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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

22

23

24

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.

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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.