stat 151 final exam 2014 summer university of alberta version abinzou/stat141/stat 151 final...
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STAT 151 FINAL EXAM
2014 SUMMER
University of Alberta
Version A
Last Name: _______________________ First Name: _______________________
Student ID: _______________________ Signature:
Instructions:
1) The exam consists of 45 multiple choice questions worth 1 mark each.
2) This is a closed book exam. You may use the STAT 151 formula sheet and the table
provided, and a non-programmable calculator.
3) All your personal items must be placed on the floor prior to the start of the exam.
4) Make sure your cell phones, and all other electronic devices are turned off.
5) No communications with other students are allowed.
6) You must mark all your answers on the Answer sheet (Second page of the exam
booklet). Otherwise, NO marks will be given.
7) Choose exactly one answer for each question.
8) Your score will be based on the number of questions you answer correctly.
9) There are no penalties for incorrect answers.
10) No credit will be given for omitted answers.
11) You must mark all your answers on the answer sheet during the allotted time. No
additional time will be allowed at the end of the session for this purpose.
12) You must return the exam booklet, the tables and the formula sheet to the instructor
after you finish the exam.
13) Copying questions or answers on paper to take from the exam room is prohibited.
14) You have 90 minutes to complete the exam.
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Answer Sheet You must write your answers on this answer sheet!
No credit will be given if answers are written elsewhere!
Question Answer Question Answer
1 26
2 27
3 28
4 29
5 30
6 31
7 32
8 33
9 34
10 35
11 36
12 37
13 38
14 39
15 40
16 41
17 42
18 43
19 44
20 45
21
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25
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1. A small company employs a supervisor at $1,100 a week, an inventory manager at
$700 a week, 6 stock boys at $400 a week, and 4 drivers at $600 a week. Which
measure of spread, would best describe the payroll?
A) IQR, because the distribution is symmetric.
B) Range, because it would be least sensitive to the outlier at $1,100.
C) Standard deviation, because it would be least sensitive to the outlier at $1,100.
D) IQR, because it would be least sensitive to the outliers at $700 and $1,100.
E) IQR, because it would be least sensitive to the outlier at $1,100.
2. The local basketball team averages 65% from the free-throw line. A player who
makes 72.5% of his free-throws has a z-score of 1.5. Find the standard deviation.
A) 5%
B) 15%
C) 3%
D) 1%
E) 10%
3. Based on the Normal model for snowfall in a certain town, N(57, 8), how many cm
of snow would represent the 75th percentile?
A) 62.4 cm
B) 49 cm
C) 65 cm
D) 51.6 cm
E) 42.8 cm
4. The relationship between the price of yachts (y) and their length (x) is analyzed.
The mean length was 41 meters with a standard deviation of 11. The mean price
was $84,000 with a standard deviation of $14,000. The correlation between the
price and the length was 0.41. Find the linear regression equation.
A) 𝑝𝑟𝑖𝑐�̂� = 31,800 + 1,270 𝑙𝑒𝑛𝑔𝑡ℎ
B) 𝑝𝑟𝑖𝑐�̂� = 70,800 + 0.0322 𝑙𝑒𝑛𝑔𝑡ℎ
C) 𝑝𝑟𝑖𝑐�̂� = −962,800 + 547 𝑙𝑒𝑛𝑔𝑡ℎ
D) 𝑝𝑟𝑖𝑐�̂� = −4,040,000 + 622 𝑙𝑒𝑛𝑔𝑡ℎ
E) 𝑝𝑟𝑖𝑐�̂� = 62,605 + 522 𝑙𝑒𝑛𝑔𝑡ℎ
5. Assume that 10% of people are left-handed. Suppose 10 people are selected at
random. What is the probability that they are not all right-handed?
A) 0.100
B) 0.651
C) 1.000
D) 0.900
E) 0.013
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6. According to a survey, 10% of students at a college are left handed, 53% are female,
and 5.3% are both female and left handed. Is being left handed independent of
gender? Explain.
A) Yes, P (left handed) = P (left handed and female)
B) No, because P(L and F) ≠ P(L) · P(F)
C) Yes, 10% of all students are left handed and 10% of female students are left
handed.
D) No, 5.3% of students are both female and left handed
E) No, 10% of all students are left handed but 5.3% of female students are left h
anded. These are not equal.
7. The table shows the political affiliation of voters in one city and their positions on
raising taxes.
Favor Oppose
NDP 0.12 0.28
Conservative 0.26 0.11
Other 0.14 0.09
What is the probability that a voter is a NDP given that he/she favors raising taxes?
A) 0.120
B) 0.520
C) 0.400
D) 0.231
E) 0.300
Answer questions 8-9 using the following information.
Your roll a die, winning nothing if the result is an odd number, $1 for a 2 or a 4, and
$16 for a 6.
8. What is the expected value and the standard deviation of your prospective winnings?
A) E(X) = $3.00, SD(X) = $5.83
B) E(X) = $6.00, SD(X) = $6.00
C) E(X) = $1.83, SD(X) = $18.20
D) E(X) = $1.00, SD(X) = $14.85
E) E(X) = $8.50, SD(X) = $4.19
9. Assume you play 30 times. What is the probability that you win at least $120?
A) 0.05
B) 0.17
C) 0.95
D) 0.83
E) 0.50
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10. A candy company claims that its jelly bean mix contains 15% blue jelly beans.
Suppose that the candies are packaged at random in small bags containing about
200 jelly beans. What is the probability that a bag will contain more than 20% blue
jelly beans?
A) 0.9761
B) 0.0239
C) 0.0422
D) 0.0478
E) 0.9578
11. A survey of 865 people in one city reveals that 408 have visited a Tim Hortons
location in the past week. Construct a 95% confidence interval for the percentage
of all people in this city who have visited a Tim Hortons location in the last week.
A) (43.8%, 50.5%)
B) (44.4%, 50.0%)
C) (46.9%, 47.5%)
D) (42.3%, 52.0%)
E) (43.1%, 51.2%)
12. A researcher wishes to estimate the proportion of fish in a certain lake that is
inedible due to pollution of the lake. How large a sample should be tested in order
to be 95% confident that the true proportion of inedible fish is estimated to be
within 5%?
A) 273
B) 269
C) 385
D) 267
E) Not enough information is given.
Use the following information to ask questions 13-14.
Five years ago, a company found that 8% of its employees commuted to work by car.
A survey investigates whether the current proportion of employees who commute by
car to work is higher than it was five years ago. A test on employee commuting by car
was done on a random sample of 1000 employees, and found car commuting to be 12%.
13. What are the null and alternative hypotheses?
A) 𝐻0: 𝑝 = 0.08; 𝐻𝐴: 𝑝 > 0.08.
B) 𝐻0: 𝑝 = 0.08; 𝐻𝐴: 𝑝 ≠ 0.08.
C) 𝐻0: 𝑝 = 0.12; 𝐻𝐴: 𝑝 > 0.08.
D) 𝐻0: 𝑝 = 0.12; 𝐻𝐴: 𝑝 > 0.12.
E) 𝐻0: 𝑝 = 0.12; 𝐻𝐴: 𝑝 ≠ 0.08.
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14. What is your conclusion based on the P-value?
A) z = 4.66; P-value < 0.00001. The data shows conclusive evidence that the
proportion in car commuting is greater than 8%.
B) z = -4.66; P-value > 0.00001. The data shows conclusive evidence that the
proportion in car commuting is greater than 8%.
C) z = 4.66; P-value > 0.00001. The data shows no evidence that the proportion
in car commuting is greater than 8%.
D) z = 4.66; P-value < 0.99999. The data shows no evidence that the proportion
in car commuting is greater than 8%.
E) z = -4.66; P-value > 0.99999. The data shows no evidence that the proportion
in car commuting is greater than 8%.
15. A newspaper reports that 58% of farmers are planning to plant genetically modified
crops this year. The article adds that the poll is based on a random sample of 798
famers and has margin of error of 4.5%. What is the significance level used in this
poll?
A) 99%
B) 98%
C) 95%
D) 90%
E) Not enough information is given.
16. A survey found that 79% of a random sample of 1024 Canadian adults approved of
cloning endangered animals. For a 90% confidence level, the margin of error is
2.09%. Among the following options, which one can guarantee a smaller margin
of error?
(I) Increase the confidence level (II) Decrease the confidence level
(III) Increase the sample size (IV) Decrease the sample size
A) I, III
B) I, IV
C) II, III
D) II, IV
E) None
17. A sample of 51 statistics students at a large university had a mean final exam score
of 76 with a standard deviation of 4. Find a 95% confidence interval for the mean
final exam score for all statistics students at this university.
A) (75.6, 76.1)
B) (75.6, 76.4)
C) (79.4, 82.6)
D) (74.9, 77.1)
E) (75.9, 76.1)
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18. Refer to the previous question, based on the confidence interval, what is your
conclusion to the claim that the average mean final exam score is different from 78
at the significance level of 5%?
A) Since 76 is different from 78, there is sufficient evidence to support the claim
that the average mean final exam score is different from 78.
B) Since 76 is included in the 95% confidence interval, there is not sufficient
evidence to support the claim that the average mean final exam score is
different from 78.
C) Since 76 is not included in the 95% confidence interval, there is sufficient
evidence to support the claim that the average mean final exam score is
different from 78.
D) Since 78 is included in the 95% confidence interval, there is not sufficient
evidence to support the claim that the average mean final exam score is
different from 78.
E) Since 78 is not included in the 95% confidence interval, there is sufficient
evidence to support the claim that the average mean final exam score is
different from 78.
19. According to the 2004 Canadian Community Health Survey, 23.1% of Canadians
aged 18 or older were considered obese; that is, they had a body mass index of 30
or higher. Suppose you believe that the percentage of obese Canadians has
decreased since 2004 and wish to test your belief. Identify the Type I error.
A) You conclude that obesity is on the rise since it can only increase.
B) You conclude that the percentage of obese Canadians is on the decline, but in
fact it has stayed the same or increased.
C) You conclude that the percentage of obese Canadians has increased or decreased,
but in fact it has stayed the same
D) You conclude that the percentage of obese Canadians has stayed the same or in
creased, but it has actually decreased.
E) A Type I error is not possible in this situation.
20. You want to estimate the average gas price in your city for a liter of regular
unleaded gasoline with 95% confidence. You believe that the standard deviation of
prices is $0.05. If you wish to achieve a margin of error no larger than $0.02, how
many gas stations should you sample?
A) 25
B) 28
C) 27
D) 24
E) 26
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21. A large software company gives job applicants a test of programming ability, and
the mean for the test has been 160 in the past. Twenty-five applicants are randomly
selected from one large university and they produce a mean score of 165, with a
standard deviation of 13. At a significance level of 0.05, does this indicate that the
sample comes from a population with a mean score greater than 160?
A) Yes. With a P-value of 0.0024, we reject the null hypothesis. There is sufficient
evidence that the population mean score is greater than 160.
B) No. With a P-value of 0.0664, we fail to reject the null hypothesis. There is not
sufficient evidence that the population mean score is greater than 160.
C) No. With a P-value of 0.9336, we fail to reject the null hypothesis. There is not
sufficient evidence that the population mean score is greater than 160.
D) Yes. With a P-value of 0.0332, we reject the null hypothesis. There is sufficient
evidence that the population mean score is greater than 160.
E) No. With a P-value of 0.9668, we fail to reject the null hypothesis. There is not
sufficient evidence that the population mean score is greater than 160.
22. A new manager, hired at a large warehouse, was told to reduce the 26% employee
sick leave. The manager introduced a new incentive program for employees with
perfect attendance. The manager decides to test the new program to see if it’s better
and receives a P-value of 0.06. What is reasonable to conclude about the new
strategy using α = 0.05?
A) We can say there is a 6% chance of seeing the new program having an effect
on the employee attendance. We conclude the new program is more effective.
B) If the sick leave percentage has not changed, then there is only a 6% chance
seeing the decrease in the sick leave percentage we observed (or lower) from
natural sampling variation. At α = 0.05, we reject the null hypothesis that the
sick leave percentage has not changed and conclude that the new incentive
program is effective.
C) There is a 6% chance of the new incentive program having no effect on the
employee attendance percentage.
D) If the sick leave percentage has not changed, then there is only a 6% chance
of seeing the decrease in the sick leave percentage we observed (or lower)
from natural sampling variation. At α = 0.05, we fail to reject the null
hypothesis that the sick leave percentage has not changed.
E) There is a 94% chance of the new incentive program having no effect on the
employee attendance percentage.
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23. In a random sample of 300 Canadian women, 68% watch CBC News. In a random
sample of 200 Canadian men, 56% watch CBC News. Construct a 98% confidence
interval for the difference in the proportion of women and men who watch CBC
News.
A) (0.017, 0.234)
B) (0.033, 0.207)
C) (0.017, 0.223)
D) (0.029, 0.211)
E) (0.006, 0.234)
24. A marketing survey involves product recognition in Alberta and British Columbia.
Suppose the proportion of Alberta residents who recognized a product is p1 and the
proportion of British Columbia residents who recognized the product is p2. The
survey found a 98% confidence interval for p1 - p2 is (0.015, 0.025). Give an
interpretation of this confidence interval.
A) We know that 98% of British Columbia residents recognized the product
between 1.5% and 2.5% more than Alberta residents.
B) We know that 98% of Alberta residents recognized the product between 1.5%
and 2.5% more than British Columbia residents.
C) We are 98% confident that the proportion of British Columbia residents who
recognized the product is between 1.5% and 2.5% higher than the proportion
of Alberta residents who recognized the product.
D) We know that 98% of all random samples will show that the proportion of
British Columbia residents who knew the product is between 1.5% and 2.5%
higher than the proportion of Alberta residents who knew the product.
E) We are 98% confident that the proportion of Alberta residents who recognized
the product is between 1.5% and 2.5% higher than the proportion of British
Columbia residents who recognized the product.
25. A 2009 update on pollution in a Canadian lake system’s fish species found that
mercury contamination was the cause of six fish advisories issued for 48 various
fish-size combinations, whereas there were 10 mercury-caused fish advisories for
the same 48 fish-size combinations in 2003. Using α = 0.05, test for a significant
difference of proportions between years.
A) P-value = 0.28. There is not enough evidence to support the existence of a
significant difference between years.
B) P-value = 0.14. There is not enough evidence to support the existence of a
significant difference between years.
C) P-value = 0.50. There is sufficient evidence to support the existence of a
significant difference between years.
D) P-value = 0.05. There is sufficient evidence to support the existence of a
significant difference between years.
E) P-value = 0.19. There is not enough evidence to support the existence of a
significant difference between years.
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26. Men diagnosed with cancer were randomly assigned to either undergo surgery or
not. Among the 338 men who had the surgery, 18 eventually died of cancer,
compared with 33 of the 335 men who did not have surgery. A 90% confidence
interval for the difference in the proportion of men who did not have the surgery
and eventually died of cancer and the proportion of men who had the surgery and
eventually died of cancer is (0.012, 0.079). Based on your confidence interval, is
there evidence that surgery may be effective in preventing death from cancer?
A) Yes. Since both numbers are positive, there is not enough evidence that surgery
may be effective in preventing death from cancer.
B) Yes. Since both numbers are positive, there is evidence that surgery may be
effective in preventing death from cancer.
C) No. Since both numbers are positive, there is not enough evidence that surgery
may be effective in preventing death from cancer.
D) No. Since both numbers are positive, there is evidence that surgery may be
effective in preventing death from cancer.
E) No. Cannot make any suggestion based on just a confidence interval.
27. For a degree of freedom of 19, what is the probability that t-value is greater than -
1.729?
(P(t > -1.729) = ?)
A) 0.05
B) 0.10
C) 0.95
D) 0.90
E) Not given in the table.
28. Two types of flares are tested for their burning times (in minutes) and sample
results are given below.
Brand X Brand Y
n = 35 n = 40
�̅� = 19.4 �̅� = 15.1
𝑠𝑥 = 1.4 𝑠𝑦 = 0.8
Construct a 95% confidence interval for the difference 𝜇𝑋 − 𝜇𝑌 based on the data.
A) (3.5, 5.1)
B) (3.8, 4.8)
C) (3.6, 5.0)
D) (-4.7, -3.9)
E) (3.2, 5.4)
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29. A manufacturer has designed athletic footwear which it hopes will improve the
performance of athletes running the 100-meter sprint. It wishes to perform a
hypothesis test to compare the times of athletes at the 100 meters with these shoes
and with their usual shoes.
A) Two-sample t-test.
B) Not enough information is given.
C) Either two-sample or paired t-test would be equally accurate.
D) Pooled t-test.
E) Paired t-test.
30. A study was made to determine which taxi company gave quicker service.
Companies A and B were each called 50 randomly selected times.
Company A Company B
Mean response time 7.6 minutes 8.2 minutes
Standard deviation 2.4 minutes 1.1 minutes
Company A claims it responds quicker than Company B. State the hypotheses.
A) 𝐻0: 𝜇𝐴 − 𝜇𝐵 > 0 𝐻𝐴: 𝜇𝐴 − 𝜇𝐵 ≤ 0
B) 𝐻0: 𝜇𝐴 − 𝜇𝐵 < 0 𝐻𝐴: 𝜇𝐴 − 𝜇𝐵 = 0
C) 𝐻0: 𝜇𝐴 − 𝜇𝐵 = 0 𝐻𝐴: 𝜇𝐴 − 𝜇𝐵 > 0
D) 𝐻0: 𝜇𝐴 − 𝜇𝐵 > 0 𝐻𝐴: 𝜇𝐴 − 𝜇𝐵 = 0
E) 𝐻0: 𝜇𝐴 − 𝜇𝐵 = 0 𝐻𝐴: 𝜇𝐴 − 𝜇𝐵 < 0
31. Refer to the previous question, at 5% level of significance, test the claim that
Company A responds to customer calls quicker than Company B.
A) P-value is less than 0.005, so there is sufficient evidence to support the claim.
B) P-value is between 0.005 and 0.01, so there is sufficient evidence to support
the claim.
C) P-value is between 0.01 and 0.025, so there is NOT sufficient evidence to
support the claim.
D) P-value is between 0.025 and 0.05, so there is sufficient evidence to support
the claim.
E) P-value is greater than 0.05, so there is NOT sufficient evidence to support the
claim.
32. You wish to construct a 95% confidence interval to compare the mean
measurement for two groups. A small pilot study yields sample standard deviation
of 20 and 30 for Group 1 and Group 2, respectively. If we wish to obtain a margin
of error of at most 10, what sample size should we take from each group? Assume
equal sample size and non-pooled case. Hint: use Z-table to find the critical value.
A) 49
B) 47
C) 50
D) 48
E) 51
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Using the following information to answer questions 33-35.
A coach uses a new technique in training middle distance runners. The time for 9
different athletes to run 800 meters before and after this training is recorded below.
Athlete A B C D E F G H I Mean SD
Before 115.2 120.9 108.0 112.4 107.5 119.1 121.3 110.8 122.3 115.28 5.848
After 116.0 119.1 105.1 111.9 109.1 115.2 118.5 110.7 120.9 114.06 5.218
Difference -0.8 1.8 2.9 0.5 -1.6 3.9 2.8 0.1 1.4 1.222 1.826
33. Construct a 99% confidence interval for the mean difference of the “before” minus
“after” time.
A) (-0.54, 2.98)
B) (-6.42, 8.87)
C) (-0.85, 3.29)
D) (-0.76, 3.20)
E) (-0.82, 3.26)
34. A coach wants to test whether this new training method is effective. What test is
appropriate? Calculate the corresponding test statistic.
A) One-sample t-test. t = 0.63.
B) Two-sample t-test (non-pooled). t = 0.47.
C) Two-sample t-test (pooled). t = 0.33.
D) Two-sample paired test. t = 2.01.
E) Two-sample z-test. z=1.96.
35. What is the degree of the freedom? What is the right conclusion given the
significance level is 5%?
A) df = 8. Since P-value > 5%, there is sufficient evidence that the new training
method is effective.
B) df = 8. Since P-value < 5%, there is not sufficient evidence that the new
training method is effective.
C) df = 8. Since P-value < 5%, there is sufficient evidence that the new training
method is effective.
D) df = 16. Since P-value > 5%, there is not sufficient evidence that the new
training method is effective.
E) df = 16. Since P-value < 5%, there is sufficient evidence that the new training
method is effective.
Page 13 of 15
Use the following information to answer questions 36-38.
Tests for adverse reaction to a new drug yielded the results given in the table. The data
will be analyzed to determine if there is sufficient evidence to conclude that an
association exists between the treatment (drug or placebo) and the reaction (whether or
not headaches were experienced).
Drug Placebo
Headaches 11 7
No headaches 73 91
36. Which test would be appropriate for the given situation?
A) Independence Chi-square test
B) Goodness-of-fit Chi-square test
C) Homogeneity Chi-square test
D) ANOVA test
E) Two-sample t-test
37. What is the expected number of people who took drugs but still experienced
headaches?
A) 18
B) 8
C) 84
D) 16
E) 11
38. Assume all conditions for an appropriate test are satisfied, and the test statistic is
1.7951. Find the degree of freedom and make your conclusion using α = 10%.
A) df = 2. Reject null hypothesis since P-value < 10%.
B) df = 2. Do not reject null hypothesis since 5% < P-value < 10%.
C) df = 1. Reject null hypothesis since P-value < 10%.
D) df = 1. Reject null hypothesis since P-value > 10%.
E) df = 1. Do not reject null hypothesis since P-value > 10%.
39. Suspecting that a die may be unfair, you want to investigate. To check, you roll it
54 times, recording the number of times each face appears.
Face 1 2 3 4 5 6
Count 9 15 6 12 6 6
Find 𝑋2 value.
A) 1.000
B) 2.000
C) 4.000
D) 8.000
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E) 10.000
40. Refer to the previous question, what is the degree of freedom? What is the P-value
of the test?
A) df = 6; 0.1 < P-value
B) df = 6; 0.025 < P-value < 0.05
C) df = 5; 0.1 < P-value
D) df = 5; 0.05 < P-value < 0.1
E) df = 5; 0.025 < P-value < 0.05
41. A researcher performed a study to determine whether political party affiliations are
different for different income brackets. State the hypotheses in the test.
A) 𝐻0: There is a relationship between political party affiliation and income.
𝐻𝐴: Income and political party affiliation have no relationship.
B) 𝐻0: Political party affiliation is uniformly distributed over income.
𝐻𝐴: Political party affiliation is not uniformly distributed over income.
C) 𝐻0: Income and political party affiliation are independent.
𝐻𝐴: Income and political party affiliation are dependent.
D) 𝐻0: Political party affiliation does not have the same distribution for each
income bracket.
𝐻𝐴: Political party affiliation has the same distribution for each income
bracket.
E) 𝐻0: Political party affiliation has the same distribution for each income
bracket.
𝐻𝐴: Political party affiliation does not have the same distribution for each
income bracket.
Use the following information to answer questions 42-44.
An industrial engineer wants to test the effect of three different ways of assembling a
part on the total assembly time. Five people are randomly assigned to each of the three
assembly methods, and the total assembly time (in seconds) is recorded.
Source df Sum-of-Squares Mean-Squares
Treatment 985535.907
Error 276844.275
42. State the null and alternative hypotheses of the ANOVA test.
A) 𝐻0: 𝜇1 = 𝜇2 = 𝜇3 𝐻𝐴: 𝜇1 ≠ 𝜇2 ≠ 𝜇3
B) 𝐻0: 𝜇1 ≠ 𝜇2 ≠ 𝜇3 𝐻𝐴: 𝜇1 = 𝜇2 = 𝜇3
C) 𝐻0: 𝜇1 = 𝜇2 = 𝜇3 𝐻𝐴: mean assembly time of five people are not equal
D) 𝐻0: 𝜇1 = 𝜇2 = 𝜇3 𝐻𝐴: mean assembly time of three methods are not equal
E) 𝐻0: 𝜇1 = 𝜇2 = 𝜇3 𝐻𝐴: not all mean assembly time of three methods are equal
Page 15 of 15
43. How many degrees of freedom does the treatment sum of squares have? How about
the error sum of squares?
A) df-treatment = 3; df-error = 5
B) df-treatment = 2; df-error = 5
C) df-treatment = 3; df-error = 15
D) df-treatment = 2; df-error = 12
E) df-treatment = 3; df-error = 12
44. Calculate the F-ratio.
A) 24.919
B) 21.359
C) 14.240
D) 16.613
E) 23.139
45. A hunting supply store is interested in knowing whether three different brands of
goose decoys are equally effective in attracting geese. Assume all the conditions
for ANOVA test are satisfied, the calculation found P-value = 0.065. What would
you conclude?
A) Reject null hypothesis using α = 0.05.
B) Reject null hypothesis using α = 0.01.
C) Reject null hypothesis using α = 0.06.
D) Reject null hypothesis using α = 0.1.
E) Reject null hypothesis at any level.