copyright © 2007 pearson education, inc. publishing as pearson addison-wesley classroom notes...

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Classroom Notes

Chapter 23 Answers


Chapter 23Inference About Means

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 23- 3

Getting Started

1. To create confidence intervals and test hypotheses about means , we will base both on the sampling distribution model.

2. The Central Limit Theorem told us that the sampling distribution model for means is Normal with mean μ and standard deviation


3. All we need is a random sample of quantitative data.

4. And the true population standard deviation, σ. Well, that’s a problem…

Getting Started (cont.)


Getting Started (cont.)5. Proportions have a link between the proportion value and the standard deviation of the sample proportion. 6. This is not the case with means—knowing the sample mean tells us nothing about

= = 7. We’ll do the best we can: estimate the population parameter σ with the sample statistic s.• Our resulting standard error is


Getting Started (cont.)8. We now have extra variation in our standard error from s, the sample standard deviation.

9.We need to allow for the extra variation so that it does not mess up the margin of error and P-value, especially for a small sample.

And, the shape of the sampling model changes—the model is no longer Normal. So, what is the sampling model?


Gosset’s t10. William S. Gosset, an employee of the Guinness Brewery in Dublin, Ireland, worked long and hard to find out what the sampling model was. The sampling model that Gosset found has been known

as Student’s t. The Student’s t-models form a whole family of related

distributions that depend on a parameter known as degrees of freedom.

11. We often denote degrees of freedom as df, and the model as tdf.


What Does This Mean for Means?

12. A practical sampling distribution model for means:When the conditions are met, the standardized sample mean

follows a Student’s t-model with n – 1 degrees of freedom. We estimate the standard error with

ytSE y

sSE yn


What Does This Mean for Means? (cont.)

13. When Gosset corrected the model for the extra uncertainty, the margin of error got bigger.

Your confidence intervals will be just a bit wider and your P-values just a bit larger than they were with the Normal model.

By using the t-model, you’ve compensated for the extra variability in precisely the right way.



14. Student’s t-models are unimodal, symmetric, and bell shaped, just like the Normal model.

• But t-models with only a few degrees of freedom have much fatter tails than the Normal model.



15. As the degrees of freedom increase, the t-models look more and more like the Normal model. In fact, the t-model with infinite degrees of freedom

is exactly Normal.


Finding t-Values By Hand

16. The Student’s t-model is different for each value of degrees of freedom. Because of this,

Statistics books usually have one table of t-model critical values for selected confidence levels.


Finding t-Values By Hand (cont.)

17. Alternatively, we could use technology to find t critical values for any number of degrees of freedom and any confidence level you need.

We’ll look at this soon.


Assumptions and Conditions

18. Gosset found the t-model by simulation. Years later, when Sir Ronald A. Fisher showed

mathematically that Gosset was right, he needed to make some assumptions to make the proof work.

We will use these assumptions when working with Student’s t.


Assumptions and Conditions (cont.)19. Independence Assumption:

a. Randomization Condition: The data arise from a random sample or suitably randomized experiment. Randomly sampled data (particularly from an SRS) are ideal.

b. 10% Condition: When a sample is drawn without replacement, the sample should be no more than 10% of the population.


20. Normal Population Assumption:We can never be certain that the data are from a population that follows a Normal model, but we can check the …a. Nearly Normal Condition: The data come from a distribution that is symmetric and without outliers.

• Check this condition by making a boxplot.• For smaller sample sizes (n < 15 or so), If severe

skewness or outliers are present, don’t use Student’s t.• For moderate sample sizes (n between 15 and 40 or

so), If outliers exist, don’t use Student’s t. • For larger sample sizes(n > 40 or so), the t methods are

safe to use even if the data are clearly skewed… Because of CLT


One-Sample t-Interval21. When the conditions are met, we are ready to find the confidence interval for the population mean, μ.

The confidence interval is

where the standard error of the mean is

The critical value depends on the particular confidence level, C, that you specify and on the number of degrees of freedom, n – 1, which we get from the sample size.

1ny t SE y

sSE yn


TI-84

Slide 23- 18

22. Type the speeds of the 23 Triphammer cars into :

29 34 34 38 30 29 38 31 2934 32 31 27 37 29 26 24 3436 31 34 36 21

-Set up a STATPLOT to create a boxplot of the data so you can check the Normal condition.-Under STAT TESTS choose 8:TInterval-Use Inpt: Data because we know the data values-For this data Frequency is 1. -Choose the confidence level you want.. Let’s try 90% for this example. THAT’s .90-CALCULATE the interval.


TI-84

Slide 23- 19

23. You should get the following results:T Interval (29.892, 33.064) = 31.47826087 = 4.429955522 n = 23

T Interval (29.563 , 33.394) = 31.47826087 = 4.429955522 n = 23

24. What would it be for a 95% confidence interval?

Interpret Confidence Interval I am 90% confident, that the true mean speed of the vehicles on Triphammer is between 29.9mph and 33.1mph.

Interpret Confidence Interval I am 95% confident, that the true mean speed of the vehicles on Triphammer is between 29.6mph and 33.4mph.


25. Interpret a Single Confidence LevelSaying that we are “#% confident” means that with this data, if many intervals were constructed with this method, we would expect approximately #% of them to contain the true (or population) mean of ________ (context).

Slide 23- 20

26. For the 95% CI (29.563 , 33.394) above,

If many intervals were constructed with this method, we would expect approximately 95% of them to contain the true mean of speed of the vehicles on Triphammer.


MORE TI-84

Slide 23- 23

27. Sometimes instead of the original data you just have the summary statistics. For instance, suppose a random sample of 53 lengths of fishing line had a mean strength of 83 pounds and standard deviation of 4 pounds. Let’s make a 95% confidence interval for the mean strength of this kind of fishing line.

Without the data you can’t check the boxplot, but a sample size of 53 is large enough that we don’t need to worry…. But you need to mention: Assuming no outliers, CLT indicates that the sample mean will be approximately Normal for a sample this large.


MORE TI-84

Slide 23- 24

27. Sometimes instead of the original data you just have the summary statistics. For instance, suppose a random sample of 53 lengths of fishing line had a mean strength of 83 pounds and standard deviation of 4 pounds. Let’s make a 95% confidence interval for the mean strength of this kind of fishing line.

28. Go back to STAT TESTS and choose 8:Tinterval.For Inpt: choose Stats.Specify the sample mean, standard deviation, and sample size.C-Level: should already be at .95Choose Calculate


TI-84

Slide 23- 25

29. You should get the following results:T Interval (81.897, 84.103) = 83 = 4 n = 53

T Interval (81.531 , 84.469) = 83 = 4 n = 53

30. What would it be for a 99% confidence interval?

I am 95% confident, that the true mean strength of the fishing line is between 81.9 lbs and 84.1 lbs.

I am 99% confident, that the true mean strength of the fishing line is between 81.5 lbs and 84.5 lbs.

Interpret the confidence interval

Interpret the confidence interval


BOXPLOT REVIEW

Slide 23- 27

Low

er F

ence

Upp

er F

ence

- 1.5(IQR) + 1.5(IQR)

Lowest Value inside fences

Highest Value inside fences

𝑸𝟏 𝑸𝟑𝑴𝒆𝒅𝒊𝒂𝒏

Locations of all upper

outliers plotted

(if any exist)

Locations of all lower

outliers plotted

(if any exist)

Length of boxrepresents the

IQRIQR =


31. IMPORTANT NOTE: If the population standard deviation is known,

then you should use the Normal model instead of the Student’s t model.

Slide 23- 33

By now you should be well versed in the 8 parts of doing any hypothesis test:

Remember PHANTOMS Now it is time to formalize what you need to

show for all Confidence intervals:P arameter of interestA ssumptions and conditions

I ntervalC onclusion in context

N ame the interval


Confidence Interval for Sample Mean

Slide 23- 35

32. A nutrition laboratory tests 40 “reduced sodium” hot dogs, finding that the mean sodium content is 310mg, with a standard deviation of 36mg. Find a 95% confidence interval for the mean sodium content of this brand of hot dog. REMEMBER PANIC

Parameter of Interest

The parameter of interest is the mean sodium content for a certain brand of hot dog.



Slide 23- 36

32. A nutrition laboratory tests 40 “reduced sodium” hot dogs, finding that the mean sodium content is 310mg, with a standard deviation of 36mg. Find a 95% confidence interval for the mean sodium content of this brand of hot dog. REMEMBER PANICAssumptions and conditions

Independence is reasonable, becausethe hot dogs used can be representative of all hotdogs of this brand and 40 hot dogs is less than 10% of all hot dogs of this brand.



Slide 23- 38

32. A nutrition laboratory tests 40 “reduced sodium” hot dogs, finding that the mean sodium content is 310mg, with a standard deviation of 36mg. Find a 95% confidence interval for the mean sodium content of this brand of hot dog. REMEMBER PANICAssumptions and conditions

Nearly Normal assumption:We don’t have data to check with a boxplot. However, assuming there are not any outliers, CLT indicates that the sample mean will be approximately normal for a sample this large.



Slide 23- 39

32. A nutrition laboratory tests 40 “reduced sodium” hot dogs, finding that the mean sodium content is 310mg, with a standard deviation of 36mg. Find a 95% confidence interval for the mean sodium content of this brand of hot dog. REMEMBER PANICName the interval

Since the conditions are met, we can create a one-sample t-interval.



Slide 23- 40

32. A nutrition laboratory tests 40 “reduced sodium” hot dogs, finding that the mean sodium content is 310mg, with a standard deviation of 36mg. Find a 95% confidence interval for the mean sodium content of this brand of hot dog. REMEMBER PANICInterval

𝟐 .𝟎𝟐𝟑

invT(.025,39)100 - .95 = 0.05



Slide 23- 41


≈5.6921

𝟐 .𝟎𝟐𝟑



Slide 23- 42


𝑆𝐸 (𝑦 )=5.6921

¿11.5¿ (2.023 )(5.6921)

𝟐 .𝟎𝟐𝟑

𝟑𝟏𝟎±𝟏𝟏 .𝟓𝒎𝒈



Slide 23- 43

32. A nutrition laboratory tests 40 “reduced sodium” hot dogs, finding that the mean sodium content is 310mg, with a standard deviation of 36mg. Find a 95% confidence interval for the mean sodium content of this brand of hot dog. REMEMBER PANICInterval 310±11.5𝑚𝑔

(𝟐𝟗𝟖 .𝟓 ,𝟑𝟐𝟏 .𝟓)



Slide 23- 44

32. A nutrition laboratory tests 40 “reduced sodium” hot dogs, finding that the mean sodium content is 310mg, with a standard deviation of 36mg. Find a 95% confidence interval for the mean sodium content of this brand of hot dog. REMEMBER PANICConclusion in context

I am 95% confident that the true mean sodium content for this brand of hot dog isbetween 298.5mg and 321.5mg.

(298.5 ,321.5)


Testing a hypothesis for mean

Slide 23- 45


One-Sample t-test for the Mean1. The conditions for the one-sample t-test for the mean are the same as for the one-sample t-interval. We test the hypothesis H0: = 0 using the statistic

The standard error of the sample mean is

When the conditions are met and the null hypothesis is true, this statistic follows a Student’s t model with n – 1 df. We use that model to obtain a P-value.

0

1nytSE y

sSE yn


Significance and Importance

2. Remember that “statistically significant” does not mean “actually important” or “meaningful.”

Because of this, it’s always a good idea when we test a hypothesis to check the confidence interval and think about likely values for the mean.


Intervals and Tests3. Confidence intervals and hypothesis tests are built from the same calculations.

• In fact, they are complementary ways of looking at the same question.

• The confidence interval contains all the null hypothesis values we can’t reject with these data.


Intervals and Tests (cont.)4. More precisely, a level C confidence interval contains all of the possible null hypothesis values that would not be rejected by a two-sided hypothesis test at alpha level 1 – C.

So a 95% confidence interval matches a 0.05 level test for these data.

Confidence intervals are naturally two-sided, so they match exactly with two-sided hypothesis tests. When the hypothesis is one sided, the

corresponding alpha level is (1 – C)/2.


Sample Size5. To find the sample size needed for a particular confidence level with a particular margin of error (ME), solve this equation for n:

But if we haven’t done the sample we don’t know the sample standard deviation or n-1.

We can use s from a small pilot study.We can use z* in place of the necessary t value.

1nsME tn

√𝒏=𝒕∗𝒏−𝟏𝒔

𝑴𝑬n

n


Sample Size (cont.)6. Sample size calculations are never exact.• The margin of error you find after collecting the

data won’t match exactly the one you used to find n.

The sample size formula depends on quantities you won’t have until you collect the data, but using it is an important first step.

Before you collect data, it’s always a good idea to know whether the sample size is large enough to give you a good chance of being able to tell you what you want to know.


What Can Go Wrong?7. Ways to Not Be Normal: Beware multimodality.

The Nearly Normal Condition clearly fails if a histogram of the data has two or more modes.

Beware skewed data. If the data are very skewed, try re-expressing

the variable.Set outliers aside—but remember to report on

these outliers individually.


What Can Go Wrong? (cont.)…And of Course:8. Watch out for bias—we can never overcome the problems of a biased sample. Make sure data are independent.

Check for random sampling and the 10% Condition. Make sure that data are from an appropriately randomized

sample. Interpret your confidence interval correctly.

Many statements that sound tempting are, in fact, misinterpretations of a confidence interval for a mean.

A confidence interval is about the mean of the population, not about the means of samples, individuals in samples, or individuals in the population.


What have we learned?9. Statistical inference for means relies on the same concepts as for proportions—only the mechanics and the model have changed. The reasoning of inference, the need to verify that

the appropriate assumptions are met, and the proper interpretation of confidence intervals and P-values all remain the same regardless of whether we are investigating means or proportions.


10. TI-84

Slide 23- 55

-Type the speeds of the 22 Triphammer Road cars into :

29 34 34 38 30 29 38 31 2934 32 31 27 37 29 26 24 3436 31 34 36

-Let’s see if the mean speed is significantly higher than the speed limit of 30mph. -Under STAT TESTS choose 2:T-Test-Use Inpt: Data because we know the data values

-For this data Frequency is 1. -Choose the -CALCULATE the test.

-Use 30 for .


TI-84

Slide 23- 56

You should get the following results:T-Test t = 2.359762364P = .01402172 = 31.95454545 = 3.884980514 n = 22

You will still need to go through all of PHANTOMS.


TI-84

Slide 23- 57

T-Test t = 2.359762364P = .01402172 = 31.95454545 = 3.884980514 n = 22


11. Parameter of interest

We are interested in the mean speed of cars on Triphammer street, .


TI-84

Slide 23- 58

T-Test t = 2.359762364P = .01402172 = 31.95454545 = 3.884980514 n = 22


12.Hypotheses = 30 > 30


TI-84

Slide 23- 59

T-Test t = 2.359762364P = .01402172 = 31.95454545 = 3.884980514 n = 22


13. Assumptions & Conditions• Independence is reasonable, because the sampled

cars can be representative of all cars on Triphammer and 22 cars is less than 10% of all cars on Triphammer.


TI-84

Slide 23- 60

T-Test t = 2.359762364P = .01402172 = 31.95454545 = 3.884980514 n = 22


13. Assumptions & Conditions

20 25 30 35 40The sample size is moderate and the boxplot of the data is fairly symmetric with no outliers so the sample distribution will be nearly Normal.


TI-84

Slide 23- 61

T-Test t = 2.359762364P = .01402172 = 31.95454545 = 3.884980514 n = 22


14. Name the TestBecause the conditions are satisfied, I will create a one-sample t-test for mean.


TI-84

Slide 23- 62

T-Test t = 2.359762364P = .01402172 = 31.95454545 = 3.884980514 n = 22


15. Test statistic

= = = 0.828 = 2.36


TI-84

Slide 23- 63

T-Test t = 2.359762364P = .01402172 = 31.95454545 = 3.884980514 n = 22


16. Obtain P-value

2.36) =

𝐭𝐜𝐝𝐟 (𝟐 .𝟑𝟔 ,𝟗𝟗 ,𝟐𝟏)

1.4%


TI-84

Slide 23- 64

T-Test t = 2.359762364P = .01402172 = 31.95454545 = 3.884980514 n = 22


17. Make a decision

With a P-value below 5%, we reject the


TI-84

Slide 23- 65

T-Test t = 2.359762364P = .01402172 = 31.95454545 = 3.884980514 n = 22


18. State the conclusion in contextThere is sufficient evidence to indicate that the mean speed on Triphammer road is greater than 30 mph.


19. IMPORTANT NOTE:

If the population standard deviation is known, then you should use the Normal model instead of the Student’s t model. This rarely is true.

Slide 23- 66


Which came first?

Slide 23- 67

Egg Sizing StandardsSize Minimum

Weight (ounces)

Minimum Weight (grams)

Mean Weight (grams)

Jumbo 2.5 71 74.5Extra Large 2.25 64 67.5Large 2 57 60.5Medium 1.75 50 53.5Small 1.5 43 46.5Peewee 1.25 36 39.5

40

Egg Weight (g)

50 60 70 80JumboXLLM

Chapter 23 Assignments

Slide 23- 70

Part I: Pages 541-545 #6, 12, 14, 18, 28

Part II: Pages 542-546 #8, 22, 24, 36, 34

6 12 14 18 28

8 22 24 36 34

Chapter 23 #6

Slide 23- 71

Software analysis of the salaries of a random sample of 288 Nevada teachers produced the confidence interval shown below. Which conclusion is correct? What’s wrong with the others?

With 90.00% Confidence,t-interval for : 38944 < < 42893

a) If we took many random samples of Nevada teachers, about 9 out of 10 of them would produce this confidence interval.

Chapter 23 #6

Slide 23- 72



b) If we took many random samples of Nevada teachers, about 9 out of 10 of them would produce a confidence interval that contained the mean salary of all Nevada teachers.

Chapter 23 #6

Slide 23- 73



c) About 9 out of 10 Nevada teachers earn between $38,944 and $42,893.

Chapter 23 #6

Slide 23- 74



d) About 9 out of 10 of the teacher surveyed earn between $38,944 and $42,893.

Chapter 23 #6

Slide 23- 75



e) We are 90% confident that the average teacher salary in the United States is between $38,944 and $42,893.

Chapter 23 #12

Slide 23- 76

Hoping to lure more shoppers downtown, a city builds a new public parking garage in the central business district. The city plans to pay for the structure through parking fees. During a two-month period (44 weekdays), daily fees collected averaged $126, with a standard deviation of $15. a) What assumptions must you make in order to use these statistics for inference?

Independence is reasonable, becausethe weekdays used can be representative of all weekdays andthe 44 weekdays are less than 10% of all weekdays.

We don’t have the actual data, but since the sample of 44 is fairly large by CLT the sampling distribution will be nearly Normal.

Chapter 23 #12

Slide 23- 77

Hoping to lure more shoppers downtown, a city builds a new public parking garage in the central business district. The city plans to pay for the structure through parking fees. During a two-month period (44 weekdays), daily fees collected averaged $126, with a standard deviation of $15. b) Write a 90% confidence interval for the mean daily income this parking garage will generate.

𝑦 ± 𝑡∗𝑛− 1( 𝑠√𝑛 )

= 126 df = 43

= 1.681InvT(0.05,43)

SE()= = = 2.261

ME = (1.681)(2.261) = 3.801263.8

(122.2 , 129.8)

Chapter 23 #12

Slide 23- 78

Hoping to lure more shoppers downtown, a city builds a new public parking garage in the central business district. The city plans to pay for the structure through parking fees. During a two-month period (44 weekdays), daily fees collected averaged $126, with a standard deviation of $15. c) Explain in context what this confidence interval means.

We are 90% confident that the true mean daily income of the parking garage is between $122.20 and $129.80.

Chapter 23 #12

Slide 23- 79

Hoping to lure more shoppers downtown, a city builds a new public parking garage in the central business district. The city plans to pay for the structure through parking fees. During a two-month period (44 weekdays), daily fees collected averaged $126, with a standard deviation of $15. d) Explain what “90% confidence” means in this context.

Chapter 23 #12

Slide 23- 80

Hoping to lure more shoppers downtown, a city builds a new public parking garage in the central business district. The city plans to pay for the structure through parking fees. During a two-month period (44 weekdays), daily fees collected averaged $126, with a standard deviation of $15. e) The consultant who advised the city on this project predicted that parking revenues would average $130 per day. Based on your confidence interval, do you think the consultant could have been correct? Why?

Chapter 23 #14

Slide 23- 81

Suppose that for budget planning purposes the city in Exercise 12 needs a better estimate of the mean daily income from parking fees.a) Someone suggests that the city use its data to create a 95% confidence interval instead of the 90% interval first created. How would this interval be better for the city? (You need not actually create the new interval.)

Chapter 23 #14

Slide 23- 82

Suppose that for budget planning purposes the city in Exercise 12 needs a better estimate of the mean daily income from parking fees.b) How would the 95% interval be worse for the planners?

Chapter 23 #14

Slide 23- 83

Suppose that for budget planning purposes the city in Exercise 12 needs a better estimate of the mean daily income from parking fees.c) How could they achieve an interval estimate that would better serve their planning needs?

Chapter 23 #14

Slide 23- 84

Suppose that for budget planning purposes the city in Exercise 12 needs a better estimate of the mean daily income from parking fees.d) How many days’ worth of data must they collect to have 95% confidence of estimating the true mean to within $3?

Estimate with

3 = 1

n = = 96.04

Use a sample size of 97.

Chapter 23 #18

Slide 23- 85

After his first attempt to determine the speed of light, Michelson conducted an “improved” experiment. In 1897 he reported results of 100 trials with a mean of 852.4 and a standard deviation of 79.0.a) What is the standard error of the mean for these data?

Chapter 23 #18

Slide 23- 86

After his first attempt to determine the speed of light, Michelson conducted an “improved” experiment. In 1897 he reported results of 100 trials with a mean of 852.4 and a standard deviation of 79.0.b) Without computing it, how would you expect a 95% confidence interval for the second experiment to differ from the 95% confidence interval for his first experiment in which he reported the results of 23 trials with a mean of 756.22 and a standard deviation of 107.12.

i) mean (describe its effect on 2nd interval)

Since the mean is different, there will be a different center.

Chapter 23 #18

Slide 23- 87


ii) standard deviation (describe its effect on 2nd SE)

Since the standard deviation is now smaller, the standard error will be smaller and thus the margin of error will be smaller.

Chapter 23 #18

Slide 23- 88


iii) sample size (describe its effect on 2nd SE)

Since the sample size was larger, the Standard Error will be smaller and thus the margin of error will be smaller.

Chapter 23 #18

Slide 23- 89

After his first attempt to determine the speed of light, Michelson conducted an “improved” experiment. In 1897 he reported results of 100 trials with a mean of 852.4 and a standard deviation of 79.0.b) Without computing it, how would you expect a 95% confidence interval for the second experiment to differ from the 95% confidence interval for his first experiment in which he reported the results of 23 trials with a mean of 756.22 and a standard deviation of 107.12. iv) sample size (describe its effect on the 2nd critical value)

Since the sample size was larger, the degrees of freedom will be larger, so the critical value will be smaller and thus the margin of error will be smaller.

Chapter 23 #18

Slide 23- 90

After his first attempt to determine the speed of light, Michelson conducted an “improved” experiment. In 1897 he reported results of 100 trials with a mean of 852.4 and a standard deviation of 79.0.c) Using Michelson’s scale, the true (mean) speed of light is 710.5. What does this indicate about Michelson’s experiments? Explain, using Michelson’s two intervals.

1st 95% CI: (709.90 km/sec, 802.54 km/sec) 2nd 95% CI: (836.72 km/sec, 868.08 km/sec)

Chapter 23 #28

Slide 23- 91

Some students checked 6 bags of Doritos marked with a net weight of 28.3 grams. They carefully weighed the contents of each bag, recording the following weights (in grams): 29.2, 28.5, 28.7, 28.9, 29.1, 29.5 a) Do these data satisfy the assumptions for inference? Explain.

Independence is reasonable, becausethese bags can be representative of all Doritos bags and 6 bags is less than 10% of all Doritos bags

Chapter 23 #28

Slide 23- 92


Nearly Normal Condition: The boxplot shows only some skew to the right without outliers. The data is nearly Normal.

28.0 28.2 28.4 28.6 28.8 29.0 29.2 29.4 29.6

Dorito Bag weight (g)

Chapter 23 #28

Slide 23- 93


Since the conditions are met, I could use the Student’s t model with 5 degrees of freedom for inference on the sample.

Chapter 23 #28

Slide 23- 94

Some students checked 6 bags of Doritos marked with a net weight of 28.3 grams. They carefully weighed the contents of each bag, recording the following weights (in grams): 29.2, 28.5, 28.7, 28.9, 29.1, 29.5 b) Find the mean and standard deviation of the observed weights.

Chapter 23 #28

Slide 23- 95

Some students checked 6 bags of Doritos marked with a net weight of 28.3 grams. They carefully weighed the contents of each bag, recording the following weights (in grams): 29.2, 28.5, 28.7, 28.9, 29.1, 29.5 c) Create a 95% confidence interval for the mean weight of such bags of chips.

𝒚 ± 𝒕∗𝒏−𝟏❑ ( 𝒔√𝒏) 𝑴𝑬=(𝟐.𝟓𝟕𝟏)(𝟎 .𝟑𝟔√𝟔 )

0.378

InvT(0.025, 5)

¿𝟎 .𝟑𝟕𝟖

(𝟐𝟖 .𝟔𝟎 ,𝟐𝟗 .𝟑𝟔)

Chapter 23 #28

Slide 23- 96

Some students checked 6 bags of Doritos marked with a net weight of 28.3 grams. They carefully weighed the contents of each bag, recording the following weights (in grams): 29.2, 28.5, 28.7, 28.9, 29.1, 29.5 d) Explain in context what your interval means.

Chapter 23 #28

Slide 23- 97

Some students checked 6 bags of Doritos marked with a net weight of 28.3 grams. They carefully weighed the contents of each bag, recording the following weights (in grams): 29.2, 28.5, 28.7, 28.9, 29.1, 29.5 e) Comment on the company’s stated net weight of 28.3 grams.

Chapter 23 #28

Slide 23- 98

Some students checked 6 bags of Doritos marked with a net weight of 28.3 grams. They carefully weighed the contents of each bag, recording the following weights (in grams): 29.2, 28.5, 28.7, 28.9, 29.1, 29.5 e) Comment on the company’s stated net weight of 28.3 grams.

Chapter 23 #8

Slide 23- 99

Based on meteorological data for the past century, a local TV weatherman estimates that the region’s average winter snowfall is 23”, with a margin of error of 2 inches. Assuming he used a 95% confidence interval, how should viewers interpret this news? Comment on each of these statements. a) During 95 of the last 100 winters, the region got between 21” and 25” of snow.

Chapter 23 #8

Slide 23- 100

Based on meteorological data for the past century, a local TV weatherman estimates that the region’s average winter snowfall is 23”, with a margin of error of 2 inches. Assuming he used a 95% confidence interval, how should viewers interpret this news? Comment on each of these statements. b) There’s a 95% chance the region will get between 21” and 25” of snow this winter.

Chapter 23 #8

Slide 23- 101

Based on meteorological data for the past century, a local TV weatherman estimates that the region’s average winter snowfall is 23”, with a margin of error of 2 inches. Assuming he used a 95% confidence interval, how should viewers interpret this news? Comment on each of these statements. c) There will be between 21” and 25” of snow on the ground for 95% of the winter days.

Chapter 23 #8

Slide 23- 102

Based on meteorological data for the past century, a local TV weatherman estimates that the region’s average winter snowfall is 23”, with a margin of error of 2 inches. Assuming he used a 95% confidence interval, how should viewers interpret this news? Comment on each of these statements. d) Residents can be 95% sure that the area’s average snowfall is between 21” and 25”.

Chapter 23 #8

Slide 23- 103

Based on meteorological data for the past century, a local TV weatherman estimates that the region’s average winter snowfall is 23”, with a margin of error of 2 inches. Assuming he used a 95% confidence interval, how should viewers interpret this news? Comment on each of these statements. e) Residents can be 95% confident that the average snowfall during the last century was between 21” and 25” per winter.

Chapter 23 #22

Slide 23- 104

During an angiogram, heart problems can be examined via a small tube (a catheter) threaded into the heart from a vein in the patient’s leg. It’s important that the company who manufactures the catheter maintain a diameter of 2.00 mm. (The standard deviation is quite small.) Each day, quality control personnel make several measurements to test at a significance level of If they discover a problem, they will stop the manufacturing process until it is corrected.a) Is this a one-sided or two-sided test? In the context of the problem, why do you think this is important?

Chapter 23 #22

Slide 23- 105

During an angiogram, heart problems can be examined via a small tube (a catheter) threaded into the heart from a vein in the patient’s leg. It’s important that the company who manufactures the catheter maintain a diameter of 2.00 mm. (The standard deviation is quite small.) Each day, quality control personnel make several measurements to test at a significance level of If they discover a problem, they will stop the manufacturing process until it is corrected.b) Explain in this context what happens if the quality control people commit a Type I error.

Chapter 23 #22

Slide 23- 106

During an angiogram, heart problems can be examined via a small tube (a catheter) threaded into the heart from a vein in the patient’s leg. It’s important that the company who manufactures the catheter maintain a diameter of 2.00 mm. (The standard deviation is quite small.) Each day, quality control personnel make several measurements to test at a significance level of If they discover a problem, they will stop the manufacturing process until it is corrected.c) Explain in this context what happens if the quality control people commit a Type II error.

Chapter 23 #24

Slide 23- 107

The catheter company in Exercise 22 is reviewing its testing procedure.

a) Suppose the significance level is changed to . Will the probability of Type II error increase, decrease, or remain the same?

Chapter 23 #24

Slide 23- 108


b) What is meant by the power of the test the company conducts?

Chapter 23 #24

Slide 23- 109


c) Suppose the manufacturing process is slipping out of proper adjustment. As the actual mean diameter of the catheters produced gets farther and farther above the desired 2.00 mm, will the power of the quality control test increase, decrease, or remain the same?

Chapter 23 #24

Slide 23- 110


d) What could they do to improve the power of the test?

Chapter 23 #36

Slide 23- 111

36. A tire manufacturer is considering a newly designed tread pattern for its all-weather tires. Tests have indicated that these tires will provide better gas mileage and longer treadlife. The last remaining test is for braking effectiveness. The company hopes the tire will allow a car traveling at 60 mph to come to a complete stop within an average of 125 feet after the brakes are applied. They will adopt the new tread pattern unless there is strong evidence that the tires do not meet this objective. The distances (in feet) for 10 stops on a test track were 129, 128, 130, 132, 135, 123, 102, 125, 128, and 130. Should the company adopt the new tread pattern? Test an appropriate hypothesis and state your conclusion. Explain how you dealt with the outlier, and why you made the recommendation you did.

Chapter 23 #36

Slide 23- 112


The parameter of interest is the mean breaking distance, .

:

:

Chapter 23 #36

Slide 23- 113


Chapter 23 #36

Slide 23- 114


Nearly Normal Condition: The breaking distance of 102 is an outlier. Once it is removed boxplot shows reasonable symmetry. The data is nearly Normal w/o the outlier.

120 122 124 126 128 130 132 134 136

Chapter 23 #36

Slide 23- 115


w/o outlier 128.889s = 3.551 = 3.285

= 1.184

Since the conditions are met, we will perform a one-sample t-test with 8 degrees of freedom.

Chapter 23 #36

Slide 23- 116

36. A tire manufacturer is considering a newly designed tread pattern for its all-weather tires. Tests have indicated that these tires will provide better gas mileage and longer treadlife. The last remaining test is for braking effectiveness. The company hopes the tire will allow a car traveling at 60 mph to come to a complete stop within an average of 125 feet after the brakes are applied. They will adopt the new tread pattern unless there is strong evidence that the tires do not meet this objective. The distances (in feet) for 10 stops on a test track were 129, 128, 130, 132, 135, 123, 102, 125, 128, and 130. Should the company adopt the new tread pattern? Test an appropriate hypothesis and state your conclusion. Explain how you dealt with the outlier, and why you made the recommendation you did.w/o outlier 128.889s = 3.551 = 3.285

= 1.184

P-value = P( > 3.285)≈ 0.006

Chapter 23 #36

Slide 23- 117


P-value = P( > 3.285) ≈ 0.006With a P-value below 5% we will reject .

There is sufficient evidence that the mean breaking distance is greater than 125 feet. The new tread pattern should not be adopted.

Chapter 23 #34

Slide 23- 118

Consumer Reports tested 14 brands of vanilla yogurt and found the following numbers of calories per serving:

160 200 220 230 120 180 140 130 170 190 80 120 100 170 a) Check the assumptions and conditions for inference.

Independence is reasonable, because calorie amounts between brands should be independent of each other.

Chapter 23 #34

Slide 23- 119


160 200 220 230 120 180 140 130 170 190 80 120 100 170 a) Check the assumptions and conditions for inference.

Nearly Normal Condition: The boxplot shows reasonable symmetry without any outliers. The data is nearly Normal.

60 80 100 120 140 160 180 200 220 240

Chapter 23 #34

Slide 23- 120


160 200 220 230 120 180 140 130 170 190 80 120 100 170 b) Create a 95% confidence interval for the average calorie content of vanilla yogurt.

157.857s = 44.752

= 1.270

Since the conditions are met, we will perform a one-sample t-test with 13 degrees of freedom.

Chapter 23 #34

Slide 23- 121

Consumer Reports tested 14 brands of vanilla yogurt and found the following numbers of calories per serving: 160 200 220 230 120 180 140 130 170 190 80 120 100 170 b) Create a 95% confidence interval for the average calorie content of vanilla yogurt. 157.857 s = 44.752

𝒚 ± 𝒕∗𝒏−𝟏❑ ( 𝒔√𝒏)

𝑴𝑬=(𝟐.𝟏𝟔𝟎)(𝟒𝟒 .𝟕𝟓𝟐√𝟏𝟒 ) 25.835

InvT(0.025, 13)

¿𝟐𝟓 .𝟖𝟑𝟓

(𝟏𝟑𝟐 .𝟎𝟐𝟐 ,𝟏𝟖𝟑 .𝟔𝟗𝟐)

Chapter 23 #34

Slide 23- 122

Consumer Reports tested 14 brands of vanilla yogurt and found the following numbers of calories per serving: 160 200 220 230 120 180 140 130 170 190 80 120 100 170 c) A diet guide claims that you will get 120 calories from a serving of vanilla yogurt. What does this evidence indicate? Use your confidence interval to test a n appropriate hypothesis and state your conclusion.

(𝟏𝟑𝟐 .𝟎𝟐𝟐 ,𝟏𝟖𝟑 .𝟔𝟗𝟐)We are 95% confident that the mean calorie content in a serving of vanilla yogurt is between 132.0 and 183.7 calories.There is evidence that the estimate of 120 calories in the diet guide is too low. The confidence interval is well above 120 calories.

Chapter 23 #34

Slide 23- 123

Consumer Reports tested 14 brands of vanilla yogurt and found the following numbers of calories per serving: 160 200 220 230 120 180 140 130 170 190 80 120 100 170 c) A diet guide claims that you will get 120 calories from a serving of vanilla yogurt. What does this evidence indicate? Use your confidence interval to test a n appropriate hypothesis and state your conclusion.

(𝟏𝟑𝟐 .𝟎𝟐𝟐 ,𝟏𝟖𝟑 .𝟔𝟗𝟐)We are 95% confident that the mean calorie content in a serving of vanilla yogurt is between 132.0 and 183.7 calories.There is evidence that the estimate of 120 calories in the diet guide is too low. The confidence interval is well above 120 calories.

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