conclude there is a difference in the two proportions.

8.76* From Megastat: (−.124, .027). Because zero falls in the interval we cannot conclude there is a difference in the two proportions.

90

Chapter 9

One-Sample Hypothesis Tests

9.1 Graphs should show a normal distribution with a mean of 80. a. Rejection region in the lower tail. b. Rejection region in both tails. c. Rejection region in the upper tail.

9.2 a. .05*1000 = 50 timesb. .01*1000 = 10 timesc. .001*1000 = 1 time

9.3 a. Null hypothesis the man is not having a heart attack. Type I error: I admit him when he does not have a heart attack. Type II error: I fail to admit him, he does have a heart attack, and dies. Better to make a Type I than Type II error, since a type II error is fatal.

b. Type I error: I reject the null and let them land even though they could have stayed up for 15 minutes (or more). Type II Don’t let the plane land and the plane runs out of fuel. It is more costly to make a Type II error.

c. Type I error: I reject the null and rush out to Staples, get caught in the snow and fail to finish the report (when if I had stayed I would have finished it). Type II error: I run out of ink and can’t finish the report. Better to stay and try to finish the report, in fact better to print out some of it than none of it.

9.4 Costly improvements may be too small to be noticed by customers and they may be unwilling to pay for the improvement.

9.5 a. Null hypothesis: Employee is not using illegal drugs.Alternative hypothesis: Employee is using illegal drugs.

b. Type I error: Test is positive for drugs when no drugs are being used by the individualType II error: Test is negative for drug use when the person is using drugs.

c. I might dismiss or discipline someone who is a non-drug user (Type I error). They could sue for wrongful damages. I might keep on someone who should be dismissed and they cause serious injury via a work related accident to themselves or others (Type II error). A Type II error could have more serious consequences than a Type I error.

9.6 a. Null hypothesis: There is no fire.Alternative hypothesis: There is a fire.

b. Type I error: A smoke detector sounds an alarm when there is no fire.Type II error: A smoke detector does not sound an alarm when there is a fire.

c. Consequence of making a type I error is that some guests will be inconvenienced by a false alarm and there is the cost of having the fire department summoned. Consequence of making a type II error is that the hotel will burn down and perhaps kill or injure many.

d. Reducing β risk would increase the likelihood making a Type I error, increasing the likelihood of a false alarm. Guests of the hotel and perhaps the fire department would be affected.

9.7 a.

.25 .22.0

.2*.8

100

z−

= =. From Appendix C: p-value = .046.

b. z = 1.90, p-value = .9713c. z = 1.14, p-value = .1271

91

9.8 a.

.7 .61.83

.6*.4

80

z−

= =. From Appendix C: p-value = .0339.

b. z = 2.07, p-value = .0384c. z = −2.33, p-value = .0098

9.9 a. .30*20 = 6 < 10 and .7*20 = 14 > 10 (cannot assume normality)b. .05*50 = 2.5 < 10 and .95*50 = 47.5 >10 (cannot assume normality)c. .10*400 = 40 > 10 and .9*400 = 360 > 10 (normality can be assumed)

9.10 H0: π = .10 versus H1: π > .10. z = 2.67, p-value = .0038 so reject H0 and conclude that the proportion who believe Pepsi is concerned with consumers’ health has increased.

9.11 a. H0: π = .997 versus H1: π < .997 Reject the null hypothesis if z < −1.645.z = −1.83 so reject H0.

b. Yes, .997*2880 = 2866 and .003*2880 = 14, both are greater than 10.c. A Type I error would sending back an acceptable shipment. This could be a problem if the hospital runs

low on insulin syringes. A Type II error would be accepting a bad shipment. The will be a problem if a defective syringe is used for an insulin injection.

d. p-value = .034

e. Reducing α would result in an increase in Type II errors. This type of error would most likely be the one

with the more severe consequence therefore reducing α is not a good idea.

9.12 a. H0: π ≤ .20 versus H1: π > .20 Reject the null hypothesis if z > 1.645. z = 2.52 so reject H0. The proportion of stores that sell cigarettes to under age teens is higher than the goal.

b. 95% CI: (.2084, .3041)

c. A two-tailed test uses α/2 to determine the critical value of the test statistic. A CI uses α/2 to determine the value of the test statistic used in the margin of error calculation. Because these are the same values, a CI can be used to make a conclusion about a two-tailed hypothesis test.

9.13 a. H0: π ≤ .50 versus H1: π > .50 Reject the null hypothesis if z > 1.645z = 2.0 so reject H0. The proportion of calls lasting more than 2 minutes is greater than .5.

b. p-value = .0228c. Yes, a difference of 12.5 % is important.

9.14 a. H0: π ≥ .052 versus H1: π < .052. Reject the null hypothesis if z < −2.33. z = −1.72 therefore fail to reject the null hypothesis.

b. p-value = .0427

c. nπ0 = 15.6 > 10 so normality is justified.

9.15 H0: π ≤ .50 versus H1: π > .50. Reject the null hypothesis if z > 1.645. z = 6.87 so reject H0.

9.16* a. Normality is not justified. nπ = 6 < 10.b. Using the normal assumption, z = 2.31 and the p-value = .0105.c. Using the binomial distribution P(X ≥ 10) = .0193. d. The normal probability is only an approximation to the binomial probability.

9.17* a. Using the binomial distribution, P(X ≥ 4 | n = 2000, π = .001) = .143. The standard is being met. b. Because there are only 4 defective items observed we cannot assume normality.

9.18* a. P(X ≥ 2 | n = 20, π = .0005) = .00005. The students are Julliard do differ from the national norm.b. Normality cannot be assumed because there were less than 10 students in the sample with perfect pitch.

92

9.19 a. z =

63 601.5

8

16

−=

, p-value = .1336

b. z = −2.0, p-value = .0228.c. z = 3.75, p-value = .0001.

9.20 a. p-value = P(Z > 1.34) = .0901b. p-value = P(Z < −2.07) = .0192c. p-value = 2*P(Z < −1.69) = .091

9.21 H0: µ = 2.035 oz versus H1: µ > 2.035 oz. z = 2.50 and p-value = .0062. Reject H0. The mean weight is heavier than it is supposed to be.

9.22 a. H0: µ ≤ 195 flights/hour versus H1: µ > 195 flights/hour. Reject H0 if z > 1.96.b. z = 2.11 so reject the null hypothesis and conclude that the average number of arrivals has increased. If

we had used an α = .01, we would have failed to reject the null hypothesis.c. We have assumed a normal population or at least one that is not badly skewed.

9.23 a. H0: µ =10 oz versus H1: µ ≠ 10 oz. Reject H0 if z > 1.96 or z < −1.96.b. z = 0.7835 so we fail to reject the null hypothesis (p = .4333). c. We assume the population is normally distributed.

9.24 a. Using Excel: TDIST(1.677, 12,1) = .0597, fail to reject H0 at α = .05.

b. TDIST(2.107,4,1) = .0514, fail to reject H0 at α = .05.

c. TDIST(1.865,33,2) = .0711, fail to reject H at α = .050.

9.25 a.

203 2001.5

8

16

t−

= =, p-value = .1544. Fail to reject the null hypothesis.

b. t = −2.0, p-value = .0285. Reject the null hypothesis.c. t = 3.75, p-value = .0003. Reject the null hypothesis.

9.26 a. H0: µ ≥ 530 bags/hour versus H1: µ < 530 bags/hour. Reject H0 if t < −1.753. t = −1.60 so fail to reject the null hypothesis.

b. When problems arise there could be an inordinately low number of bags processed. This would create a skewed distribution.

9.27 a. H0: µ ≥ 400 sf/gal versus H1: µ < 400 sf/gal. Reject H0 if t < −1.476. t = −1.98 therefore reject H0.

b. Yes, if α were less than or equal to .05 our decision would be different. c. p-value = .0525. Because .0525 < .10, we would reject the null hypothesis.d. A significant result in a hypothesis does not always translate to a practically important difference. In this

case, if the painter plans his or her paint purchase based on coverage of 400 square feet per gallon, but in reality the paint covers 5% less, the painter may run short on large jobs. A difference of 5% may not matter on a small job.

9.28 a. H0: µ ≥ 18 oz versus H1: µ < 18 oz. Reject H0 if t < −1.74. t = −2.28 therefore reject H0.

b. Yes, if α =.01, we would have failed to reject the null hypothesis.c. p-value = .018. Because the p-value < .05 we reject the null hypothesis.

9.29 a. H0: µ ≥ 19 minutes versus H1: µ < 19 minutes. Reject H0 if t < −1.729. t = −2.555 therefore reject H0.d. p-value = .0097. Because the p-value < .05 we reject the null hypothesis.

93

9.30 a. H0: µ ≤ 30,000 miles versus H1: µ > 30,000 miles. Reject H0 if t > 1.325. t = 1.53 therefore reject H0. This dealer shows a significantly greater mean number of miles than the national average for two year leases.

9.31 a. H0: µ = 3.25 versus H1: µ ≠ 3.25. Reject H0 if t > 2.11 or t <−2.11. t = 1.70 therefore fail to reject H0.b. The 95% confidence interval is: (3.2257, 3.4743). Because this interval does contain 3.25 we would fail

to reject the null hypothesis.c. We constructed the 95% confidence interval using the t statistic associated with .025 in the tail areas.

This is the same area we used to determine the critical value for the t statistic. A two-tailed test and a

confidence should always result in the same conclusion as long as α is the same for both.

9.32* a. Power = .4622b. Power = .7974c. Power = .9459


When you decrease α the power will decrease as shown above.


Using Learning Stats Excel file 09-08 PowerCurvesDIY.xls:

Results from Learning Stats:

.

94

BetaLeft

9.35 a. Power = .2595b. Power = .6388c. Power = .9123

Results from Learning Stats:

BetaLeft

9.36* H0: σ2 ≥ 24 versus H1: σ2 < 24. Reject the null hypothesis if χ2 > 16.92 or χ2 < 3.325. χ2 = 6.0 therefore we fail to reject H0.

9.37* H0: σ2 ≤ 1.21 versus H1: σ2 > 1.21. Reject the null hypothesis if χ2 > 28.87. χ2 = 29.16 therefore we reject H0.

9.38* H0: σ2 = 0.01 versus H1: σ2 ≠ 0.01. Reject the null hypothesis if χ2 > 27.49 or χ2 < 6.262. χ2 = 14.65 therefore we fail to reject H0.

9.39* H0: σ2 = 625 versus H1: σ2 ≠ 625. Reject the null hypothesis if χ2 > 21.92 or χ2 < 3.816. χ2 = 8.26 therefore we fail to reject H0.

9.40 a. P(Type II error) = 0.b. This is bad policy because the chance of making a Type I error is uncontrolled.

9.41 a. P(Type I error) = 0.b. This is bad policy because the chance of making a Type II error is uncontrolled.

9.42 a. H0: µ ≤ 90 versus H1: µ > 90.b. Type I error occurs when the physician concludes a patient has high blood pressure when they do not.

Type II error occurs when the physician concludes that a patient’s blood pressure if OK when it is too high.

c. A Type II error would have a more serious consequence. The patient could have severe health problems if high blood pressure is undiagnosed.

9.43 a. H0: User is authorized versus H1: User is unauthorized. b. Type I error occurs when the scanner fails to admit an authorized user. Type II error occurs when the

scanner admits an unauthorized user.c. A Type II error has a more serious consequence. Allowing entry to an unauthorized user could result in

damage to the plant or possibly even a terrorist attack.

95

9.44 P(Type II error) = 0. We’ve rejected the null hypothesis therefore it is impossible to make a Type II error.

9.45 P(Type I error) = 0. There can be no Type I error if we fail to reject the null hypothesis.

9.46 a. H0: A patient is does not have cancerous cells versus H1: A patient has cancerous cells. A false negative is a Type II error and means that the test shows no cancerous cells are present when in fact there are. A false positive is a Type I error and means that the test shows cancerous cells are present when they are not.

b. In this case “null” stands for absence.c. The patient bears the cost of a false negative. If their health problems are not diagnosed early they will

not seek treatment. The insurance company bears the costs of a false positive. Typically more tests will need to be done to check the results.

9.47 a. H0: A patient does not have an infected appendix versus H1: A patient does have an infected appendix. A Type I error occurs when a healthy appendix is removed. A Type II error occurs when an infected appendix goes undetected. The consequences of a Type I error include all the risks one is subjected to when undergoing surgery as well as the cost of an unnecessary operation. The consequences of a Type II error include a ruptured appendix which can cause serious health issues.

b. Type II error rates are high because diagnosing appendicitis is actually quite difficult. Type I error rates are high because the consequences of not removing an infected appendix are very serious.

9.48 a. Type I error: You should have been accepted, but the scanner rejected you. Type II error: You should

have been rejected, but the scanner accepted you.b. The consequence of falsely rejecting someone is not as severe as falsely accepting someone. Or it could

be that the scanner is dirty and cannot read the fingerprint accurately.

9.49 The likelihood of the PSA test result showing positive for cancer is 25%. The patient who is told he has cancer as well as his family is affected. Most likely, with an error rate this high, the physician would perform a second test to verify the results.

9.50 This is the probability of making a type I error. This means that half of the women who test positive are initially told that they do, not that half of the women tested do not have cancer.

9.51 a A two-tailed test would be used. You would not want to overfill or under-fill the can. b. Overfilling costs you money and under-filling cheats the customer. c. Because the weight is normally distributed and the population standard deviation is known the sample

mean will have a normal distribution.d. Reject the null hypothesis if z > 2.575 or z < −2.575.

9.52 a. Because the population distribution is normal and you know the population standard deviation, you should use the normal distribution for the sampling distribution on the sample mean.

b. H0: µ = 520 versus H1: µ ≠ 520. Reject H0 if z > 1.96 or z < −1.96. c. z = −5.0 therefore reject the null hypothesis. The sample result is highly significant showing there is a

difference in the mean fill.

9.53 a. H0: µ ≥ 90 versus H1: µ < 90.

b.0

Xt =

s

n

− µ

Reject H0 if t < −2.998.

c.

88.375 90

4.984

8

=-0.92 t = −

Because −0.92 < −2.998 we fail to reject the null hypothesis. The sample does not

give enough evidence to reject Bob’s claim that he is a 90+ student.

96

d. We assume that the population distribution in normal.e. The p-value = .1936. Because .1936 > .01 we fail to reject the null hypothesis.

9.54 a. H0: µ ≤ 10 pages versus H1: µ > 10 pages. Reject H0 if t > 2.441. t = 5.90 so reject the null hypothesis and conclude that the true mean is greater than 10 pages.

b. The p-value ≈ 0 so we would reject the null hypothesis.

9.55 a. H0: µ ≥ 2.268 grams versus H1: µ < 2.268 grams. Reject H0 if t < −1.761. t = −1.79 so reject the null hypothesis and conclude that the true mean is less than 2.268 grams.

b. With use, the metal could erode slightly so that the average weight is less than the newly minted dimes.

9.56 a. H0: π ≤ .50 versus H1: π > .50. Reject H0 if z > 1.282. z = 2.07 so reject the null hypothesis and conclude that the true proportion is greater than .5.

b. p-value = .0194 so we would reject the null hypothesis. The coin is biased towards heads.

9.57 a. H0: π ≤ .10 versus H1: π > .10. Reject H0 if z > 1.645. z = 2.00 so reject the null hypothesis and conclude that the true proportion is greater than .1.

b. Yes, if α were less than .0228, our decision would be different.c. p-value = .0228. Conclude that more than 10% of all one-dollar bills have something extra written on

them.

9.58 a. H0: π ≤ .25 versus H1: π > .25. Reject H0 if z > 1.645. z = 1.39 so fail to reject the null hypothesis.b. This is not a close decision.

c. We assume a normal distribution on the sample statistic, p. This makes sense because both nπ > 10 and

n(1−π) > 10.

9.59 a. H0: π ≤ .05 versus H1: π > .05. Reject H0 if z > 1.96. z = 1.95 so we fail to reject the null hypothesis at the .025 level of significance. The standard is not being violated.

b. p-value = .0258. .0258 > .025 therefore fail to reject the null hypothesis. This decision is very close.

9.60 a. H0: µ ≤ 30 years versus H1: µ > 30 years. Reject H0 if t > 1.796. t = 3.10 so reject the null hypothesis and conclude that the true mean age is greater than 30 years.

b. The sample mean was 33.92. This difference is probably unimportant.c. p-value = .0051 which is much smaller than .05 so the result is statistically significant.

9.61 a. H0: π ≤ .10 versus H1: π > .10. Reject H0 if z > 1.645. z = 1.11 so fail to reject the null hypothesis. We do not have strong evidence to conclude that more than 10% of all flights have contaminated drinking water.

b. p-value = .1327.

9.62 H0: π ≤ .95 versus H1: π > .95. Reject H0 if z > 1.96. z = 2.05 so reject the null hypothesis and conclude that the true proportion is greater than .95. The company is exceeding its goal.

9.63 a. H0: π ≥ .50 versus H1: π < .50. Reject H0 if z < −1.645. z = −2.07 so reject the null hypothesis and conclude that the true proportion is less than .5.

b. p-value = .0193. .0193 < .05 therefore we would reject the null hypothesis.c. The sample proportion was .46. This is a difference of 4%. This is an important difference. There are

thousands of college athletes in the US. Increasing the graduation rate for college athletes is a goal that many universities are striving for today.

9.64 a. H0: µ ≤ $250 versus H1: µ > $250. Reject H0 if t > 1.711. t = 1.64 so we fail to reject the null hypothesis. It does not appear that the average out of pocket expense is greater than $250.

b. The decision if fairly close.

97

9.65 a. 95% CI (.173, .2684)b. This sample is consistent with the hypothesis that no more than 25% of hams are underweight.

H0: π ≤ .25 versus H1: π > .25. However, if the goal were stated as having less than 25% of the hams

underweight, the set of hypotheses would be: H0: π ≥ .25 versus H1: π < .25. In this case, the sample would not support the goal.

c. A confidence interval is equivalent to a two-tailed test because the critical value of the test statistic used in the hypothesis test is the same value used to calculate the margin of error in the confidence interval.

9.66 H0: µ ≤ 5 days versus H1: µ > 5 days. Reject H0 if t > 1.796. t = 0.10 so we fail to reject the null hypothesis. It does not appear that the average repair time is longer than 5 days so the goal is being met.

9.67 a. H0: µ ≤ 300 rebounds versus H1: µ > 300 rebounds. Reject H0 if t > 2.201. t = 0.204 so we fail to reject the null hypothesis. It does not appear that the average number of rebounds is greater than 300.

b. There may be outliers in the population of NBA players.

9.68 a. H0: µ = 1.223 kg versus H1: µ ≠ 1.223 kg. Reject H0 if t > 2.201 or t < −2.201. b. t = −0.33 so we fail to reject the null hypothesis. It does not appear that the mean weight is different from

1.223 kg.

9.69* a. P(X ≥ 3 | n = 100, π = .01) = .0794. Because .0794 > .025 we fail to reject the null hypothesis.b. The p-value is .0794. This sample does not contradict the automaker’s claim.

9.70* a. P(X ≥ 2 | n = 36, π = .02) = .1618. Because .1618 > .10 we fail to reject the null hypothesis.b. p-value = .1618. This sample does not show that the standard is exceeded.

9.71* H0: π ≤ .50 versus H1: π > .50. Let n = 16 and x = 10. Find P(X ≥ 10 | n = 16, π = .5) = .2272. Because .2272 > .1, we cannot conclude that more than 50% feel better with the experimental medication.

9.72 H0: π ≥ .10 versus H1: π < .10. P(X = 0 | n = 31, π = .1) = .0382. Because .0382 < .10, we can reject the null hypothesis. It appears that the on-time percentage has fallen.

9.73* a. From MINITAB: 95% confidence interval is (0, .0154).

b. A binomial distribution should be used because nπ =0 which is less than 10.c. Yes, this sample shows that the proportion of patients who experience restenosis is less than 5%.

9.74 a. The p-value is .042. A sample proportion as extreme would occur by chance about 42 times in 1,000 samples if in fact the null hypothesis were true. This is fairly convincing evidence that the drug is effective.

b. A p-value of .087 is approximately twice .042. This sample is less convincing of the effectiveness of the drug.

9.75 a. The p-value tells us the chance of making this particular sample observation if in fact the null hypothesis is true. A small p-value says that there is a very small chance of making this sample observation assuming the null hypothesis is true therefore our assumption about the null hypothesis must be false.

98

9.76* Using the worksheet 09-08 PowerCurvesDIY.xls:

BetaLeft

9.77* Using the worksheet 09-08 PowerCurvesDIY.xls:

BetaLeft

99

BetaLeft

9.78* H0: σ2 = 64 versus H1: σ2 ≠ 64. Reject the null hypothesis if χ2 > 39.36 or χ2 < 12.40. χ2 = 24.68 therefore we fail to reject H0.

9.79* a. H0: µ ≤ 106 versus H1: µ > 106. Reject the null hypothesis if t > 2.807. t = 131.04 so reject the null hypothesis. The mean brightness is considerably greater than 106.

b. H0: σ2 ≥ .0025 versus H1: σ2 < .0025. Reject the null hypothesis if χ2 < 9.26. χ2 = 12.77 therefore we would fail to reject the null hypothesis. This sample does not provide evidence that the variance is less than .0025.

9.80 Answers will vary but should consider the following points:a. The null hypothesis is that the patient’s cholesterol is less than the threshold of treatable

hypercholesterolemia. The alternative is that the patient’s cholesterol is greater than the threshold of treatable hypercholesterolemia. A Type I error is a false positive; we rejected the null when it is true. A type II error is a false negative; we fail to reject the null hypothesis when the null is false.

b. Discussion should focus on the costs borne by the doctor for a false negative vs. costs borne by patient for living with a false positive (both financial as well as psychological.)

c. Patient wants to minimize a Type I error. Doctor or HMO want to minimize a Type II error.d. Discussion could include proper diet, American fast food culture, the movie “Supersize Me”, choice of

food, individual’s right to eat what they want, responsibility (or not) of businesses to offer alternative foods to help lower cholesterol, responsibility of individuals with respect to food choices.

100

Chapter 10

Two-Sample Hypothesis Tests

10.1 For each problem, the following formulas were used:

1 21 2

calc c

1 2

c c

1 2

combined number of successesp p x x = where p z

combined sample sizen n1 1p (1 - p ) +

n n

− += =

+

a. Standard error: .0987Z Test Statistic: −2.43p-value: 0.0075Z Critical: -2.3263Decision is not close: reject H0

b. Standard error: .0884 Z Test Statistic: 2.26 p-value: .0237Z Critical: +/- 1.645Decision is not close: reject H0

c. Standard error: .07033Z Test Statistic: −1.7063

p-value: 0.0440Z Critical: -1.645

Decision is close: reject H0

10.2 For each problem, the following formulas are used:

1 21 2

calc c

1 2

c c

1 2

combined number of successesp p x x = where p z

combined sample sizen n1 1p (1 - p ) +

n n

− += =

+

a. Standard error: .0555Z Test Statistic: 1.4825p-value: 0.1382Z Critical: +/- 1.9600Decision is not close: fail to reject H0

b. Standard error: .0618Z Test Statistic: −2.162p-value: .0153Z Critical: −2.3263Decision is not close: reject H0

c. Standard error: .01526Z Test Statistic: 1.638p-value: .0507Z Critical: 1.645Decision is close: fail to reject H0

101

10.3 a. Define π1 = proportion of shoppers that paid by debit card in 1999. Define π2 = proportion of shoppers that paid by debit card in 2004.

H0: π1 = π2 versus H1: π1 < π2. This is a left-tailed test. Reject the null hypothesis if z < −2.33.b. z = −2.28 so we fail to reject the null hypothesis (although the decision is close.) The sample does not

provide strong enough evidence to conclude that there is a difference in the two proportions.c. p-value = .0113.d. Normality is assumed since n1p1 > 10 and n2p2 > 10.

10.4 a. Define π1 = proportion of loyal mayonnaise purchasers. Define π2 = proportion of loyal soap purchasers.

H0: π1 = π2 versus H1: π1 ≠ π2. This is a two-tailed test. Reject the null hypothesis if z < −1.96 or z > 1.96. z = 1.725 therefore we fail to reject the null hypothesis. The sample evidence does not show a significant difference in the two proportions.

b. 95% confidence interval: (−.015, .255). Yes, the interval does contain zero.

10.5 a. Define π1 = proportion of respondents in first group (the group given the gift certificate.) Define π2 = proportion of respondents in the second group.

H0: π1 = π2 versus H1: π1 ≠ π2. This is a two-tailed test. Reject the null hypothesis if z < −1.96 or z > 1.96. z = 2.021 therefore we reject the null hypothesis. The sample shows a significant difference in response rates.

b. 95% confidence interval: (.0013, .0787). No, the interval does not contain zero. We estimate that the response rate for the group given the gift certificate is higher than the group that did not receive the gift certificate.

10.6 a. Define π1 = proportion of flights with contaminated water in August and September 2004. Define π2 = proportion of flights with contaminated water in November and December 2004.

H0: π1 = π2 versus H1: π1 < π2. Reject the null hypothesis if z < −1.645. z = −1.1397 so we fail to reject the null hypothesis. The level of contamination was not lower in the first sample.

b. p-value: 0.1272c. From the public health perception, importance outweighs significance. Our sample information did not

allow us to conclude that the contamination proportion has gone down after sanitation improvements.d. Yes, normality is assumed because both n1p1 > 10 and n2p2 > 10.

10.7 a. Survival rates: 28/39 = .72 and 50/53 = .94, respectively. Reject the null hypothesis that the survival rates are equal if z < −1.28. z = −2.975 so we reject the null hypothesis. The survival rate for people with pets is higher than for those without pets.

b. In the second sample n2 (1−p2) < 10.c. It is not clear that owning a pet is the direct cause of longer survival. There may be underlying causes

that contribute to longer survival that were not identified in the study.

10.8 a. H0: πM = πW versus H1: πM ≠ πW. Reject the null hypothesis if z < −1.645 or z > 1.645. b. pM = .60 and pW = .6875c. z = −.69, p-value = .492. The sample does not show a significant difference in proportions.d. Normality can be assumed because both n1p1 > 10 and n2p2 > 10.

10.9 a. H0: πB = πC versus H1: πB ≠ πC. Reject the null hypothesis if z < −1.96 or z > 1.96. z = −0.669 so we fail to reject the null hypothesis. This sample does not give enough evidence to conclude that the proportions are different.

b. Normality cannot be assumed because n2p2 < 10.

10.10 a. H0: π2 − π1 ≤ .05 versus H1: π2 − π1 > .05. Reject the null hypothesis if z > 1.28. b. z = 1.14 and the p-value = .1272 so we fail to reject the null hypothesis. The percentage of shoppers

paying by debit card did not increase by 5%.

102

10.11 a. H0: π1 − π2 ≤ .10 versus H1: π1 − π2 > 10. Reject the null hypothesis if z > 1.645. b. z = .63 and the p-value = .2644 so we fail to reject the null hypothesis. The proportion of calls lasting at

least five minutes has not decreased by 10%.

10.12 a. H0: π1 − π2 ≤ .20 versus H1: π1 − π2 > .20. Reject the null hypothesis if z > 1.96. b. z = 3.006 and the p-value = .0013 so we reject the null hypothesis. The response rate did increase by at

least 20%.

10.13 Use the following formula for each test in (a) – (c), substituting the appropriate values:

calc2 2

1 21 2

1 2 1 2

1 2 = t

( 1) + ( 1) 1 1n ns s +

( 1) + ( 1)n n n n

X X

− −

− −

−

a. d.f.: 28Standard error: 0.0931t-calculated: -2.1483p-value: 0.0202t-critical: -2.0484Decision: RejectFormula for p-value: =TDIST(ABS(-2.1483),28,2)

b. d.f.: 39Standard error: 1.8811t-calculated: -1.5948p-value: .1188t-critical: +/- 2.0227Decision: Not RejectFormula for p-value: =TDIST(ABS(-1.5948),39,2)

c. d.f.: 27Standard error: 1.0335t-calculated: 1.9351p-value: 0.0318t-critical: 1.7033Decision: RejectFormula for p-value: =TDIST(ABS(1.9351),27,1)

10.14 Use the following formulas for each test in (a) – (c), substituting the appropriate values:

calc2 2

1 2

1 2

1 2 = t

s s +

n n

X X−

with ( ) ( )

22 2

1 1 2 2

2 22 2

1 1 2 2

1

s n s nd.f .

s n s n

n 1 n2 1

+=

+− −

a. d.f.: 24Standard error: 0.0931t-calculated: -2.1483p-value: 0.0210t-critical: -2.0639Decision: RejectFormula for p-value: =TDIST(ABS(-2.0639),24,2)

103

b. d.f.: 32Standard error: 1.9275t-calculated: -1.5564p-value: .1294t-critical: +/- 2.0369Decision: Not RejectFormula for p-value: =TDIST(ABS(-1.5564),32,2)

c. d.f.: 23Standard error: 1.0403t-calculated: 1.9226p-value: 0.0335t-critical: 1.7139Decision: RejectFormula for p-value: =TDIST(ABS(1.9226),23,1)

10.15 a. H0: µ1 = µ2 versus H1: µ1 ≠ µ2. Reject the null hypothesis if t < −1.677 or t > 1.677 (48 df). t = −0.7981 so we fail to reject the null hypothesis. There is no difference in the average length of stay between men and women pneumonia patients.

b. The p-value = .4288.

10.16 a. H0: µexped = µexplo versus H1: µexped < µexplo. Reject the null hypothesis if t < −2.552 (18 df). t = −3.704 so we reject the null hypothesis. The average MPG is lower for the Expedition than the Explorer.


10.17 a. H0: µ1 ≤ µ2 versus H1: µ1 > µ2. Reject the null hypothesis if t >2.462 (29 df). t = 1.902 so we fail to reject the null hypothesis. The average amount of purchases when the music is slow is not less than when the music is fast.


10.18 H0: µ1 ≤ µ2 versus H1: µ1 > µ2. Reject the null hypothesis if t >2.074 (22 df). t = 2.26 so we reject the null hypothesis. The average shoe size appears to have increased.

10.19 H0: µ1 ≥ µ2 versus H1: µ1 < µ2. Reject the null hypothesis if t < −2.145. t = −2.22 so we reject the null hypothesis. The sample data show a significant decrease in the average number of migraines each month when using Topiramate.

10.20 a. Define the difference as New – Old. H0: µd ≤ 0 versus H0: µd > 0. Reject the null hypothesis if t > 1.833 (9 df.) t = 2.03 so we reject the null hypothesis. The new battery shows significantly greater average hours of charge.

b. The decision is not close. The p-value is .0363 which is less than .05.c. Yes, this is an important difference. The sample showed a difference of 5 hours.

10.21 a. Define the difference as Daughter’s Height – Mother’s Height. H0: µd ≤ 0 versus H0: µd > 0. Reject the null hypothesis if t > 1.943 (6 df.) t = 1.93 so we fail to reject the null hypothesis. There is not a significant difference in height between mothers and daughters.

b. The decision is close. The p-value is .0509 which is slightly greater than .05.c. A daughter’s height is affected by her father’s height as well as her grandparents. Nutrition also plays a

role in a person’s development.

10.22 a. Define the difference as Old – New. H0: µd ≤ 0 versus H0: µd > 0. Reject the null hypothesis if t > 2.132 (4 df.) t = 2.64 so we reject the null hypothesis. The new method shows a significantly faster average.

b. The decision is not close. The p-value is .0287 which is less than .05.

104

10.23 a. Define the difference as No Late Fee – late Fee. H0: µd ≤ 0 versus H0: µd > 0. Reject the null hypothesis if t > 1.383 (9 df.) t = 2.86 so we reject the null hypothesis. The average number of rentals has increased.

b. The decision is not close. The p-value is .0094 which is less than .10.c. Yes, this is an important difference. The sample showed an average increase of 2 rentals per month

which is a 20% increase. This means more revenue for the store.

10.24 a. Define the difference as Daughter – Mother. H0: µd ≤ 0 versus H0: µd > 0. Reject the null hypothesis if t > 2.718. t = 3.17 so we reject the null hypothesis. The average shoe size of a daughter is greater than her mother.

b. The decision is not close. The p-value is .0045 which is less than .01.c. Not sure if this is an important distinction. The sample showed a difference of less than a whole shoe

size. d. In general, adults are showing a trend of increasing size.

10.25 Define the difference as Entry − Exit. H0: µd = 0 versus H0: µd ≠ 0. Reject the null hypothesis if t > 3.499 or t < −3.499 (7 df.) t = −1.71 so we fail to reject the null hypothesis. There is no difference between the number of entry failures and exit failures. The decision is not close. The p-value is .1307 which is much greater than .01.

10.26 a. H0: σ12 = σ2

2 versus σ12 ≠ σ2

2. Reject H0 if F > 4.76 or F < .253. (ν1 = 10, ν2 = 7.) F = 2.54 so we fail to reject the null hypothesis.

b. H0: σ12 = σ2

2 versus σ12 < σ2

2. Reject H0 if F < .264. (ν1 = 7, ν2 = 7.) F = .247 so we reject the null hypothesis.

c. H0: σ12 = σ2

2 versus σ12 > σ2

2. Reject H0 if F > 2.80 (ν1 = 9, ν2 = 12.) F = 19.95 so we reject the null hypothesis.

10.27 a. H0: µ1 ≥ µ2 versus H1: µ1 < µ2. Reject the null hypothesis if t < −1.86 (8 df.) t = −4.29 so we reject the null hypothesis. The sample provides evidence that the mean sound level has been reduced with the new flooring.

b. H0: σ12 = σ2


2. Reject H0 if F > 9.60 or F < .104. (ν1 = 4, ν2 = 4.) F = .6837 so we fail to reject the null hypothesis. The variance has not changed.

10.28 H0: σ12 = σ2

2 versus σ12 < σ2

2. Reject H0 if F < .3549. (ν1 = 11, ν2 = 11.) F = .103 so we reject the null hypothesis. The new drill has a reduced variance.

10.29 a. H0: µ1 ≤ µ2 versus H1: µ1 > µ2. Reject the null hypothesis if t > 1.714 (11 df.) t = 3.163 so we reject the null hypothesis. The sample provides evidence that the mean weight of an international bag is greater than a domestic bag.

b. H0: σ12 ≤ σ2

2 versus σ12 > σ2

2. Reject H0 if F > 2.65 or F < .33. (ν1 = 9, ν2 = 14.) F = 6.74 so we reject the null hypothesis. The variance of international bag weight is greater than domestic bag weight.

10.30 a. H0: π1 = π2 versus H1: π1 > π2. b. Reject the null hypothesis if z > 2.326. c. p1 = .0001304, p2 = .0000296, z = 2.961d. We reject the null hypothesis. The sample evidence shows a significant difference in the two proportions.e. The p-value = .0031. This result is not due to chance.f. Yes, accidents have severe consequences therefore even small reductions make a difference.g. The normality assumption is questionable because there were only 4 accidents observed with the yellow

fire trucks.

10.31 a. H0: π1 = π2 versus H1: π1 > π2. b. Reject the null hypothesis if z > 1.645. c. p1 = .980, p2 = .93514, z = 4.507d. We reject the null hypothesis. The sample evidence shows a significant difference in the two proportions.e. The p-value ≈ .0000. This result is not due to chance.

105

f. Normality assumption is valid because both n1p1 > 10 and n2p2 > 10.

106

10.32 a. H0: π1 = π2 versus H1: π1 > π2. Reject the null hypothesis if z > 1.645.b. p1 = .169, p2 = .1360.c. z = 2.98, p-value = .003. Reject the null hypothesis. d. The increase is most likely due to an increase in women executives and an increased awareness of the

benefit of having more diverse boards.

10.33 a. p1 = .17822, p2 = .143.b. z = 1.282, p-value = .2000. Because the p-value is greater than .05, we fail to reject the null hypothesis.

There is not enough evidence in this sample to conclude that there is a difference in the proportion of minority men (out of all males) and minority women (out of all females) on Fortune 100 boards.

10.34 a. H0: π1 = π2 versus H1: π1 > π2. Reject the null hypothesis if z > 1.28. p1 = .40, p2 = .3333. z = .4839, p-value = .3142. We fail to reject the null hypothesis.

b. Yes, the normality assumption is valid.c. Early finishers might know the material better and finish faster. On the other hand, if a student has not

studied they might quickly right an answer down and turn in their exam just to get it over with.

10.35 a. H0: π1 = π2 versus H1: π1 ≠ π2. Reject the null hypothesis if z > 2.576 or z < −2.576. z = −2.506 so we fail to reject the null hypothesis. The decision is very close.

b. The p-value = .0122.c. Normality assumption is valid because both n1p1 > 10 and n2p2 > 10.d. Gender differences may imply different marketing strategies.

10.36 a. H0: π1 = π2 versus H1: π1 > π2. b. z = 9.65, p-value ≈ 0.c. Normality assumption is valid.d. Yes, this difference is quite important because the safety of children is involved.

10.37 a. H0: π1 ≥ π2 versus H1: π1 < π2. Reject the null hypothesis if z < 2.33. z = −8.003 so we reject the null hypothesis.

b. p-value = .0000. This is less than .01 so the difference is quite significant.c. Normality can be assumed.

10.38. a. pE = .1842, pW = .2580.

b. H0: πE = πW versus H1: πE ≠ πW. Reject the null hypothesis if z > 1.96 or z < −1.96. z = −2.46 so we reject the null hypothesis and conclude that there is a greater proportion of large gloves sold on the west side of Vail.

c. There could be a different type of skier on the east side of Vail, perhaps more children ski on the east side as opposed to the west side.

10.39 a. H0: π1 ≥ π2 versus H1: π1 > π2. Reject the null hypothesis if z > 2.326.b. z = 2.932, p-value = .0017. The p-value is less than .01 so we would reject the null hypothesis.c. Normality assumption is valid.d. While the difference may seem small on paper, breast cancer has very serious consequences. Small

reductions are important.e. Were diet, smoking, exercise, hereditary factors considered?

10.40 H0: πP = πX versus H1: πP ≠ πX. Reject the null hypothesis if z > 1.645 or z < −1.645. z = −1.222 so we fail to reject the null hypothesis.

10.41 a. H0: π1 ≥ π2 versus H1: π1 > π2. Reject the null hypothesis if z > 2.326.b. z = 8.254, p-value = .0000.c. Normality assumption is valid.d. Yes, the difference is important because the risk is almost three times greater for those with a family

history of heart disease.

107

10.42 a. Group 1: (.300, .700)Group 2: (−.007, .257)Group 3: (−.015, .098)

b. H0: π1 = π2 versus H1: π1 ≠ π2. z = 2.803, p-value = .0051. Reject the null hypothesis.c. While the confidence intervals may be more appealing, the procedure in part b is more appropriate.d. Normality is questionable because there were fewer than 10 observations in groups 2 and 3.

10.43 H0: µ1 ≤ µ2 versus H1: µ1 > µ2. Reject the null hypothesis if t > 2.374 (84 df.) t = 4.089 so we reject the null hypothesis and conclude that the virtual team mean is higher.

10.44 a. H0: π1 = π2 versus H1: π1 < π2. b. z = −3.987, p-value = .0000. Reject the null hypothesis.c. Normality can be assumed.d. Yes, differences are important.e. Physicians should ask about exercise habits, nutrition, weight, smoking etc.

10.45 a. H0: π1 = π2 versus H1: π1 < π2. b. z = −2.7765, p-value = .0027. Reject the null hypothesis.c. Normality can be assumed.d. Yes, differences are important.e. Exercise habits, nutrition, weight, smoking etc., might influence the decision. Also, many people cannot

afford them and lack insurance to pay the costs.

10.46 a. H0: µ1 = µ2 versus H1: µ1 > µ2. b. t = 1.221, p-value = .1141c. The results of the sample are not statistically significant although students might think 8 points is an

important difference.d. Yes, the sample standard deviations appear similar.e. F = .620. FL = 0.488, FR = 2.05. Fail to reject the null hypothesis. This sample does not provide evidence

that the variances are different.

10.47 a. H0: µ1 ≤ µ2 versus H1: µ1 > µ2. Assuming unequal variances, t = 1.718. The p-value is .0525. We fail to reject the null hypothesis.

b. A paired sample test may have made more sense. By comparing the costs from one year to the next for the same 10 companies we would have eliminated a source of variation due to different businesses.

10.48 a. H0: µ1 = µ2 versus H1: µ1 > µ2.b. t = 2.640, p-value = .0050. Because the p-value > .01 we would reject the null hypothesis.c. The distribution could be skewed to the right by one or two extremely long calls. A heavily skewed

distribution could make the t distribution an unwise choice.

10.49 a.

New Bumper:

Control Group:

108

b. H0: µ1 ≥ µ2 versus H1: µ1 < µ2.c. Assuming equal variances, reject H0 if t < −1.729 with df = 19.d. t = −1.63.e. Fail to reject the null hypothesis.f. The p-value = .0600. This decision was close.g. A sample difference of approximately 3 days downtime would be considered important but the variation

in the downtimes is large enough that we cannot conclude the true means are different.

10.50 a. H0: µN ≥ µS versus H1: µN < µS. Reject the null hypothesis if t < −2.650 using ν* = 13.b. t = −5.29.c. This sample provides strong evidence that the average spending in the northern region is much higher

than average spending in the southern region.d. Folks in the south may use services differently or may be older.

10.51 a. Use a two-tailed test comparing two means assuming unequal variances.

b. H0: µ1 = µ2 versus H1: µ1 ≠ µ2.

c. t = 2.651 with df = 86. Because the p-value is .0096, we easily reject the null hypothesis at α = .05. Although the sample difference isn’t large, large samples have high power.

d. Students might be more alert in the morning.e. Yes, the standard deviations are similar.

f. H0: σ12 = σ2


2. Reject H0 if F < .53 or F > 1.88. (ν1 = 41, ν2 = 45.) F = .1.39 so we fail to reject the null hypothesis.

10.52 a. H0: µ1 = µ2 versus H1: µ1 ≠ µ2.b. Reject the null hypothesis if t < −1.686 or t > 1.686.c. t = −1.549d. Because t > −1.686, we fail to reject the null hypothesis.e. The p-value = .130. Because the p-value > .10, we fail to reject the null hypothesis.

10.53 a. Dot plots suggest that the mean differ and variances differ. Note the outlier in men’s salaries.

b. H0: µ1 ≤ µ2 versus H1: µ1 > µ2.c. Reject the null hypothesis if t > 2.438 with df = 35.d. Assuming equal variances, t = 4.742.

e. Reject the null hypothesis at α = .05. Men are paid more on average.f. p-value = .0000. This shows that the sample result would be unlikely if H0 were true.g. Yes, the large difference suggests gender discrimination.

10.54 a. H0: µ1 = µ2 versus H1: µ1 ≠ µ2.b. Reject the null hypothesis if t > 2.045 or t < −2.045.c. t = −1.623d. Fail to reject the null hypothesis. There appears to be no difference in the average order size between

Friday and Saturday night.e. p-value = .1154

109

10.55 a. The distributions appear skewed to the right.

b. H0: µ1 = µ2 versus H1: µ1 ≠ µ2. Assume equal variances.c. Reject the null hypothesis t > 2.663 or if t < −2.663.d. t = .017.e. We fail to reject the null hypothesis. It does not appear that the means are different.f. The p-value = .9886. This indicates that the sample result shows no significant difference.

10.56 H0: σ12 = σ2


2. Fcalculated = 1.991. The p-value = .0981. We cannot reject the null hypothesis

at α = .05. The variances are not different.

10.57 a. H 0: µ1 = µ2 versus H1: µ1 ≠ µ2. Assume equal variances.b. Reject the null hypothesis if t < −1.673 or t > 1.673. Use 55 df.c. Since t = −3.162 and the p-value = .0025, we reject the null hypothesis. Mean sales are lower on the east

side.

10.58 a. H 0: µd = 0 versus H1: µd ≠ 0.b. Reject the null hypothesis if t < −2.776 or t > 2.776.c. t = −1.31. Fail to reject the null hypothesis. The average sales appear to be the same.

10.59 H0: σ12 = σ2


2. df1 = 30, df2 = 29. Reject the null hypothesis if F > 2.09 or F < .47. Fcalculated

= .76 so we fail to reject the null hypothesis. The variances are not different.

10.60 a. H 0: µd = 0 versus H1: µd ≠ 0.b. Reject the null hypothesis if t > 2.045 or t < −2.045. c. t = −1.256. d. We fail to reject the null hypothesis. e. The p-value = .2193. There is no evidence that the heart rates are different before and after a class break.

10.61 Assume independent samples. H 0: µ1 = µ2 versus H1: µ1 ≠ µ2. Assume equal variances. Reject the null hypothesis if p-value < .01. t = −.05 and the p-value = .9622 (two tailed test) so we fail to reject the null hypothesis. The average assessed value from the company’s assessor and the employee’s assessor are the same.

10.62 H 0: µ1 = µ2 versus H1: µ1 ≠ µ2. Reject the null hypothesis if the p-value is less than .10. t = −1.336 and the p-value = .2004 (two tailed test) so we fail to reject the null hypothesis. The average size of the homes in the two neighborhoods are the same.

10.63 H 0: µ1 = µ2 versus H1: µ1 ≠ µ2. Assume unequal variances. t = 1.212 with p-value = .2433. Fail to

reject the null hypothesis. The average defect rates appear to be the same. It is questionable whether the normal assumption applies because of the very low incidence of bad pixels. Perhaps the Poisson distribution should be used.

10.64 H 0: µd = 0 versus H1: µd ≠ 0. Reject the null hypothesis if the p-value < .10. t = −1.76 and the p-value = .1054. We fail to reject the null hypothesis but the decision is quite close.

10.65 a. H0: σA2 = σB

2 versus σA2 > σB

2. df1 = 11, df2 = 11. Reject the null hypothesis if F > 3.53. Fcalculated = 9.86 so we reject the null hypothesis. Portfolio A has greater variance than portfolio B.

110

b. H 0: µ1 = µ2 versus H1: µ1 ≠ µ2. Assume unequal variances (from part a.) t = .49 with a p-value = .6326. We fail to reject the null hypothesis. The portfolio means are equal.

Chapter 11

Analysis of Variance

11.1 a. The hypotheses to be tested are:

H0: µA = µB = µC � mean scrap rates are the same

H1: Not all the means are equal � at least one mean is different

b. One factor, F = 5.31 and critical value for α = .05 is F2,12 = 3.89.

c. We reject the null hypothesis since the test statistic exceeds the critical value.

d. The p-value of .0223 is less than .05. At least one mean scrap rate differs from the others.

e. From the dot plot, we see Plant B above the overall mean and Plant C below the overall mean.

Mean n Std. Dev Treatment

12.30 5 1.573 Plant A

13.96 5 2.077 Plant B

9.58 5 2.651 Plant C

11.95 15 2.728 Total

One-Factor ANOVA

Source SS df MS F p-value

Treatment 48.897 2 24.4487 5.31 .0223

Error 55.260 12 4.6050

Total 104.157 14


H0: µ1 = µ2 = µ3 = µ4 � physician means are the same



c. We reject the null hypothesis since the test statistic exceeds the critical value (close).

d. The p-value of .0310 is less than .05. At least one physician mean differs from the others.

e. From the dot plot, we see Physician 1 and Physician 3 below the overall mean and Physician 2 above the

overall mean.

111


28.3 7 4.89 Physician 1

34.2 6 4.12 Physician 2

27.3 8 4.62 Physician 3

32.0 7 4.24 Physician 4

30.2 28 5.08 Total

One-Factor ANOVA


Treatment 212.35 3 70.782 3.50 .0310

Error 485.76 24 20.240

Total 698.11 27


H0: µ1 = µ2 = µ3 = µ4 � mean GPAs are the same



c. We reject the null hypothesis since the test statistic exceeds the critical value (close).

d. The p-value of .0304 is less than .05. At least one GPA mean differs from the others.

e. From the dot plot, we see the GPA for Accounting below the overall mean and Human Resources and

Marketing above the overall mean.


2.834 7 0.5053 Accounting

3.024 7 0.1776 Finance

3.241 7 0.3077 Human Resources

3.371 7 0.2575 Marketing

3.118 28 0.3785 Total

One-Factor ANOVA


Treatment 1.1812 3 0.39372 3.52 .0304

Error 2.6867 24 0.11195

Total 3.8679 27

112


H0: µ1 = µ2 = µ3 = µ4 � mean sales are the same



c. We reject the null hypothesis since the test statistic exceeds the critical value.

d. The p-value of .0153 is less than .05. At least one mean differs from the others.

e. From the dot plot, we see the weekly sales for Stores 2 and 3 below the overall mean and Store 1 above

the overall mean.


108.0 5 5.34 Store 1

87.4 5 10.83 Store 2

91.0 5 11.11 Store 3

101.0 5 10.30 Store 4

96.9 20 12.20 Total

One-Factor ANOVA


Treatment 1,325.35 3 441.783 4.71 .0153

Error 1,501.20 16 93.825

Total 2,826.55 19

113

11.5 Using Tukey simultaneous comparison t-values, Plant B and Plant C differ. Using the pairwise t-tests, Plant

B and Plant C differ.

Post hoc analysis

Tukey simultaneous comparison t-values (d.f. = 12)

Plant C Plant A Plant B

9.58 12.30 13.96

Plant C 9.58

Plant A 12.30 2.00

Plant B 13.96 3.23 1.22

critical values for experimentwise error rate:

0.05 2.67

0.01 3.56

p-values for pairwise t-tests

Plant C Plant A Plant B

9.58 12.30 13.96

Plant C 9.58

Plant A 12.30 .0682

Plant B 13.96 .0073 .2448

11.6 Using Tukey simultaneous comparison t-values, Physicians 2 and 3 differ. Using the pairwise t-tests,

Physicians 2 and 3 are one pair and Physicians 1 and 2 are another pair.

Post hoc analysis


Physician 3Physician

1Physician

4 Physician 2

27.3 28.3 32.0 34.2

Physician 3 27.3

Physician 1 28.3 0.44

Physician 4 32.0 2.04 1.54

Physician 2 34.2 2.85 2.35 0.87

critical values for experiment wise error rate:

0.05 2.76

0.01 3.47


Physician 3Physician

1Physician

4 Physician 2

27.3 28.3 32.0 34.2

Physician 3 27.3

Physician 1 28.3 .6604

Physician 4 32.0 .0525 .1355

Physician 2 34.2 .0089 .0274 .3953

114

11.7 Using Tukey simultaneous comparison t-values, Marketing and Accounting differ. Using the pairwise t-

tests, Marketing and Accounting are one pair and Human Resources and Accounting are another pair.

Post hoc analysis


Accounting Finance Human Resources Marketing

2.834 3.024 3.241 3.371

Accounting2.83

4

Finance3.02

4 1.06

Human Resources3.24

1 2.28 1.21

Marketing3.37

1 3.00 1.94 0.73


0.05 2.76

0.01 3.47


Accounting Finance Human Resources Marketing

2.834 3.024 3.241 3.371

Accounting2.83

4

Finance3.02

4 .2986

Human Resources3.24

1 .0320 .2365

Marketing3.37

1 .0062 .0641 .4743

11.8 Using Tukey simultaneous comparison t-values, Store 1 and Store 2 differ. Using the pairwise t-tests, Store

1 and Store 2 are one pair, Store 4 and Store 2 are another pair, and Store 1 and Store 3 are a third.

Post hoc analysis


Store 2 Store 3 Store 4 Store 1

87.4 91.0 101.0 108.0

Store 2 87.4

Store 3 91.0 0.59

Store 4 101.0 2.22 1.63

Store 1 108.0 3.36 2.77 1.14


0.05 2.86

0.01 3.67

115


Store 2 Store 3 Store 4 Store 1

87.4 91.0 101.0 108.0

Store 2 87.4

Store 3 91.0 .5650

Store 4 101.0 .0412 .1221

Store 1 108.0 .0040 .0135 .2700

For Exercises 11.9 through 11.12 The hypotheses to be tested are:

H0: σ12 = σ2

2 = ... = σc2

H1: Not all the σj2 are equal

where c = the number of groups. The test statistic is:2

maxmax 2

min

sF

s= .

Critical values of Fmax may be found in Table 11.5 using degrees of freedom given by:

Numerator ν1 = c

Denominator ν2 = n/c–1 (round down to next lower integer if necessary).

11.9* Critical value from Table 11. 5 is 15.5 (df1 = c = 3, df2 = n/c−1 = 4). We fail to reject the null hypothesis of

variance homogeneity since the text statistic Fmax = 7.027/2.475 = 2.84 is less than the critical value. This

result agrees with Levene’s test (p-value = .843) and the confidence intervals overlap.

Mean n Std. DevVarianc

eTreatmen

t

12.30 5 1.573 2.475 Plant A

13.96 5 2.077 4.313 Plant B

9.58 5 2.651 7.027 Plant C

11.95 15 2.728 Total

Plant

95% Bonferroni Confidence Intervals for StDevs

Plant C

Plant B

Plant A

1086420

Bartlett's Test

0.843

Test Statistic 0.95

P-Value 0.622

Lev en e's Test

Test Statistic 0.17

P-Value

Test for Equal Variances for Scrap Rate

116

11.10* Critical value from Table 11.5 is 10.4 (df1 = c = 4, df2 = n/c−1 = 6). We fail to reject the null hypothesis of


result agrees with Levene’s test (p-value = .885) and the confidence intervals overlap..

Mean n Std. Dev Variance Treatment

28.3 7 4.89 23.90Physician

1

34.2 6 4.12 16.97Physician

2

27.3 8 4.62 21.36Physician

3

32.0 7 4.24 18.00Physician

4

30.2 28 5.08 Total

Physician


Physician 4

Physician 3

Physician 2

Physician 1

1412108642

Bartlett's Test

0.885

Test Statistic 0.20

P-Value 0.978

Lev en e's Test

Test Statistic 0.21

P-Value

Test for Equal Variances for Wait Time


variance homogeneity since the text statistic Fmax = (0.2553)/(0.0315) = 8.10 is less than the critical value.

This result agrees with Levene’s test (p-value = .145). However, both tests are closer than in Exercises 11.9

and 11.9 (the high variance in accounting is striking, even though the confidence intervals do overlap.).

Mean n Std. Dev Variance Treatment

2.834 7 0.5053 0.2553 Accounting3.024 7 0.1776 0.0315 Finance

3.241 7 0.3077 0.0947 Human Resources3.371 7 0.2575 0.0663 Marketing

3.118 28 0.3785 Total

Major


Marketing

Human Resources

Finance

Accounting

1.61.41.21.00.80.60.40.20.0

Bartlett's Test

0.145

Test Statistic 6.36

P-Value 0.095

Lev en e's Test

Test Statistic 1.98

P-Value

Test for Equal Variances for GPA

117



result agrees with Levene’s test (p-value = .810) and the confidence intervals overlap..

Mean n Std. DevVarianc

e Treatment

108.0 5 5.34 28.50 Store 1

87.4 5 10.83 117.30 Store 2

91.0 5 11.11 123.50 Store 3

101.0 5 10.30 106.00 Store 4

96.9 20 12.20 TotalStore


Store 4

Store 3

Store 2

Store 1

50403020100

Bartlett's Test

0.810

Test Statistic 2.07

P-Value 0.558

Lev en e's Test

Test Statistic 0.32

P-Value

Test for Equal Variances for Sales

11.13 a. Date is the blocking factor and Plant is the treatment or research interest.Rows (Date):

H0: A1 = A2 = A3

H1: Not all the Aj are equal to zero

Columns: (Plant)

H0: B1 = B2 = B3 = B4 = 0

H1: Not all the Bk are equal to zerob. See tables.c. αPlant means differ at = .05, F = 41.19, p-value = .0002. Blocking factor (date) also significant, F = 8.62,

p-value = .0172.d. A test statistic of this magnitude would arise about 2 times in 10,000 samples if the null were true.e. Plot suggests that Plants 1 and 2 are below overall mean, Plants 3 and 4 above.

ANOVA table: Two factor without replication


Treatments (Plant) 216.25 3 72.083 41.19 .0002

Blocks (Date) 30.17 2 15.083 8.62 .0172

Error 10.50 6 1.750

Total 256.92 11

118

Mean n Std. Dev Factor Level

20.333 3 1.528 Plant 1

18 3 2 Plant 2

29 3 2.646 Plant 3

25 3 2.646 Plant 4

21.5 4 4.041 Mar 4

25.25 4 5.377 Mar 11

22.5 4 5.508 Mar 18

Post hoc analysis


Plant 2 Plant 1 Plant 4 Plant 3

18.000 20.333 25.000 29.000

Plant 2 18.000

Plant 1 20.333 2.16

Plant 4 25.000 6.48 4.32

Plant 3 29.000 10.18 8.02 3.70


0.05 3.46

0.01 4.97


Plant 2 Plant 1 Plant 4 Plant 3

18.000 20.333 25.000 29.000

Plant 2 18.000

Plant 1 20.333 .0741

Plant 4 25.000 .0006 .0050

Plant 3 29.000 .0001 .0002 .0100

119

11.14 a. Vehicle Size is the blocking factor and Fuel Type is the treatment or research interest.

Rows (Vehicle Size):

H0: A1 = A2 = A3 = A4


Columns (Fuel Type)

H0: B1 = B2 = B3 = B4 = 0

H1: Not all the Bk are equal to zerob. See tables.c. αFuel type means differ at = .05, F = 6.94, p-value = .0039. Blocking factor (Vehicle Size) also significant,

F = 34.52, p-value = .0000.d. A test statistic of this magnitude would arise about 39 times in 10,000 samples if the null were true.e. Plot suggests that 89 Octane and 91 Octane are somewhat above the overall mean. The Tukey tests show a

significant difference in fuel economy between Ethanol 10% and 89 Octane, Ethanol 10% and 91 Octane, and 87 Octane and 91 Octane. The pairwise t-tests confirm this plus a couple of weaker differences.



Treatments (Fuel Type) 54.065 4 13.5163 6.94 .0039

Blocks (Vehicle Size) 201.612 3 67.2040 34.52 0.0000

Error 23.363 12 1.9469

Total 279.040 19

Mean n Std. Dev Group

22.5750 4 3.5575 87 Octane

25.5500 4 3.2254 89 Octane

25.8000 4 4.2716 91 Octane

22.7500 4 4.5625 Ethanol 5%

21.8250 4 3.5874 Ethanol 10%

28.0200 5 2.0130 Compact

25.4200 5 2.3392 Mid-Size

21.1600 5 1.4293 Full-Size

20.2000 5 2.7911 SUV

23.7000 20 3.8323 Total

Post hoc analysis

Tukey simultaneous comparison t-values (d.f. = 12)Ethanol

10% 87 Octane Ethanol 5% 89 Octane 91 Octane

21.8250 22.5750 22.7500 25.5500 25.8000

Ethanol 10%21.825

0

87 Octane22.575

0 0.76

Ethanol 5% 22.750 0.94 0.18

120

0

89 Octane25.550

0 3.78 3.02 2.84

91 Octane25.800

0 4.03 3.27 3.09 0.25


0.05 3.19

0.01 4.13

p-values for pairwise t-testsEthanol

10% 87 Octane Ethanol 5% 89 Octane 91 Octane

21.8250 22.5750 22.7500 25.5500 25.8000

Ethanol 10%21.825

0

87 Octane22.575

0 .4618

Ethanol 5%22.750

0 .3670 .8622

89 Octane25.550

0 .0026 .0108 .0150

91 Octane25.800

0 .0017 .0067 .0093 .8043

121

11.15 a. Exam is the blocking factor and Professor is the treatment or research interest.

Rows (Exam):

H0: A1 = A2 = A3 = A4


Columns (Professor)

H0: B1 = B2 = B3 = B4 = B5= 0

H1: Not all the Bk are equal to zerob. See tables.c. αProfessor means are on the borderline at = .05, F = 3.26, p-value = .0500. Blocking factor (Exam) is not

significant, F = 1.11, p-value = .3824.d. A test statistic of this magnitude would arise about 5 times in 100 samples if the null were true.e. Plot shows no consistent differences in means for professors. The Tukey tests and the pairwise tests are not

calculated since the treatments do not significantly affect the exam scores.



Treatments (Professors) 134.403 4 33.6008 3.26 .0500

Blocks (Exams) 34.404 3 11.4680 1.11 .3824

Error 123.721 12 10.3101

Total 292.528 19

Mean nStd. Dev Group

76.325 4 4.8321 Prof. Argand

75.225 4 2.3977 Prof. Blague

80.250 4 3.1859 Prof. Clagmire

76.700 4 3.1591 Prof. Dross

72.200 4 1.8655 Prof. Ennuyeux

78.040 5 6.1064 Exam 1

75.120 5 3.0971 Exam 2

76.660 5 2.5530 Exam 3

74.740 5 3.3366 Final

76.140 20 3.9238 Total

122

11.16 a. Qtr is the blocking factor and Store is the treatment or research interest.

Rows (Qtr):

H0: A1 = A2 = A3 = A4


Columns (Store)

H0: B1 = B2 = B3 = 0

H1: Not all the Bk are equal to zerob. See tables.c. αStore means do not differ at = .05, F = 1.60, p-value = .2770. Blocking factor (Qtr) is significant, F =

15.58, p-value = .0031.d. A test statistic of this magnitude would arise about 28 times in 100 samples if the null were true.e. Plot shows no consistent differences in means for stores. The Tukey tests and the pairwise tests are not

calculated since the treatments do not significantly affect the sales.



Treatments (Store) 41,138.67 2 20,569.333 1.60 .2779

Blocks (Qtr) 601,990.92 3 200,663.639 15.58 .0031

Error 77,277.33 6 12,879.556

Total 720,406.92 11

Mean n Std. Dev

1,456.250 4 231.073 Store 1

1,375.250 4 261.153 Store 2

1,518.250 4 323.770 Store 3

1,509.000 3 205.263 Qtr 1

1,423.333 3 59.878 Qtr 2

1,120.667 3 63.760 Qtr 3

1,746.667 3 97.079 Qtr 4

1,449.917 12 255.913 Total

123

11.17 Factor A: Row Effect (Year)

H0: A1 = A2 = A3 = 0 � year means are the same

H1: Not all the Aj are equal to zero � year means differ

Factor B: Column Effect (Portfolio Type)

H0: B1 = B2 = B3 = B4 = 0 �stock portfolio type means are the same

H1: Not all the Bk are equal to zero � stock portfolio type means differ

Interaction Effect (Year×Portfolio)

H0: All the ABjk are equal to zero � there is no interaction effect

H1: Not all ABjk are equal to zero � there is an interaction effectb. See tables.c. αYears differ at = .05, F = 66.82, p α-value < .0001. Portfolios differ at = .05, F = 5.48, p-value = .0026.

αInteraction is significant at = .05, F = 4.96, p-value = .0005.d. The small p-values indicate that the sample would be unlikely if the null were true.e. The interaction plot lines do cross and support the interaction found and reported above. The visual

indications of interaction are strong for the portfolio returns data.

Table of Means

Factor 2 (Portfolio Type)

Factor 1 (Year) Health Energy Retail Leisure

2004 15.74 22.20 18.36 18.52 18.7

1

2005 22.84 27.98 23.92 25.46 25.0

5

2006 13.24 12.62 19.90 10.98 14.1

9

17.27 20.93 20.73 18.32 19.3

1

Two-Factor ANOVA with Replication


Factor 1 (Year) 1,191.584 2 595.7922 66.82 1.34E-14

Factor 2 (Portfolio) 146.553 3 48.8511 5.48 .0026

Interaction 265.192 6 44.1986 4.96 .0005

Error 427.980 48 8.9162

Total 2,031.309 59

Post hoc analysis for Factor 1


Row 3 Row 1 Row 2

14.19 18.71 25.05

Row 3 14.19

Row 1 18.71 4.79

Row 2 25.05 11.51 6.72


0.05 2.42

0.01 3.07


124


Health Leisure Retail Energy

17.27 18.32 20.73 20.93

Health 17.27

Leisure 18.32 0.96

Retail 20.73 3.17 2.21

Energy 20.93 3.36 2.40 0.19


0.05 2.66

0.01 3.29

11.18 Factor A: Row Effect (Year)

H0: A1 = A2 = A3 = 0 � year means are the same

H1: Not all the Aj are equal to zero � year means differ

Factor B: Column Effect (Department)

H0: B1 = B2 = B3 = 0 �department means are the same

H1: Not all the Bk are equal to zero � department type means differ

Interaction Effect (Year×Department)


H1: Not all ABjk are equal to zero � there is an interaction effectb. See tables.c. αYears do not differ at = .05, F = 0.64, p α-value = .5365. Departments differ at = .05, F = 12.66, p-value =

α.0004. Interaction is not significant at = .05, F = 2.38, p-value = .0899.d. The p-values range from highly significant (Department) to insignificant (Year). The interaction effect, if

any, is weak since about 9 samples in 100 would show an F statistic this large in the absence of interaction.e. The interaction plot lines do cross for Department, but are approximately parallel for Year and support

the lack of interaction found and reported above. The visual indications of interaction are, therefore, non-existent for the team ratings.

Table of Means

Factor 2 (Department)

Factor 1 (Year) Marketing Engineering Finance

2004 84.7 73.0 89.3 82.3

2005 79.0 77.0 89.7 81.9

2006 88.7 79.7 84.7 84.3

84.1 76.6 87.9 82.9

Two factor ANOVA with Replication


125

Factor 1 (Year) 30.52 2 15.259 0.64 .5365

Factor 2 (Department) 599.41 2 299.704 12.66 .0004

Interaction 225.48 4 56.370 2.38 .0899

Error 426.00 18 23.667

Total 1,281.41 26



Engineering Marketing Finance

76.6 84.1 87.9

Engineering 76.6

Marketing 84.1 3.29

Finance 87.9 4.94 1.65


0.05 2.55

0.01 3.32

11.19 Factor A: Row Effect (Age Group)

H0: A1 = A2 = A3 = A4 = 0 � age group means are the same

H1: Not all the Aj are equal to zero � age group means differ

Factor B: Column Effect (Region)

H0: B1 = B2 = B3 = B4= 0 �region means are the same

H1: Not all the Bk are equal to zero � region means differ

Interaction Effect (Age Group×Region)


H1: Not all ABjk are equal to zero � there is an interaction effectb. See tables.c. αAge groups differ at = .05, F = 36.96, p α-value < .0001. Regions do not differ at = .05, F = 0.55, p-value

α= .6493. Interaction is significant at = .05, F = 3.66, p-value = .0010.d. The p-values range from highly significant (Age Group) to insignificant (Region). The interaction effect is

significant since only about 1 sample in 1000 would show an F statistic this large in the absence of interaction.

e. The interaction plot lines do cross (e.g., MidWest crosses the others by age group) but visually there is not a strong indication of interaction. This is perhaps because the data range is not large (data appear to be rounded to nearest .1 so there is only 2-digit accuracy).

126

Table of Means

Factor 2 (Region)

Factor 1 (Age Group) Northeast Southeast Midwest West

Youth (under 18) 4.00 4.12 3.68 4.06 3.97

College (18-25) 3.86 3.70 3.88 3.78 3.81

Adult (25-64) 3.50 3.42 3.76 3.52 3.55

Senior (65 +) 3.42 3.52 3.18 3.36 3.37

3.70 3.69 3.63 3.68 3.67



Factor 1 (Age Group) 4.193 3 1.3975 36.96 5.56E-14

Factor 2 (Region) 0.062 3 0.0208 0.55 .6493

Interaction 1.245 9 0.1383 3.66 .0010

Error 2.420 64 0.0378

Total 7.920 79


Tukey simultaneous comparison t-values (d.f. = 64)Senior(65 +)

Adult(25-64)

College(18-25)

Youth(Under 18)

3.37 3.55 3.81 3.97

Senior (65 +) 3.37

Adult (25-64) 3.55 2.93

College (18-25) 3.81 7.07 4.15

Youth (under 18) 3.97 9.68 6.75 2.60


0.05 2.64

0.01 3.25

127

11.20 Factor A: Row Effect (Quarter)

H0: A1 = A2 = A3 = A4 = 0 � quarter means are the same

H1: Not all the Aj are equal to zero � quarter means differ

Factor B: Column Effect (Supplier)

H0: B1 = B2 = B3 = 0 �supplier means are the same

H1: Not all the Bk are equal to zero � supplier means differ

Interaction Effect (Quarter×Supplier)


H1: Not all ABjk are equal to zero � there is an interaction effectb. See tables.c. αQuarters differ at = .05, F = 6.01, p α-value < .0020. Suppliers differ at = .05, F = 4.30, p-value = .0211.

αInteraction is significant at = .05, F = 0.44, p-value = .8446.d. The p-values indicate that both main effects are significant. The interaction effect is not significant, since

about 84 samples in 100 would show an F statistic this large in the absence of interaction.e. The interaction plot lines do not cross to a noticeable degree, so we see no evidence of interaction.

Table of Means

Factor 2 (Supplier)

Factor 1 (Quarter) Supplier 1 Supplier 2 Supplier 3

Qtr 1 12.3 10.8 14.3 12.4

Qtr 2 12.3 11.0 12.3 11.8

Qtr 3 10.3 8.5 10.0 9.6

Qtr 4 10.5 9.5 10.8 10.3

11.3 9.9 11.8 11.0


Source SS df MS F p-

value

Factor 1 (Quarter) 63.23 3 21.076 6.01 .0020

Factor 2 (Supplier) 30.17 2 15.083 4.30 .0211

Interaction 9.33 6 1.556 0.44 .8446

Error 126.25 36 3.507

Total 228.98 47



Qtr 3 Qtr 4 Qtr 2 Qtr 1

9.6 10.3 11.8 12.4

Qtr 3 9.6

Qtr 4 10.3 0.87

Qtr 2 11.8 2.94 2.07

Qtr 1 12.4 3.71 2.83 0.76


0.05 2.70

0.01 3.35

128



Supplier 2 Supplier 1 Supplier 3

9.9 11.3 11.8

Supplier 2 9.9

Supplier 1 11.3 2.08

Supplier 3 11.8 2.83 0.76


0.05 2.45

0.01 3.11

11.21 We fail to reject the null hypothesis of equal means. The p-value (.1000) exceeds .05. There is no significant difference among the GPAs. We ignore importance, since the results are not significant. The dot plot comparison confirms that differences are not strong. For tests of homogeneity of variances, the critical

value of Hartley’s Fmax statistic from Table 11.5 is 13.7 (df1 = c = 4, df2 = n/c−1 = 25/4 − 1 = 5). We fail to reject the null hypothesis of homogeneous variances since the text statistic Fmax = (0.3926)/(.0799) = 4.91 is less than the critical value. This result agrees with Levene’s test (p-value = .290) and the confidence intervals overlap.

Mean n Std. Dev Variance Group

2.484 5 0.6240 0.3894 Freshman

2.916 7 0.6265 0.3926 Sophomore

3.227 7 0.2826 0.0799 Junior

3.130 6 0.4447 0.1978 Senior

2.968 25 0.5477 Total

One factor ANOVA


value

Treatment 1.8180 3 0.60599 2.36 .1000

Error 5.3812 21 0.25625

Total 7.1992 24

129

Class


Sophomore

Senior

Junior

Freshman

2.52.01.51.00.50.0

Bartlett's Test

0.606

Test Statistic 3.75

P-Value 0.290

Lev en e's Test

Test Statistic 0.63

P-Value

Test for Equal Variances for GPA

11.22 We fail to reject the null hypothesis of equal means. The p-value (.0523) exceeds .05, although it is a very close decision. We ignore importance, since the results are not significant. The dot plot does suggest that differences exist. A larger sample might be in order. For tests of homogeneity of variances, the critical value

of Hartley’s Fmax statistic from Table 11.5 is 8.38 (df1 = c = 3, df2 = n/c−1 = 23/3 − 1 = 6). We fail to reject the null hypothesis of homogeneous variances since the text statistic Fmax = (451.11)/(89.41) = 5.05 is less than the critical value. This result agrees with Levene’s test (p-value = .092) and the confidence intervals

overlap. However, the variances would have been judged unequal had we used α = .10.

Mean n Std. DevVariance

s Group

261.2 5 11.95 142.70 Budgets

238.0 10 21.24 451.11 Payables

244.4 8 9.46 89.41 Pricing

245.3 23 17.91 Total

One factor ANOVA


value

Treatment 1,803.76 2 901.880 3.43 .0523

Error 5,256.68 20 262.834

Total 7,060.43 22

130

Dept


Pricing

Payables

Budgets

50403020100

Bartlett's Test

0.140

Test Statistic 4.77

P-Value 0.092

Lev en e's Test

Test Statistic 2.17

P-Value

Test for Equal Variances for Days

11.23 We reject the null hypothesis of equal means. The p-value (.0022) is less than .05. Even a small difference in output could be important in a large array of solar cells. The dot plot does suggest that differences exist. Cell Type C is above the overall mean, while Cell Type B is below the overall mean. For tests of homogeneity of variances, the critical value of Hartley’s Fmax statistic from Table 11.5 is 10.8 (df1 = c = 3, df2

= n/c−1 = 18/3 − 1 = 5). We fail to reject the null hypothesis of homogeneous variances since the text statistic Fmax = (4.57)/(4.00) = 1.14 is less than the critical value. This result agrees with Levene’s test (p-value = .975) and the confidence intervals overlap. Tukey tests show that C differs from A and B.

Mean n Std. Dev Variances Group

123.8 6 2.04 4.17 Cell Type A

123.0 6 2.00 4.00 Cell Type B

127.8 6 2.14 4.57 Cell Type C

124.9 18 2.91 Total

One factor ANOVA


value

Treatment 80.11 2 40.056 9.44 .0022

Error 63.67 15 4.244

Total143.7

8 17

131

Post hoc analysis

Tukey simultaneous comparison t-values (d.f. = 15)Cell Type

BCell Type

ACell Type

C

123.0 123.8 127.8

Cell Type B 123.0

Cell Type A 123.8 0.70

Cell Type C 127.8 4.06 3.36


0.05 2.60

0.01 3.42

p-values for pairwise t-testsCell Type

BCell Type

ACell Type

C

123.0 123.8 127.8

Cell Type B 123.0

Cell Type A 123.8 .4943

Cell Type C 127.8 .0010 .0043

Cell Type


Cell Type C

Cell Type B

Cell Type A

7654321

Bartlett's Test

0.975

Test Statistic 0.02

P-Value 0.989

Lev en e's Test

Test Statistic 0.03

P-Value

Test for Equal Variances for Watts

132

11.24 We cannot reject the null hypothesis of equal means. The p-value (.4188) exceeds .05. Since the means do not differ significantly, the issue of importance is moot. The dot plot does not suggest any differences. For tests of homogeneity of variances, the critical value of Hartley’s Fmax statistic from Table 11.5 is 15.5 (df1 = c

= 3, df2 = n/c−1 = 15/3 − 1 = 4). We fail to reject the null hypothesis of homogeneous variances since the text statistic Fmax = (246,324)/(103,581) = 2.38 is less than the critical value. This result agrees with Levene’s test (p-value = .715) and the confidence intervals overlap.


1,282.0 5 496.31 246,324 Goliath

1,376.0 5 321.84 103,581 Varmint

1,638.0 5 441.78 195,170 Weasel

1,432.0 15 424.32 Total

One factor ANOVA

Source SS d

f MS F p-

value

Treatment 340,360.00 2 170,180.000 0.94 .4188

Error 2,180,280.00 12 181,690.000

Total 2,520,640.00 14

Vehicle


Weasel

Varmint

Goliath

2000150010005000

Bartlett's Test

0.569

Test Statistic 0.67

P-Value 0.715

Lev en e's Test

Test Statistic 0.59

P-Value

Test for Equal Variances for Damage

133

11.25 We cannot reject the null hypothesis of equal means. The p-value (.1857) exceeds .05. Since the means do

not differ significantly, the issue of importance is moot. The dot plot does not suggest any differences. For tests of

homogeneity of variances, the critical value of Hartley’s Fmax statistic from Table 11.5 is 20.6 (df1 = c = 4, df2 = n/c−1 =

22/4 − 1 = 4). We fail to reject the null hypothesis of homogeneous variances since the text statistic Fmax = (141.610)/

(45.428) = 3.12 is less than the critical value. This result agrees with Levene’s test (p-value = .739) and the confidence

intervals overlap.

Mean n Std. Dev Variances Group

14.2 5 8.29 68.724 Hospital A

21.5 4 11.90 141.610 Hospital B

16.9 7 7.95 63.203 Hospital C

9.3 6 6.74 45.428 Hospital D

15.0 22 8.98 Total

One factor ANOVA


Treatment 388.96 3 129.655 1.79 .1857

Error 1,305.99 18 72.555

Total 1,694.95 21

Hospital


Hospital D

Hospital C

Hospital B

Hospital A

80706050403020100

Bartlett's Test

0.645

Test Statistic 1.26

P-Value 0.739

Lev en e's Test

Test Statistic 0.56

P-Value

Test for Equal Variances for Wait

135

11.26 We reject the null hypothesis of equal means. The p-value (.0029) is less than .05. Productivity differences could be important in a competitive market, and might signal a need for additional worker training. The dot plot suggests that Plant B has lower productivity. For tests of homogeneity of variances, the critical value of

Hartley’s Fmax statistic from Table 11.5 is 6.94 (df1 = c = 3, df2 = n/c−1 = 25/3 − 1 = 7). We fail to reject the null hypothesis of homogeneous variances since the text statistic Fmax = (2.9791)/(0.68558) = 4.35 is less than the critical value. This result agrees with Levene’s test (p-value = .122) and the confidence intervals overlap.


3.97 9 0.828 0.685584 Plant A

3.02 6 1.094 1.196836 Plant B

5.57 10 1.726 2.979076 Plant C

4.38 25 1.647 Total

One factor ANOVA


value

Treatment 26.851 2 13.4253 7.72 .0029

Error 38.269 22 1.7395

Total 65.120 24

Post hoc analysis


Plant B Plant A Plant C

3.02 3.97 5.57

Plant B 3.02

Plant A 3.97 1.37

Plant C 5.57 3.75 2.65


0.05 2.52

0.01 3.25


Plant B Plant A Plant C

3.02 3.97 5.57

Plant B 3.02

Plant A 3.97 .1855

Plant C 5.57 .0011 .0148

136

Plant


Plant C

Plant B

Plant A

4.03.53.02.52.01.51.00.5

Bartlett's Test

0.208

Test Statistic 4.21

P-Value 0.122

Lev en e's Test

Test Statistic 1.69

P-Value

Test for Equal Variances for Output

11.27 It appears that the researcher is not treating this as a randomized block, since both factors appear to be of research interest. Hence, this will be referred to as a two-factor ANOVA without replication.Factor A (Method):

H0: A1 = A2 = A3


Factor B: (Road Condition)

H0: B1 = B2 = B3

H1: Not all the Bk are equal to zeroMean stopping distance is significantly affected by surface (p = 0.0002) but not by road condition (p = 0.5387). Tukey tests show significant differences between Ice and the other two surfaces. To test for homogeneous variances the critical value of Hartley’s statistic is F3,2 = 87.5. Since Fmax = 1.37 (for Method) and Fmax = 14.5 (for Surface) we cannot reject the hypothesis of equal variances.

Table of Means


452.000 3 9.849 Ice

184.667 3 37.528 Split Traction

154.000 3 11.358 Packed Snow

271.000 3 151.803 Pumping

249.667 3 177.827 Locked

270.000 3 164.739 ABS

263.556 9 143.388 Total

137

Two-Factor ANOVA Without Replication


Column (Surface) 161,211.56 2 80,605.778 134.39 .0002

Row (Method) 869.56 2 434.778 0.72 .5387

Error 2,399.11 4 599.778

Total 164,480.22 8

Post hoc analysis


Packed Snow Split Traction Ice

154.000 184.667 452.000

Packed Snow 154.000

Split Traction 184.667 1.53

Ice 452.000 14.90 13.37


0.05 3.56

0.01 5.74


Packed Snow Split Traction Ice

154.000 184.667 452.000

Packed Snow 154.000

Split Traction 184.667 .1999

Ice 452.000 .0001 .0002

11.28 We cannot reject the null hypothesis of equal means. The p-value (.3744) exceeds .05. The dot plot does not show large differences among manufacturers. For tests of homogeneity of variances, the critical value of

Hartley’s Fmax statistic from Table 11.5 is 333 (df1 = c = 7, df2 = n/c−1 = 22/7 − 1 = 2). We fail to reject the null hypothesis of homogeneous variances since the text statistic Fmax = (.000372567)/(0.000013815) = 27.0 is less than the critical value. This result agrees with Levene’s test (p-value = .315) and the confidence intervals overlap (although they are rather strange in appearance).

138

Table of Means


0.03255 2 0.004455 0.000019847 Aunt Millie's

0.03648 5 0.015563 0.000242207 Brownberry

0.02950 2 0.008768 0.000076878 Compass Food

0.03943 3 0.004277 0.000018293 Interstate Brand Co.

0.02007 3 0.019302 0.000372567 Koepplinger's Bakery

0.03437 3 0.003717 0.000013816 Metz Baking Co.

0.04223 4 0.009982 0.000099640 Pepperidge Farm

0.03441 22 0.012320 Total

One Factor ANOVA

Source SS d

f MS F p-value

Manufacturer0.00101

4 6 0.0001690 1.17 .3744

Error0.00217

4 15 0.0001449

Total0.00318

8 21

Mfgr


Pepperidge Farm

Metz Baking Co.

Koepplinger's Bakery

Interstate Brand Co.

Compass Food

Brownberry

Aunt Millie's

2.01.51.00.50.0

Bartlett's Test

0.729

Test Statistic 7.07

P-Value 0.315

Lev en e's Test

Test Statistic 0.60

P-Value

Test for Equal Variances for Fat

139

11.29 We cannot reject the null hypothesis of equal means. The p-value (.8166) exceeds .05. The dot plot does not show large differences among groups, although the fourth quintile seems to have smaller variance. For tests of homogeneity of variances, the critical value of Hartley’s Fmax statistic from Table 11.5 is 7.11 (df1 = c = 5,

df2 = n/c−1 = 50/5 − 1 = 9). We reject the null hypothesis of homogeneous variances since the text statistic Fmax = 112.04/14.13 = 7.93 exceeds the critical value. This result agrees with Levene’s test (p-value = .036) even though the confidence intervals do overlap.

Table of Means


30.00 10 9.548 Quintile 1

29.12 10 10.213 Quintile 2

31.17 10 10.585 Quintile 3

28.71 10 3.759 Quintile 4

26.66 10 6.305 Quintile 5

29.13 50 8.286 Total

One Factor ANOVA


Treatment 111.959 4 27.9897 0.39 .8166

Error 3,252.570 45 72.2793

Total 3,364.529 49

Quintile


Quintile 5

Quintile 4

Quintile 3

Quintile 2

Quintile 1

2520151050

Bartlett's Test

0.018

Test Statistic 10.28

P-Value 0.036

Lev en e's Test

Test Statistic 3.33

P-Value

Test for Equal Variances for Dropout

140

11.30 This is a replicated experiment with two factors and interaction. Based on the p-values, we conclude that means differ for Angle (p = .0088) and for Vehicle(p = .0007). However, there is no significant interaction

for Angle×Vehicle (p = .6661). The interaction plots support this conclusion, as the lines do not cross. The Tukey tests say that pairwise means differ for Rear End and Slant, and that Goliath differs from Varmint and Weasel.

Table of Means

Factor 2 (Vehicle)

Factor 1 (Angle) Goliath Varmint Weasel

Head-On 983.3 1,660.0 1,896.7 1,513.3

Slant 1,470.0 1,733.3 1,996.7 1,733.3

Rear end 973.3 1,220.0 1,513.3 1,235.6

1,142.2 1,537.8 1,802.2 1,494.1

Two Factor ANOVA with Replication


value

Factor 1 (Angle) 1,120,029.63 2 560,014.815 6.22 .0088

Factor 2 (Vehicle) 1,985,985.19 2 992,992.593 11.04 .0007

Interaction 216,637.04 4 54,159.259 0.60 .6661

Error 1,619,400.00 18 89,966.667

Total 4,942,051.85 26



Rear end Head-On Slant

1,235.6 1,513.3 1,733.3

Rear end 1,235.6

Head-On 1,513.3 1.96

Slant 1,733.3 3.52 1.56


0.05 2.55

0.01 3.32



Goliath Varmint Weasel

1,142.2 1,537.8 1,802.2

Goliath 1,142.2

Varmint 1,537.8 2.80

Weasel 1,802.2 4.67 1.87


0.05 2.55

0.01 3.32

141

11.31 This is a replicated experiment with two factors and interaction. The only difference between this experiment and the previous one is that the sample size is doubled, which raises the F statistics and reduces the p-values. Based on the p-values, we conclude that means differ for Crash Type (p < .0001) and for

Vehicle(p < .0001). However, there is no significant interaction for Crash Type×Vehicle (p = .2168). Notice, however, that the interaction p-value is smaller than in the previous experiment, showing that larger sample size alone (ceteris paribus) can make an effect more “significant.” The interaction plots support the conclusion of no interaction, as the lines do not cross to any major extent. The Tukey tests suggest that pairwise means differ for Rear End and Slant, and that Goliath differs from Varmint and Weasel.

Table of Means

Factor 2 (Vehicle)

Factor 1 (Angle) Goliath Varmint Weasel

Head On 983.3 1,660.0 1,896.7 1,513.3

Slant 1,470.0 1,733.3 1,996.7 1,733.3

Rear end 973.3 1,220.0 1,513.3 1,235.6

1,142.2 1,537.8 1,802.2 1,494.1



value

Factor 1 (Angle) 2,240,059.26 2 1,120,029.630 15.56 7.30E-06

Factor 2 (Vehicle) 3,971,970.37 2 1,985,985.185 27.59 1.51E-08

Interaction 433,274.07 4 108,318.519 1.50 .2168

Error 3,238,800.00 45 71,973.333

Total 9,884,103.70 53



Rear end Head On Slant

1,235.6 1,513.3 1,733.3

Rear end 1,235.6

Head On 1,513.3 3.11

Slant 1,733.3 5.57 2.46


0.05 2.42

0.01 3.08

142



Goliath Varmint Weasel

1,142.2 1,537.8 1,802.2

Goliath 1,142.2

Varmint 1,537.8 4.42

Weasel 1,802.2 7.38 2.96


0.05 2.42

0.01 3.08

11.32 This is a replicated experiment with two factors and interaction. Based on the p-values, the means differ for Temperature (p = .0000) and for PVC Type(p = .0013). However, there is no interaction for

Temperature×PVC Type (p = .9100). We conclude that the burst strength is affected by temperature, by PVC type, but not by the interaction between temperature and PVC type. The dot plots suggest that PVC2 is the best brand. The pairwise Tukey tests indicate that there is a difference between PVC2 and PVC3, but no difference between PVC1 and PVC2 or PVC1 and PVC3.

Table of Means

Factor 2 (PVC Typr)

Factor 1 (Temperature) PVC1 PVC2 PVC3

Hot (70 Degrees C) 268.0 287.0 258.0 271.0

Warm (40 Degrees C) 314.0 334.3 306.0 318.1

Cool (10 Degrees C)) 354.0 361.3 335.3 350.2

312.0 327.6 299.8 313.1



Factor 1 (Temperature) 28,580.22 2 14,290.111 81.04 0.0000

Factor 2 (PVC Type) 3,488.89 2 1,744.444 9.89 .0013

Interaction 171.56 4 42.889 0.24 .9100

Error 3,174.00 18 176.333

Total 35,414.67 26

143



Hot (70o C) Warm (40o C) Cool (10o C)

271.0 318.1 350.2

Hot (70o C) 271.0

Warm (40o C) 318.1 7.53

Cool (10o C)) 350.2 12.66 5.13


0.05 2.55

0.01 3.32



PVC3 PVC1 PVC2

299.8 312.0 327.6

PVC3 299.8

PVC1 312.0 1.95

PVC2 327.6 4.44 2.48


0.05 2.55

0.01 3.32

11.33 This is a two-factor ANOVA without replication. We conclude that tax audit rate is not significantly affected by year (p = 0.6153) but is significantly affected by taxpayer class (p < .0001). There is no interaction as there is no replication. MegaStat calls the column factor the “treatment” but the problem wording suggests that both factors are of research interest.

144

Table of Means


2.21500 10 1.55302 1990

2.31800 10 1.64241 1991

2.08000 10 1.53937 1992

1.85700 10 1.30518 1993

1.99800 10 1.36745 1993

2.41400 10 1.50986 1994

1.00167 6 0.49906 1040 A TPI

0.95000 6 0.20995 1040 TPI < $25,000

0.74167 6 0.17360 1040 TPI $25,000-50,000

1.06167 6 0.23853 1040 TPI $50,000-100,000

4.37167 6 1.30310 1040 TPI > $100,000

3.02000 6 1.69680 C-GR < $25,000

2.55833 6 0.38301 C-GR $25,000-100,000

3.81333 6 0.25524 C-GR > $100,000

1.32167 6 0.23216 F-GR < $100,000

2.63000 6 0.80230 F-GR > $100,000

2.14700 60 1.43886 Total

Two Factor ANOVA without Replication


Column (Year) 2.1594 5 0.43189 0.72 .6153

Row (Taxpayer Class) 92.8146 9 10.31274 17.08 8.46E-12

Error 27.1744 45 0.60388

Total 122.1485 59

145

11.34 This is a two-factor ANOVA with replication and interaction. Based on the p-values, we conclude that the means differ by Weight (p = .0009) and by Medication (p = .0119). There is no significant interaction effect

Weight×Medication (p = .9798).

Table of Means

Means: Factor 2 (Medication)

Factor 1 (Weight) Med 1 Med 2 Med 3 Med 4

1.1 or Less 133.0 141.0 136.0 127.5 134.4

1.1 to 1.3 140.5 141.5 140.5 132.0 138.6

1.3 to 1.5 148.5 153.0 148.5 140.0 147.5

140.7 145.2 141.7 133.2 140.2


Factor 1 (Weight) 717.58 2 358.792 13.25 .0009

Factor 2 (Medication) 459.00 3 153.000 5.65 .0119

Interaction 27.75 6 4.625 0.17 .9798

Error 325.00 12 27.083

Total 1,529.33 23



1.1 or Less 1.1 to 1.3 1.3 to 1.5

134.4 138.6 147.5

1.1 or Less 134.4

1.1 to 1.3 138.6 1.63

1.3 to 1.5 147.5 5.04 3.41


0.05 2.67

0.01 3.56



Med 4 Med 1 Med 3 Med 2

133.2 140.7 141.7 145.2

Med 4 133.2

Med 1 140.7 2.50

Med 3 141.7 2.83 0.33

Med 2 145.2 3.99 1.50 1.16


0.05 2.97

0.01 3.89

146

11.35 This is a two-factor ANOVA with replication and interaction. We conclude that means do not differ by Instructor Gender (p = .43) or by Student Gender (p = .24) but there is an interaction effect between the two

factors Instructor Gender×Student Gender (p = .03). The sample size is very large, so it is unlikely that any effect was overlooked (the test should have excellent power).

11.36 This is an unreplicated two-factor ANOVA. Although MegaStat calls it a randomized block ANOVA, the wording of the problem suggests that both factors are of research interest. We conclude that texture is not significantly affected by age group (p = 0.2999) or by surface type (p = 0.2907). The dot plots support these conclusions, since there are no strong or consistent differences in the groups. No interaction is estimated since there is no replication.

Table of Means

Mean n Std. Dev

5.1500 4 1.2261 Shiny

5.3750 4 0.8846 Satin

6.1250 4 0.4924 Pebbled

4.9500 4 0.8851 Pattern

5.7750 4 1.1236 Youth (Under 21)

5.7250 4 0.4031 Adult (21 to 39)

5.4500 4 0.9292 Middle-Age (40 to 61)

4.6500 4 0.9983 Senior (62 and over)

5.4000 16 0.9345 Total



value

Columns (Surface) 3.165 3 1.0550 1.42 .2999

Rows (Age Group) 3.245 3 1.0817 1.46 .2907

Error 6.690 9 0.7433

Total 13.100 15

147

11.37 This is an unreplicated two-factor ANOVA. Although MegaStat calls it a randomized block ANOVA, the wording of the problem suggests that both factors are of research interest. Call waiting time is not significantly affected by day of the week (p = 0.1760) but is significantly affected by time of day (p = 0.0001) as indicated in the bar chart of means. No interaction is estimated since there is no replication.

Table of Means

Mean n Std. Dev

49.077 26 25.575 Mon

60.269 26 35.629 Tue

53.692 26 28.369 Wed

49.577 26 28.365 Thu

44.808 26 17.253 Fri

43.200 5 15.786 6:00

62.400 5 16.502 6:30

61.800 5 32.874 7:00

68.400 5 28.789 7:30

65.800 5 11.883 8:00

65.200 5 20.042 8:30

57.800 5 22.061 9:00

60.800 5 36.224 9:30

60.000 5 18.371 10:00

88.200 5 45.779 10:30

45.600 5 19.256 11:00

34.400 5 5.727 11:30

70.200 5 29.811 12:00

53.000 5 16.598 12:30

47.000 5 14.577 13:00

69.600 5 41.107 13:30

86.800 5 35.024 14:00

38.000 5 3.674 14:30

35.200 5 1.095 15:00

43.200 5 11.167 15:30

28.800 5 19.045 16:00

27.400 5 10.922 16:30

32.600 5 4.506 17:00

26.800 5 3.271 17:30

42.400 5 41.283 18:00

24.000 5 17.436 18:30

148

51.485 130 27.745 Total



Columns (Day of Week) 3,537.58 4 884.396 1.62 .1760

Rows (Time of Day) 41,046.47 25 1,641.859 3.00 .0001

Error 54,716.42 100 547.164

Total 99,300.47 129

0

10

20

30

40

50

6:00

7:00

8:00

9:00

10:00

11:00

12:00

13:00

14:00

15:00

16:00

17:00

18:00

Time of Day

Mean

Call

Vo

lum

e

11.38 (a) This is a two Factor ANOVA. (b) There are 4 friends since df = 3 (df = r−1) and 3 months since df =2

(df = c−1). The total number of observations is 36 since df = 35 (df = n−1). Thus, since the data matrix is

4×3 (12 cells) there must have been 36/12 = 3 observations per cell (i.e., 3 bowling scores per friend per

month). (c) Based on the p-values, we see Month (p = .0002) is significant at α = .01, Friend (p < .0001) is

significant at α = .01, and there is only a weak interaction since Month×Friend (p = .0786) is only significant

at α = .10. We conclude that mean bowling scores are influenced by the month, friend and possibly by an interaction between the month (time of year) and the bowler.

149

11.39 (a) This is a randomized block (unreplicated two-factor) ANOVA. (b) Based on the p-values, air pollution is significantly affected by car type (p < .0001) and time of day (p < .0001). (c) Variances may appear to be unequal. Equal variances are important because analysis of variance assumes that observations on the response variable are from normally distributed populations that have the same variance. However, we cannot rely on our eyes alone to judge variances, and we should do a test for homogeneity. (d) In Hartley’s test, for freeway, we get Fmax = (14333.7)/(2926.7) = 4.90 which is less than the critical value from Table 11.5 is F4, 4 = 20.6, and so we fail to reject the hypothesis of equal variances. Similarly, we fail to reject the null of equal variances for time of day, since Fmax = (14333.6)/872.9 = 16.4 is less than the critical value from Table 11.5 F5, 3 = 50.7.

11.40 (a) This is a two factor ANOVA with replication. (b) There are 5 suppliers since df = 4 (df = r−1) and 4

quarters since df =3 (df = c−1). The total number of observations is 100 since df = 99 (df = n−1). Therefore,

we have a 5×4 data matrix (20 cells) which implies 100/20 = 5 observations per cell (i.e., 5 observations per supplier per quarter). (c) Based on the p-values Quarter (p = .0009) and Supplier (p < .0001), we conclude

that both main effects are significant at α = .01. However, there is also a very strong interaction effect

Quarter×Supplier (p = .0073). We conclude that shipment times are influenced by the quarter, supplier and the interaction between the quarter (time of year) and the supplier. However, in view of the interaction effect, the main effects may be problematic.

11.41 (a) This is a one Factor ANOVA. The number of bowlers is 5 since d.f. = 4 (df = c−1). That is, there were 5

data columns. The sample size is 67 since df = 66 (df = n−1). (c) Based on the p-value from the ANOVA table (p < .0001) we reject the null hypothesis of no difference between the mean scores and conclude there is a difference. (d) The sample variances range from 77.067 to 200.797. To test the hypothesis of homogeneity, we compare Hartley’s critical value F5,12 = 5.30 with the sample statistic Fmax = (200.797)/(77.067) = 2.61, and fail to reject the hypothesis of equal variances.

11.42 (a) This is a one Factor ANOVA (b) Based on the p-value from the ANOVA table (essentially zero) we strongly reject the null hypothesis of no difference in mean profit/asset ratios. (c) The plots indicate that company size (as measured by employees) does affect profitability per dollar of assets. There are possible outliers in several of the groups. (d) Variances may be unequal, based on the dot plots and possible outliers. (e) To test the hypothesis of homogeneity, we compare Hartley’s critical value F4,123 = 1.96 with the sample statistic Fmax = (34.351)/(8.108) = 4.24 and reject the hypothesis of equal variances. There isn’t anything we can do about it, though. (f) Specifically, the Tukey tests show that small companies differ significantly from

medium large, and huge companies (although the latter three categories are not different at α = .05).

150

Chapter 12

Bivariate Regression

12.1 For each sample: H0: ρ = 0 versus H1: ρ ≠ 0.

Summary Table

Sample df r t tα rα Decision

a 18 .45 2.138 2.101 .444 Reject

b 28 −.35 −1.977 1.701 .306 Reject

c 5 .6 1.677 2.015 .669 Fail to Reject

d 59 −.3 −2.416 2.39 .297 Reject

12.2 a. The scatter plot shows a positive correlation between hours worked and weekly pay.

b.

Hours Worked (X) Weekly Pay (Y)2( )ix x− 2( )iy y− ( )( )i ix x y y− −

10 93 100 7056 840

15 171 25 36 30

20 204 0 729 0

20 156 0 441 0

35 261 225 7056 1260

20 177 350 15318 2130

x y SSxx SSyy SSxy

2130.9199

350 15318r = =

c. t.025 = 3.182

d. 2

5 2.9199 4.063

1 (.9199)t

−= =

−. We reject the null hypothesis of zero correlation.

e. p-value = .0269.

151

12.3 a. The scatter plot shows a negative correlation between operators and wait time.

b.

Operators (X) Wait (Y)2( )ix x− 2( )iy y− ( )( )i ix x y y− −

4 385 4 1444 −76

5 335 1 144 12

6 383 0 1296 0

7 344 1 9 −3

8 288 4 3481 −118

6 347 10 6374 −185

x y SSxx SSyy SSxy

185.7328

10 6374r

−= = −

c. t.025 = ±3.183

d. 2

5 2.7328 1.865

1 ( .7328)t

−= − = −

− −. We fail to reject the null hypothesis of zero correlation.

e. p-value = .159.

12.4 a. The scatter plot shows little correlation between age and amount spent.

b. rcalculated = −.292c. t.025 = ±2.306

d. 2

10 2.292 .864

1 ( .292)t

−= − = −

− −

e. critical2

2.306.692

2.306 10 2r = ± = ±

+ −.

f. Because rcalculated (−.292) > −.692, we fail to reject the null hypothesis of zero correlation.

152

12.5 a. The scatter plot shows a positive correlation between returns from last year and returns from this year. b. rcalculated = .5313c. t.025 = ±2.131

d. 2

17 2.5313 2.429

1 (.5313)t

−= =

−

e. critical2

2.131.482

2.131 17 2r = ± = ±

+ −f. Because rcalculated (.5313) > .482, we reject the null hypothesis of zero correlation.

12.6 a. The scatter plot shows a positive correlation between orders and ship cost.

b. rcalculated = .820c. t.025 = ±2.228

d. 2

12 2.820 4.530

1 (.820)t

−= =

−

e. critical2

2.228.576

2.228 12 2r = ± = ±

+ −f. Because rcalculated (.820) > .576, we reject the null hypothesis of zero correlation.

12.7 a. Correlation Matrix

1-Year 10-Year

1-Year 1.000

3-Year -.095

5-Year .014

10-Year .341 1.000

12 sample size

± .576 critical value .05 (two-tail)


d. There were positive correlations between years 3 and 5 and years 5 and 10. Higher returns in Year 3 lead to higher returns in Year 5 and also in Year 10.

12.8 a. An increase in the price of $1, reduces its expected sales by 37.5 units. b. Sales = 842 – (20)*37.5 = 92c. From a practical point of view no. A zero price is unrealistic.

153

12.9 a. Increasing the size by 1 square foot raises the price by $150.b. HomePrice = 125000 + 150*(2000) = $425,000c. No, the intercept has no meaning.

12.10 a. Increasing the average revenue by 1 million dollars raises the net income by $30,700.b. If revenue is zero, then net income is 2277 millions dollars., suggests that the firm has net income when

revenue is zero. Does not seem to be meaningful.c. Revenue = 2277 + .0307*(1000) = 2307.7 million dollars

12.11 a. Increasing the median income by $1,000 raises the median home price by $2610.b. If median income is zero, then the model suggests that median home price is $51,300. While it does not

seem logical that the median family income for any city is zero, it is unclear what the lower bound would be.

c. HomePrice = 51.3 + 2.61*(50) = $181,800Homeprice = 51.3 + 2.61*(100) = $312,300

12.12 a. Increasing the number of hours worked per week by 1 hour reduces the expected number of credits by .07.

b. Yes, the intercept makes sense in this situation. It is possible that a student does not have a job outside of school.

c. Credits = 15.4 + .07*(0) = 15.4 creditsCredits = 15.4 + .07*(40) = 12.6 creditsThe more hours a student works, the less credits (courses) he will take on average.

12.13 a. Chevy Blazer: a one year increase in vehicle age reduces the price by $1050. Chevy Silverado: a one year increase in vehicle age reduces the price by $1339.

b. Chevy Blazer: If age = 0 then price = $16,189. This could be the price of a new Blazer.Chevy Silverado: If age = 0 then price = $22,951. This could be the price of a new Silverado.

c. 16,189 – 1,050*5 = $10,939

22,951 −1,339*5 = $16,256

12.14 a. Tips = 20+ 10*Hours (Answers will vary.)b. One hour of work yields on average $10 in tips.c. The intercept has no meaning in this case.

12.15 a. Units Sold = 300 − 150*Price (Answers will vary.)b. One dollar reduction in price increases units sold by 150 on average.c. If price is zero, then units sold = 300. This is not meaningful, price is never zero.

12.16 a.

Hours Worked (X) Weekly Pay (Y)2( )ix x− 2( )iy y− ( )( )i ix x y y− −

10 93 100 7056 840

15 171 25 36 30

20 204 0 729 0

20 156 0 441 0

35 261 225 7056 1260

20 177 350 15318 2130

x y SSxx SSyy SSxy

b. 1

21306.086

350b = = , 0 177 6.086(20) 55.286b = − = , y = 55.286 + 6.086X

154

c.

Hours Worked (xi) Weekly Pay (yi)

Estimated

Pay ( ˆiy ) ˆ

i iy y− 2ˆ( )i iy y− 2ˆ( )iy y− 2( )iy y−

10 93 116.146 -23.146 535.7373 3703.209 7056

15 171 146.576 24.424 596.5318 925.6198 36

20 204 177.006 26.994 728.676 3.6E-05 729

20 156 177.006 -21.006 441.252 3.6E-05 441

35 261 268.296 -7.296 53.23162 8334.96 7056

20 177 177.006 -0.006 3.6E-05 3.6E-05 0

20 177 2355.429 12963.79 15318

x y SSE SSR SST

d.2 12,963

.846215,318

R = =

e.

12.17 a.

Operators (X) Wait (Y)2( )ix x− 2( )iy y− ( )( )i ix x y y− −

4 385 4 1444 −76

5 335 1 144 12

6 383 0 1296 0

7 344 1 9 −3

8 288 4 3481 −118

6 347 10 6374 −185

x y SSxx SSyy SSxy

b. 1

18518.5

10b

−= = − , 0 347 18.5(6) 458b = + = , y = 458 − 18.5X

155

c.

Operators (xi) Wait Time (yi)

Estimated

Time ( ˆiy ) ˆ

i iy y− 2ˆ( )i iy y− 2ˆ( )iy y− 2( )iy y−

4 385 384 1 1 1369 1444

5 335 365.5 -30.5 930.25 342.25 144

6 383 347 36 1296 0 1296

7 344 328.5 15.5 240.25 342.25 9

8 288 310 -22 484 1369 3481

6 347 2951.5 3422.5 6374

x y SSE SSR SST

d.2 3, 422.5

.53696,374.0

R = =

e.

12.18 a. and b.

c. An increase of 1% in last year’s return leads to an increase, on average, of .458% for this year’s return.d. If last year’s return is zero, this year’s return is 11.155%. Yes, this is meaningful, returns can be zero.e. R2 = .2823 Only 28.23% of the variation in this year’s return is explained by last year’s return.

156

12.19 a. and b.

c. An increase of 100 orders leads to an average increase in shipping cost of $493.22. d. The intercept is not meaningful in this case.

e. R2 = .6717 67.17% of the variation in shipping costs is explained by number of orders.

12.20 a. and b.

c. An increase in age of 10 years leads to an average decrease in spending of $5.30. d. The intercept is not meaningful in this case.e. R2 = .0851 8.51% of the variation in spending is due to the variation in age. Age of the consumer has

little impact on the amount spent.

12.21 a. Y = 557.4511 + 3.0047*Xb. The 95% confidence interval is 3.0047 ± 2.042(0.8820) or (1.203, 4.806).

c. H0: β1 ≤ 0 versus H1: β1 > 0. Reject the null hypothesis if t > 1.697. t = 3.407 so we reject the null hypothesis.

d. p-value = .000944 so we reject the null hypothesis. The slope is positive. Increased debt is correlated with increased NFL team value.

12.22 a. Y = 7.6425 + 0.9467*Xb. The 95% confidence interval is 0.9467 ± 2.145(0.0936) or (0.7460, 1.1473).

c. H0: β1 ≤ 0 versus H1: β1 > 0. Reject the null hypothesis if t > 1.761. t = 10.118 so we reject the null hypothesis.

d. p-value = .000 so we reject the null hypothesis. The slope is positive. Increased revenue is correlated with increased expenses.

157

12.23 a. Y = 1.8064 + .0039*Xb. Intercept: t = 1.8064/0.6116 = 2.954, Slope: t = 0.0039/0.0014 = 2.786 (Excel output may be different

due to internal rounding.)c. df = 10, t.025 = 2.228.d. Intercept: p-value = .0144. Slope: p-value = .0167.e. (2.869)2 = 8.23f. This model fits the data quite fairly well. The F statistic is highly significant. Also, R2 = .452 indicating

almost half of the variation in annual taxes is explained by home price.

12.24 a. Y = 614.930 − 1.09.11*Xb. Intercept: t = 614.930/51.2343 = 12.002. Slope: t = −109.112/51.3623 = −2.124.c. df = 18, t.025 = 2.101.d. Intercept: p-value = .0000, Slope: p-value = .0478e. (−2.124)2

= 4.51f. This model has a poor fit. The F statistic is barely significant at a level of .05 and R2 = .2. Only 20% of

the variation in units sold can be explained by average price.

12.25 a.

SUMMARY OUTPUT

Regression Statistics

Multiple R 0.5313237

R Square 0.2823049

Adj R Square 0.2344586

Standard Error 4.3346058

Observations 17

ANOVA

df SS MS FSignificance

F

Regression 1 110.8584768 110.85848 5.9002402 0.028176778

Residual 15 281.8321114 18.788807

Total 16 392.6905882

CoefficientsStandard

Error t Stat P-value Lower 95% Upper 95%

Intercept 11.15488 2.190744205 5.091822 0.0001326 6.485418897 15.82434033

Last Year 0.4579749 0.188541436 2.429041 0.0281768 0.056108323 0.859841435

b. (0.05611, 0.85984) This interval does not contain zero therefore we can conclude that the slope is greater than zero.

c. The t statistic is 2.429 and the p-value is 0.02828. Because the p-value is less than 0.05, we can conclude that the slope is positive.

d. F = 5.90 with a p-value = .0282. This indicates that the model does provide some fit to the data.e. The p-values match. (2.429)2 = 5.90.f. This model provides modest fit to the data. Although the F statistic is significant, R2 shows that only 28%

of the variation in this year’s return is explained by last year’s return.

158

12.26 a.

SUMMARY OUTPUT


Multiple R 0.819555843

R Square 0.671671781

Adj R Square 0.638838959


Observations 12

ANOVA


F

Regression 1 7340819.551 7340819.55 20.457327 0.001103268

Residual 10 3588357.115 358835.7115

Total 11 10929176.67



Intercept -31.18952293 1059.8678 -0.0294277 0.9771025 -2392.7222 2330.343177

Orders 4.932152105 1.0904657 4.52297768 0.0011033 2.5024431 7.361861124

b. (2.502, 7.362) This interval does not contain zero therefore we can conclude that the slope is greater than zero.

c. The t statistic is 4.523 and the p-value is 0.0011. Because the p-value is less than 0.05, we can conclude that the slope is positive.

d. F = 20.46 with a p-value = .0011. This indicates that the model does provide some fit to the data.e. The p-values match. (4.523)2 = 20.46.f. This model provides a good fit to the data. The F statistic is highly significant and R2 shows that 67% of

the variation in shipping cost is explained by number of orders.

12.27 a.

SUMMARY OUTPUT


Multiple R 0.291791

R Square 0.085142

Adj R Square -0.0292153


Observations 10

ANOVA


F

Regression 1 3.372666112 3.3726661 0.7445265 0.413332705

Residual 8 36.23958389 4.529948

Total 9 39.61225



159

Intercept 6.9609354 2.088494459 3.332992 0.0103383 2.144858565 11.777012

Age -0.0529794 0.061399756 -0.8628595 0.4133327 -0.194567453 0.0886087

b. (−0.1946, 0.0886) This interval does contain zero therefore we cannot conclude that the slope is greater than zero.

c. The t statistic is −0.863 and the p-value is ..4133. Because the p-value is greater than 0.05, we cannot conclude that the slope is positive.

d. F = 0.745 with a p-value = .4133. This indicates that the model does not fit the data.e. The p-values match. (−0.863)2 = 0.745.f. This model does not fit the data. The F statistic is not significant and R2 shows that only 8.5% of the

variation in dollar’s spent is explained by the moviegoer’s age.

12.28 For only two of the data sets, F and I, are the data time series data. The rest are cross-sectional data.



12.31 For A, B, D, F, H, and I should expect a positive sign for slope.

12.32 A: positive relationship between income and home priceB: positive relationship between employees and revenueC: positive relationship between ELOS and ALOSD: positive relationship between HP and Cruising SpeedE: inverse relationship between years in circulation and weight of a nickelF: no relationship between changes in the money supply and changes in the CPIG: inverse relationship between weight of a car and the gas mileage it gets in the city.H: positive relationship between fat calories per gram and calories per gramI: positive relationship between usage of electricity and monthly expenditure.

Data Set A Data Set B

Data Set C Data Set D

160

Data Set E Data Set F

Data Set G Data Set H

Data Set I

12.33 A: YesB: YesC: YesD: YesE: YesF: NoG: YesH: YesI: Yes

161

12.34 A: An increase in median income of $1000 , increases home price by $2,609.8 . No, the intercept does not have meaning. B: An increase in the number of employees by 1 unit increases revenue by .304 units. Yes, the intercept does have meaning. It is possible that the revenue is zero.C: An increase in ELOS of 1month increases ALOS by 1.03months. No, the intercept does not have meaning.D: An increase in one unit of horsepower, increases cruise speed by .1931 mph. No, the intercept does not have meaning.E: An increase in age of 1 year, reduces the weight by 0.004 grams.F: An increase in M1 of 1%in the prior year, increases the CPI by .1993% in the current year.G: An increase in the weight of a car by 1 pound, reduces its city mpg by 0.0045 mpg.H: An increase in the fat calories per gram by 1, increases total calories per gram by 2.2179.I: An increase in 1 kwh of usage, increases the monthly expenditure by $0.1037.

For 12.35 through 12.43, filling out the MegaStat Regression Dialog Box as displayed below will provide the information required for these questions. The dialog box displayed is for the Data Set A.

12.36 a. A: No, it means that the slope is different from zero.B: No, it means that the slope is different from zero.C: No, it means that the slope is different from zero.D: No, it means that the slope is different from zero.E: No, it means that the slope is different from zero.F: Yes, it means that the slope is not different from zero.G: No, it means that the slope is different from zero.H: No, it means that the slope is different from zero.I: No, it means that the slope is different from zero.

b. The hypothesis for each data set is: H0: β1 = 0 versus H1: β1 ≠ 0.A:.DF = 32, t-critical = 2.037; (e) Yes.B: DF = 22, t-critical = 2.074; (e) Yes.C: DF = 32, t-critical = 2.145; (e) Yes.D: DF = 50, t-critical = 2.009; (e) Yes.E: DF = 29, t-critical = 2.045; (e) Yes.F: DF = 39, t-critical = 2.023; (e) No.G: DF = 41, t-critical = 2.020; (e) Yes.H: DF = 18, t-critical = 2.101; (e) Yes.I: DF = 22, t-critical = 2.074; (e) Yes.

162

c. The p-value measures the chance of making this sample observation if the null hypothesis were true. Small p-values tell us that the null is false

d. The p-value approach is easier since the p-value is reported as part of the regression output and can easily be compared to the level of significance.

e. See part b above.

12.37 A: (a) Good; (b) see 12.36 (c); (c) Yes.B: (a) Very Good; (b) see 12.36 (c); (c) Yes.C: (a) Very Good; (b) see 12.36 (c); (c) Yes.D: (a) Good; (b) see 12.36 (c); (c) Yes.E: (a) Good; (b) see 12.36 (c); (c) Yes.F: (a) Very Poor; (b) see 12.36 (c); (c) No.G: (a) Good; (b) see 12.36 (c); (c) Yes.H: (a) Very Good; (b) see 12.36 (c); (c) Yes.I: (a) Excellent; (b) see 12.36 (c); (c) Yes.

12.38 A:. observations 20 and 29 have unusual residuals; no outliers.B: observations 6 and 21 have unusual residuals; no outliers.C: no observations have unusual residuals or outliers.D: observations 42 has an unusual residual; observation 28 is an outlier.E: observations 5, 8, and 28 have unusual residuals.F: observations 14, 19, and 20 have unusual residuals.G: observation 42 is an outlier. There are no unusual residuals.H: There are no unusual residuals or outliers.I: observation 14 is an unusual residual and 16 is an outlier.

12.39* Assumption of normal errors violated for: G and I

12.40* Heteroscedasticity a problem for: None

12.41* Durbin-Watson test appropriate only for: F, the value is .58. Indicates that autocorrelation is present. I the value is 1.95, indicating that autocorrelation is not present.

12.42* Answers will vary.

12.43* A: observation 8 has high leverage.B: observations 2 and 8 have high leverage.C: observations 4 and 10 have high leverage.D: observations 2, 3, 4, 14, 16, and 17 have high leverage.E: observations 5 and 25 have high leverage.F: observations 27, 33, 37, and 41 have high leverage.G: observations 12, 13 and 22 have high leverage.H: observations 6 and 11 have high leverage.I: observations 9 and 13 have high leverage.

12.44 No, r measures the strength and direction of the linear relationship, but not the amount of variation explained by the explanatory variable. R2.

12.25 H0: β1 = 0 versus H1: β1 ≠ 0. tcritical = 2.3069. t = 2.3256 so we reject the null hypothesis. The correlation is not zero.

12.46 The correlation coefficient is only .13, indicating that there exists a very weak positive correlation between prices on successive days. The fact that it is a highly significant result stems from a large sample size which increases power of the test. This means that very small correlations will show statistical significance even though the correlation is not truly important.

163

12.47 a. Y = 55.2 +.73*2000 = 1515.2 total free throws expected.b. No, the intercept is not meaningful. You can’t make free throws without attempting them.

c. Quick rule: 2ˆ

i n yxy t s−± 1515.2 + t(27)*53.2 = 1515.2 +/- 2.052*53.2 (1406.03, 1624.37)

12.48 a. Y = 30.7963+ .0343*X (R2 = .202, syx = 6.816)b. DF = 33, t.025 = 2.035c. t = 2.889 so we will reject the null hypothesis that the slope is zero.d. We are 95% confident that the slope is contained in the interval .0101 to .0584. This CI does not contain

zero, hence, there is a relationship between the weekly pay and the income tax withheld.e. (2.889)2 = 8.3463 f. The value of R-squared assigns only 20% of the variation in income withholding to the weekly pay.

While the F statistic is significant, the fit is only a modest fit. 12.49 a. Y = 1743.57 − 1.2163*X (R2 = .370, syx = 286.793)

b. DF = 13, t.025 = 2.160c. t = −2.764 so we will reject the null hypothesis that the slope is zero.d. We are 95% confident that the slope is contained in the interval −2.1617 to −0.2656. This CI does not

contain zero, hence, there is a relationship between the weekly pay and monthly machine downtime.e. (−2.764)2 = 7.639696 f. The value of R-squared assigns only 37% of the variation in monthly machine downtime to the monthly

maintenance spending (dollars). Thus, throwing more “money” at the problem of downtime will not completely resolve the issue. Indicates that there are most likely other reasons why machines have the amount of downtime incurred.

12.50 a. Y = 6.5763 +0.0452*X (R2 = .519, syx = 6.977)b. DF= 62, t.025 = 2.00 (using DF = 60)c. t = 8.183 so we will reject the null hypothesis that the slope is zero.

d. We are 95% confident that the slope is contained in the interval 0.0342 to 0.0563. This CI does not contain zero, hence, there is a relationship between the total assets (billions) and total revenue (billions).

e. (8.183)2 = 66.96 f. The value of R-squared assigns 51.9% of the variation in total revenue (billions) to the total assets

(billions). Thus, increasing assets will lead to an increase in income. However, the results also indicate that there are most likely other reasons why companies earn the revenue they do.

12.51 a. r = .677 b. The critical values for α =.01 are ± .393. The correlation coefficient of .677 is outside of these limits, so

we reject the hypothesis of no correlation and the sample evidence supports the notion of positive correlation.

c. The scatterplot shows a positive correlation between IBM and EDS stock prices.

164

12.52 a.

b. r = .792. This shows a fairly strong positive linear relationship between gestation and longevity.

c. At α = .01, the correlation coefficient of .792 is outside the critical range ± .537. We reject the hypothesis of no correlation. There is significant correlation.

12.53 a. The scatter plot indicates that there is a negative correlation between life expectancy and fertility.

b. r = −.846. There is a strong negative linear relationship between a nation’s life expectancy and their fertility rate.

c. At α = .01, the correlation coefficient of =.846 is outside the critical range ± .463. We reject the hypothesis of no correlation. There is a negative correlation between life expectancy and fertility rates.

12.54 a. The scatter plot shows almost no pattern.

165

b. r = −.105. At α = .05, the correlation coefficient of = −.105 is not outside the critical range ± .381. We fail to reject the hypothesis of no correlation. It appears there is very little relationship between price and accuracy rating of speakers.

12.55 For each of these, the scatter plot will contain the answers to (a), (b), and (d) with respect to the fitted equation.

c. Salary: The fit is good. Assessed: The fit is excellent. HomePrice2: The fit is good.d. Salary: An increase in the age by 1 year increases salary by $1447.4.

Assessed: An increase in 1 sq. ft. of floor space increases assessed value by $313.30.HomePrice2: An increase in 1 sq. ft. of home size increases the selling price by $209.20.

e. The intercept is not meaningful for any of these data sets as a zero value for any of X’s respectively cannot realistically result in a positive Y value.

12.56 a. estimated slope

standard errort =

Dependent Variable Estimated SlopeStd Error Differ from 0?

Highest grade achieved -0.027 0.009 −3.00 0.008 Yes

Reading grade equivalent

-0.07 0.018 −3.89 0 Yes

Class standing -0.006 0.003 −2.00 0.048 No

Absence from school 4.8 1.7 2.82 0.006 Yes

Grammatical reasoning 0.159 0.062 2.57 0.012 No

Vocabulary -0.124 0.032 −3.88 0 Yes

Hand-eye coordination 0.041 0.018 2.28 0.02 No

Reaction time 11.8 6.66 1.77 0.08 No

Minor antisocial behavior

-0.639 0.36 −1.77 0.082 No

c. It would be inappropriate to assume cause and effect without a better understanding of how the study was conducted.

166

12.57 a.

c. The fit of this regression is weak as given by R2 = 0.2474. 24% of the variation in % Operating Margin is explained by % Equity Financing.

12.58 a.

c. The fit of this regression is very good as given by the r-squared value 0.8216. The regression line does show a strong positive linear relationship between molecular wt and retention time, indicating that the greater the molecular wt the greater is the retention time.

12.59 a. Based on both the r-squared = 0 and the p-value > .10, there is no relationship between the class size and teacher ratings.

b. Given that r-squared = 0, we have not “explained” teacher ratings in this bivariate model. Other factors would be students expected GPA, years teaching, core class, age of student, gender of student, gender of instructor, etc. Answers will vary with respect to other teachers.

167

12.60 a. The scatter plot shows a positive relationship.

c. The fit of this regression is very good as given by the r-squared value .8206. The regression line shows a strong positive linear relationship between revenue and profit, indicating that the greater the revenue the greater is the profit.

12.61 a. The slope of each model indicates the impact of an additional year of the vehicle has on the price. This relationship for each model is negative indicating that an additional year of age reduces the asking price. This impact ranges from a low for the Taurus (an additional year reduces the asking price by $906) to a high for the Ford Explorer (an additional year reduces the asking price by $2,452).

b. The intercepts could indicate the price of a new vehicle.c. Based on the R-squared values: The fit is very good for the Explorer, the F-150 Pickup and the Taurus.

The fit is weak for the Mustang. One reason for the seemingly poor fit for the Mustang is the fact that this is a collector item (if in good condition) so that the age is less important of a factor in determining the asking price.

d. Answers will vary, but a bivariate model for 3 of the vehicles explains approximately 2/3 of the variation in asking price at a minimum. Other factors: condition of the car, collector status, proposed usage, price of a new vehicle.

12.62 a. The regression results are not significant, based on the p-value, for the 1-year holding period. The results for the 2-Year period are significant at the 5% level, while for 2 years and beyond the results are significant at the 1% level. For each regression there is an inverse relationship between P/E and the stock return. For the 8-year and 10-year period the relationship is approximately -1. The R-squared increases as the holding period increases. This indicates that P/E ratio explains a greater portion of the variation in stock return, the longer the stock is held.

b. Yes, given the data are time series, the potential for autocorrelation is present. Also, it is commonly recognized that stock returns do exhibit a high degree of autocorrelation, as do most financial series.

12.63 a. Using Father’s Height: My Predicted Height = 71+2.5 = 73.5” My actual height = 73” Using Average of Parent’s Height: My Predicted Height = 68+2.5 = 70.5”

b. Fairly accurate within 0.5” when using my father’s height, within 2.5” when using average parent height. May be there is improved accuracy using only father’s height for males.

c. Regression analysis of samples of daughters and sons, with respective average height of parents.

Separate samples of each.

168

Chapter 13

Multiple Regression

13.1 a. Y = 4.31 − 0.082*ShipCost + 2.265*PrintAds + 2.498*WebAds + 16.7*Rebate%b. The coefficient of ShipCost says that each additional $1 of shipping cost reduces about$ 0.082 from net

revenue.The coefficient of PrintAds says that each additional $1000 of printed ads adds about $2,265 to net revenue.The coefficient of WebAds says that each additional $1000 of printed ads adds about $2,498 to net revenue.The coefficient of Rebate% says that each additional percentage in the rebate rate adds about $16,700 to net revenue.

c. The intercept is meaningless. You have to supply some product, so shipping cost can’t be zero. You don’t have to have a rebate or ads, they can be zero.

d. NetRevenue = $467,160.

13.2 a. Y = 1225 + 11.52*FloorSpace − 6.935*CompetingAds − .1496*Price b. The coefficient of FloorSpace says that each additional square foot of floor space adds about 11.52 to

sales (in thousand of dollars).The coefficient of CompetingAds says that each additional $1000 of CompetingAds reduces about 6.935 from sales (in thousand of dollars).The coefficient of Price says that each additional $1 of Advertised Price reduces about .1496 from net revenue (in thousand of dollars).

c. No. If all of these variables are zero, you wouldn’t sell a bike (no one will advertise a bike for zero).d. Sales = $48.6 thousand

13.3 a. DF are 4, 45b. F.05 = 2.61, using df = 4, 40.c. F = 64,853/4990 = 12.997. Yes, the overall regression is significant.

H0: All the coefficients are zero (β1 = β2 = β3 = 0)H1: At least one coefficients is non-zero

d. R2 = 259,412/483,951 = .536 R2adj = 1 − (1 − .536)(49/45) = .4948

13.4 a. DF are 3, 26b. F.05 = 2.61c. F = 398802/14590 = 27.334. Yes, the overall regression is significant.

H0: All the coefficients are zero (β1 = β2 = β3 = 0)H1: At least one coefficients is non-zero

d. R2 = 1196410/1575741 = .759 R2adj = 1 − (1 − .759)(29/265) = .731

13.5 a.

Predictor t-value

Intercept 0.06085850.9517414

ShipCost -0.01752890.9860922

PrintAds 2.15714290.0363725

WebAds 2.95376610.0049772

Rebate% 4.6770308

b. t.005 = 2.69. Web Ads and Rebate% differ significantly from zero (p-value < .01 and t-value>2.69.)c. See table in part a.

169

13.6 a.

Predictor t-value

Intercept 3.08431920.0034816

FloorSpace 8.6631579

CmpetingAds -1.77592830.0825069

Price-0.14955 -1.67525480.1008207

Critical Value of t

b. t.005 = 2.779. Only Floor Space differs significantly from zero (p-value < .01 and t-value>2.779.c. See above table

13.7 Use formulas in text: 13.11b and 13.12b and t with 34 df and .025 in the upper tail:

îy ± tn−k−1SE = ˆ

iy ± 2.032*3620 = îy ± 7355.84

Using the quick rule: îy ± 2SE ˆ

iy ± 2*(3620) = îy ± 7240

Yes, the quick rule gives similar results. 13.8 Use formulas 13.11b and 13.12b and t with 20 df and .025 in the upper tail:

îy ± tn−k−1SE = ˆ

iy ± 2.086*1.17 = îy ± 2.44062

Using the quick rule îy ± 2SE ˆ

iy ± 2*(1.17) = v ± 2.34

Yes, the quick rule gives similar results.

13.9 All are cross-sectional data except for Data Set D which is time series.

13.10 Answers will vary. Casual observation indicates that X and Y data for each data set are well conditioned.

13.11 Answers will vary. Sample Answers based on selection of independent variables:A: Length (-) Width (-) Weight (-) Japan (+)B: Price (-) Shelf (+) C: Floor (+) Offices (+)Entrances (+) Age (-)Freeway (+)D: CapUtil (+) ChgM1 (+) ChgM2 (+) ChgM3 (+)E: Dropout (-) EdSpend (+) Urban (+) Age (-) FemLab (+) Neast (+) Seast (+) West (+)F: TotalHP (+) NumBlades (+) Turbo (+)G: MW(+) BP (-) RI (+) H1 (+) H2 (+) H3 (+) H4 (+) H5 (+)H: Height (+) Line (+) LB (+) DB (+) RB (+)I: Age (+) Weight (+) Height (+) Neck (+) Chest (+) Abdomen (+) Hip (+) Thigh (+)J: Age (-) Car (+) Truck (+) SUV (+)

13.12 Evans’ Rule : A, B, D, F, HDoane’s Rule: C, E, G, I

13.13 Data Set A: y = 43.9932 − 0.0039length − 0.1064width − 0.0041weight − 1.3228Japan. Data Set B: y = 87.1968 − 0.0016Price − 1.3881Shelf.Data Set C: y = −59.3894 + 0.2509Floor + 97.7927Offices + 72.8405Entrances − 0.4570Age +

116.1786Freeway.Data Set D: y = −21.6531 + 0.2745CapUtil + 0.2703ChgM1− 0.2012ChgM2 + 0.4630ChgM3.

Data Set E: y = −2.1471 − 0.0258Dropout + 0.0006EdSpend + 0.0891Urban − 0.2685Age + 0.3516Femlab + 3.9749Neast + 1.4456Seast + 1.8117.

Data Set F: y= −696.9390 + 0.3927Year + 0.1787TotalHP + 8.8269NumBlades + 15.9752Turbo.Data Set G: y = 51.3827 − 0.1772MW + 1.4901BP − 13.1620RI − 13.8067H1 − 6.4334H2 − 12.2297H3 −

0.5823H4.Data Set H: y = −12.0098 + 2.8141Height + 69.0801Line + 23.7299LB − 5.3320DB.

170

Data Set I: y = −35.4309Age − 0.1928Weight − 0.0642Height − 0.3348Neck.Data Set J: y = 15,340.7233 − 693.9768Age − 533.5731Car + 5,748.1799Truck + 3,897.5375SUV.

The Regression Analysis Output for each data set follows. Please refer to the output for answers to questions 13.14 – 13.17.

Data Set A

R² 0.703 Adjusted

R² 0.671 n 43

R 0.838 k 4

Std. Error 2.505 Dep. Var. City

ANOVA table


Regression 563.9264 4 140.9816 22.46 1.40E-09

Residual 238.5387 38 6.2773

Total 802.4651 42

variablescoefficient

s std. error t

(df=38) p-value 95% lower 95% upper VIF

Intercept 43.9932 8.4767 5.190 7.33E-06 26.8330 61.1534

Length -0.0039 0.0445 -0.087 .9311 -0.0939 0.0862 2.672

Width -0.1064 0.1395 -0.763 .4501 -0.3888 0.1759 2.746

Weight -0.0041 0.0008 -4.955 1.53E-05 -0.0058 -0.0024 2.907

Japan -1.3228 0.8146 -1.624 .1127 -2.9718 0.3262 1.106

Mean VIF 2.358

Data Set B

R² 0.034

Adjusted R² 0.000 n 27

R 0.185 k 2

Std. Error 4.060 Dep. Var. Accuracy

ANOVA table


Regression 14.0006 2 7.0003 0.42 .6588

Residual 395.6290 24 16.4845

Total 409.6296 26

variables coefficients std. error t (df=24) p-value 95% lower 95% upper VIF

Intercept 87.1968 2.4030 36.286 1.76E-22 82.2372 92.1564

Price -0.0016 0.0047 -0.338 .7382 -0.0113 0.0081 1.05

4

Shelf 1.3881 1.8307 0.758 .4557 -2.3903 5.1666 1.05

4

Data Set C

R² 0.967 Adjusted R² 0.961 n 32

171

R 0.983 k 5 Std. Error 90.189 Dep. Var. Assessed

ANOVA table


Regression6,225,261.256

1 5 1,245,052.25 153.07 2.01E-18Residual 211,486.6189 26 8,134.11

Total 6,436,747.875

0 31

variables coefficientsstd.

error t (df=26) p-value 95% lower 95% upper VIF

Intercept -59.3894 71.9826 -0.825 .4168 -207.3520 88.5731 Floor 0.2509 0.0218 11.494 1.08E-11 0.2060 0.2957 3.757

Offices 97.7927 30.8056 3.175 .0038 34.4708 161.1146 3.267 Entrances 72.8405 38.7501 1.880 .0714 -6.8115 152.4924 1.638

Age -0.4570 1.2011 -0.380 .7067 -2.9258 2.0118 1.169 Freeway 116.1786 34.7721 3.341 .0025 44.7035 187.6536 1.185

Mean VIF 2.203

Data Set D

R² 0.347


R 0.589 k 4

Std. Error 2.672 Dep. Var. ChgCPI

ANOVA table


Regression 136.8772 4 34.2193 4.79 .0033

Residual 257.0584 36 7.1405

Total 393.9356 40

Regression output confidence interval


error t (df=36) p-value 95% lower 95% upper VIF

Intercept -21.6531 9.5228 -2.274 .0290 -40.9662 -2.3399

CapUtil 0.2745 0.1130 2.429 .0203 0.0453 0.5038 1.205

ChgM1 0.2703 0.1069 2.530 .0159 0.0536 0.4870 1.193

ChgM2 -0.2012 0.2981 -0.675 .5040 -0.8058 0.4034 5.017

ChgM3 0.4630 0.2463 1.879 .0683 -0.0366 0.9626 4.489

Mean VIF 2.976

Data Set E

R² 0.729


R 0.854 k 8

172

Std. Error 2.128 Dep. Var. ColGrad%

ANOVA table


Regression 500.5063 8 62.5633 13.82 1.78E-09

Residual 185.6579 41 4.5282

Total 686.1642 49

variables coefficients std. error t

(df=41) p-value95%

lower95%

upper VIF

Intercept -2.1471 11.3532 -0.189 .8509 -25.0753 20.7811

Dropout -0.0258 0.0564 -0.458 .6495 -0.1398 0.0881 2.189

EdSpend 0.0006 0.00036045 1.568 .1245 -0.0002 0.0013 2.343

Urban 0.0891 0.0253 3.520 .0011 0.0380 0.1402 1.492

Age -0.2685 0.2517 -1.067 .2923 -0.7769 0.2398 1.640

Femlab 0.3516 0.0894 3.935 .0003 0.1711 0.5321 1.652

Neast 3.9749 1.0908 3.644 .0007 1.7720 6.1778 2.254

Seast 1.4456 1.2430 1.163 .2516 -1.0647 3.9559 3.112

West 1.8117 0.9069 1.998 .0524 -0.0198 3.6432 1.831

Mean VIF 2.064

Data Set F

R² 0.768


R 0.876 k 4

Std. Error 18.097 Dep. Var. Cruise

ANOVA table


Regression 54,232.9050 4 13,558.23 41.40 2.75E-15

Residual 16,375.2041 50 327.50

Total 70,608.1091 54

Regression output confidence interval

variables coefficients std. error t (df=50) p-value 95% lower95%

upper VIF

Intercept -696.9390 393.3465 -1.772 .0825 -1,486.9990 93.1209

Year 0.3927 0.1991 1.972 .0541 -0.0073 0.7927 1.131

TotalHP 0.1787 0.0195 9.167 2.76E-12 0.1396 0.2179 1.459

NumBlades 8.8269 5.7530 1.534 .1313 -2.7284 20.3823 1.716

Turbo 15.9752 6.2959 2.537 .0143 3.3296 28.6208 1.201

Mean VIF 1.377

Data Set G

R² 0.987


173

R 0.993 k 7

Std. Error 8.571 Dep. Var. Ret

ANOVA table


Regression 146,878.2005 7 20,982.6001 285.64 1.27E-23

Residual 1,983.3648 27 73.4580

Total 148,861.5653 34

variables coefficients std. error t (df=27)p-

value95%

lower95%

upper VIF

Intercept 51.3827 162.7418 0.316 .7546 -282.535

385.3010

MW -0.1772 0.3083 -0.575 .5701 -0.8097 0.4553 21.409

BP 1.4901 0.1831 8.139 9.64E-

09 1.1144 1.8657 31.113

RI -13.1620 107.2293 -0.123 .9032 -233.178

206.8542 13.115

H1 -13.8067 9.7452 -1.417 .1680 -33.8022 6.1888 9.235

H2 -6.4334 8.6848 -0.741 .4652 -24.2531 11.3863 2.816

H3 -12.2297 8.1138 -1.507 .1434 -28.8779 4.4184 2.458

H4 -0.5823 4.8499 -0.120 .9053 -10.5335 9.3689 1.793

Mean VIF 11.706

Data Set H

R² 0.806 Adjusted R² 0.789 n 50

R 0.898 k 4 Std. Error 19.256 Dep. Var. Weight

ANOVA table


Regression 69,245.0001 4 17,311.25 46.69 1.84E-15Residual 16,685.0799 45 370.78

Total 85,930.0800 49

variables coefficients std. error t (df=45) p-value95%

lower95%

upper VIF

Intercept -12.0098 118.3477 -0.101 .9196 -250.3743 226.354

7 Height 2.8141 1.6495 1.706 .0949 -0.5083 6.1364 2.257 Line 69.0801 10.1884 6.780 2.16E-08 48.5597 89.6006 3.141 LB 23.7299 8.9644 2.647 .0111 5.6748 41.7851 1.734 DB -5.3320 8.0565 -0.662 .5115 -21.5587 10.8947 1.502

Mean VIF 2.158

Data Set I

R² 0.841


R 0.917 k 8

Std. Error 3.957 Dep. Var. Fat%

174

ANOVA table


Regression 3,399.1446 8 424.893

1 27.144.82E-

14

Residual 641.8882 41 15.6558

Total 4,041.0328 49


error t

(df=41) p-value95%

lower95%

upper VIF

Intercept -35.4309 24.9040 -1.423 .1624-

85.7256 14.8639

Age 0.0905 0.0880 1.028 .3099 -0.0872 0.2682 1.712

Weight -0.1928 0.0783 -2.462 .0181 -0.3510 -0.0346 31.111

Height -0.0642 0.1160 -0.554 .5827 -0.2984 0.1700 1.689

Neck -0.3348 0.4023 -0.832 .4100 -1.1472 0.4776 5.472

Chest 0.0239 0.1788 0.133 .8945 -0.3373 0.3850 11.275

Abdomen 0.9132 0.1640 5.570 1.77E-

06 0.5821 1.2444 17.714

Hip -0.3107 0.2749 -1.130 .2649 -0.8658 0.2445 25.899

Thigh 0.7787 0.2907 2.678 .0106 0.1915 1.3658 11.931

Mean VIF 13.350

Data Set J

R² 0.139 Adjusted R² 0.134 n 637

R 0.373 k 4 Std. Error 8573.178 Dep. Var. Price

ANOVA table


Regression 7,512,691,866 4 1,878,172,966 25.55 1.20E-19

Residual46,451,606,04

7 632 73,499,377

Total53,964,297,91

3 636

variables coefficients std. error t (df=632) p-value 95% lower 95% upper VIF

Intercept 15,340.7233 1,239.0560 12.381 1.12E-3112,907.556

3 17,773.890

3 Age -693.9768 117.6801 -5.897 6.02E-09 -925.0682 -462.8853 1.017 Car -533.5731 1,225.8598 -0.435 .6635 -2,940.8263 1,873.6802 3.201 Truck 5,748.1799 1,318.6111 4.359 1.52E-05 3,158.7885 8,337.5713 2.662 SUV 3,897.5375 1,315.4861 2.963 .0032 1,314.2828 6,480.7923 2.749

Mean VIF 2.407

175

13.14 Answers will vary by dataset, see output . The main conclusion is that if the 95 percent confidence interval contains the value 0, the predictor coefficient is not significantly different from zero. Predictor coefficients that are shaded in yellow do not include the value zero. These predictors are those that do have an impact on the dependent variable.

176

13.15 The hypothesis for each data set is: H0: βi = 0 versus H1: βi ≠ 0For each output provided, see yellow shaded cells. These are the predictor variables for which the null

hypothesis is rejected and are the same ones that did not include zero in the 95 percent confidence interval from 13.14 .

A:.DF = 38, t-critical = 2.024B: DF = 24, t-critical = 2.064C: DF = 26, t-critical = 2.056D: DF = 36, t-critical = 2.028E: DF = 41, t-critical = 2.020F: DF = 50, t-critical = 2.009G: DF = 27, t-critical = 2.052H: DF = 45, t-critical = 2.014I: DF = 41, t-critical = 2.020J: DF = 632, t-critical = 1.96

13.16 a. For full model results see shaded answers in output provided. These are the predictors whose p-values are less than 0.05.

b. Yes, the predictors that were found to have significant coefficients from the t-tests are the same ones that are significant from using the p-values.

c. Most prefer the p-value approach, easier to check for significance.

13.17 A:. Very GoodB: Very PoorC: ExcellentD: PoorE: Very GoodF: Very Good.G: ExcellentH: Very GoodI: Very GoodJ: Poor

13.18 Std errors are calculated for each full model. Use equation 13.11b to construct the prediction intervals.

A:. îy ± tn−k−1 SE = ˆ

iy ± 2.024*2.505 = îy ± 5.07012

B: îy ± tn−k−1 SE = ˆ

iy ± 2.064*176.291 = îy ± 363.864624

C: îy ± tn−k−1 SE = ˆ

iy ± 2.056*90.189 = îy ± 185.428584

D: îy ± tn−k−1 SE = ˆ

iy ± 2.028*2.672 = îy ± 5.563104

E: îy ± tn−k−1 SE = ˆ

iy ± 2.020*2.128 = îy ± 4.29856

F: îy ± tn−k−1 SE = ˆ

iy ± 2.0009*18.097 = îy ± 36.356873

G: îy ± tn−k−1 SE = ˆ

iy ± 2.052*8.571 = îy ± 17.587692

H: îy ± tn−k−1 SE = ˆ

iy ± 2.014*19.256 = îy ± 38.781584

I: îy ± tn−k−1 SE = ˆ

iy ± 2.020*3.957 = îy ± 7.99314

J: îy ± tn−k−1 SE = ˆ

iy ± 1.96*8573.178 = îy ± 16.80342888

177

13.19 a.Data Set A Correlation Matrix

Length Width Weight Japan

Length 1.000

Width .720 1.000

Weight .753 .739 1.000

Japan -.160 -.267 -.093 1.000

43 sample size± .301 critical value .05 (two-tail)± .389 critical value .01 (two-tail)Yes, width and height are correlated with other.

Data Set B Correlation Matrix

Price Shelf

Price 1.000

Shelf -.227 1.000

27 sample size± .381 critical value .05 (two-tail)± .487 critical value .01 (two-tail)No correlation found.

Data Set C Correlation Matrix

Offices Entrances Freeway

Floor

Offices 1.000

Entrances .444 1.000

Age -.241 .136

Freeway -.368 -.082 1.000

32 sample size± .349 critical value .05 (two-tail)± .449 critical value .01 (two-tail)Offices and Entrances correlated with each other.

Data Set D Correlation Matrix

CapUtil ChgM1 ChgM2 ChgM3

CapUtil 1.000

ChgM1 -.241 1.000

ChgM2 -.265 .266 1.000

ChgM3 -.071 .080 .857 1.000

41 sample size± .308 critical value .05 (two-tail)± .398 critical value .01 (two-tail)M2 and M3 are highly correlated.

178

Data Set E Correlation Matrix

Dropout EdSpend Urban Age Femlab Neast Seast WestMidwes

t

Dropout 1.000

EdSpend -.047 1.000

Urban .096 .260 1.000

Age -.067 .340 -.099 1.000

Femlab -.445 .258 .101 -.226 1.000

Neast -.009 .667 .080 .316 .169 1.000

Seast .550 -.394 -.380 .089 -.495 -.298 1.000

West -.059 -.135 .352 -.428 .138 -.331 -.350 1.000

Midwest -.466 -.108 -.066 .053 .182 -.315 -.333 -.370 1.000

50 sample size± .279 critical value .05 (two-tail)± .361 critical value .01 (two-tail)

Regional differences correlated with other predictor variables. This is to be expected as regional differences influence college graduation rate as well as the factors that influence those rates.

Data Set F Correlation Matrix

TotalHP NumBlades

Year

TotalHP 1.000

NumBlades .491 1.000

Turbo .096 .388


Number of Blades is correlated with both Turbo and Total HP

Data Set G Correlation Matrix

35 sample size± .334 critical value .05 (two-tail)± .430 critical value .01 (two-tail)BP correlated with MW, RI and H1RI correlated with H5 H5 correlated with H1

179

Data Set H Correlation Matrix

Height Line LB DB RB

Height 1.000

Line .683 1.000

LB .032 -.359 1.00

0

DB -.351 -.381 -.266 1.00

0

RB -.447 -.403 -.281 -.298 1.000

50 sample size± .279 critical value .05 (two-tail)± .361 critical value .01 (two-tail)Line, DB and RB correlated with HeightDB and Line correlatedRB and Line correlatedThese correlations make sense. Each position is specialized, so if you are fit for one, chances are you are not fit for any other.

Data Set I Correlation Matrix

Age Weight Height Neck ChestAbdome

n Hip Thigh

Age 1.00

0

Weight .265 1.000

Height -.276 .109 1.000

Neck .176 .882 .201 1.000

Chest .376 .912 .014 .820 1.000

Abdomen .442 .915 -.052 .781 .942 1.000

Hip .314 .959 -.045 .804 .911 .942 1.000

Thigh .219 .937 -.037 .823 .859 .890 .938 1.000


Weight is correlated with the body parts given. Other body parts are correlated with each other. This is not unexpected.

Data Set J Correlation Matrix

Age Car Truck SUV Van

Age 1.000

Car .003 1.000

Truck .017 -.478

1.000

SUV -.092 -.495 -.308 1.00

0

Van .106 -.283 -.176 -.182 1.000

637 sample size



637 sample size± .078 critical value .05 (two-tail)

180

± .102 critical value .01 (two-tail)SUV correlated with Age, Car, and TruckVan is correlated with all variablesCar is correlated with Truck, SUV and VanSuch a large n reduces the critical values so even small r is significant.

181

13.20 a. See output.b. Multicollinearity is a potential problem if the VIF is greater than 10 (rule of thumb): G has potential

multicollinearity problems based on the VIFs

13.21 A: observation 42 is an outlier, no unusual residuals.B: observation 11 has an unusual residual; no outliers.C: no observations have unusual residuals or outliers.D: observations 19 and 20 have unusual residuals; no outliers.E: observation 6 has an unusual residual; there are no outliers.F: observations 23, 39 and 46 have unusual residuals; no outliers.G: observations 15, 17 and 25 are unusual residuals; no outliers.H: observations 1, 6, 26, 48 are unusual residuals, no outliers.I: no unusual residuals or outliers.J: observations 246, 397 and 631-632 are unusual residuals; 212, 342, 502 are outliers.

13.22 A: observations 2, 8, 13, and 21 have high leverage.B: observation 18 has high leverage.C: No Leverage effects present.D: observations 16 and 33 have high leverage.E: observations 2, 44 and 48 have high leverage.F: observations 43 and 46 have high leverage.G: observation 24 has high leverage.H: observations 20 and 44 have high leverage.I: observations 5, 15, 36, 39, and 42 have high leverage.J: observations 1-4, 52, 77, 92-101, 116, 126, 178-181, 184, 270-272, 298, 493, 502, 522, 554, 556-564, 611-

624, 627-628 have high leverage.

13.23 Normality is a problem for J.

13.24 Heteroscedasticity is a concern for J.

13.25 Durbin-Watson for D: 0.74, indicating autocorrelation could be a potential problem.

13.26 a. Each slope measures the additional revenue earned by selling one more unit (one more car, truck, or SUV, respectively.)

b. The intercept is not meaningful. Ford has to sell at least one car, truck or SUVto earn revenue. No sales mean no revenue.

c. The predictor variables are highly correlated to each other (multicollinearity problem), as well as related to “missing variables” that influence their sales as well as revenue.

13.27 The sample size is too small relative to the number of predictors. Using the following:

Evans’ Rule (conservative) n/k ≥ 10 (at least 10 observations per predictor)

Doane’s Rule (relaxed) n/k ≥ 5 (at least 5 observations per predictor)

A researcher would have to either reduce the number of predictors or increase the size of the sample. With 8 predictors, one needs a minimum of 40 observations using Doane’s Rule or 80 using Evans’ Rule. If increasing the sample size is not feasible, then pairwise t-tests on group means could be performed by recalculating the groupings of the proposed binaries.

13.28 a. One binary must be omitted to prevent perfect multicollinearity.b. Same reasoning as in (a). The rule is if you use the intercept then you must use one less binary than the

total number of binaries as predictors. The interpretation of the intercept is the missing binary. c. Monday: 11.2 + 1.19 = 12.39

d. Shift 3 11.2, Shift 1 and Shift 2 have lower AvgOccupancy given that they have negative coefficients.e. The intercept represents the AvgOccupancy on Sundays during Shift 3.

182

f. The fit is poor.

183

13.29 Main points:1. The regression as a whole is not significant based on the p-value of .37102. R-squared is 0.117 indicating a very poor fit.3. Examination of the individual regression coefficients indicates that the two binary variables are not significantly different from zero, p-values >.10.4. Conclusion: cost per average load does not differ based on whether or not it is a top-load washer or whether or not powder was used. No apparent cost savings based on washer type or detergent type.

13.30 Main points:1. The best model in terms of fit as measured by the R-squared is model (NVAR=3), although only a small improvement over model (NVAR=2). No gain if fit is achieved by adding LifeExp and Density.2. Examination of the individual regression coefficients indicates that the InfMort and Literate have p-values < .01 and GDPCap has a p-value < 0.05. 3. Conclusion: Infant mortality and Literate have the greatest impact on birth rates.

13.31 a. Yes, the coefficients make sense, except for TrnOvr. One would think that turnovers would actually reduce the number of wins, not increase them.

b. No. It is negative and the number of games won is limited to zero or greater. You can’t win games with all of the presented values = 0.

c. One needs either 5 observations or 10 observations per predictor. Here we have 6, so we need 30 observations minimum for Doane’s rule, but we only have 23. Yes, the sample size is a problem.

d. Rebounds and points highly correlated. We don’t need both of them and the variance of rebounds is increased, which increases the denominator of the test-statistic biasing it towards the non-rejection of the null hypothesis.

13.32 Main points:1. The regression as a whole indicates a very strong fit. 2. R-squared is .81. The predictor variables as a group explain 81.1% of the variation in Salary.3. Examination of the individual regression coefficients indicates that all of the variables are significantly different from zero, p-values <0.014. Conclusion: The ethnicity of a professor does matter. A professor who is African-American earns on average $2,093 less than one who is not. Assistant professors earn on average $6,438 less than other professors. New hires earn less than those who have been there for some time. The finding that those who have been there longer have higher salaries is a fact of rank and the tenure system. The finding that after controlling for this, race does matter does support that racial discrimination is present.

13.33 a. Both men and women who had prior marathon experience had lower times on average than those who were running for the first time.

b. No the intercept does not have any meaning. If all predictor/binary variables were 0 then you wouldn’t have an individual racer.

c. It is suspected non-linearity is present among age, weight, and height. In this model we see increases in age decreases times, but at an increasing rate, increases in weight decreases time, but at an increasing rate and increasing height increases time, but at a decreasing rate.

d. The model predicted that I would run the marathon in about 12 and ½ hours. And that could be right. I

can walk 4 mph so it would take at least 6 to 7 hours miminum!

184

Chapter 14

Time Series Analysis

Note: For questions that deal with “best model” the student should consider four criteria:

• Occam’s Razor Would a simpler model suffice?

• Overall fit How does the trend fit the past data?

• Believability Does the extrapolated trend “look right?”

• Fit to recent data Does the fitted trend match the last few data points

14-1 a. See graphs.b. Diesel fuel may be cheaper and may offer better MPG.c. See graphs.d. Exponential model is simple and looks like a good fit.e. Linear seems too conservative (about 41,000 by 2006). Quadratic is reasonable (about 61,000 by 2006).

Exponential is aggressive (about 84,000 by 2006).

t Linear Exponential Quadratic

12 33,980 49,723 43,895

13 37,256 64,778 52,130

14 40,533 84,391 61,127

y = 3276.5x - 5338.6

R2 = 0.8338

0

5,000

10,000

15,000

20,000

25,000

30,000

35,000

40,000

45,000

1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003

y = 381.37x2 - 1299.9x + 4576.9

R2 = 0.9219

0

10,000

20,000

30,000

40,000

50,000

60,000

70,000

1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003

185

y = 2080.2e0.2645x

R2 = 0.9372

0

10,000

20,000

30,000

40,000

50,000

60,000

70,000

80,000

90,000

1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003

14-2 a. Non-linear (decline, then rise).b. Rise of internet, increase in high speed connections.c. Quadratic is by far the best fit.d. Quadratic forecast is about 10.0 for 2007.e. Linear and exponential forecasts are flat, but could be right

t LinearExponentia

lQuadrati

c

8 7.23 7.20 9.71

y = 0.0357x + 6.9429

R2 = 0.009

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

2000 2001 2002 2003 2004 2005 2006

y = 0.2071x2 - 1.6214x + 9.4286

R2 = 0.9219

0.0

2.0

4.0

6.0

8.0

10.0

12.0

14.0

16.0

2000 2001 2002 2003 2004 2005 2006

186

y = 6.8934e0.0055x

R2 = 0.0106

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

2000 2001 2002 2003 2004 2005 2006

14.3 a. Somewhat linear, but maybe slowing.b. Rise of internet, higher speed connections, increased PC use, hacker cult c. See graphs.d. Use criteria from text to assess. Students may like the quadratic because it has a good fit, but also

because its projections look more reasonable than the other two models.e. Quadratic forecasts about 120. But, in hindsight, exponential (almost 240) may have come closer to

foretelling the virus explosion of the first decade of the 2000s.

t LinearExponentia

lQuadrati

c

8 135.7 236.92 119.4

y = 18.143x - 9.4286

R2 = 0.9244

0

20

40

60

80

100

120

140

160

180

200

1996 1997 1998 1999 2000 2001 2002

y = -1.3571x2 + 29x - 25.714

R2 = 0.9399

0

20

40

60

80

100

120

140

1996 1997 1998 1999 2000 2001 2002

187

y = 9.4401e0.4028x

R2 = 0.8785

0

100

200

300

400

500

600

1996 1997 1998 1999 2000 2001 2002

14.4 a. Fairly linear increase.b. Needs of family, expenses incurred requires more incomec. See graphs.d. Each has merit. Simplicity favors the linear, but quadratic captures the slowing.e. Linear gives 3,770 hours, quadratic only about 3,693 hours.

t LinearExponentia

lQuadrati

c

19 3,770 3,782 3,693

y = 29.611x + 3206.9

R2 = 0.9232

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y = -1.2051x2 + 52.508x + 3130.5

R2 = 0.9559

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y = 3211.8e0.0086x

R2 = 0.9139

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14.5 a. Quite linear.b. Rise of health care concerns, change to healthy dietc. yt = 581.73 + 25.55t.d. Increased capacity needed for production and distribution.e. Using t = 6 we get y6 = 581.73 + 25.55(6) = 735.

y = 25.55x + 581.73

R2 = 0.9898

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1980 1985 1990 1995 2000

y = -0.7786x2 + 30.221x + 576.28

R2 = 0.9911

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189

y = 584.92e0.0389x

R2 = 0.9886

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1980 1985 1990 1995 2000

14.6 a. See graphs. A cyclical pattern is observed (positive autocorrelation).b. As m increases (i.e., more smoothing) the trendline does not describe the data as well. m = 2 gives the

best “fit” while m = 5 gives the most smoothing. A larger m is actually helpful if we are trying to reduce the impact of day-to-day fluctuations (if you want to track the data exactly, why smooth at all).

c. Can help anticipate the next day’s rate. Trend models aren’t much help.

190

14.7 a. Yields seem to exhibit a cyclical pattern that is not like any standard trend model (looks like positive autocorrelation) so exponential smoothing seems le a good choice for making a one-period forecast.

b-f. MegaStat output is lengthy. Here is a summary table:

α = .10 α = .20 α = .30

Mean Squared Error 0.039 0.028 0.021

Mean Absolute Percent Error 3.8% 3.1% 2.7%

Percent Positive Errors 42.3% 46.2% 51.9%

Forecast for Period 53 4.3 4.37 4.43

14.8 a. Seasonality is present as well as a positive trend.

Calculation of Seasonal Indexes

1 2 3 4

1 0.923 1.170

2 0.858 0.982 0.935 1.141

3 0.875 1.027 1.030 1.208

4 0.788 0.974 1.007 1.160

5 0.847 0.981 1.002 1.160

6 0.866 0.979

mean: 0.847 0.988 0.979 1.168 3.982

adjusted: 0.850 0.993 0.984 1.173 4.000

191

b. Time and seasonal binaries are significant.

Regression Analysis

R² 0.842


R 0.918 k 4

Std. Error 528.304 Dep. Var. Revenue

variables coefficients std. error t (df=19) p-value

Intercept 5,845.3500 308.8056 18.929 8.64E-14

Qtr1 -1,835.2060 308.6711 -5.946 1.01E-05

Qtr2 -1,111.0262 306.6461 -3.623 .0018

Qtr3 -1,145.5131 305.4247 -3.751 .0014

t 111.1536 15.7861 7.041 1.06E-06

c. Forecasts for 2005:

Period Forecasts

2005 Q1 6788.98

2005 Q2 7624.32

2005 Q3 7700.98

2005 Q4 8957.65

14.9 a. No trend present, but there is seasonality. It is important to emphasize that 1.00 is the key reference point for a multiplicative seasonal index. The creative student might plot the seasonal indexes. A chart can show that May, Jun, Sep, Oct, and Nov are above average (seasonal index exceeding 1.00). MegaStat’s trend fitted to deseasonalized data (shown below) indicates a slight downward trend, but without a t-statistic we cannot tell if it is significant.

Calculation of Seasonal Indexes

1 2 3 4 5 6 7 8 9 10 11 12

1 0.981 0.611 1.087 1.127 1.066 0.925

2 0.907 0.869 1.062 1.111 0.991 1.003 0.890 0.711 0.980 1.288 1.684 1.060

3 0.597 0.869 1.175 0.663 0.794 1.154 1.017 1.108 1.136 1.168 1.128 0.749

4 1.006 0.527 0.636 1.065 2.484 1.000 0.957

5

mean: 0.836 0.755 0.958 0.946 1.423 1.052 0.961 0.810 1.068 1.194 1.293 0.911

adjusted: 0.822 0.742 0.941 0.930 1.399 1.034 0.945 0.796 1.049 1.174 1.271 0.896

0.00

0.20

0.40

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0.80

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1.20

1.40

1.60

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Sea

so

nal

Ind

ex

192

b. Fit of the multiple regression model is not very good (R2 = .282, R2adj = .036) and only April (t = 2.296,

p = .0278) shows significant seasonality at α = .05. This is reasonable, since spring might cause a spike in Corvette sales. The trend term (Time) is negative, but is not significant.

R² 0.282


R 0.531 k 12

Std. Error 802.557 Dep. Var. Sales


Intercept 2,391.5417 477.6169 5.007 1.57E-05

Jan -242.5153 575.3861 -0.421 .6760

Feb 410.7361 574.0241 0.716 .4790

Mar 523.9875 572.7890 0.915 .3666

Apr 1,312.4889 571.6817 2.296 .0278

May 546.4903 570.7028 0.958 .3448

Jun 298.2417 569.8531 0.523 .6040

Jul -62.7569 569.1332 -0.110 .9128

Aug 470.7444 568.5434 0.828 .4133

Sep 599.9958 568.0843 1.056 .2981

Oct 910.2472 567.7561 1.603 .1179

Nov 87.2486 567.5592 0.154 .8787

Time -7.5014 8.6341 -0.869 .3909

c. Forecasts for 2004 are shown. These forecasts involve plugging the correct binary value (0, 1) for each month into the fitted regression equation. It is simpler than it seems at first glance, since all binaries are zero except the month being forecast. Nonetheless, it will be a challenge for most students.

Period Forecast Period Forecast

January 1,781.46 July 1,916.21

February 2,427.21 August 2,442.21

March 2,532.96 September 2,563.96

April 3,313.96 October 2,866.71

May 2,540.46 November 2,036.21

June 2,284.71 December 1,941.46

193

14.10 a. See graphs for each seriesb. Answers will vary. All have positive upward trend. Spirit is an aggressive new airline. To some extent,

its growth is more rapid because it is starting from a smaller base.c. See graphs.d. For all three variables, the linear model gives a good fit, and by Occam’s Razor we might choose it for

short-term forecasts. However, in many business situations, the exponential model is also a viable model. In this case, it gives somewhat more aggressive forecasts.

e. Forecasts for period t = 6 using the linear model:Revenue: y6 = 67.1(6) + 83.9 = 486.5Aircraft: y6 = 3.5(6) + 12.1 = 33.1Employees: y6 = 362.4(6) + 606.4 = 2781

Revenue

Aircraft

195

Employees

14.11 a. Dual scale graph is helpful, due to differing magnitudes (or separate graphs).

b. Electronic sales are large, but have a declining trend (yt = 31,578 −1615.7t, R2 = .9449). Mechanical sales are small, but with a rising trend (yt = 2467 + 40.543t, R2 = .7456).

c. Electronic sales are falling at 6.27% (yt = 32091e−0.0627t, R2 = .9413) while mechanical are rising at 1.54% (yt = 2470.6e0.0154t, R2 = .7465).

d. Fascination with electronic gadgets may be waning and/or competitors may be moving in on the Swiss watch industry. They may have a stronger specialty niche in mechanical watches.

e. Electronic forecast for 2004 is about 20,268, mechanical about 2,750.

Electronic: y7 = −1615.7(7) + 31578= 20,268Mechanical: y7 = 40.543(7) + 2466.9 = 2,750

2,400

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1998 1999 2000 2001 2002 2003

0

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10,000

15,000

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25,000

30,000

35,000

Mechanical

Electronic

14.12 a. See graph. Trend is steadily upwards.b. People have more leisure time and use it watching the tube.c. See graph. Linear gives a good fit.

196

d. The linear trend should give good results based on overall fit and fit to recent data.e. y12 = 18.609(12) + 256.98 = 480.3 minutesf. Yes, hours in a day are finite (24 hours) and people also must work and sleep.

14.13 a. See graph. Trend is downwards.b. Possible reasons for voter apathy might be disillusionment with politicians.c. See graph. Linear gives a reasonable fit. Quadratic is not shown (Occam’s Razor).

d. The linear trend should give good results based on overall fit and fit to recent data.

e. y19 = −0.3672(19) +59.144 = 52.2 percent.f. The Committee for the Study of the American Electorate reported that more than 122 million people

voted in the 2004 Presidential election, the highest turnout (60.7 percent) since 1968. Source: http://www.washingtonpost.com. This is much higher than any of the forecasting models would have predicted. This shows that past trends may not be a guide to the future.

14.14 a. See graph.b. Steady upward trend since 1996.

197

c. Fitted trend equations:Linear: yt = 11.599+ .8125 t (R² = 0.6192)Exponential: yt = 12.551e.00424 t (R² = 0.5972)

Quadratic: yt = 18.396 −1.736 t + .1699 t2 (R² = 0.9658)

d. Quadratic is the best model based on overall fit and fit to recent data.Quadratic forecast: y15 = 18.396 – 1.736(15) +.1699(152) = 30.58

14.15 a. See graph. All three are increasing steadily.b. Linear growth due to convenience of credit and flexibility in when to pay the bills.c. Linear model seems appropriate (Occam’s Razor, good fit to recent data).d. Forecasts for 2004 (t = 10):

Total: y10 = 994.86+112.45(10) = 2119.36Non-revolving: y10 = 583.58+73.75(10) = 1321.1Revolving: y10 = 411.17 + 38.76(10) = 798.8

y = 112.45x + 994.86

y = 73.75x + 583.58

y = 38.7x + 411.17

0

500

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2,000

2,500

1995 1996 1997 1998 1999 2000 2001 2002 2003 2004

Total

Revolving

Non-revolving

14.16 a. See graph.b. Answers will vary. Airplanes are capital goods, affected by business cycle (stronger demand expected

during periods of growth), interest rates (credit availability for financing), and foreign demand (exchange rates, economic conditions abroad, foreign competition). There appears to be a structural break after the 1979 peak (cyclical prior to 1979).

c. No fitted trend can capture the whole pattern of aircraft sales.d See graph. Even this subset of the data has a problematic downturn that all of the standard trend models

will miss. This may be related to the 2001 attack on the U.S. World Trade Center.

U.S. Manufactured General Aviation Shipments, 1966-2003

y = 721.53e0.1159x

R2 = 0.808

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3500

1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004

Year

Pla

ne

s

e. Exponential trend is yt = 721.53e.1159t (R2 = .805).f. Exponential forecast for 2004 is y13 = 721.53e.1159(13) = 3,255. This forecast is aggressive, considering the

recent past. A judgment forecast might be better, just by eyeballing the most recent data, e.g., y13 = 2,500 (better than 2002-2003, worse than 2000-2001). It is well to remind students that equations do not always give better forecasts than judgment, especially when patterns are erratic.

198

14.17 Answers depend on which series is chosen. Soft drinks (both regular and diet) are increasing, Whole milk is decreasing, while reduced fat milk increased and then leveled off. Beer and wine are flat, while liquor is decreased over this period. Tastes in beverages are influenced by prices, fads, perceived health benefits, calories, changing age demographics, advertising, and many other factors that students will think of. In the absence of strong trends (all but soft drinks) a judgment forecast will be reasonable (2005 same as 2000). For soft drinks, a linear forecast would probably make sense.

U.S. Beverage Consumption

0

5

10

15

20

25

30

35

40

1980 1985 1990 1995 2000

Gallons P

er

Capita . Whole milk

Reduced-fat milk

Diet soft drinks

Regular soft drinks

Fruit juices

Beer

Wine

Liquor

14.18 a. See graphs.b. Federal taxes and spending are political variables that are affected by perceived societal needs, the

balance of political power in Congress, and leadership from the executive branch. Economic growth, inflation, business cycles, and economic policies underlie changes in the GDP and federal debt.

199

c. Until 2000, receipts were growing faster than outlays (5.13% versus .3.95% in exponential model) leading to a rising budget surplus. Since then, the opposite is true. Federal debt was growing more slowly (4.93%) than the GDP (5.91%) over this entire period, though since 2000 it appears that federal debt is increasing more rapidly. It would be useful to look up recent data in the Economic Report of the President (www. gpoaccess.gov/eop/) to see what has been happening since 2004.

d. Answers will vary. The main difficulty is deciding what trend (if any) can be used to fit budget receipts, since there has been a structural change since 2000.

e. Answers will vary. Relevant groups are Congress, Federal Reserve, President, businesses, households.

14.19 a. Both have slight downward trends.b. The CD data file goes back to 1970, while the textbook data sets starts at 1980. Using the textbook data,

we fit trends to 1980-2005 and equate the women’s trend to men’s trend (148.66 − 0.2186t = 129.96 − .0145t). Solving for t gives t = 92 or year 2137 (recall that 2005 is t = 26). Whether this will actually happen depends on human physiology and its limits, as well as issues such as performance enhancing drugs. If a student uses the entire data set 1970 to 2005, the equated trends (173.62 – 1.0916t = 153.75

− .1531t) converge at t = 42.5 or about 2012 (in the whole data set, 2005 is t = 36). Bothe graphs are shown below.

c. Moving average gives a good fit.d. Yes, it is reasonable since trend is slight.

Data from 1980-2005 Only

y = -0.2186x + 148.66

y = -0.0145x + 129.96

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Women

200

Data from 1970-2005

y = -1.0916x + 173.64

y = -0.1531x + 133.75

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Women

3-Period Moving Average

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14.20 a. See graphs.b. See graphs.c. Answers will vary depending on the example chosen. Commercial banks main have a downward trend

while branches have and upward trend. Rise of online banking, ATM machines, direct deposit are examples of why branch use has increased. For saving institutions, both series have downward trends.

d. See graphse. Answers will varyf. There is little difference between the two models. Both yield similar, almost identical results. The linear

model is therefore used since it is easier to work the linear trend model.

y = 1407.2x + 55024

R2 = 0.9812

y = -273.47x + 10052

R2 = 0.9741

0

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1995 1996 1997 1998 1999 2000 2001 2002 2003

Banks-Main

Banks-Branches

201

y = -181.87x + 13745

R2 = 0.8545

y = -73.967x + 2044.4

R2 = 0.9639

0

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1995 1996 1997 1998 1999 2000 2001 2002 2003

Savings-Main

Savings-Branches

14.21 a. See graph. Trend is rapid growth.b. Answers will vary, but there has been a dramatic increase in shares, need to have quick access to travel,

unevenness of commercial airplane travel, burden of security procedures for commercial air travel, convenience of smaller airports (closer to ultimate destination than main hubs), and rise of a highly compensated executive class with access to company expense accounts.

c. Possibly, if underlying causes remain unchanged, such as continued crowding and deterioration of quality in commercial air travel (e.g., bankrupt airlines, labor disputes). However, the exponential model’s fit to recent data is poor.

d. The exponential forecast for 2003 is so aggressive as to be unconvincing (y18 = 3.8256e.4512(18) = 12,879. Perhaps a judgment forecast would be better (e.g., somewhere around 8,000).

Fractional Aircraft Ownership

y = 3.8256e0.4512x

R2 = 0.9708

0

2000

4000

6000

8000

10000

12000

14000

1986 1988 1990 1992 1994 1996 1998 2000 2002

14.22 a. See graph.b. See graph.c. Answers will vary, use criteria from above and in text to assess, answers will vary, but all exhibit a small,

positive upward trend. Linear trend model works best for cars, least for light trucks, based on overall fit and fit to recent data.

d. Purchasers of new vehicles, producers of new vehicles, governments, oil companies and refiners

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14.23 a. See graph. Except for spike in 2001 (possible ralated to 9-11 attacks) there is little trend.b. Answers will vary, but the fitted trend, while not strong, is positive. c. A linear trend does not fit the data. It is tempting to try a higher-order polynomial, but such a model

would be useless for forecasting.d. Would expect it to be around 140 (judgment based on last 4 years excluding 2001).

U.S. Law Enforcement Officers Killed, 1994-2002

y = 3.9167x + 119.64

R2 = 0.1136

100

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240

1994 1995 1996 1997 1998 1999 2000 2001 2002 2003

Year

Kille

d

14.24 a. See graph.b. Answers will vary, but the trend, while not strong, has been negative and declining in the numbers killed.

Awareness of how to avoid lightning (despite popularity of outdoor sports like golf, climbing, boating).c. The exponential model is a good overall fit and is a simple way to describe the observed trend in a

believable way.

d. See graph. The 2005 forecast is y14 = 329.01e−.1415(14) = 45.4 deaths.

203

y = 329.01e-0.1415x

R2 = 0.9253

0

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14.25 a. See graph.b. Answers will vary. A linear trend does not fit the data well. The quadratic does a good job of capturing

the pattern, and its forecasts look reasonable (similar to what someone might get using a judgment forecast based on the most recent data).

c. See graph. The quadratic forecast is y11 = 62.72 (11)2 - 771.68 (11) + 11384 = 10,485.

y = 62.72x2 - 771.68x + 11384

R2 = 0.8741

0

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12000

1993 1994 1995 1996 1997 1998 1999 2000 2001 2002

14.26 a. See graph.b. See graph.c. Yes, but will be beyond 2040.d. Trend equations are given in the graph.

e. Issue is of interest to athletes, trainers, ergonomists, and sports physiologists. Set 11.967 −0.0804 x = 10.541 – 0.0402 x and solve for x = 37.14 (round up to 38). Year 38 is 2084. It would be conceivable if trends stay the same. However, if men get faster as women get faster, and if the trends are nonlinear, it will take even longer for convergence to happen.

204

14.27 Answers will vary based on series selected. It is clear that the oil consumption in the US is increasing. This is a good opportunity for students to access the internet (e.g., the Energy Information Agency www.eia.doe.gov) to get more information to augment their preconceived ideas and facts.

14.28 Answers will vary based on series selected. The trend for these series has been positive and increasing, but at a decreasing rate. Results indicate rising costs of prison interment, since all series show a projected increase. The legal, demographic, and sociological roots of these trends are worth exploring. This is a good opportunity for students to access the internet (e.g., www.ojp.usdoj.gov/bjs) to get more information to augment their preconceived ideas and facts.

0

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3,000

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7,000

8,000

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1990

1992

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2000

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2004

2006

2008

2010

Total

Probation

Jail

Prison

Parole

14.29 Answers will vary based on series selected. There are different trends in males vs. females on verbal as well as mathematical portion of exam. There are implications for learning styles by gender, college admissions, and perhaps even career choices. This is a good opportunity for students to access the Web (e.g., www.collegeboard.com) to get more information to augment their preconceived ideas and facts.

14.30 a. See graph.b. Answers will vary. Students will see that m = 2 (shown below) gives the best “fit.” However, m = 2

offers little smoothing.. Being heavily influenced by the most recent data values could be a disadvantage in making a future projection.

c. Yes., based on overall fit, fit to recent data, and the inapplicability of any of the usual trend models.

205

14.31 a. See graphs.

b. See graphs. The degree of smoothing varies dramatically as α is increased.

c. For this data α = .20 seems to track the recent data well, yet provides enough smoothing to iron out the “blips.” It gives enough weight to recent data to bring its forecasts above 1.80 (lagging but reflecting the

recent rise in rates). In contrast, α = .05 or α = .10 are not responsive to recent data (too much

smoothing) so they give a forecast below 1.80. While α = .50 gives a good “fit” it does not smooth the

data very much. Forecasters generally use smaller α values (the default of α = .20 is a common choice).d. Yes, based on graphs and fitted trendline.

α = .05 α = .10

α = .20 α = .50

206

14.32 a. See graph.b. Yes. Seasonality is quite apparent. There are highs and lows, depending on the month. There is also an

upward trend. The seasonal swings seem to increase in amplitude.c. See output below. It is important to note that 1.00 is the key reference point for a multiplicative seasonal

index. The creative student might plot the seasonal indexes to show that Nov through Mar (the winter months) are above 1.00, Apr is near 1.00, and the summer and fall months are below 1.00.

d. December is the highest, August is the lowest. This is logical, based on the seasons’ impact on heating degree-days (presumably the residence is in a northern climate).

e. There is some upward trend. Logical, since generally, fossil fuels are getting more expensive.f. We used December as our norm. All binaries are significant and negative (although Jan and Feb are

significant only at rather large α) meaning that December is the highest month (all other months reduce the cost below December). The next two highest months are February and January.

R² 0.803


R 0.896 k 11

Std. Error 22.411 Dep. Var. Cost

ANOVA table


Regression73,856.453

5 11 6,714.2230 13.37 1.23E-09

Residual18,081.219

1 36 502.2561

Regression output

207


Intercept 133.1025 11.2055 11.878 5.16E-14

Jan -19.4400 15.8470 -1.227 .2279

Feb -16.4225 15.8470 -1.036 .3070

Mar -44.1600 15.8470 -2.787 .0084

Apr -62.6500 15.8470 -3.953 .0003

May -95.5400 15.8470 -6.029 6.36E-07

Jun -113.1775 15.8470 -7.142 2.14E-08

Jul -96.8575 15.8470 -6.112 4.92E-07

Aug -114.0950 15.8470 -7.200 1.80E-08

Sep -108.0000 15.8470 -6.815 5.74E-08

Oct -77.4050 15.8470 -4.885 2.14E-05

Nov -38.8175 15.8470 -2.450 .0193

14.33 a. See graph. There is no apparent trend.b. Yes, one sees the demand for permits start to increase before the prime building months of the summer

season. Seasonality is apparent in the recurring cycles.c. See output and graph. Note that 1.00 is the key reference point for a multiplicative seasonal index. The

creative student might plot the seasonal indexes to show that April through October seasonal indexes exceed 1.00, while the other months are below 1.00.

d. April, May, June have the most; while January, December, February have the least. Yes, this is logical, based on weather patterns.

e. No there is no trend (see graph).

208

14.34 a. See graph. There is a pattern, but it is cyclical and obviously not easily modeled.b. Yes, but not easily detected. Shipments are slower in the first quarter and stronger in the fourth quarter.c. See output and graph. Yes, there is an upward trend, but with a heavy cyclical pattern.

209

14.35 a. See graph.b. See output and graph. Yes, there is an upward trend in the deseasonalized data. This is logical, given

consumers’ dependence on credit. There is a hint of a level jump in 2004.c. Highest: November, December, January (holiday buying and bill-paying). Lowest: October, September,

July. Yes this is logical. Credit increases due to the Christmas buying season. Drops off during the month of July (vacation), and September and October (kids back to school, waiting for Christmas spending season). Most of the months that are below 1.00 are only slightly below.

14.36 a. See graph.b. See output. Yes there is a trend (a little hard to see because the scale is magnified by the seasonal

“spike” every December.c. December is the highest (greatly different from the other months) and January is the lowest. Yes, based

on the Christmas retail season makes sense. The other months are not too far from 1.00 (the all-important reference point that would indicate an “average” month in a multiplicative model). There is a trend in the deseasonalized data.

210

d. See output. The seasonal binaries are constructed with the highest month, December as the base. The seasonal binaries measure the amount of sales for the month that are less than December. All seasonal binaries are negative, indicating that December is indeed the highest month. All coefficients, including Time, are significant in the regression model. The model can be used to estimate monthly Jewelry Sales in the U.S.

R² 0.967 Adjusted R² 0.960 n 72

R 0.983 k 12

Std. Error 149.970 Dep. Var. Sales

211


Intercept 3,455.0500 71.1368 48.569 2.81E-49Jan -2,798.6837 87.1031 -32.131 4.43E-39Feb -2,575.2579 87.0134 -29.596 4.27E-37Mar -2,665.1655 86.9322 -30.658 6.05E-38Apr -2,604.2397 86.8595 -29.982 2.08E-37May -2,343.4806 86.7952 -27.000 6.66E-35Jun -2,470.0548 86.7395 -28.477 3.59E-36Jul -2,534.4623 86.6923 -29.235 8.42E-37Aug -2,479.3698 86.6537 -28.612 2.76E-36Sep -2,570.6107 86.6237 -29.676 3.68E-37Oct -2,503.1849 86.6022 -28.904 1.58E-36Nov -2,165.5925 86.5893 -25.010 4.27E-33Time 6.2409 0.8624 7.237 1.08E-09

14.37 a. See graph.b. See output. No strong trend is found. There is only slight seasonality, with December demand for cash

apparently being about 2.3% higher than the other months (seasonal index 1.023) and the other months having seasonal indexes very near 1.00. Only February is noticeably below 1.00 (seasonal index is 0.989), perhaps because it is a shortened month.

c. See output. December was omitted (to serve as a reference point). All the other months have negative coefficients, indicating that December was, on average, the highest month in terms of M1, February, May, and October are the only months with somewhat significant seasonality in the regression.

212

R² 0.121


R 0.348 k 12

Std. Error 31.417 Dep. Var. M1


Intercept 1,132.0405 13.7115 82.561 2.79E-72

Jan -21.5584 16.8664 -1.278 .2053

Feb -42.1388 16.8537 -2.500 .0147

Mar -31.9335 16.8422 -1.896 .0620

Apr -17.6567 16.8319 -1.049 .2977

May -38.0086 16.8228 -2.259 .0269

Jun -31.4176 16.8150 -1.868 .0658

Jul -29.2837 16.8083 -1.742 .0858

Aug -33.2498 16.8028 -1.979 .0517

Sep -29.5302 16.7986 -1.758 .0831

Oct -35.4392 16.7955 -2.110 .0384

Nov -24.2767 16.7937 -1.446 .1527

Time 0.0375 0.1428 0.263 .7934

Using the fitted regression equation, here are forecasts for 2002:

Period Forecast

85 1,089.90

86 1,100.11

87 1,114.38

88 1,094.03

89 1,100.62

90 1,102.76

91 1,098.79

92 1,102.51

93 1,096.60

94 1,107.76

95 1,132.08

96 1,132.04

213

14.38 a. See graph. The trend is positive and increasing. The MegaStat seasonal index show that the 2nd quarter is the highest (1.085 or 8.5% above normal) and the 1st quarter is the lowest (.877 or about 12% below normal). This is consistent with temperature and the seasons in the U.S. as they might affect sales of Coca Cola products.

b. All seasonal binaries are significant. Each measures departure from quarter 4 (i.e., Qtr4 was the omitted seasonal binary). Results confirm quarter 1 is below quarter 4 and quarters 2 and 3 are above it. The time coefficient is also significant confirming the presence of a trend.

c. See output.

R² 0.866


R 0.930 k 4

Std. Error 197.012 Dep. Var. Revenue


Intercept 4,713.1250 115.1581 40.927 5.40E-20

Qtr1 -443.7292 115.1079 -3.855 .0011

Qtr2 596.6250 114.3528 5.217 4.90E-05

Qtr3 487.8125 113.8973 4.283 .0004

time 20.3125 5.8869 3.450 .0027

214

Here are the forecasts for 2005 from fitted regression equation:

Period Forecast

25 4772

26 5833

27 5738

28 5277

14.39 a. See output, the regression is significant. All binaries and time are significant, except for January and February. Each seasonal binary measures the distance from December (i.e., December is the omitted binary).

R² 0.871


R 0.933 k 12

Std. Error 488.245 Dep. Var. Issued


Intercept 5,485.9750 231.5951 23.688 7.91E-32

Jan 575.5780 283.5755 2.030 .0469

Feb 580.1012 283.2834 2.048 .0450

Mar 1,343.2911 283.0189 4.746 1.36E-05

Apr 1,344.6476 282.7821 4.755 1.32E-05

May 1,586.8375 282.5729 5.616 5.56E-07

Jun 2,525.5274 282.3915 8.943 1.42E-12

Jul 2,665.3839 282.2380 9.444 2.10E-13

Aug 2,933.4071 282.1123 10.398 5.84E-15

Sep 2,386.9304 282.0144 8.464 9.03E-12

Oct 1,910.4536 281.9445 6.776 6.47E-09

Nov 1,215.1435 281.9026 4.311 .0001

Time -36.5232 2.8077 -13.008 5.57E-19

b. Applying the fitted regression equation, here are the forecasts for 1997. No web site could be found to check these forecasts.

Period Forecast

73 3,395.36

74 3,363.36

75 4,090.03

76 4,054.86

77 4,260.53

78 5,162.69

79 5,266.03

80 5,497.53

81 4,914.53

82 4,401.53

83 3,669.69

84 2,418.03

215

14.40* Translate each equation into the form yt = y0(1+r)t. For each equation, we have y0 = a and r = eb−1. We then verify that both forms yield the same result for t = 3 (an arbitrary choice). Students may not think of doing this kind of confirmation, but it makes the equivalency clear.

a. yt = 456(1+.1309)t

b. yt = 228(1+..0779)t

c. yt = 456(1-0.0373)t

t=3 t=3

y0 r y0(1 + r)t yt = a ebt

a. 456 0.1309 659.5071 659.5071

b. 228 0.0779 285.5296 285.5296

c. 456 -0.0373 406.8696 406.8696

14.41* Translate each of the following fitted compound interest trend models into an exponential model of the form yt = a ebt. For each equation, we set a = y0 and b = ln(1 + r). We then verify that both forms yield the same result for t = 3. Students may not think of doing this kind of confirmation, but it makes the equivalency clear.

a. yt = 123 e.0853t

b. yt = 654 e.1964t

c. yt = 308e-0.0598 t

t=3 t=3

a b y0(1 + r)t yt = a ebt

a. 123 0.853 158.8506 158.8506

b. 654 0.1964 1178.825 1178.825

c. 308 -0.0598 257.4562 257.4562

216

Chapter 15

Chi-Square Tests

15.1 a. H0: Earnings and Approach are independent.

b. Degrees of Freedom = (r−1)(c−1) = (4−1)(3−1) = 6c. CHIINV(.01,6) = 16.81 and test statistic = 127.57.

d. Test statistic is 127.57 (p < .0001) so reject the null at α = .01.e. No Clear Effect and Business Combinations contributes the most.f. All expected frequencies exceed 5.g. p-value is near zero (observed difference not due to chance).

Increase Decrease No Effect Total

Expenses and Losses Observed 133 113 23 269 Expected 142.07 83.05 43.88 269.00 (O - E)² / E 0.58 10.80 9.93 21.31

Revenue and Gains Observed 86 20 8 114 Expected 60.21 35.20 18.59 114.00 (O - E)² / E 11.05 6.56 6.04 23.64

Business Combinations Observed 12 22 33 67 Expected 35.39 20.69 10.93 67.00 (O - E)² / E 15.46 0.08 44.58 60.12

Other Approaches Observed 41 4 20 65 Expected 34.33 20.07 10.60 65.00 (O - E)² / E 1.30 12.87 8.33 22.49

Total Observed 272 159 84 515 Expected 272.00 159.00 84.00 515.00 (O - E)² / E 28.38 30.31 68.88 127.57

127.57 chi-square6 df

p-value

15.2 a. H0: Age Group and Ownership are independent.

b. Degrees of Freedom = (r−1)(c−1) = (2−1)(4−1) = 3c. CHIINV(.01,3) = 11.34 and test statistic = 19.31.d. Since the p-value (.0002) is less than .01, we reject the null and find dependence.e. Adults and Europe and Adults and Latin America contribute the most.f. All expected frequencies exceed 5.g. The p-value from MegaStat shows that observed difference would arise by chance only 2 times in 10,000

samples if the two variables really were independent.

217

U.S. Europe Asia Latin. America Total

Observed 80 89 69 65 303

Expected 75.75 75.75 75.75 75.75 303.00

(O - E)² / E 0.24 2.32 0.60 1.53 4.68

Observed 20 11 31 35 97

Expected 24.25 24.25 24.25 24.25 97.00

(O - E)² / E 0.74 7.24 1.88 4.77 14.63

Observed 100 100 100 100 400

Expected 100.00 100.00 100.00 100.00 400.00

(O - E)² / E 0.98 9.56 2.48 6.29 19.31

19.31 chi-square

3 df

.0002 p-value

15.3 a. H0: Verbal and Quantitative are independent

b. Degrees of Freedom = (r−1) (c−1) = (3−1)(3−1) = 4c. CHIINV(.005,4) = 14.86 and test statistic = 55.88.

d. Test statistic is 55.88 (p < .0001) so we reject the null at α = .005.e. The upper left cell (Under 25 and Under 25) contributes the most.f. Expected frequency is less than 5 in two cells.g. p-value is nearly zero (observed difference not due to chance).

Under 25 25 to 35 35 or More Total

Under 25 Observed 25 9 1 35

Expected 10.50 14.00 10.50 35.00

(O - E)² / E 20.02 1.79 8.60 30.40

25 to 35 Observed 4 28 18 50

Expected 15.00 20.00 15.00 50.00

(O - E)² / E 8.07 3.20 0.60 11.87

35 or More Observed 1 3 11 15

Expected 4.50 6.00 4.50 15.00

(O - E)² / E 2.72 1.50 9.39 13.61

Total Observed 30 40 30 100

Expected 30.00 40.00 30.00 100.00

(O - E)² / E 30.81 6.49 18.58 55.88

55.88 chi-square

4 df

.0000 p-value

15.4 a. H0: Privilege Level and Disciplinary Action are independent.

b. Degrees of Freedom = (r−1)(c−1) = (3−1)(2−1) = 2c. CHIINV(.01,2) = 9.210 and test statistic = 13.77.d. Since the p-value (.0010) is less than .01, we reject the null and find dependence.e. Not Disciplined and Low contributes the most.f. All expected frequencies exceed 5 except Low and Not Disciplined.g. The p-value from MegaStat shows that observed difference would arise by chance only 1 time in 1000


218

Disciplined Not Disciplined Total

Observed 20 11 31

Expected 26.29 4.71 31.00

(O - E)² / E 1.51 8.42 9.93

Medium Observed 42 3 45

Expected 38.17 6.83 45.00

(O - E)² / E 0.38 2.15 2.53

Observed 33 3 36

Expected 30.54 5.46 36.00

(O - E)² / E 0.20 1.11 1.31

Observed 95 17 112

Expected 95.00 17.00 112.00

(O - E)² / E 2.09 11.68 13.77

13.77 chi-square

2 df

.0010 p-value

15.5 a. H0: Return Rate and Notification are independent.

b. Degrees of Freedom = (r−1)(c−1) = (2−1)(2−1) = 1c. CHIINV(.025,1) = 5.024 and test statistic = 5.42 (close decision).d. Since the p-value (.0199) is less than .025, we reject the null and find dependence.e. Returned and No, Returned and Yes contribute the most.f. All expected frequencies exceed 5.g. The p-value from MegaStat shows that observed difference would arise by chance only 20 times in 1000


Returned Not Returned Total

Observed 39 155 194

Expected 30.66 163.34 194.00

(O - E)² / E 2.27 0.43 2.70

Observed 22 170 192

Expected 30.34 161.66 192.00

(O - E)² / E 2.29 0.43 2.72

Observed 61 325 386

Expected 61.00 325.00 386.00

(O - E)² / E 4.56 0.86 5.42

5.42 chi-square

1 df

.0199 p-value

Hypothesis test for two independent proportions

p1 p2 pc

0.6393 0.4769 0.5026 p (as decimal)

39/61 155/325 194/386 p (as fraction)

39 155 194 X

61 325 386 n

0.1624 difference

219

0. hypothesized difference

0.0698 std. error

2.33 z

.0199 p-value (two-tailed)

5.419795 z-squared

0.0302 confidence interval 95.% lower

0.2946 confidence interval 95.% upper

0.1322 half-width

h. In a two-tailed two-sample z test for π1 = π2 we verify that z2 is the same as the chi-square statistic, as presented in the table above. The p-value is the same (.0199) and z2 = 5.42 (after rounding) is equal to the chi-square value from the previous table.

15.6 Most purchasers are 18-24 years of age. Fewest bought by those 65 and over. At α = .01, this sample does contradict the assumption that readership is uniformly distributed among these six age groups, since the p-value is less than .01.

Goodness of Fit Test

observed expected O - E (O - E)² / E % of chisq

38 20.000 18.000 16.200 51.76

28 20.000 8.000 3.200 10.22

19 20.000 -1.000 0.050 0.16

16 20.000 -4.000 0.800 2.56

10 20.000 -10.000 5.000 15.97

9 20.000 -11.000 6.050 19.33

120 120.000 0.000 31.300 100.00

31.30 chi-square

5 df

8.17E-06 p-value

15.7 Vanilla and Mocha are the leading flavors. Coffee least favorite as measured by sales. However, at α = .05, this sample does not contradict the assumption that sales are the same for each beverage, since the p-value (.8358) is greater than .05.

220



18 21.000 -3.000 0.429 50.00

23 21.000 2.000 0.190 22.22

23 21.000 2.000 0.190 22.22

20 21.000 -1.000 0.048 5.56

84 84.000 0.000 0.857 100.00

.86 chi-square

3 df

.8358 p-value

15.8 Graph reveals that 0 and 8 are occurring the most frequently, while 1 has the smallest occurrence rate. At

α = .05, we cannot reject the hypothesis that the digits are from a uniform population, since the p-value (.5643) is greater than .05.

221



33 27.000 6.000 1.333 17.31

17 27.000 -10.000 3.704 48.08

25 27.000 -2.000 0.148 1.92

30 27.000 3.000 0.333 4.33

31 27.000 4.000 0.593 7.69

28 27.000 1.000 0.037 0.48

24 27.000 -3.000 0.333 4.33

25 27.000 -2.000 0.148 1.92

32 27.000 5.000 0.926 12.02

25 27.000 -2.000 0.148 1.92

270 270.000 0.000 7.704 100.00

7.70 chi-square

9 df

.5643 p-value

15.9 At α = .05, we cannot reject the hypothesis that the movie goers are from a uniform population, since the p-value (.1247) is greater than .05


Age observed expected O - E (O - E)² / E % of chisq

10 < 20 5 8.000 -3.000 1.125 11.25

20 < 30 6 8.000 -2.000 0.500 5.00

30 < 40 10 8.000 2.000 0.500 5.00

40 < 50 3 8.000 -5.000 3.125 31.25

50 < 60 14 8.000 6.000 4.500 45.00

60 < 70 9 8.000 1.000 0.125 1.25

70 < 80 9 8.000 1.000 0.125 1.25

56 56.000 0.000 10.000 100.00

10.00 chi-square

6 df

.1247 p-value

15.10 a. The sample mean and standard deviation are close to those used to generate the values.

Sample Generated

5.20 5.00

2.4075 2.2361

222

b. and c. See table below. Since the p-value (.7853) is greater than 0.05, we don’t reject the null hypothesis that the observations were coming from a Poisson distribution. Note that the end categories were collapsed so that their expected frequencies would be at least 5. A common error that students make is to fail to check that their probabilities sum to 1 and that the expected frequencies sum to n. If these sum to less than expected, it is an indication that they forgot the Poisson probabilities beyond the highest observed value (X = 12, 13, 14, ... etc.).

X P(X) observed expected O - E (O - E)² / E

2 or less 0.12465 6 6.2326 -0.2326 0.0087

3 0.14037 7 7.0187 -0.0187 0.0000

4 0.17547 7 8.7734 -1.7734 0.3585

5 0.17547 8 8.7734 -0.7734 0.0682

6 0.14622 10 7.3111 2.6889 0.9889

7 0.10444 3 5.2222 -2.2222 0.9456

8 or more 0.13337 9 6.6686 2.3314 0.8151

1.00000 50 50.0000 0.0000 3.1850

3.185chi-square

6df

.7853p-value

d. If we use λ = 5.2 instead of λ = 5, the test statistic changes and we lose one degree of freedom (because λ is being estimated from the sample). However, in this case, the p-value (.7610) is about the same, so we still fail to reject the hypothesis of a Poisson distribution.

X P(X) observed expected O - E (O - E)² / E

2 or less 0.10879 6 5.4393 0.5607 0.0578

3 0.12928 7 6.4639 0.5361 0.0445

4 0.16806 7 8.4031 -1.4031 0.2343

5 0.17479 8 8.7393 -0.7393 0.0625

6 0.15148 10 7.5740 2.4260 0.7771

7 0.11253 3 5.6264 -2.6264 1.2260

8 or more 0.15508 9 7.7539 1.2461 0.2002

1.00000 50 50.0000 0.0000 2.602

2.602chi-square

5df

.7610p-value

223

15.11 Using sample mean λ = 4.948717949 the test statistic is 3.483 (p-value = .4805) with d.f. = 6−1−1 = 4.

The critical value for α = .05 is 9.488 so we cannot reject the hypothesis of a Poisson distribution. Note that the end categories were collapsed so that their expected frequencies would be at least 5. A common error that students make is to fail to check that their probabilities sum to 1 and that the expected frequencies sum to n. If these sum to less than expected, it is an indication that they forgot the Poisson probabilities beyond the highest observed value (X = 11, 12, 13, ... etc.).

X P(X) Obs Exp O−E (O−E)² / E

2 or less 0.12904 3 5.032 -2.032 0.821

3 0.14326 5 5.587 -0.587 0.062

4 0.17724 9 6.912 2.088 0.631

5 0.17542 10 6.841 3.159 1.458

6 0.14468 5 5.643 -0.643 0.073

7 or more 0.23036 7 8.984 -1.984 0.438

1.00000 39 39.000 0.000 3.483

15.12 At α = .05, you cannot reject the hypothesis that truck arrivals per day follow a Poisson process, since

the p-value (.2064) is greater than .05. For this test, we use the estimated sample mean λ = 2.6. Note that the top categories were collapsed so that their expected frequencies would be at least 5. A common error that students make is to fail to check that their probabilities sum to 1 and that the expected frequencies sum to n. If these sum to less than expected, it is an indication that they forgot the Poisson probabilities beyond the highest observed value (X = 8, 9, 10, ... etc.).

X Days P(X) Exp Obs-Exp Chi-Square

0 4 0.07427 7.4274 -3.4274 1.5816

1 23 0.19311 19.3111 3.6889 0.7047

2 28 0.25104 25.1045 2.8955 0.3340

3 22 0.21757 21.7572 0.2428 0.0027

4 8 0.14142 14.1422 -6.1422 2.6677

5 or more 15 0.12258 12.2577 2.7423 0.6135

Total 100 1.00000 100.0000 0.0000 5.9041

df 4

p-value 0.2064

15.13 From sample, x = 75.375 and s = 8.943376. Set ej = 40/8 = 5. Students might form categories somewhat differently, so results may vary slightly depending on rounding. Using Visual Statistics with 8 classes with class limits to ensure equal expected frequencies (the “optimal expected frequencies” option)

the test statistic is 6.000 (p-value = .3062) using d.f. = 8−2−1 = 5. The critical value for α = .05 is 11.07 so we cannot reject the hypothesis of a normal distribution. Visual Statistics helpful because you can adjust for expected frequencies less than 5 easily and quickly.

Score Obs Exp Obs−Exp Chi-Square

Under 65.09 5 5.000 0.000 0.000

65.09 < 69.34 3 5.000 -2.000 0.800

69.34 < 72.53 5 5.000 0.000 0.000

72.53 < 75.38 3 5.000 -2.000 0.800

75.38 < 78.22 9 5.000 4.000 3.200

78.22 < 81.41 7 5.000 2.000 0.800

81.41 < 85.66 4 5.000 -1.000 0.200

85.66 or more 4 5.000 -1.000 0.200

Total 40 40.000 0.000 6.000

224

15.14 For this test, we use the estimated sample mean 31.1512 and standard deviation 9.890436. Set ej = 42/8 = 5.25 Students might form categories somewhat differently, so results may vary slightly depending on rounding. Results shown below are from Visual Statistics using the option for equal (optimal) expected

frequencies. At α = .025, you cannot reject the hypothesis that carry-out orders follow a normal population, since the p-value (.7074) is greater than .025.

Cost of Order Obs Exp Obs-Exp Chi-Square

Under 19.77 6 5.2500 0.7500 0.1070

19.77 < 24.48 6 5.2500 0.7500 0.1070

24.48 < 28.00 6 5.2500 0.7500 0.1070

28.00 < 31.15 3 5.2500 -2.2500 0.9640

31.15 < 34.30 5 5.2500 -0.2500 0.0120

34.30 < 37.82 6 5.2500 0.7500 0.1070

37.82 < 42.53 3 5.2500 -2.2500 0.9640

42.53 or more 7 5.2500 1.7500 0.5830

Total 42 42.0000 0.0000 2.9520

d.f. 5

p-value .7074

15.15* The probability plot looks rather linear, yet the p-value (.033) for the Anderson-Darling test is less than α = .05. This tends to contradict the chi-square test used in Exercise 15.13. However, the Kolmogorov-Smirnov test (DMax = .158) = does not reject normality (p > .20). The data are a borderline case, having some characteristics of a normal distribution. If we have to choose one test, the A-D is the most powerful.

Exam Score

Percent

1009080706050

99

95

90

80

70

60

50

40

30

20

10

5

1

Mean

0.033

75.38

StDev 8.943

N 40

AD 0.811

P-Value

Probability Plot of Exam ScoreNormal

225

15.16* The probability plot looks linear and the p-value (.404) for the Anderson-Darling test exceeds α = .05. The Kolmogorov-Smirnov test (DMax = .085) = does not reject normality (p > .20). Therefore, we cannot reject the hypothesis of normality.

Cost of Order

Percent

605040302010

99

95

90

80

70

60

50

40

30

20

10

5

1

Mean

0.404

31.15

StDev 9.890

N 42

AD 0.373

P-Value

Probability Plot of Cost of OrderNormal

226

15.17 a. H0: Pay Category and Job Satisfaction are independent.

b. Degrees of Freedom = (r−1)(c−1) = (2−1)(3−1) = 2c. CHIINV(.05,2) = 5.991 and test statistic = 4.54.d. Since the p-value (.1032) is greater than .05, we cannot reject the null and find independence.e. Highlighted cells contribute the most — see table and (O - E)2 / E.f. No small expected frequenciesg. The p-value from MegaStat shows that observed difference would arise by chance only 103 times in

1000 samples if the two variables really were independent.

Satisfied Neutral Dissatisfied Total

Observed 20 13 2 35

Expected 15.28 13.80 5.92 35.00

(O - E)² / E 1.46 0.05 2.59 4.10

Observed 135 127 58 320

Expected 139.72 126.20 54.08 320.00

(O - E)² / E 0.16 0.01 0.28 0.45

Observed 155 140 60 355

Expected 155.00 140.00 60.00 355.00

(O - E)² / E 1.62 0.05 2.88 4.54

4.54 chi-square

2 df

.1032 p-value

15.18 a. H0: Credits Earned and Certainty of Major are independent.

b. Degrees of Freedom = (r−1)(c−1) = (3−1)(3−1) = 4c. CHIINV(.01,4) = 13.28d. Since the p-value (.0052) is less than .01, we can reject the null and conclude dependence.e. Highlighted cells contribute the most (see table).f. No small expected frequenciesg. The p-value from MegaStat shows that observed difference would arise by chance only 5 times in 1000


VeryUncertain

SomewhatCertain

VeryCertain Total

0 to 9 Observed 12 8 3 23

Expected 7.55 6.83 8.63 23.00

(O - E)² / E 2.63 0.20 3.67 6.50

10 to 59 Observed 8 4 10 22

Expected 7.22 6.53 8.25 22.00

(O - E)² / E 0.08 0.98 0.37 1.44

60 or more Observed 1 7 11 19

Expected 6.23 5.64 7.13 19.00

(O - E)² / E 4.39 0.33 2.11 6.83


Expected 21.00 19.00 24.00 64.00

(O - E)² / E 7.11 1.51 6.15 14.76

14.76 chi-square

4 df

.0052 p-value

227

15.19 a. H0: Order Handed In and Grade are independent.

b. Degrees of Freedom = (r−1)(c−1) = (2−1)(2−1) = 1c. CHIINV(.1,1) = 2.706 and test statistic = 0.23.d. Since the p-value is greater than .10, we cannot reject the null and find independence.e. Highlighted cells contribute the most—see table and (O - E)2 / E.f. No small expected frequenciesg. The p-value from MegaStat shows that observed difference would arise by chance 628 times in 1000

samples if the two variables really were independent, so the sample result is not convincing.h. See table below. The z2 does equal the chi-squared value and gives the same two-tailed p-value.

EarlierHand-In

LaterHand-In Total

“B” or better Observed 10 8 18

Expected 9.18 8.82 18.00

(O - E)² / E 0.07 0.08 0.15

“C” or worse Observed 15 16 31

Expected 15.82 15.18 31.00

(O - E)² / E 0.04 0.04 0.09

Total Observed 25 24 49

Expected 25.00 24.00 49.00

(O - E)² / E 0.11 0.12 0.23

.23 chi-square

1 df

.6284 p-value


p1 p2 pc

0.4 0.3333 0.3673 p (as decimal)

10/25 8/24 18/49 p (as fraction)

10. 8. 18. X

25 24 49 n

0.0667 difference


0.1378 std. error

0.48

0.2304 z-squared


228

15.20 a. H0: Type of Planning and Competition are independent.

b. Degrees of Freedom = (r−1)(c−1) = (3−1)(3−1) = 4c. CHIINV(.05,4) = 9.488 and test statistic = 24.59.d. Since the p-value (.0001) is less than .05, we can reject the null, we find dependence.e. Highlighted cells contribute the most—see table and (O - E)2 / E.f. No small expected frequenciesg. The p-value from MegaStat shows that observed difference would arise by chance only 1 time in 10,000


Limited Constituency Comprehensive Total

Observed 11 25 33 69

Expected 23.87 23.87 21.26 69.00

(O - E)² / E 6.94 0.05 6.49 13.48

Moderate Observed 19 23 15 57

Expected 19.72 19.72 17.56 57.00

(O - E)² / E 0.03 0.55 0.37 0.94


Expected 29.41 29.41 26.18 85.00

(O - E)² / E 6.28 0.66 3.22 10.16

Observed 73 73 65 211

Expected 73.00 73.00 65.00 211.00

(O - E)² / E 13.25 1.26 10.08 24.59

24.59 chi-square

4 df

.0001 p-value

15.21 a. H0: Graduation and Sport are independent. H1: Graduation and Sport are not independent.

b. Degrees of Freedom = (r−1)(c−1) = (12−1)(2−1) = 11c. CHIINV(.01,11) = 24.73 and test statistic = 82.73.d. Since the p-value (less than .0001) is less than .01, we can reject the null, we find dependence.e. Highlighted cells contribute the most—see table and (O - E)2 / E.f. No small expected frequenciesg. The tiny p-value from MegaStat shows that observed difference would not arise by chance if the two

variables really were independent. Point out to students that the large sample size could make almost any deviation from independence “significant.”

229

Grad in6 Years

Not Grad in6 years Total

Tennis Observed 42 16 58

Expected 29.61 28.39 58.00

(O - E)² / E 5.18 5.40 10.58

Swimming Observed 116 51 167

Expected 85.27 81.73 167.00

(O - E)² / E 11.07 11.55 22.63

Soccer Observed 35 17 52

Expected 26.55 25.45 52.00

(O - E)² / E 2.69 2.80 5.49

Gymnastics Observed 40 23 63

Expected 32.17 30.83 63.00

(O - E)² / E 1.91 1.99 3.90

Golf Observed 30 21 51

Expected 26.04 24.96 51.00

(O - E)² / E 0.60 0.63 1.23

Track Observed 97 69 166

Expected 84.76 81.24 166.00

(O - E)² / E 1.77 1.84 3.61

Football Observed 267 317 584

Expected 298.19 285.81 584.00

(O - E)² / E 3.26 3.40 6.67

Wrestling Observed 70 87 157

Expected 80.16 76.84 157.00

(O - E)² / E 1.29 1.34 2.63

Baseball Observed 77 98 175

Expected 89.36 85.64 175.00

(O - E)² / E 1.71 1.78 3.49

Hockey Observed 39 66 105

Expected 53.61 51.39 105.00

(O - E)² / E 3.98 4.16 8.14

Basketball Observed 36 61 97

Expected 49.53 47.47 97.00

(O - E)² / E 3.70 3.86 7.55

Other Observed 18 5 23

Expected 11.74 11.26 23.00

(O - E)² / E 3.33 3.48 6.81

Total Observed 867 831 1698

Expected 867.00 831.00 1698.00

(O - E)² / E 40.49 42.24 82.73

82.73chi-square

11df

4.36E-13p-value

230

15.22 a. H0: Vehicle Type and Mall Location are independent.

b. Degrees of Freedom = (r−1)(c−1) = (5−1)(4−1) = 12c. CHIINV(.05,12) = 21.03 and test statistic = 24.53.d. Since the p-value (.0172) is less than .05, we can reject the null, we find dependence.e. Highlighted cells contribute the most—see table and (O - E)2 / E.f. Small expected frequencies in the full size van row.g. The p-value (.0172) from MegaStat shows that observed difference would arise by chance about 17 times

in 1,000 samples if the two variables really were independent.

Somerset Oakland Great Lakes Jamestown Total

Car Observed 44 49 36 64 193

Expected 48.25 48.25 48.25 48.25 193.00

O - E -4.25 0.75 -12.25 15.75 0.00

(O - E)² / E 0.37 0.01 3.11 5.14 8.64

Minivan Observed 21 15 18 13 67

Expected 16.75 16.75 16.75 16.75 67.00

O - E 4.25 -1.75 1.25 -3.75 0.00

(O - E)² / E 1.08 0.18 0.09 0.84 2.19

Full-size Van Observed 2 3 3 2 10

Expected 2.50 2.50 2.50 2.50 10.00

O - E -0.50 0.50 0.50 -0.50 0.00

(O - E)² / E 0.10 0.10 0.10 0.10 0.40

SUV Observed 19 27 26 12 84

Expected 21.00 21.00 21.00 21.00 84.00

O - E -2.00 6.00 5.00 -9.00 0.00

(O - E)² / E 0.19 1.71 1.19 3.86 6.95

Truck Observed 14 6 17 9 46

Expected 11.50 11.50 11.50 11.50 46.00

O - E 2.50 -5.50 5.50 -2.50 0.00

(O - E)² / E 0.54 2.63 2.63 0.54 6.35

Total Observed 100 100 100 100 400

Expected 100.00 100.00 100.00 100.00 400.00

O - E 0.00 0.00 0.00 0.00 0.00

(O - E)² / E 2.29 4.64 7.12 10.48 24.53

24.53chi-square

12df

.0172p-value

15.23 a. H0: Smoking and Race are independent.

b. Degrees of Freedom = (r−1)(c−1) = (2−1)(2−1) = 1c. CHIINV(.005,1) = 7.879 and test statistic = 5.84 (for males) and 14.79 (for females).d. For males, the p-value (.0157) is not less than .005, so we cannot reject the hypothesis of independence.

However, for females, the p-value (.0001) is less than .005 so we conclude dependence.e. Highlighted cells contribute the most—see table and (O - E)2 / E.f. No small expected frequencies.g. The p-value for males is just within the chance level, while the female p-value indicates significance.

231

Males

Smoker Nonsmoker Total

Observed 145 280 425

Expected 136.00 289.00 425.00

(O - E)² / E 0.60 0.28 0.88

Observed 15 60 75

Expected 24.00 51.00 75.00

(O - E)² / E 3.38 1.59 4.96


Expected 160.00 340.00 500.00

(O - E)² / E 3.97 1.87 5.84

5.84 chi-square

1 df

.0157 p-value

Females

Smoker Nonsmoker Total


Expected 102.09 312.91 415.00

(O - E)² / E 1.90 0.62 2.51

Observed 7 78 85

Expected 20.91 64.09 85.00

(O - E)² / E 9.25 3.02 12.27


Expected 123.00 377.00 500.00

(O - E)² / E 11.15 3.64 14.79

14.79chi-square

1df

.0001p-value

15.24 a. H0: Cockpit Noise Level and Flight Phase are independent.

b. Degrees of Freedom = (r−1)(c−1) = (3−1)(3−1) = 4c. CHIINV(.05,4) = 9.488 and test statistic = 15.16.d. Since the p-value (.0044) is less than .05, we can reject the null, i.e., we find dependence.e. Highlighted cells contribute the most—see table and (O - E)2 / E.f. Small expected frequencies in the Cruise column.g. The p-value (.0044) from MegaStat shows that observed difference would arise by chance about 44 times

in 1,000 samples if the two variables really were independent.

232

Climb Cruise Descent Total

Observed 6 2 6 14

Expected 5.74 1.84 6.43 14.00

(O - E)² / E 0.01 0.01 0.03 0.05

Medium Observed 18 3 8 29

Expected 11.89 3.80 13.31 29.00

(O - E)² / E 3.15 0.17 2.12 5.43

Observed 1 3 14 18

Expected 7.38 2.36 8.26 18.00

(O - E)² / E 5.51 0.17 3.98 9.67

Observed 25 8 28 61

Expected 25.00 8.00 28.00 61.00

(O - E)² / E 8.67 0.36 6.13 15.16

15.16 chi-square

4 df

.0044 p-value

15.25 a. H0: Actual Change and Forecasted Change are independent.

b. Degrees of Freedom = (r−1)(c−1) = (2−1)(2−1) = 1c. CHIINV(.10,1) = 2.706 and test statistic = 1.80.d. Since the p-value (.1792) exceeds .10, we cannot reject the null, i.e., we find independence.e. Highlighted cells contribute the most—see table and (O - E)2 / E.f. No small expected frequencies.g. The p-value (.1792) from MegaStat shows that observed difference would arise by chance about 18 times

in 100 samples if the two variables really were independent.

Decline Rise Total

Observed 7 12 19

Expected 8.94 10.06 19.00

(O - E)² / E 0.42 0.37 0.80

Observed 9 6 15

Expected 7.06 7.94 15.00

(O - E)² / E 0.53 0.47 1.01

Observed 16 18 34

Expected 16.00 18.00 34.00

(O - E)² / E 0.96 0.85 1.80

1.80 chi-square

1 df

.1792 p-value

15.26 a. H0: Smoking and Education Level are independent.

b. Degrees of Freedom = (r−1)(c−1) = (4−1)(4−1) = 6c. CHIINV(.005,6) = 18.55 and test statistic = 227.78.d. Since the p-value (less than .0001) is smaller than .005, we reject the null, i.e., we find dependence.e. First and fourth rows contribute the most—see table and (O - E)2 / E.f. No small expected frequencies.g. The tiny p-value from MegaStat is highly significant. Point out to students that this is partly an artifact

due to the huge sample size (i.e., in large samples, just about any deviation from independence would be significant).

233

No Smoking < 1/2 Pack >= 1/2 Pack Total

< High School Observed 641 196 196 1033

Expected 764.76 139.74 128.50 1033.00

(O - E)² / E 20.03 22.65 35.46 78.14

High School Observed 1370 290 270 1930

Expected 1428.83 261.09 240.08 1930.00

(O - E)² / E 2.42 3.20 3.73 9.35

Some College Observed 635 68 53 756

Expected 559.69 102.27 94.04 756.00

(O - E)² / E 10.13 11.48 17.91 39.53

College Observed 550 30 18 598

Expected 442.72 80.90 74.39 598.00

(O - E)² / E 26.00 32.02 42.74 100.76


Expected 3196.00 584.00 537.00 4317.00

(O - E)² / E 58.58 69.36 99.84 227.78

227.78 chi-square

6 df

2.28E-46 p-value

15.27 a. H0: ROI and Sales Growth are independent.

b. For 2×2 table: Degrees of Freedom = (r−1)(c−1) = (2−1)(2−1) = 1

For 3×3 table: Degrees of Freedom = (r−1)(c−1) = (3−1)(3−1) = 4

c. For 2×2 table: CHIINV(.05,1) = 3.841 and test statistic = 7.15.

For 3×3 table: CHIINV(.05,4) = 9.488 and test statistic = 12.30.

d. For 2×2 table: Conclude dependence since p-value = .0075 is smaller than .05.

For 3×3 table: Conclude dependence since p-value = .0153 is smaller than .05.e. First column contributes the most—see table and (O - E)2 / E.f. No small expected frequencies.

g. The tables agree. Both p-values are significant at α = .05.

2×2 Cross-Tabulation of Companies

Low High Total

Observed 24 16 40

Expected 17.88 22.12 40.00

(O - E)² / E 2.09 1.69 3.78

Observed 14 31 45

Expected 20.12 24.88 45.00

(O - E)² / E 1.86 1.50 3.36

Observed 38 47 85

Expected 38.00 47.00 85.00

(O - E)² / E 3.95 3.20 7.15

7.15 chi-square

1 df

.0075 p-value

234

3×3 Cross-Tabulation of Companies

Low Medium High Total

Observed 9 12 7 28

Expected 5.27 12.52 10.21 28.00

(O - E)² / E 2.64 0.02 1.01 3.67

Medium Observed 6 14 7 27

Expected 5.08 12.07 9.85 27.00

(O - E)² / E 0.17 0.31 0.82 1.30

Observed 1 12 17 30

Expected 5.65 13.41 10.94 30.00

(O - E)² / E 3.82 0.15 3.36 7.33


Expected 16.00 38.00 31.00 85.00

(O - E)² / E 6.63 0.48 5.19 12.30

12.30 chi-square

4 df

.0153 p-value

15.28 a. H0: Type of Cola Drinker and Correct Response are independent.

b. Degrees of Freedom = (r−1)(c−1) = (2−1)(2−1) = 1c. CHIINV(.05,1) = 3.841 and test statistic = 0.63.d. Since the p-value (.4282) exceeds .05, we cannot reject the null, i.e., we find independence.e. Highlighted cells contribute the most—see table and (O - E)2 / E.f. No small expected frequencies.g. The p-value shows that observed difference would arise by chance about 43 times in 100 samples if the

two variables really were independent. We get the same p-value result using a two-tailed test of two proportions, and z2 = 0.792 = .63—the same as the chi-square test statistic (except for rounding).

Regular Cola Diet Cola Total

Observed 7 7 14

Expected 5.78 8.22 14.00

(O - E)² / E 0.26 0.18 0.44

Observed 12 20 32

Expected 13.22 18.78 32.00

(O - E)² / E 0.11 0.08 0.19

Observed 19 27 46

Expected 19.00 27.00 46.00

(O - E)² / E 0.37 0.26 0.63

.63 chi-square

1 df

.4282 p-value

235


p1 p2 pc

0.3684 0.2593 0.3043 p (as decimal)

7/19 7/27 14/46 p (as fraction)

7. 7. 14. X

19 27 46 n

0.1092 difference


0.1378 std. error

0.79 z

0.6277


15.29 a. H0: Student Category and Reason for Choosing are independent.

b. Degrees of Freedom = (r−1)(c−1) = (3−1)(3−1) = 4c. CHIINV(.01,4) = 13.28 and test statistic = 54.18.d. Since the p-value (less than .0001) is smaller than .01, we reject the null, i.e., we find dependence.e. No consistent pattern—see table and (O - E)2 / E.f. No small expected frequencies.g. Tiny p-value indicates that the variables are not independent.

Tuition Location Reputation Total

Freshmen Observed 50 30 35 115

Expected 30.49 34.41 50.09 115.00

O - E 19.51 -4.41 -15.09 0.00

(O - E)² / E 12.48 0.57 4.55 17.59

Transfers Observed 15 29 20 64

Expected 16.97 19.15 27.88 64.00

O - E -1.97 9.85 -7.88 0.00

(O - E)² / E 0.23 5.06 2.23 7.52

MBAs Observed 5 20 60 85

Expected 22.54 25.44 37.03 85.00

O - E -17.54 -5.44 22.97 0.00

(O - E)² / E 13.65 1.16 14.25 29.06


Expected 70.00 79.00 115.00 264.00

O - E 0.00 0.00 0.00 0.00

(O - E)² / E 26.36 6.79 21.03 54.18

54.18chi-square

4df

4.83E-11p-value

236

15.30 a. H0: Dominance of Parent and Favoring Legalizing Marijuana are independent.

b. Degrees of Freedom = (r−1)(c−1) = (2−1)(3−1) = 2c. CHIINV(.10,2) = 4.605 and test statistic = 4.23.d. Since the p-value (.1204) exceeds .10, we cannot reject the null, i.e., we find independence.e. No consistent pattern—see table and (O - E)2 / E.f. One expected frequency is below 5 (Father and Yes).g. This is a close decision. The test statistic does not quite exceed the critical value, but is close.

Mother Neither Father Total

Observed 9 13 12 34

Expected 12.24 11.56 10.20 34.00

(O - E)² / E 0.86 0.18 0.32 1.35

Observed 9 4 3 16

Expected 5.76 5.44 4.80 16.00

(O - E)² / E 1.82 0.38 0.68 2.88


Expected 18.00 17.00 15.00 50.00

(O - E)² / E 2.68 0.56 0.99 4.23

4.23 chi-square

2 df

.1204 p-value

15.31 At α = .10, this sample does not contradict the assumption that President’s deaths is uniformly distributed among the four seasons, since the p-value (.6695) is greater than .10. No parameters are estimated, so

d.f. = c−1−m = 4−1−0 = 3.


11 9.000 2.000 0.444 28.57

9 9.000 0.000 0.000 0.00

10 9.000 1.000 0.111 7.14

6 9.000 -3.000 1.000 64.29

36 36.000 0.000 1.556 100.00

1.56 chi-square

3 df

.6695 p-value

15.32 At α = .05, this sample does not contradict the assumption that the 50 answers are uniformly distributed

since the p-value (.6268) is greater than .05. No parameters are estimated, so d.f. = c−1−m = 5−1−0 = 4.


8 10.000 -2.000 0.400 15.38

8 10.000 -2.000 0.400 15.38

9 10.000 -1.000 0.100 3.85

11 10.000 1.000 0.100 3.85

14 10.000 4.000 1.600 61.54

50 50.000 0.000 2.600 100.00

2.60 chi-square

4 df

.6268 p-value

237

15.33 To obtain expected values, multiply the U.S. proportions by 50. At α = .05, Oxnard employees do not differ significantly from the national distribution, since the p-value (.1095) exceeds .05. No parameters

are estimated, so d.f. = c−1−m = 4−1−0 = 3. A common error that students may make is to treat percentages as if they were frequencies (i.e., to convert the Oxnard frequencies to percentages). Doing so is a serious error because it doubles the sample size.


4 8.250 -4.250 2.189 36.22

20 22.900 -2.900 0.367 6.08

15 12.200 2.800 0.643 10.63

11 6.650 4.350 2.845 47.07

50 50.000 0.000 6.045 100.00

6.045 chi-square

3 df

.1095 p-value

15.34 At α = .01, you cannot reject the hypothesis that the digits are from a uniform population since the p-

value (.6570) is greater than .01. There are 356 occurrences since 89×4 = 356. No parameters are estimated, so

d.f. = c−1−m = 10−1−0 = 9.


39 35.600 3.400 0.325 4.77

27 35.600 -8.600 2.078 30.51

35 35.600 -0.600 0.010 0.15

39 35.600 3.400 0.325 4.77

35 35.600 -0.600 0.010 0.15

35 35.600 -0.600 0.010 0.15

27 35.600 -8.600 2.078 30.51

42 35.600 6.400 1.151 16.90

36 35.600 0.400 0.004 0.07

41 35.600 5.400 0.819 12.03

356 356.000 0.000 6.809 100.00

6.81 chi-square

9 df

.6570 p-value

15.35 At α = .10, you cannot reject the hypothesis that the die is fair, since the p-value (.4934) is greater than .

10. No parameters are estimated, so d.f. = c−1−m = 6−1−0 = 5.


7 10.000 -3.000 0.900 20.45

14 10.000 4.000 1.600 36.36

9 10.000 -1.000 0.100 2.27

13 10.000 3.000 0.900 20.45

7 10.000 -3.000 0.900 20.45

10 10.000 0.000 0.000 0.00

60 60.000 0.000 4.400 100.00

4.40chi-square

5df

.4934p-value

238

15.36 At α = .025, you cannot reject the hypothesis that goals per game follow a Poisson process, since the p-

value (.9293) is greater than .025. One parameter is estimated, so d.f. = c−1−m = 7−1−1 = 5. A common error that students may make is to fail to define the top category as open ended (X = 6, 7, 8, ...) so that the

last entry in the P(X) column actually is P(X ≥ 6) = 1−P(X ≤ 5). If this error is made, the probabilities will sum to less than 1 and the expected frequencies will sum to less than 232. Another common mistake

is not combining end categories to enlarge expected frequencies (e.g., Cochran’s rule requires ej ≥ 5).

Goals fj P(X) ej fj-ej (fj-ej)2 (fj-ej)

2/ej

0 19 0.08387 19.4586 -0.4586 0.21031 0.010811 49 0.20788 48.2271 0.7729 0.59733 0.01239

2 60 0.25760 59.7642 0.2358 0.05559 0.00093

3 47 0.21282 49.3742 -2.3742 5.63674 0.11416

4 32 0.13187 30.5928 1.4072 1.98010 0.06472

5 18 0.06536 15.1646 2.8354 8.03976 0.53017

6 or more 7 0.04060 9.4185 -2.4185 5.84900 0.62101

Total games 232 1.00000 232 0.0000 1.35419

Total goals 575

Mean goals/game 2.47844828

df 5

p-value 0.92926

15.37* Estimated mean is λ = 1.06666667. For d.f. = c−1−m = 4−1−1 = 2 the critical value is CHIINV(.025,2) = 7.378, test statistic is 4.947 (p= .0943) so we can’t reject the hypothesis of a Poisson distribution A common error that students may make is to fail to define the top category as open ended (X = 3, 4, 5, ...)

so that the last entry in the P(X) column actually is P(X ≥ 3) = 1−P(X ≤ 2). If this error is made, the probabilities will sum to less than 1 and the expected frequencies will sum to less than 60. Another common mistake is not combining the top categories to enlarge expected frequencies (e.g., Cochran’s

rule requires ej ≥ 5).

X fj P(X) ej fj-ej (fj-ej)2/ej

0 25 0.344154 20.64923 4.35077 0.9171 18 0.367097 22.02584 -4.02584 0.7362 8 0.195785 11.74712 -3.74712 1.195

3 or more 9 0.092964 5.57781 3.42219 2.100Total 60 1.000000 60.00000 0.00000 4.947

15.38 Results may vary, depending on which software package was used, how the categories were defined, and which options were selected (e.g., equal expected frequencies versus equal class widths). Results are shown for Visual Statistics (chi-square test with equal expected frequency option) and MINITAB (histogram with fitted normal curve and Anderson-Darling test). For the chi-square test, we use d.f. =

c−3 since two parameters are estimated,. i.e. c−1−m = c−1−2 = c−3. Note that the chi-square test’s p-value may not agree with the A-D test’s p-value. Point out to students that the chi-square test is based on grouped frequencies, whereas the A-D test is based on individual data values, and hence they may disagree. The A-D test is more powerful, but its methods are less intuitive for most students.

239

Data Set A Kentucky Derby Winning Times, 1950-2005 (n = 56)

Time Obs Exp Obs-Exp Chi-Square

Under 120.7 6 5.6 0.4 0.029

120.7 < 121.2 7 5.6 1.4 0.35

121.2 < 121.6 0 5.6 -5.6 5.6

121.6 < 121.9 7 5.6 1.4 0.35

121.9 < 122.1 6 5.6 0.4 0.029

122.1 < 122.4 14 5.6 8.4 12.6

122.4 < 122.7 1 5.6 -4.6 3.779

122.7 < 123.1 5 5.6 -0.6 0.064

123.1 < 123.5 5 5.6 -0.6 0.064

123.5 or more 5 5.6 -0.6 0.064

Total 56 56 0 22.929

Parameters from sample d.f. = 7 p < 0.002

Derby Time

Percent

125124123122121120119

99

95

90

80

70

60

50

40

30

20

10

5

1

Mean

0.223

122.1

StDev 1.100

N 56

A D 0.482

P-Value

Probability Plot of Derby TimeNormal

Data Set B National League Runs Scored Leader, 1900-2004 (n = 105)

Runs Obs Exp Obs-Exp Chi-Square

Under 103.7 11 13.13 -2.13 0.344

103.7 < 110.9 12 13.13 -1.13 0.096

110.9 < 116.3 16 13.13 2.88 0.630

116.3 < 121.2 15 13.13 1.88 0.268

121.2 < 126.0 14 13.13 0.88 0.058

126.0 < 131.4 11 13.13 -2.13 0.344

131.4 < 138.6 13 13.13 -0.13 0.001

138.6 or more 13 13.13 -0.13 0.001

Total 105 105 0 1.743


240

Derby Time

Frequency

125124123122121120

18

16

14

12

10

8

6

4

2

0

Mean 122.1

StDev 1.100

N 56

Histogram of Derby TimeNormal

Runs

Percent

18016014012010080

99.9

99

95

90

80

7060504030

20

10

5

1

0.1

Mean

0.561

121.2

StDev 15.13

N 105

A D 0.306

P-Value

Probability Plot of RunsNormal

Data Set C Weights (in grams) of Pieces of Halloween Candy (n = 78)

Weight (gm) Obs Exp Obs-Exp Chi-Square

Under 1.120 9 11.14 -2.14 0.412

1.120 < 1.269 9 11.14 -2.14 0.412

1.269 < 1.385 14 11.14 2.86 0.733

1.385 < 1.492 13 11.14 1.86 0.31

1.492 < 1.607 8 11.14 -3.14 0.886

1.607 < 1.757 17 11.14 5.86 3.079

1.757 or more 8 11.14 -3.14 0.886

Total 78 78 0 6.718


Candy Wt (gm)

Percent

2.52.01.51.00.5

99.9

99

95

90

80

7060504030

20

10

5

1

0.1

Mean

0.148

1.438

StDev 0.2985

N 78

A D 0.555

P-Valu e

Probability Plot of Candy Wt (gm)Normal

Data Set D Price-Earnings Ratios for Specialty Retailers (n = 58)

PE Ratio Obs Exp Obs-Exp Chi-Square

Under 10.35 4 8.29 -4.29 2.217

10.35 < 15.19 11 8.29 2.71 0.889

15.19 < 18.92 13 8.29 4.71 2.682

18.92 < 22.39 13 8.29 4.71 2.682

22.39 < 26.12 6 8.29 -2.29 0.631

26.12 < 30.96 6 8.29 -2.29 0.631

30.96 or more 5 8.29 -3.29 1.303

Total 58 58 0 11.034


241

Candy Wt (gm)

Frequency

2.11.81.51.20.90.6

14

12

10

8

6

4

2

0

Mean 1.438

StDev 0.2985

N 78

Histogram of Candy Wt (gm)Normal

Runs

Frequency

15013512010590

20

15

10

5

0

Mean 121.2

StDev 15.13

N 105

Histogram of RunsNormal

PE Ratio

Percent

80706050403020100

99

95

90

80

70

60

50

40

30

20

10

5

1

M ean

<0.005

20.66

S tDev 9.651

N 58

A D 2.307

P -Valu e

Probability Plot of PE RatioNormal

Data Set E U.S. Presidents’ Ages at Inauguration (n = 43)

Age Obs Exp Obs-Exp Chi-Square

Under 48.88 6 7.17 -1.17 0.19

48.88 < 52.20 10 7.17 2.83 1.12

52.20 < 54.86 5 7.17 -2.17 0.655

54.86 < 57.52 11 7.17 3.83 2.05

57.52 < 60.84 2 7.17 -5.17 3.725

60.84 or more 9 7.17 1.83 0.469

Total 43 43 0 8.209


Age

Percent

70656055504540

99

95

90

80

70

60

50

40

30

20

10

5

1

Mean

0.556

54.86

StDev 6.186

N 43

A D 0.304

P-Value

Probability Plot of AgeNormal

Data Set F Weights of 31 Randomly-Chosen Circulated Nickels (n = 31) � Nickels

Weight (gm) Obs Exp Obs-Exp Chi-Square

Under 4.908 4 5.17 -1.17 0.263

4.908 < 4.943 4 5.17 -1.17 0.263

4.943 < 4.972 6 5.17 0.83 0.134

4.972 < 5.000 4 5.17 -1.17 0.263

5.000 < 5.036 8 5.17 2.83 1.554

5.036 or more 5 5.17 -0.17 0.005

Total 31 31 0 2.484


242

Age

Frequency

6864605652484440

18

16

14

12

10

8

6

4

2

0

Mean 54.86

StDev 6.186

N 43

Histogram of AgeNormal

PE Ratio

Frequency

644832160

30

25

20

15

10

5

0

Mean 20.66

StDev 9.651

N 58

Histogram of PE RatioNormal

Nickel Wt (gm)

Percent

5.155.105.055.004.954.904.854.80

99

95

90

80

70

60

50

40

30

20

10

5

1

M ean

0.021

4.972

S tDev 0.06623

N 31

A D 0.881

P -Value

Probability Plot of Nickel Wt (gm)Normal

15.39* In this problem, the estimated Poisson mean is λ = 0.702479339 runs/inning. For d.f. = c−m−1 = 4−1−1 = 2, the critical value is CHIINV(.05,2) = 5.991. The test statistic is 64.02 (p-value less than .0001) so we reject the hypothesis that runs per inning are Poisson. A common error that students may make is to fail to define the top category as open ended (X = 3, 4, 5, ...) so that the last entry in the P(X) column

actually is P(X ≥ 3) = 1−P(X ≤ 2). If this error is made, the probabilities will sum to less than 1 and the expected frequencies will sum to less than 121. Another common mistake is not combining the top

categories to enlarge expected frequencies (e.g., Cochran’s rule requires ej ≥ 5). Why is the Poisson fit poor? Perhaps because runs in baseball are conditional on arrival at another base (usually) and waiting at that base before proceeding to a run scored following someone else’s hit (so the runner can arrive at home plate) whereas in hockey, a goal is scored or not when you arrive at the opponent’s goal.

Runs fj P(X) ej fj-ej (fj-ej)2/ej

0 83 0.495356 59.93803 23.06197 8.873405

1 15 0.347977 42.10523 -27.10523 17.44898

2 7 0.122223 14.78903 -7.78903 4.102294

3 or more 16 0.034444 4.16771 11.83229 33.59226

Total 121 1.000000 121.00000 0.00000 64.01695

15.40* Different pitchers are faced. It is not the same pitcher throughout the game. Hitters also vary by inning. Some innings have the team’s best hitters up, others do not. Also, batters face different degrees of

pressure in different innings, depending on how close the game is, how many are out, and so on. At α = .01, you can reject the hypothesis that the runs scored per inning are from a uniform population since the

p-value (.0084) is less than .01. We use d.f. = c−m−1 = 9−1−0 = 8,

Runs by Inning - Test for Uniform Distribution

Inning fj ej fj-ej (fj-ej)2 (fj-ej)

2/ej

1 9 9.44 -0.44 0.19753 0.02092

2 15 9.44 5.56 30.86420 3.26797

3 10 9.44 0.56 0.30864 0.03268

4 4 9.44 -5.44 29.64198 3.13856

5 17 9.44 7.56 57.08642 6.04444

6 10 9.44 0.56 0.30864 0.03268

7 8 9.44 -1.44 2.08642 0.22092

8 11 9.44 1.56 2.41975 0.25621

9 1 9.44 -8.44 71.30864 7.55033

Total 85 20.56471

df 8

p-value 0.008398

243

Nickel Wt (gm)

Frequency

5.125.044.964.884.80

9

8

7

6

5

4

3

2

1

0

Mean 4.972

S tDev 0.06623

N 31

Histogram of Nickel Wt (gm)Normal

15.41* Results will vary, but should be close to the intended distribution in a chi-square test or z-test for the intended mean. However, students often are surprised at how much a “normal” sample can differ from a perfect bell-shape, and even the mean and standard deviation may not be “on target.” That is the nature of random sampling, as we learned in Chapters 8 and 9. The following sample was created using Excel’s function =NORMINV(RAND(),0,1). The histogram is reasonably bell-shaped. The test statistic for zero mean is z = (10)(0.031808) = 0.3181 (p-value = .7504) so the mean is not significantly different from 0. The standard deviation is a little larger than 1. The chi-square test from Visual Statistics (using equal class widths) gives a p-value of .834 so the hypothesis of normality should not be rejected.


count 100

mean 0.031808

sample std dev 1.117752

minimum -2.35182753

maximum 3.337159939

1st quartile -0.672417

median -0.005821

3rd quartile 0.824195

Classes Obs Exp Obs-Exp Chi-Square

Under -1.641 8 6.73 1.27 0.24

-1.641 < -0.930 10 12.76 -2.76 0.596

-0.930 < -0.218 24 21.66 2.34 0.254

-0.218 < 0.493 23 24.85 -1.85 0.138

0.493 < 1.204 21 19.29 1.71 0.152

1.204 < 1.915 10 10.12 -0.12 0.001

1.915 or more 4 4.6 -0.6 0.079

Total 100 100 0 1.461

d.f. = 4 p < 0.834

244

15.42* Results will vary, but should be close to the intended distribution in a chi-square test or test for the desired mean. However, students often are surprised at how much a “uniform” sample can differ from a perfect uniform shape, and even the mean and standard deviation may not be “on target.” That is the nature of random sampling, as we learned in Chapters 8 and 9. The following sample was created using

Excel’s function =RAND(). The histogram does not look quite uniform. The test statistic for µ = 0.50000

is z = (0.533437−0.500000)/.0288675 = 1.1583 (p-value = .2467) so the mean is not significantly different from 0.5000. The standard deviation (.29142) is a little larger than .28868. However, the chi-square test from Visual Statistics (dividing the observed range into 8 classes) does givesa p-value less than .05 (mainly due to the surfeit of observations in the sixth class) so the hypothesis of uniformity might be rejected.

Histogram

0

2

4

6

8

10

12

14

16

18

0.00

00

0.10

00

0.20

00

0.30

00

0.40

00

0.50

00

0.60

00

0.70

00

0.80

00

0.90

00

1.00

00

uniform

Perc

ent


count 100

mean 0.533437

sample variance 0.084926


minimum 0.017795

maximum 0.998479

1st quartile 0.275343

median 0.620489

3rd quartile 0.772453

Classes Obs Exp Obs-Exp Chi-Square

Under 0.1404 12 12.5 -0.5 0.02

0.1404 < 0.2630 10 12.5 -2.5 0.5

0.2630 < 0.3856 14 12.5 1.5 0.18

0.3856 < 0.5081 9 12.5 -3.5 0.98

0.5081 < 0.6307 6 12.5 -6.5 3.38

0.6307 < 0.7533 23 12.5 10.5 8.82

0.7533 < 0.8759 12 12.5 -0.5 0.02

0.8759 or more 14 12.5 1.5 0.18

Total 100 100 0 14.08

Parameters from user d.f. = 7 p < 0.050

15.43* Results will vary, but should be close to the intended distribution in a chi-square test or z-test for the intended mean. However, students often are surprised at how much a “Poisson” sample can differ from a

245

expected shape, and even the mean and standard deviation may not be “on target.” That is the nature of random sampling, as we learned in Chapters 8 and 9. The following sample was created using Excel’s

Tools > Data Analysis > Random Number Generation with a mean of λ = 4. The histogram looks fine.

The test statistic for µ = 4 is z =(5)(4.17)−20 = 0.85 (p-value = .3953) so the mean doe not differ

significantly from 4. The standard deviation (2.09) is very close to 2 (the square root of λ = 4). The chi-square test from Visual Statistics gives a p-value of .722 so the hypothesis of normality should not be rejected.

Histogram

0

5

10

15

20

25

30

1 2 3 4 5 6 7 8 9 10

11

12

Data

Perc

ent


count 100

mean 4.17


minimum 1

maximum 11

1st quartile 3.00

median 4.00

3rd quartile 5.00

interquartile range 2.00

mode 4.00

X Values Obs Exp Obs-Exp Chi-Square

1 or less 9 9.16 -0.16 0.003

2 11 14.65 -3.65 0.91

3 20 19.54 0.46 0.011

4 24 19.54 4.46 1.02

5 14 15.63 -1.63 0.17

6 8 10.42 -2.42 0.562

7 7 5.95 1.05 0.184

8 3 2.98 0.02 0

9 or more 4 2.13 1.87 1.631

Total 100 100 0 4.49

d.f. = 7 p < 0.722

Chapter 16

246

Nonparametric Tests

16.1 Since the p-value (from MegaStat) is greater than .05 we fail to reject the null hypothesis of randomness.

Runs Test for Random Sequence

n runs

12 7 A

15 7 B

27 14 total

14.333 expected value

2.515 standard deviation

-0.133 z test statistic

-0.066 z (with continuity correction)


Note: MegaStat uses a continuity correction (subtracting 0.5 from the difference in the numerator when R is below its expected value) which will lead to different z values and p-values than if the textbook formula is used. MegaStat’s p-value shown.

16.2 Since the p-value (from MegaStat) is greater than .10 we fail to reject the null hypothesis of randomness.


n runs

10 6 F

14 6 T

24 12 total




0.072 z (with continuity correction)


Note: MegaStat uses a continuity correction (subtracting 0.5 from the difference in the numerator when R is below its expected value) which will lead to different z values and p-values than if the textbook formula is used. MegaStat’s p-value shown.

247

16.3 a. At α = .10, the population median does not differ from 50 (p-value = .4732). The worksheet and test statistic calculation are shown.

Student xi xi-50 | xi-50 | Rank R+ R-

1 74 24 24 14.5 14.5

2 5 -45 45 24 24

3 87 37 37 20 20

4 26 -24 24 14.5 14.5

5 60 10 10 5.5 5.5

6 99 49 49 27 27

7 37 -13 13 9 9

8 45 -5 5 3 3

9 7 -43 43 22 22

10 78 28 28 17 17

11 70 20 20 13 13

12 84 34 34 19 19

13 97 47 47 25 25

14 93 43 43 22 22

15 54 4 4 2 2

16 24 -26 26 16 16

17 62 12 12 7.5 7.5

18 32 -18 18 12 12

19 60 10 10 5.5 5.5

20 66 16 16 11 11

21 2 -48 48 26 26

22 43 -7 7 4 4

23 62 12 12 7.5 7.5

24 7 -43 43 22 22

25 100 50 50 28 28

26 64 14 14 10 10

27 17 -33 33 18 18

28 48 -2 2 1 1

406 234.5 171.5

Test Statistic:

( 1) 28(28 1)234.5

234.5 2034 40.7173

43.9147( 1)(2 1) 28(28 1)(56 1)

24 24

+ +− −

−= = = =

+ + + +

n nW

zn n n

b. The histogram appears platykurtic, but the A-D test statistic (p = .468) indicates that the hypothesis of normality should not be rejected.

249

100806040200

Median

Mean

70656055504540

A nderson-Darling Normality Test

Variance 917.041

Skewness -0.199844

Kurtosis -0.988643

N 28

Minimum 2.000

A-Squared

1st Quartile 27.500

Median 60.000

3rd Quartile 77.000

Maximum 100.000

95% C onfidence Interv al for Mean

41.936

0.34

65.421

95% Confidence Interv al for Median

39.690 68.207

95% Confidence Interv al for StDev

23.942 41.219

P-V alue 0.468

Mean 53.679

StDev 30.283

95% Confidence Intervals

Summary for Score

16.4 a. In the Wilcoxon/Mann-Whitney test at α = .05, there is a difference in the population median scores on the two exams (p-value = .00234). The worksheet is shown.

Student Exam 1 Exam 2 d |d| Rank R+ R-

9 52 53 -1 1 1.5 1.5

12 95 96 -1 1 1.5 1.5

8 71 69 2 2 3 3

20 54 58 -4 4 4 4

10 79 84 -5 5 5.5 5.5

14 81 76 5 5 5.5 5.5

3 65 59 6 6 7 7

16 54 47 7 7 8 8

4 60 68 -8 8 9.5 9.5

18 92 100 -8 8 9.5 9.5

15 59 68 -9 9 11.5 11.5

17 75 84 -9 9 11.5 11.5

7 72 82 -10 10 13 13

1 70 81 -11 11 14.5 14.5

19 70 81 -11 11 14.5 14.5

5 63 75 -12 12 16.5 16.5

11 84 96 -12 12 16.5 16.5

2 74 89 -15 15 18 18

13 83 99 -16 16 19 19

6 58 77 -19 19 20 20

210 23.5 186.5

Test Statistic:

( 1) 20(20 1)23.5

23.5 1054 43.04

26.7862( 1)(2 1) 20(20 1)(40 1)

24 24

+ +− −

−= = = = −

+ + + +

n nW

zn n n

b. In a t-test at α = .05, there is no difference in the population mean scores on the two exams since the p-value is greater than .05. The two tests reveal different results. Samples are too small for a meaningful test for normality. The MegaStat results are shown.

250

Hypothesis Test: Independent Groups (t-test, pooled variance)

Exam 1 Exam 2

70.55 77.10 mean

12.55 15.26 std. dev.

20 20 n

38 df

-6.550 difference (Exam 1 - Exam 2)

195.178 pooled variance

13.971 pooled std. dev.

4.418 standard error of difference

0 hypothesized difference

-1.48 t


16.5 a. At α = .05, there is no difference in the medians, since the p-value is greater than .05. MegaStat uses a correction for ties, so students may get different z values and p-values. The calculations and p-value shown are from MegaStat, on the assumption that students will use MegaStat for the calculations.

Wilcoxon - Mann/Whitney Test

n sum of ranks

10 135 Bob's Portfolio

12 118 Tom’s Portfolio

22 253



0.96 z, uncorrected

1.29 z, corrected for ties


b. MegaStat’s results are shown, assuming equal variances (t = 1.62, p = .0606). At α = .05, there is no difference in the means, since the p-value is greater than .05. If you assume unequal variances, the result is similar (t = 1.661, p = .0565). Both tests lead to the same decision. Samples are too small for a meaningful test for normality.

Hypothesis Test: Independent Groups (t-test, pooled variance)Bob's Portfolio Tom’s Portfolio

6.040 4.100 mean

2.352 3.119 std. dev.

10 12 n

20 df1.9400 difference (Bob's - Tom’s Portfolio)7.8392 pooled variance2.7999 pooled std. dev.1.1988 standard error of diff

0 hypothesized difference1.618 t.0606 p-value (one-tail upper)

251

16.6 a. We fail to reject the null hypothesis that there is a difference in the medians since the p-value is greater than .05. MegaStat’s results are shown.


n sum of ranks

9 125 Old Bumper

12 106 New Bumper

21 231



1.81


b. We fail to reject the null hypothesis that there is a difference in the means since the p-value is greater than .05. MegaStat’s results are shown. We have the same decision as in (a). Samples are too small for a meaningful test for normality.

Hypothesis Test: Independent Groups (t-test, pooled variance)Old Bumper New Bumper

1,766.11 1,101.42 mean

837.62 696.20 std. dev.

9 12 n

19 df664.694 difference (Old - New)

576,031.463 pooled variance

758.967 pooled std. dev.334.673 standard error of difference

0 hypothesized difference1.99 t


.0308 p-value (one-tailed)

16.7 MegaStat results are shown. At α = .05, there is no difference in median volatility in these four portfolios (p = .0892). The ANOVA test gives the same conclusion, but the decision is very close (p = .

0552). Had we used α = .10, the difference would have been significant in either test.

Kruskal-Wallis Test

Median n Avg. Rank

16.20 15 20.03 Health

22.70 12 35.13 Energy

21.05 14 29.71 Retail

18.10 13 26.69 Leisure

19.65 54 Total

6.511 H (corrected for ties)

3 d.f.

.0892 p-value

252

One factor ANOVA

Mean n Std. Dev

19.92037037 17.34 15 4.630 Health

19.92037037 23.18 12 6.311 Energy

19.92037037 20.62 14 4.032 Retail

19.92037037 19.14 13 6.711 Leisure

19.92 54 5.716 Total

ANOVA table


Treatment 241.815 3 80.6049 2.71 .0552

Error 1,489.913 50 29.7983

Total 1,731.728 53

Based on the four individual histograms, we would doubt normality. However, each sample is rather small for a normality test. Pooling the samples, we get a p-value of .490 for MINITAB’s Anderson-Darling test statistic, so normality can’t be rejected.

253

302418126

Median

Mean

222120191817


Variance 32.674

Skewness 0.014614

Kurtosis -0.137501

N 54

Minimum 4.900

A-Squared

1st Quartile 15.250

Median 19.650

3rd Quartile 24.275

Maximum 32.500


18.360

0.34

21.481


17.636 22.064


4.805 7.057

P-V alue 0.490

Mean 19.920

StDev 5.716


Summary for Volatility

16.8 a. At α = .05, there is a difference in median productivity since the p-value is less than .05.

Kruskal-Wallis Test

Mediann Avg. Rank

4.10 9 11.61 Station A

2.90 6 6.67 Station B

5.40 10 18.05 Station C

4.50 25 Total


2 d.f.

.0087 p-value

multiple comparison values for avg. ranks

8.63 10.58

b. At α = .05, there is a difference in median productivity since the p-value is less than .05.

One factor ANOVA

Mean n Std. Dev

4.38 3.97 9 0.828 Station A

4.38 3.02 6 1.094 Station B

4.38 5.57 10 1.726 Station C

4.38 25 1.647 Total

ANOVA table


Treatment 26.851 2 13.4253 7.72 .0029

Error 38.269 22 1.7395

Total 65.120 24

The samples are rather small for a normality test. Pooling the samples them, we get a p-value of .392 for MINITAB’s Anderson-Darling test statistic, so normality can’t be rejected.

254

8765432

Median

Mean

5.04.54.03.53.0


Variance 2.7133

Skewness 0.660729

Kurtosis 0.072543

N 25

Minimum 1.9000

A-Squared

1st Quartile 3.0000

Median 4.5000

3rd Quartile 5.2500

Maximum 8.4000


3.7001

0.37

5.0599


3.2396 4.9802


1.2862 2.2915

P-V alue 0.392

Mean 4.3800

StDev 1.6472


Summary for Units Per Hour

16.9 The median ratings of surfaces do not differ at α = .05 since the p-value is greater than .05.

Friedman Test

Sum of Ranks Avg. Rank

9.00 2.25 Shiny

10.00 2.50 Satin

17.50 4.38 Pebbled

10.00 2.50 Pattern

13.50 3.38 Embossed

60.00 3.00 Total

4 n

5.013 chi-square (corrected for ties)

4 d.f.

.2860 p-value


3.14 (.05) 3.68 (.01)

255

16.10 The median sales of coffee sizes do not differ at α = .05, since the p-value is greater than .05

Friedman Test


10.00 2.00

10.00 2.00 Medium

10.00 2.00

30.00 2.00

5


2

1.0000


1.51 1.86

16.11 a. Worksheet is shown for rank correlation.

Profit in year: Rank in year:

Obs Company 2004 2005 2004 2005

1 Campbell Soup 595 647 6 7

2 ConAgra Foods 775 880 5 5

3 Dean Foods 356 285 10 10

4 Del Monte Foods 134 165 13 14

5 Dole Food 105 134 15 15

6 Flowers Foods 15 51 19 18

7 General Mills 917 1055 3 3

8 H. J. Heinz 566 804 7 6

9 Hershey Foods 458 591 8 8

10 Hormel Foods 186 232 12 11

11 Interstate Bakeries 27 -26 17 20

12 J. M. Schmucker 96 111 16 16

13 Kellogg 787 891 4 4

14 Land O'Lakes 107 21 14 19

15 McCormick 211 215 11 13

16 Pepsico 3568 4212 1 1

17 Ralcorp Holdings 7 65 20 17

18 Sara Lee 1221 1272 2 2

19 Smithfield Foods 26 227 18 12

20 Wm. Wrigley, Jr. 446 493 9 9

Rank sum:210 210

b. Spearman rank correlation found by using the Excel function CORREL on the rank columns is 0.9338. c. t-statistic for Spearman rank correlation is 11.706. Clearly, we can reject the hypothesis of no correlation

at any of the customary α levels.

256

2 2

0.933811.076

1 1 0.9338

20 22

= = =− −

−−

s

s

rt

r

n

Critical values:

α = 0.025 t0.025 = 2.093 Reject H0

α = 0.01 t0.01 = 2.539 Reject H0

α = 0.005 t0.005 = 2.861 Reject H0

d. MegaStat’s calculations are shown.

Spearman Coefficient of Rank Correlation

2004 2005

2004 1.000

2005 .934 1.000

20 sample size



e. Calculated using the CORREL function on the actual data (not the ranks) we get r = 0.9960:f. In this example, there is no strong argument for the Spearman test since the data are ratio. However, the

assumption of normality may be dubious (samples are too small for a reliable normality test).

16.12 There is a discrepancy between the textbook’s data and the student CD data. The textbook’s margin answer is based on the CD data. Students’ answers will depend on which data set they use. Calculations for each data set are shown below. This discrepancy will be corrected in future editions.

Data Set from CD: Textbook Data Set:

12-Mo 5-Yr 12-Mo 5-Yr

12-Mo 1.000 12-Mo 1.000

5-Yr .742 1.000 5-Yr .373 1.000

24 sample size 24 sample size



The worksheets for each data set are shown:

257

Data Set from CD Ranks: Textbook Data Set Ranks:

Fund 12-Mo 5-Yr Fund 12-Mo 5-Yr

1 17.5 18.5 1 17.5 18.5

2 1 5 2 1 5

3 12 11 3 12 11

4 6 4 4 6 4

5 7 1 5 7 1

6 15 14 6 15 14

7 23 22 7 23 22

8 10 7 8 10 7

9 9 10 9 9 10

10 16 21 10 16 21

11 11 9 11 14 9

12 17.5 23 12 4 23

13 22 15 13 11 15

14 19 24 14 17.5 24

15 21 20 15 22 20

16 3 2 16 19 2

17 8 18.5 17 21 18.5

18 24 12 18 3 12

19 2 3 19 8 3

20 13 8 20 24 8

21 14 17 21 2 17

22 4 13 22 13 13

23 5 6 23 5 6

24 20 16 24 20 16

Rank sum: 300 Rank sum: 300

e. Pearson correlation found by using the Excel function CORREL is 0.6560 (from student CD data set) or 0.2796 (from the data printed in the textbook).

f. In this example, there is no strong argument for the Spearman test since the data are ratio. Despite the low outlier in 5-year returns, both samples pass the test for normality (p = .541 and .460 respectively). The following tests are based on the CD data set

20151050

Median

Mean

111098765


V ariance 34.8478

S kew ness -0.022436

Kurtosis -0.236113

N 24

M inimum -2.4000

A -Squared

1st Q uartile 3.5250

M edian 8.6500

3rd Q uartile 11.5000

M aximum 21.1000


5.4615

0.31

10.4469

95% C onfidence I nterv al for M edian

5.8016 11.2000

95% C onfidence Interv al for StDev

4.5880 8.2808

P -V alue 0.541

M ean 7.9542

S tDev 5.9032


Summary for 12-Mo

151050

Median

Mean

10987


V ariance 17.8220


Kurtosis 0.850848

N 24

M inimum -2.9000

A -Squared


M edian 9.1500


M aximum 14.7000


6.6049

0.34

10.1701


6.9441 10.2520


3.2811 5.9219

P -V alue 0.460

M ean 8.3875

S tDev 4.2216


Summary for 5-Yr

258

16.13 Since the p-value (.5300) is greater than .05, we fail to reject the null hypothesis of randomness. Note: MegaStat’s z-value subtracts a continuity correction of 0.5 from the numerator of the test statistic when R is less than its expected value. This will give a slightly different result than the formula shown in the textbook (see Siegel and Castellan, Nonparametric Statistics for the Behavioral Sciences, McGraw-Hill, 1988).

H0: Events follow a random patternH1: Events do not follow a random pattern


n runs

21 14 B

29 14 A

50 28 total



0.775 z test statistic



16.14 Since the p-value (.9145) is greater than .01 we fail to reject the null hypothesis of randomness. Note: MegaStat’s z-value subtracts a continuity correction of 0.5 from the numerator of the test statistic when R is less than its expected value. This will give a slightly different result than the formula shown in the textbook (see Siegel and Castellan, Nonparametric Statistics for the Behavioral Sciences, McGraw-Hill, 1988).



n runs

21 9 H

14 8 M

35 17 total







259



n runs

14 8 T

11 7 F

25 15 total









n runs

21 10 N

12 10 H

33 20 total








260


n runs

18 11 C

16 11 X

34 22 total









n runs

34 13 Up

27 13 Dn

61 26 total






Since the p-value is greater than .05 we fail to reject the null hypothesis of randomness.



261


n runs

13 5 Lo

11 4 Hi

24 9 total






16.20 MegaStat results are shown. At α = .10, the median ELOS does not differ for the two groups, since the p-value (.5720) is greater than .10. The hypotheses are:

H0: M1 = M2 (no difference in ELOS)

H1: M1 ≠ M2 (ELOS differs for the two groups)


n sum of ranks

10 124 Clinic A

12 129 Clinic B

22 253





Although the histogram is somewhat platykurtic in appearance, normality may be assumed at α = .10 based on the Anderson-Darling p-value (.147). To perform this test, the two samples were pooled. Even so, the sample is rather small for a normality test.

50403020

Median

Mean

37.535.032.530.027.525.0


Variance 119.100

Skewness 0.460007

Kurtosis -0.632645

N 22

Minimum 16.000

A-Squared

1st Quartile 20.000

Median 30.000

3rd Quartile 40.000

Maximum 52.000


25.525

0.54

35.202


23.891 36.109


8.396 15.596

P-V alue 0.147

Mean 30.364

StDev 10.913


Summary for Weeks

262

16.21 MegaStat results are shown. At α = .05, the median defect counts do not differ for the two groups, since the p-value (.4731) is greater than .05. The hypotheses are:

H0: M1 = M2 (no difference in number of bad pixels)

H1: M1 ≠ M2 (number of bad pixels differs for the two groups)


n sum of ranks

12 162.5

12 137.5

24 300





The histogram is strongly right-skewed and the Anderson-Darling p-value is small (less than .005) so the assumption of normality is untenable. To perform this test, the two samples were pooled. Even so, the sample is rather small for a normality test.

543210

Median

Mean

2.01.51.00.50.0


V ariance 2.3460

Skewness 1.11533

Kurtosis 0.65863

N 24

Minimum 0.0000

A -Squared

1st Quartile 0.0000

Median 1.0000

3rd Quartile 2.0000

Maximum 5.0000


0.8116

1.36

2.1051


0.0000 2.0000


1.1904 2.1486

P-V alue < 0.005

Mean 1.4583

StDev 1.5317


Summary for Bad Pixels

16.22 MegaStat results are shown. At α = .05, the median weights do not differ for the two groups, since the p-value (.1513) is greater than .01. The hypotheses are:

H0: M1 = M2 (no difference in weight of linemen)

H1: M1 ≠ M2 (Weight of linemen differs for the two teams)

263


n sum of ranks

14 226 WeightD

13 152 WeightP

27 378





The histogram is bell-shaped, and normality may be assumed at any common α based on the Anderson-Darling p-value (.360). To perform this test, the two samples were pooled.

340320300280

Median

Mean

305300295290


Variance 241.95

Skewness 0.642175

Kurtosis 0.197856

N 27

Minimum 274.00

A-Squared

1st Quartile 288.00

Median 300.00

3rd Quartile 308.00

Maximum 338.00


292.74

0.39

305.04


288.00 305.00


12.25 21.32

P-V alue 0.360

Mean 298.89

StDev 15.55


Summary for Pounds

16.23 MegaStat results are shown. At α = .05, the median difference in heart rates does not differ from zero, since the p-value (.0923) is greater than .05. Note: The 5th observation is omitted because its difference is zero (heart rate of 82 before and after) which leaves only n = 29. The hypotheses are:

H0: Md = 0 � the median difference in pulse rate is zero

H1: Md ≠ 0 � the median difference in pulse rate is not zero

Wilcoxon Signed Rank Test

Paired Data: :Before − After

149sum of positive ranks

286sum of negative ranks

29 n



-1.68 z, corrected for ties

.0923 p-value (two-tailed)The worksheet is shown:

264

Student Before After d | d | Rank R+ R-

1 60 62 -2 2 12 12

2 70 76 -6 6 24.5 24.5

3 77 78 -1 1 5 5

4 80 83 -3 3 16.5 16.5

5 82 82 0 0

6 82 83 -1 1 5 5

7 41 66 -25 25 29 29

8 65 63 2 2 12 12

9 58 60 -2 2 12 12

10 50 54 -4 4 20.5 20.5

11 82 93 -11 11 27 27

12 56 55 1 1 5 5

13 71 67 4 4 20.5 20.5

14 67 68 -1 1 5 5

15 66 75 -9 9 26 26

16 70 64 6 6 24.5 24.5

17 69 66 3 3 16.5 16.5

18 64 69 -5 5 23 23

19 70 73 -3 3 16.5 16.5

20 59 58 1 1 5 5

21 62 65 -3 3 16.5 16.5

22 66 68 -2 2 12 12

23 81 77 4 4 20.5 20.5

24 56 57 -1 1 5 5

25 64 62 2 2 12 12

26 78 79 -1 1 5 5

27 75 74 1 1 5 5

28 66 67 -1 1 5 5

29 59 63 -4 4 20.5 20.5

30 98 82 16 16 28 28

Rank sum: 435 149 286

The histograms are bell-shaped, and normality may be assumed at any common α based on the Anderson-Darling p-values (.543 and .388).

96806448

Median

Mean

7270686664

A nderson-Darling Normality T est

V ariance 131.361

S kewness 0.182087

Kurtosis 0.832718

N 30

M inimum 41.000

A -S quared


M edian 66.500


M aximum 98.000

95% C onfidence Interv al for M ean

63.854

0.31

72.413

95% C onfidence Interv al for M edian

64.000 70.771

95% C onfidence I nterv al for StDev

9.128 15.408

P -V alue 0.543

M ean 68.133

S tDev 11.461


Summary for Before

90807060

Median

Mean

75.072.570.067.565.0

A nderson-Darling Normality T est

Variance 92.102

Skewness 0.431633

Kurtosis -0.347835

N 30

Minimum 54.000

A -S quared


Median 67.500


Maximum 93.000

95% C onfidence Interv al for M ean

66.050

0.38

73.217

95% C onfidence Interv al for M edian

64.229 74.771

95% C onfidence I nterv al for StDev

7.643 12.901

P-V alue 0.388

Mean 69.633

StDev 9.597


Summary for After

265

16.24 MegaStat results are shown. At α = .05, the median downtimes (days) do not differ for the two groups, since the p-value (.1735) is greater than .05. The hypotheses are:

H0: M1 = M2 (no difference in number of repair incidents)

H1: M1 ≠ M2 (Number of repair incidents differs for the two groups)


n sum of ranks

12 112.5 New Bumper

9 118.5 Old Bumper

21 231 total





The histogram is right-skewed, but the Anderson-Darling p-value (.088) would not lead to rejection of

the hypothesis of normality at α = .05. To perform this test, the two samples were pooled. Even so, the sample is rather small for a normality test.

161284

Median

Mean

98765


Variance 18.5619

Skewness 0.938235

Kurtosis 0.530538

N 21

Minimum 1.0000

A-Squared

1st Quartile 4.0000

Median 7.0000

3rd Quartile 10.0000

Maximum 18.0000


5.2293

0.63

9.1516


4.6735 8.3265


3.2961 6.2216

P-V alue 0.088

Mean 7.1905

StDev 4.3084


Summary for Days

16.25 MegaStat results are shown. At α = .01, the median square footage does differ for the two groups, since the p-value (.0022) is less than .01. The hypotheses are:

H0: M1 = M2 (no difference in square footage)

H1: M1 ≠ M2 (square footage differs for the two groups)

266


n sum of ranks

11 79.5 Grosse Hills

11 173.5 Haut Nez Estates

22 253 total






4500400035003000

Median

Mean

3800370036003500340033003200


Variance 237208.7

Skewness 0.890787

Kurtosis 0.934743

N 22

Minimum 2800.0

A-Squared

1st Quartile 3215.0

Median 3450.0

3rd Quartile 3850.0

Maximum 4850.0


3343.1

0.39

3775.0


3220.0 3752.7


374.7 696.0

P-V alue 0.359

Mean 3559.1

StDev 487.0


Summary for Sq Ft

16.26 MegaStat results are shown. At α = .01, the median GPA does not differ among the classes, since the p-value (.1791) is greater than .01.

H0: All c population medians are the sameH1: Not all the population medians are the same

Kruskal-Wallis Test

Median n Avg. Rank

2.19 5 7.40

2.96 7 11.93

3.26 7 16.21

3.10 6 15.17

3.01 25


3

.1791


10.98 13.09

267

The histogram is rather bimodal, but normality cannot be rejected at any common α based on the Anderson-Darling p-value (.277). To perform this test, the four samples were pooled.

3.63.22.82.42.0

Median

Mean

3.33.23.13.02.92.82.7


Variance 0.3000

Skewness -0.388769

Kurtosis -0.543016

N 25

Minimum 1.9100

A-Squared

1st Quartile 2.5800

Median 3.0100

3rd Quartile 3.3250

Maximum 3.8900


2.7419

0.43

3.1941


2.8339 3.3081


0.4277 0.7619

P-V alue 0.277

Mean 2.9680

StDev 0.5477


Summary for GPA

16.27 MegaStat results are shown. At α = .01, the median crash damage does not differ among the cars, since the p-value (.4819) is greater than .01. Sample sizes are too small for a reasonable test for normality, even if the samples are combined.


Kruskal-Wallis Test

Mediann Avg. Rank

1,220.00 5 6.40

1,390.00 5 7.80

1,830.00 5 9.80

1,390.00 15

1.460

2

.4819


6.77 8.30

16.28 MegaStat results are shown. At α = .05, the median waiting time does not differ among the hospitals,

since the p-value (.1775) is greater than .05.


268

Kruskal-Wallis Test

Median n Avg. Rank

11.00 5 11.20 Hospital A

18.00 7 15.43 Hospital B

11.50 6 10.33 Hospital C

7.00 4 6.75 Hospital D

13.50 22


3

.1775


10.33 12.31

The histogram looks rather right-skewed, but normality may be assumed at any common α based on the Anderson-Darling p-value (.561). To perform this test, the four samples were pooled.

3020100

Median

Mean

201816141210


Variance 80.712

Skewness 0.557667

Kurtosis -0.076613

N 22

Minimum 0.000

A-Squared

1st Quartile 8.750

Median 13.500

3rd Quartile 21.000

Maximum 36.000


11.062

0.30

19.029


9.973 19.027


6.912 12.839

P-V alue 0.561

Mean 15.045

StDev 8.984


Summary for Minutes

269

16.29 MegaStat results are shown. At α = .05, the median output (watts) does differ among the three types of cells, since the p-value (.0104) is smaller than .05.


Kruskal-Wallis Test

Mediann Avg. Rank

123.50 6 7.75

122.00 6 6.00

128.00 6 14.75

125.00 18


2

.0104


7.38 9.05

The histogram appears right-skewed, but normality may be assumed at any common α based on the Anderson-Darling p-value (.524). To perform this test, the three samples were pooled.

132130128126124122

Median

Mean

127126125124123122


Variance 8.46

Skewness 0.446497

Kurtosis -0.585434

N 18

Minimum 121.00

A-Squared

1st Quartile 122.00

Median 125.00

3rd Quartile 127.25

Maximum 131.00


123.44

0.31

126.34


122.52 126.48


2.18 4.36

P-V alue 0.524

Mean 124.89

StDev 2.91


Summary for Watts

16.30 MegaStat results are shown. At α = .01, the median stopping distance does not differ among the types of braking method, since the p-value (.2636) is greater than .01. Normality test is not practical with only nine data points.

H0: All c populations have the same median (braking method not related to stopping distance).H1: Not all the populations have the same median (braking method related to stopping distance).

270

Friedman Test


6.00 2.00 Pumping

4.00 1.33

8.00 2.67

18.00 2.00

3


2

.2636


1.95 2.40

16.31 MegaStat results are shown. At α = .01, the median waiting time does not differ among the time of day, since the p-value (.5964) is greater than .01.

H0: All c populations have the same median (waiting time not related to time of day).H1: Not all the populations have the same median (waiting time related to time of day).

Friedman Test


76.00 2.92

87.50 3.37

78.00 3.00

79.50 3.06

69.00 2.65

390.00 3.00

26


4

.5964


1.23 1.44

All five histograms are skewed. Only one histogram (Thursday) might be considered normal, according to the Anderson-Darling test (individual tests not shown). Combining the five days (shown below) we reject normality (p < .005). There are high outliers. A non-parametric test is desirable.

271

1501209060300

Median

Mean

5550454035


V ariance 769.771

Skewness 1.20873

Kurtosis 1.35480

N 130

Minimum 0.000

A -Squared

1st Quartile 34.000

Median 42.000

3rd Quartile 63.750

Maximum 152.000


46.670

5.13

56.299


37.365 47.000


24.733 31.598

P-V alue < 0.005

Mean 51.485

StDev 27.745


Summary for Waiting Time

16.32 At α = .05, there is a significant rank correlation between gestation and on longevity since the Spearman coefficient of rank correlation is outside the critical region as given in the MegaStat output (.769 > .423).

H0: Rank correlation is zero (ρs ≤ 0) (no relationship between gestation and longevity)

H1: Rank correlation is positive (ρs > 0) (there is a relationship between gestation and longevity)


Gestation Longevity

Gestation 1.000

Longevity .769 1.000

22 sample size



Responses to analyzing the question on human beings will vary (you could do a regression and show that the prediction for humans does not fit the regression, since we have gestation of about 270 days and live,

say, 75 years on average). The histograms are right-skewed. Using α = .05, based on the Anderson-Darling test, normality is rejected for gestation (p = .012) but not for longevity (p = .068). It should be

noted that neither sample would pass the normality test at α = .10, so there is some reason to doubt normality. Thus, the nonparametric Spearman test is attractive.

6004002000

Median

Mean

30025020015010050


V ariance 31046.92

S kew ness 1.08380

Kurtosis 0.63782

N 22

M inimum 13.00

A -Squared

1st Q uartile 58.75

M edian 129.50


M aximum 660.00


116.69

0.96

272.94


62.95 286.23


135.56 251.80

P -V alue 0.012

M ean 194.82

S tDev 176.20


Summary for Gestation

3020100

Median

Mean

1614121086


Variance 58.323

Skewness 1.29820

Kurtosis 2.85056

N 22

Minimum 1.000

A -S quared

1st Q uartile 5.000

Median 12.000


Maximum 35.000


8.296

0.67

15.068


6.945 15.000


5.875 10.914

P-V alue 0.068

Mean 11.682

StDev 7.637


Summary for Longevity

272

16.33 At α = .05, there is a significant rank correlation between fertility in 1990 and fertility in 2000 since the Spearman coefficient of rank correlation is outside the critical region as given in the MegaStat output (.812 > .514).

H0: Rank correlation is zero (ρs ≤ 0) (no relationship exists)

H1: Rank correlation is positive (ρs > 0) (there is a relationship)


1990 2000

1990 1.000

2000 .812 1.000

15 sample size




2.11.81.51.2

Median

Mean

1.701.651.601.551.50


Variance 0.0567

Skewness 0.284054

Kurtosis -0.056615

N 30

Minimum 1.1000

A-Squared

1st Quartile 1.4000

Median 1.6000

3rd Quartile 1.7000

Maximum 2.1000


1.4811

0.36

1.6589


1.5000 1.6771


0.1896 0.3200

P-V alue 0.437

Mean 1.5700

StDev 0.2380


Summary for Fertility

16.34 At α = .01, there is not a significant rank correlation between calories and sodium since the Spearman coefficient of rank correlation is inside the critical region as given in the MegaStat output (.229 < .623). Samples are too small for reliable test for normality. However, there is one severe outlier in the calories (possibly a data recording error). The sodium histograms is somewhat right-skewed. All in all, the nonparametric test seems like a good idea.

H0: Rank correlation is zero (ρs ≤ 0) (no relationship between calories and sodium)

H1: Rank correlation is positive (ρs > 0) (there is a relationship between calories and sodium)


Fat (g) Calories Sodium (mg)

Fat (g) 1.000

Calories .680 1.000

Sodium (mg) .559 .229 1.000

16 sample size



273

16.35 At α = .05, there is a significant rank correlation between colon cancer rate and per capita meat consumption, since the Spearman coefficient of rank correlation is outside the critical region as given in the MegaStat output (.813 > .413).

H0: Rank correlation is zero (ρs ≤ 0) (no relationship)



Colon Cancer Rate Per Capita Meat

Colon Cancer Rate 1.000

Per Capita Meat .813 1.000

23 sample size



The colon cancer histogram is right-skewed and its Anderson-Darling p-value (.027).suggest non-

normality at α = .05. However, the meat consumption histogram appears normal and the Anderson-Darling p-value (.621) confirms this.

403020100

Median

Mean

2018161412108


Variance 100.376

Skewness 1.23068

Kurtosis 1.35479

N 23

Minimum 1.100

A -S quared

1st Q uartile 7.500

Median 12.500


Maximum 41.800


10.141

0.83

18.806


7.814 15.610


7.748 14.180

P-V alue 0.027

Mean 14.474

StDev 10.019


Summary for Colon Cancer Rate

300250200150100500

Median

Mean

160140120100


V ariance 5178.31

S kew ness 0.633858

Kurtosis 0.535117

N 23

M inimum 19.00

A -Squared

1st Q uartile 81.00

M edian 134.00


M aximum 313.00


107.19

0.28

169.42


103.00 168.10


55.65 101.85

P -V alue 0.621

M ean 138.30

S tDev 71.96


Summary for Per Capita Meat

16.36 At α = .05, there is a significant rank correlation between gas prices and carbon emission since the Spearman coefficient of rank correlation is outside the critical region as given in the MegaStat output (|

−.588| > .355).

H0: Rank correlation is zero (ρs ≤ 0) (no relationship)



Gas Price ($/L) CO2/GDP (kg/$)

Gas Price ($/L) 1.000

CO2/GDP (kg/$) -.588 1.000

31 sample size



The gas price histogram appears left-skewed, but its Anderson-Darling p-value (.169).suggests normality

at α = .05. However, the CO2 histogram is strongly right-skewed and non-normal (p < .005).

274

1.21.00.80.60.4

Median

Mean

1.000.950.900.850.800.75


V ariance 0.05581


Kurtosis -0.915642

N 31

M inimum 0.38100

A -Squared


M edian 0.87900


M aximum 1.19800


0.74983

0.52

0.92314


0.76341 0.99017


0.18878 0.31577

P -V alue 0.169

M ean 0.83648

S tDev 0.23623


Summary for Gas Price ($/L)

2.01.51.00.5

Median

Mean

0.90.80.70.60.50.40.3


Variance 0.33697

Skew ness 1.39163

Kurtosis 0.78585

N 31

Minimum 0.13000

A -Squared


Median 0.41000


Maximum 2.08000


0.47288

2.55

0.89873


0.35000 0.67949


0.46388 0.77593

P-Value < 0.005

Mean 0.68581

StDev 0.58049


Summary for CO2/GDP (kg/$)

16.37 At α = .05, there is a significant rank correlation between this week’s points and last week’s points since the Spearman coefficient of rank correlation is outside the critical region as given in the MegaStat output (. 812 > .444). We are not surprised that team rankings usually do not change much from week to week. There is no reason to expect normality since ratings do not tend toward a common mean.

H0: Rank correlation is zero (ρs ≤ 0)(no relationship)



This Week Last Week

This Week 1.000

Last Week .812 1.000

20 sample size



275

Chapter 17

Quality Management

17.1 a. See text, p 732.b. See text, p 731.c. See text p 732-733.

17.2 Common cause variation is normal and expected. Special cause variation is abnormal.

17.3 Zero variation is an asymptote of aspiration, not achievable in human endeavors.

17.4 Answers will vary. Students may see themselves as interval customers of higher education, and employers as external customers, or may refer to their place of employment or organizations like Starbuck’s or music that they like.

17.5 Answers will vary. Use Likert scales for service attributes.a. Cleanliness of vehicle, full gas tank, waiting time for sales help.b. Length of queues, friendliness of staff (Likert), interest paid on accounts.c. Price, seat comfort, picture quality (Likert scale for all).

17.6 Examples of barriers include employee fear, inadequate equipment, inadequate equipment maintenance, insufficient employee training, flawed process design, unclear task definitions, poor supervision, lack of support for employees.

17.7 Deming felt that most workers want to do a good job, but are often hampered by the work environment,

management policies, and fear of reprisal if they report problems.

17.8 Students may name Deming, Shewhart, Ishikawa, Taguchi, and others they’ve heard of.

17.9 Deming’s 14 points (abbreviated) are on p. 736. See www.deming.org for a more complete list.

17.10 Techniques of SPC (statistical process control) are a specific subset of tools of TQM (total quality management) and CQI (continuous process improvement).

17.11 Define parameters, set targets, monitor until stable, check capability, look for sources of variation or nonconformance, make changes, repeat steps.

17.12 Attribute control charts are for nominal data (e.g., proportion conforming) while variable control charts are for ratio or interval data (e.g., means).

17.13 a. Sampling frequency depends on cost and physical possibility of sampling.b. For normal data, small samples may suffice for a mean (Central Limit Theorem).c. Large samples may be needed for a proportion to get sufficient precision.

17.14 We can estimate σ using the sample standard deviation (s), or using 2R d where R is the average range and

d2 is a control chart factor from Table 17.4, or using the average of the sample standard deviations of many

samples ( s ). If the process standard deviation σ is known, we do not need to estimate σ. But a little thought

will show that σ can only be “known” from one of the preceding methods.

276

17.15 This is the Empirical Rule (see chapters 4 and 7):a. Within ± 1 standard deviations � 68.26 percent of the timeb. Within ± 2 standard deviations � 95.44 percent of the timec. Within ± 3 standard deviations � 99.73 percent of the time

17.16 Students may need to be reminded that “sigma” refers to the standard error of the mean nσ .

Rule 1. Single point outside 3 sigmaRule 2. Two of three successive points outside 2 sigma on same side of centerlineRule 3. Four of five successive points outside 1 sigma on same side of centerlineRule 4. Nine successive points on same side of centerline

17.172

0.42UCL 3 12.5 3 12.742

2.326 5

Rx

d n= + = + =

2

0.42LCL 3 12.5 3 12.258

2.326 5

Rx

d n= − = − =

17.182

5UCL 3 400 3 403.643

2.059 4

Rx

d n= + = + =

2

5LCL 3 400 3 396.357

2.059 4

Rx

d n= − = − =

17.19 Estimated σ is 2/R d = 30/2.059 = 14.572, UCL = 98.37, LCL = 54.63

1 2 9... 72.25 74.25 ... 82.25

76.59 9

x x xx

+ + + + + += = =

1 2 9... 43 31 ... 41

309 9

R R RR

+ + + + + += = =

17.20 1 2 9... 5.52 5.51 ... 5.51

5.508 8

x x xx

+ + + + + += = =

1 2 25... 0.13 0.11 ... 0.13

0.1108 8

R R RR

+ + + + + += = =

Estimate of µ = 5.50 and estimate of σ = (0.110)/(2.326) = 0.0473

17.21 R = 0.82 (centerline)

UCL = D4 R = (2.004)(0.82) = 1.64328

LCL = D3 R = (0)(0.82) = 0

17.22 R = 0.82 � centerline for R chart

UCL = D4 R = (2.574)(12) = 30.888 � upper control limit

LCL = D3 R = (0)(12) = 0 � lower control limit

17.23 By either criterion, process is within acceptable standard (Cp = 1.67, Cpk = 1.67).

Cp index:

USL LSL 725 7151.667

6 6(1)p

C− −

= = =σ

277

Cpk index:

USL

USL 725 7205.00

1z

− µ −= = =

σ and

LSL

LSL 720 7155.00

1z

µ − −= = =

σ

zmin = USL LSLmin( , )z z = min{ 5.00, 5.00} = 5.00 and so min5.00

1.6673 3

pk

ZC = = =

17.24 If the minimum capability index is 1.33, this process meets the Cp but fails on the Cpk criterion.

Cp index:

USL LSL 0.432 0.4231.50

6 6(0.001)p

C− −

= = =σ

Cpk index:

USL

USL 0.432 0.4266.00

0.001z

− µ −= = =

σ and

LSL

LSL 0.426 0.4233.00

0.001z

µ − −= = =

σ

zmin = USL LSLmin( , )z z = min{6.00, 3.00} = 3.00 and so min3.00

1.003 3

pk

ZC = = =

17.25 If the minimum capability index is 1.33, the process fails on both criteria, especially Cpk due to bad centering (Cp = 1.67, Cpk = 1.67).

Cp index:

USL LSL 55.9 55.21.1667

6 6(0.1)p

C− −

= = =σ

Cpk index:

USL

USL 55.9 55.45.00

.1

− µ −= = =

σz and

LSL

LSL 55.4 55.22.00

.1

µ − −= = =

σz

zmin = USL LSLmin( , )z z = min{5.00, 2.00} = 2.00 and so min2.00

0.6673 3

pk

ZC = = =

17.26 Yes, it is OK to assume normality since nπ = (500)(.02) = 10.

(1 )UCL 3

n

π − π= π +

(.02)(.98).02 3 .0388

500= + =

(1 )LCL 3

n

π − π= π − =

(.02)(.98).02 3 .0012

500− =

17.27 Yes, safe to assume normality since nπ = (20)(.50) = 10.

(1 )UCL 3

n

π − π= π +

(.5)(.5).50 3 .8354

20= + =

(1 )

LCL 3n

π − π= π − =

(.50)(.50).50 3 .1646

20− =

17.28 Since n(1−π) = 40(.10) =4 is not greater than 10, we can’t assume normality.

(1 )UCL 3

n

π − π= π +

(.90)(.10).90 3 1.042302

40= + = (use 1.000 since UCL cannot exceed 1).

278

(1 )

LCL 3n

π − π= π − =

(.90)(.10).90 3 .757698

40− =

17.29 Services are often assessed using percent conforming or acceptable quality, so we use p charts.

17.30 The charts and their purposes are:

a. x chart monitors a process mean for samples of n items. Requires estimates of µ and σ (or R and

control factor d2).

b. R chart monitors variation around the mean for samples of n items. Requires estimate of R or σ and

control chart factor D4.

c. p chart monitors the proportion of conforming items in samples of n items. Requires estimate of π.

d. I chart monitors individual items when inspection is continuous (n = 1). Requires estimates of µ and σ.

17.31 Answers will vary. For example:a. GPA, number of classes re-taken, faculty recommendation letters (Likert).b. Knowledge of material, enthusiasm, organization, fairness (Likert scales for all).c. Number of bounced checks, size of monthly bank balance errors, unpaid VISA balance.d. Number of print errors, clarity of graphs, useful case studies (Likert scales for last two).

17.32 Answers will vary. For example:a. Percent of time “out of range,” frequency of poor reception, perceived ease of use of menus.b. Percent of time server is unavailable, frequency of spam or “pop-up” ads.c. Customer wait in queue to pick up or drop off, rating of garment cleanliness, rating of staff courtesy.d. Waiting time in office, staff courtesy, percent of cost covered by insurance.e. Waiting time for service, perceived quality of haircut, rating of friendliness of haircutter.f. Waiting time for service, perceived quality of food, rating of staff courtesy.

17.33 Answers will vary. For example:a. MPG, repair cost.b. Frequency of jams, ink cost.c. Frequency of re-flushes, water consumption.d. Battery life, ease of use (Likert scale).e. Cost, useful life, image sharpness (Likert scale).f. Cost, useful life, watts per lumen.

17.34 a. Sampling (not cost effective to test every engine).b. 100% inspection (airlines record fuel usage and passenger load on every flight).c. 100% inspection (McDonald’s computers would have this information for each day).d. Sampling (you can’t test the life of every battery).e. Sampling (cost might prohibit hospitals from recording this in normal bookkeeping).

17.35 x is normally distributed from the Central Limit Theorem for sufficiently large values of n (i.e., symmetric distribution). However, the range and standard deviation do not follow a normal distribution (e.g., standard deviation has a chi distribution).

17.36 Answers will vary. It is because x is normally distributed from the Central Limit Theorem for sufficiently large values of n. However, some processes may not be normal, and subgroups typically are too small for the CLT to apply unless the data are at least symmetric (see Chapter 8). For small n, normality would exist if the underlying process generates normally-distributed data—a reasonable assumption for many, but not all, processes (especially in manufacturing). If non-normal, special techniques are required (beyond the scope of an introductory class in statistics).

17.37 a. Variation and chance defects are inevitable in all human endeavors.b. Some processes have very few defects (maybe zero in the short run, but not in the long run).c. Quarterbacks cannot complete all their passes, etc.

279

17.38 Answers will vary, depending on how diligent a web search is conducted.

17.39 Answers will vary (e.g., forgot to set clock, clock set incorrectly, couldn’t find backpack, stopped to charge cell phone, had to shovel snow in driveway, alarm didn’t go off, traffic, car won’t start, can’t find parking).

17.40 Answers will vary (addition or subtraction error, forgot to record a deposit or withdrawal, recorded data incorrectly e.g., $54.65 instead of $56.54, missing check number, lost debit card receipt).

17.41 Answers will vary (e.g., weather, union slowdown, pilot arrived late, crew change required, de-icing planes in winter, traffic congestion at takeoff , no arrival gate available).

17.42 a. If µ = 1.00 mils and σ = 0.05 mils, and if the minimum capability index is 1.33, this process is well below capability standards (Cp = Cpk = 0.95).

Cp index:

USL LSL 1.20 0.800.952

6 6(0.07)p

C− −

= = =σ

Cpk index:

USL

USL 1.20 1.002.857

0.07z

− µ −= = =

σ and

LSL

LSL 1.00 .802.857

0.07z

µ − −= = =

σ

zmin = USL LSL

min( , )z z = min{ 2.587, 2.857} = 2.857 and so min2.857

.9523 3

pk

ZC = = =

b. If µ = 1.00 mils and σ = 0.05 mils, and if the minimum capability index is 1.33, this process meets capability standards (Cp = Cpk = 1.33).

Cp index:

USL LSL 1.20 .80

6 6(0.05)

1.33− −

= =σ

=p

C

Cpk index:

USL

USL 1.20 1.04.00

0.05z

− µ −= = =

σ and

LSL

LSL 1.00 0.804.00

0.05z

µ − −= = =

σ

zmin = USL LSLmin( , )z z = min{4.00, 4.00} = 4.00 and so min

4.001.333

3 3pk

ZC = = =

c. The point is to show that a reduction in the process standard deviation can improve the capability index.

17.43 a. If µ = 1.00 mils and σ = 0.05 mils, and if the minimum capability index is 1.33, this process meets capability standards (Cp = Cpk = 1.33).

Cp index:

USL LSL 1.20 .80

6 6(0.05)

1.33− −

= =σ

=p

C

280

Cpk index:

USL

USL 1.20 1.004.00

0.05z

− µ −= = =

σ and

LSL

LSL 1.00 0.804.00

0.05z

µ − −= = =

σ


4.001.333

3 3pk

ZC = = =

b. If µ = 0.90 mils and σ = 0.05 mils, and if the minimum capability index is 1.33, this process meets capability standards (Cp = 1.33, Cpk = 0.67).

Cp = 1.33, Cpk = 0.67

Cp index:

USL LSL 1.20 0.80

6 6(0.05)

1.33p

C− −

= =σ

=

Cpk index:

USL

USL 1.20 0.906.00

0.05z

− µ −= = =

σand

LSL

LSL 0.90 0.802.00

0.05z

µ − −= = =

σ


2.000.667

3 3= = =

pk

ZC

c. This example shows why we need more than just the Cp index. A change in the process mean can reduce the Cpk index, even though the Cp index is unaffected.

17.44 a. If µ = 140 mg and σ = 5 mg, and if the minimum capability index is 1.33, this process meets capability standards (Cp = Cpk = 1.33).

Cp index:

USL LSL 160 120

6 6(5)

1.33− −

= =σ

=p

C

Cpk index:

USL

USL 160 1404.00

5z

− µ −= = =

σ and

LSL

LSL 140 1204.00

5z

µ − −= = =

σ

zmin = USL LSL

min( , )z z = min{4.00, 4.00} = 4.00 and so min4.00

1.3333 3

pk

ZC = = =

b. If µ = 140 mg and σ = 3 mg, and if the minimum capability index is 1.33, this process exceeds capability standards (Cp = Cpk = 2.22).

Cp index:

USL LSL 160 120

6 6(3)

2.22p

C− −

= =σ

=

Cpk index:

USL

USL 160 1406.67

3z

− µ −= = =

σ and

LSL

LSL 140 1206.67

3z

µ − −= = =

σ

zmin = USL LSLmin( , )z z = min{6.67, 6.67} = 6.67 and min

6.672.22

3 3pk

ZC = = =

281

c. The point is to show that a reduction in the process standard deviation can improve the capability of a process that already meets the requirement.

17.45 a.100

UCL 6050 3 6223.2053

= + = and 100

LCL 6050 3 5876.7953

= − =

b. Chart violates no rules.c. Process is in control.

17.46. a. Histogram is bell-shaped and probability plot is linear with one possible low outlier (the Anderson-Darling statistic has p-value = .296).

Pounds

Percent

630062006100600059005800

99

95

90

80

70

60

50

40

30

20

10

5

1

Mean

0.296

6073

StDev 85.93

N 24

A D 0.422

P-Value

Probability Plot of PoundsNormal

b. Yes, it approximates the normal distribution.c. Sample mean is 6072.625 and the sample standard deviation is 85.92505. They are both close to the

process values.

17.47 a..07

UCL 1.00 3 1.09395

= + = and .07

LCL 1.00 3 .90615

= − =

b. Chart violates no rules.c. Process is in control.

282

17.48 a. Histogram is bell-shaped and probability plot is linear with one possible high outlier (the Anderson-Darling statistic has p-value = .656).

Mils

Percent

1.201.151.101.051.000.950.900.85

99

95

90

80

70

60

50

40

30

20

10

5

1

M ean

0.656

1.006

S tDev 0.06547

N 35

A D 0.270

P -Valu e

Probability Plot of MilsNormal

b. The distribution is approximately normal.c. Sample mean is 1.006 and the sample standard deviation is 0.0655, both close to the process values.

17.49 a. Cp = 1.00 and Cpk = 0.83.

Cp index:

USL LSL 30 18

6 6(2)

1.00− −

= =σ

=p

C

Cpk index:

USL

USL 30 233.50

2z

− µ −= = =

σ and

LSL

LSL 23 182.50

2z

µ − −= = =

σ


2.500.833

3 3pk

ZC = = =

b. If the minimum capability index is 1.33, this process is well below capability standards.c. The frequency of the door of the door being opened. Door not being closed tightly.

17.50 a. 2

UCL 23.00 3 26.004

= + = and 2

LCL 23.025 3 20.004

= − =

b. Control chart suggests a downward trend but does not violate Rule 4.

283

c. The sixth mean hits the UCL, so possibly not in control.

17.51 a. The sample mean of 23.025 and the standard deviation of 2.006 are very close to the process values (µ =

23, σ = 2).b. The histogram is symmetric, though perhaps platykurtic. Probability plot is linear but Anderson-Darling

test statistic has a p-value below .005 so fails normality test.

Temperature

Percent

30.027.525.022.520.017.515.0

99.9

99

95

90

80

7060504030

20

10

5

1

0.1

M ean

<0.005

23.03

S tDev 2.006

N 80

A D 1.348

P -Valu e

Probability Plot of TemperatureNormal

17.52 a. Cp = 2.00 and Cpk = 2.00. If the minimum capability index is 1.33, this process is capable.

Cp index:

USL LSL 14.6 13.4

6 6(.10)

2.00p

C− −

= =σ

=

Cpk index:

USL

USL 14.60 14.006.00

.10z

− µ −= = =

σ and

LSL

LSL 14.00 13.406.00

.10z

µ − −= = =

σ


6.002.00

3 3pk

ZC = = =

b. Since the process is capable, there is no reason to change unless the customers can see the variation.

284

17.53 a. Histogram is arguably normal, though somewhat bimodal. However, the probability plot is linear and the Anderson-Darling test’s p-value of .795 indicates a good fit to a normal distribution.

Hours

Percent

9400920090008800860084008200

99

95

90

80

70

60

50

40

30

20

10

5

1

Mean

0.795

8785

StDev 216.1

N 20

A D 0.224

P-Value

Probability Plot of HoursNormal

b. Center line = 8760, 200

UCL 8760 3 9028.335

= + = , and 200

LCL 8760 3 8491.675

= − =

c. Center line = 8784.75, 216.1398

UCL 8784.75 3 9074.735

= + = , and 216.1398

LCL 8784.75 3 8494.775

= − =

d. The UCL and LCL from the sample differ substantially from those based on the assumed process parameters. This small sample is perhaps not a reliable basis for setting the UCL and LCL.

17.54 a. Cp = 2.00 and Cpk = 1.83.

Cp index:

USL LSL 477 453

6 6(2)

2.00p

C− −

= =σ

=

Cpk index:

USL

USL 477 4665.50

2z

− µ −= = =

σ and

LSL

LSL 466 4536.50

2z

µ − −= = =

σ


5.501.83

3 3pk

ZC = = =

b. If the minimum capability index is 1.33, this process is capable.c. Need to pack the box without crushing the Chex. Given the smallness and fragility of each Chex it

would be difficult to attain.

17.55 a.3

UCL 465 3 470.203

= + =3

LCL 465 3 459.803

= − =

b. No rules are violated. The process is in control.

285

17.56 a. The histogram and probability plot do not appear grossly non-normal, but the p-value (.042) for the

Anderson-Darling test suggests that the box fill may not be normal (our conclusion depends on α).

Grams

Percent

472470468466464462460458456

99

95

90

80

70

60

50

40

30

20

10

5

1

Mean

0.042

464.2

StDev 2.797

N 30

A D 0.761

P-Value

Probability Plot of GramsNormal

b. Sample mean is 464.2 which is very close to 465.

17.57 a. From MegaStat: UCL = 12.22095, LCL = 11.75569, centerline = 11.98832

b. Process appears to be in control.

c. Histogram approximates the normal distribution c. The histogram and probability plot (Anderson-Darling p-value = .871) suggest a normal distribution.

Weight in Grams

Percent

12.812.612.412.212.011.811.611.411.2

99.9

99

95

90

80

7060504030

20

10

5

1

0.1

Mean

0.871

11.99

StDev 0.2208

N 84

A D 0.204

P-Valu e

Probability Plot of Weight in GramsNormal

286

17.58 a.(1 )

UCL 3π − π

= π +n

(.06)(.94)

.06 3 .11038200

= + =

(1 )

LCL 3π − π

= π −n

(.06)(.94).06 3 .0096

120= − =

b. Yes, 200(.06) = 12 and 200(.94) = 188 both are greater than 10.

17.59 a.(1 )

UCL 3π − π

= π +n

(.05)(.95)

.05 3 .1154100

= + =

(1 )

LCL 3π − π

= π −n

(.05)(.95)

.05 3 .0154100

= − = − (set to .0000 since LCL can’t be negative).

b. Sample 7 hits the LCL, so the process may not be in control.

c. Samples are too small to assume normality (nπ = 5). Better to use MINITAB’s binomial option to set the control limits.

17.60 Chart A: TrendChart B: OscillationChart C: Level ShiftChart D: InstabilityChart E: NoneChart F: Cyclical

17.61 Chart A: Rule 4.Chart B: No rules violated.Chart C: Rule 4.Chart D: Rules 1, 4.Chart E: No rules violated.Chart F: Rules 1, 2.

287

17.62 Each pattern is clearly evident, except possibly instability in the third series. See charts below.

Yes, upward trend present. Yes, downward trend present

Yes, instability (Rule 1). Yes, cyclical (only 6 centerline crossings).

Maybe oscillation (14 centerline crossings).

288

17.63 Each pattern is clearly evident, except possibly instability in the third series.

In Control?

0.90

1.00

1.10

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Sample

Up Trend?

0.90

1.00

1.10

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Sample

Yes, in control (no rules violated). Yes, upward trend, but no rule violations.

Down Trend?

0.90

1.00

1.10

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Sample

Unstable?

0.90

1.00

1.10

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Sample

Yes, downward trend, but no rule violations. Yes, unstable (Rule 1).

Cycle?

0.90

1.00

1.10

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Sample

Yes, cycle (only 7 centerline crossings).

17.64 Answers will vary, sample points presented.a. p-chart since these are attributes (e.g., percent of patients who received aspirin).b. Randomness can’t be overcome completely, each doctor/nurse interprets standards differently. Also,

there may be cost constraints or issues of time in emergency situations.c. Reduction in variation raises costs: training, monitoring and evaluating

17.65 a. Type I error: disease is not present, but remove meat anyway (sample result shows process out of control when it is not). Type II error: disease is present, but fail to remove the meat (sample result shows process in control when in fact it is not).

b. 27,400,000/4,000 = 6,850 two-ton truckloads. Need to know the cost of trucks, drivers, and disposal fees. But where to put it?

c. NIMBY (not in my backyard).

289

17.66 a. Type I error: disease is not present, but remove meat anyway (sample result shows process out of control when it is not). Type II error: disease is present, but fail to remove the meat (sample result shows process in control when in fact it is not).

b. Public cannot see the salmonella, so they have a collective interest in hiring an agent on their behalf to inspect the meat (e.g., government inspectors, who presumably are unbiased and have the safety of the public in mind).

17.67 Type I error: cereal actually was safe for human consumption, but was discarded. Type II: cereal was unsafe for human consumption, but was sold. The former is a cost borne by the company (lost profit) and perhaps by consumers (higher prices), while the latter is a cost borne both by the public (possible health hazard) and by the company (possible litigation). Although both parties bear some cost, the costs are different in nature and may differ in magnitude as well. Consumers cannot see or detect pesticide, so government inspection is needed to protect the public interest and to apply the laws.

17.68 Answers will vary. It may be hard top define a “defective M.” Since sampling is easy (and tasty) It

should be possible to inspect enough M&Ms to get a reliable confidence interval for π.17.69 Answers will vary. Observers may disagree as to what constitutes a defect. Presumably, a newer car has

fewer paint defects, though it does depend on usage.

17.70 Answers will vary. It may be hard to define a “broken Cheerio.” Randomness should be attainable.

290

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