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Test 3

ECO220Y–L5101 Quantitative Methods in Economics

Monday February 6, 2017; 18:00 – 20:00

U of T e-mail: ___________________________________@mail.utoronto.ca

Surname

(last name):

Given name

(first name):

UTORID:

(e.g. lihao8)

Student Number __Solution_________________________________

x You have 2 hours to complete this test.

x This exam consists of 21 questions in 7 pages including this cover page. It is the

student’s responsibility to hand in all pages of this exam. Any missing page will get zero

mark.

x Write your answers clearly, completely and concisely in the designated space provided

immediately after each question. No extra space/pages are possible. You cannot use

blank space for other questions nor can you write answers on the Supplement. Your entire

answer must fit in the designated space provided immediately after each question.

x You are encouraged to write in pencil and to use an eraser as needed. This

way you can make sure to fit your final answer (including work and reasoning) in the

appropriate space.

x A formula sheet and the standard normal table are attached.

x This exam is worth 17% of your course grade.

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Part 1. Multiple Choice Questions. 3 points each. No part mark.

Circle the BEST answer. If there are more than one correct answer, circle the best one.

1. When the true value under the alternative hypothesis shifts closer to the value under the null

hypothesis, while the critical value stays the same,

√(a) The power of the test decreases.

(b) The power of the test increases.

(c) Probability of Type I error increases

(d) Probability of Type I error decreases

(e) Probability of Type II error decreases

2. You are to conduct hypothesis testing for a population mean. Your research assistant incorrectly

memorized the formula for sample standard errors, causing him to always calculate standard errors

to be larger than what they really are. Suppose the point estimate of the population mean is correctly

calculated. If you do not find out about his systematic mistakes, what consequences will they have

on the results of your tests?

(a) Your tests will always result in type I error.

(b) Your tests will always result in type II error.

(c) There will be no effect on the results of your tests.

√(d) You will tend to reject the null hypothesis less often than you actually should.

(e) You will tend to reject the null hypothesis more often than you actually should.

3. The standard deviation of income is $30,750 and the population is not normal. How big of a sample

should you collect if you wish to estimate with 96% confidence the average income with a margin

of error of plus or minus $1,000?

(a) 2,559 (b) 2,913 (c) 3,633 √(d) 3,994 (e) 4,600

4. Suppose you used this formula

n

X Z V

r 0.05 and correctly computed a 90% confidence interval

estimator of the population mean to be [10, 30]. Which of the following is FALSE?

(a) The sample mean is equal to 20.

(b) The significance level D is equal to 0.10.

(c) The point estimate of the population mean is 20.

√(d) About 90% of the population lies between 10 and 30.

(e) You have a sufficiently large sample size or the population is Normal.

5. For a confidence interval estimator for the population mean μ, which one of the following

determines the probability that it includes μ?

(a) sample size √(d) confidence level | (b) tabular t value | (c) tabular z value |

(e) standard deviation of the population | ||

6. | If four confidence interval estimates for the population mean are constructed with 95% confidence from four independent samples, the probability that all four intervals contain the population mean is | |

closest to √(a) 0.815 | (d) 0.93 | (e) 0.95 |

(b) 0.857 | (c) 0.903 |

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7. For a normal population, which of these probabilities would be the largest?

I. II. | PP V X P V PP V X P V where n =10. |

III. | ¸ |

P P | P |

V |

· ¹

§¨©

n

X

n

V where n =100.

(a) I.

√(b) II.

(c) III.

(d) All three probabilities (I., II. and III.) are equal.

(e) There is not enough information to answer this question.

8. In estimating the percentage of people in Vancouver who are in favour of a policy, a random sample

of 625 voters is selected from Vancouver. A 95% confidence interval for the percentage in favour is

(0.58, 0.62). If a random sample of the same size is taken in Toronto, which has a much larger

population, the results will be

√(a) About the same accuracy as those for Vancouver.

(b) Substantially less accurate than those for Vancouver.

(c) Substantially more accurate than those for Vancouver.

(d) Can’t tell without further information.

(e) None of the above.

9. Suppose a population is positively skewed and has a population mean μ and a population variance

2

V . Which of the following statements about the sample mean X for a sample of n=20

observations is true?

(a) | 2.575 0.02 / 2.575 ¸ ¸ n X V P | P |

· ¹

§¨¨©

(b) 1.96 0.05

/

1.96 ¸ ¸

· ¹

§¨¨©

n

X

P

V

P

(c) 1.645 0.10

/

1.645 ¸ ¸

· ¹

§¨¨©

n

X

P

V

P

(d) All of the above

√(e) None of the above

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Questions 10–11: Suppose a 95% confidence interval for a population mean is (14.844, 27.156). The

sample size is 28.

10. What is the standard deviation of the sample?

(a) 0.107 (b) 3.000 (c) 6.144 √(d) 15.875 (e) 16.619

11. Which one of the following statements is true?

(a) 95% of the sample fall inside the interval (14.844, 27.156).

(b) 95% of the population fall inside the interval (14.844, 27.156).

(c) 95% of the times, the sample mean falls inside the interval (14.844, 27.156).

√(d) 95% of the times, the population mean falls inside the interval (14.844, 27.156).

(e) In testing H0 : P 25 vs Ha : P z 25 at significant level 0.05, we will reject H 0.

Questions 12–13: Bank Today, the newsletter by a trade association of bankers, reported on the results

achieved by Crown Bank in improving customer satisfaction by listening to the “voice of the

customer.” A key measure of customer satisfaction is the response on a scale from 1 to 10 to the

question, “Considering all business you do with Crown Bank, what is your overall satisfaction with

Crown Bank?” The historical proportion of Crown Bank customers who responded with a 9 or a 10 is

known to be 0.48. As a market analyst, you would like to test if the customer satisfaction of Crown

Bank improved as the report suggests. Suppose that you obtained a random sample of 350 current

customers from Crown Bank and you found 195 customers giving a response of 9 to 10.

12. What is your set of hypotheses corresponding to your research question?

√(a) H0: p=0.48 H1: p>0.48

(b) H0: p=0.48 H1: p≠0.48

(c) H0: p=0.48 H1: p<0.48

(d) H0: p=0.56 H1: p ≠ 0.56

(e) H0: p=0.56 | H1: p>0.56 | |

13. What is the p-value of your test for the hypotheses that you identified in question 12? | ||

(a) 0.0000 | √(b) 0.0019 (c) 0.4404 | (d) 4800 (e) 0.5571 |

Questions 14–15: A newspaper believes that it captures over 12% of the Toronto market. The

hypotheses to be tested are H0 : p 0.12 and Ha : p ! 0.12. A random sample of 400 shows that this

newspaper captures 14.5% of the Toronto market.

14. The value of the test statistic is closest to

(a) 0.82 | √(b) 1.54 | (c) 1.645 | (d) 1.96 | (e) 2.33 |

15. The p-value is closest to | ||||

(a) 0.01 | (b) 0.05 | √(c) 0.0618 (d) 0.1 | (e) 0.1512 |

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Questions 16–18: a manager of a company quotes that the average weekly sale per office is $25,000. A

random sample of 36 offices across the country shows an average weekly sale of $23,860 with a

standard deviation of $3,800.

16. The p-value to test the manager’s quote is closest to

(a) 0.0359 (b) 0.05 √(c) 0.0718 (d) 0.142 (e) 0.3637

17. For simplicity, we use “accept” to mean “do not reject” in this question. Let D be the significance

level. The manager’s quote is

(a) rejected at D = 0.05 but accepted at D = 0.10.

(b) rejected at D = 0.05 and also rejected at D = 0.10.

(c) accepted at D = 0.05 and also accepted at D = 0.10.

√(d) accepted at D = 0.05 but rejected at D = 0.10.

(e) none of the above.

18. After the manager has reviewed the sample results, he revises his quote that the weekly sale per

office is over $22,000. Based on the same sample data, the p-value to test the manager’s revised

quote is closest to

√(a) 0.0016 (b) 0.0032 (c) 0.0064 (d) 0.0128 (e) 0.0256

Part 2 Show your work in all questions.

x Write your answers clearly, completely and concisely in the designated space provided

immediately after each question.

x Show your reasoning. Your mark depends on your solution and your reasoning. No extra

space/pages are possible.

Page 6 of 9

19. (11 points)

Suppose 1 in 10 consumers favour cola brand A. After a promotional campaign in a given sales

region, 200 cola drinkers were randomly selected from consumers in the market area and were

interviewed to determine the effectiveness of the campaign. The result of the survey showed that a

total of 26 people expressed a preference for cola brand A. Do these data present sufficient

evidence to indicate an increase in the acceptance of brand A in the region?

[Hint: (i) Write down the null and alternative hypotheses using the notations in this course.

(ii) Perform the hypotheses testing using a significance level 0.05.

(iii) Draw your conclusion. Also write your conclusion in one or two sentences in the

context of the question.]

(11 points)

Solution: Let p be the percent of consumers who favour cola brand A. Let

p

be the sample percentage of

customers favouring cola brand.

The hypotheses to be test are H0 : p 0.1, Ha : p ! 0.1.

Method 1. For D 0.05 , the decision rule is to reject H 0 if Z t1.645 , do not reject H 0 if Z 1.645 ,

where

0.1 0.9 / 200

0.1

u

p

Z .

Test statistic is 0.13

200

26

p . Hence

0.1 0.9 / 200

0.13 0.1

0.1 0.9 / 200

0.1

u

pu

Z =1.414, which falls

in the acceptance region. Do not reject H 0 and conclude that the result of the survey does not

support a preference for cola brand A.

Method 2. ForD 0.05 , let c be the critical value. The decision rule is to reject H 0 if p t c

, do not

reject H 0 if p c

. Then D 0.05 = Preject H0 when H0 is true = ¸

· ¹

P¨ © § p t c when p 0.1

=

·¸¸¸¹

§¨¨¨©

0.1 c | t | p 0.1 | = ¸ u t P Z |

u u 0.1 0.9 / 200 0.1 0.9 / 200 |

P · ¹

§¨©

0.1 0.9 / 200

c 0.1

From the Z table, 1.645

0.1 0.9 / 200

0.1

cu

, 0.1349

200

0.1 0.9

c 0.11.645 u

Therefore the decision rule is to reject H 0 if t 0.1349

p

, do not reject H 0 if 0.1349

p

The test statistic is and 0.13

200

26

p , which falls in the acceptance region. Do not reject H 0 and

conclude that the result of the survey does not support a preference for cola brand A.

Method 3. ForD 0.05 , the decision rule is:

Reject H 0 if p-value d 0.05, do not reject H 0 if p-value > 0.05

Test statistic is 0.13 200 26 p , p-value = ¸ t P p 0.13 P | t 0.13 0.1 | 0.1 p |

u u 0.1 0.9 / 200 0.1 0.9 / 200 |

·¸¸¸¹

§¨¨¨©

· ¹

¨ © §

= PZ t1.414 |0.0793, which falls in the acceptance region. Do not reject H 0 and conclude that

the result of the survey does not support a preference for cola brand A.

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20. (10 points)

Scheer Industries is considering a new computer assisted program to train maintenance employees

to do machine repairs. In order to fully evaluate the program, the director of manufacturing

requested an estimate of population mean time required for maintenance employees to complete the

computer assisted training. Assume that the time for maintenance employees to complete the

computer assisted training follows approximately a normal distribution.

(a) Suppose a large random sample of size is selected. Based on this sample, a 80% confidence

interval for the population mean time required for maintenance employees to complete the computer

assisted training is calculated as 48.3, 54.7 .

Based on this same sample, find a 90% confidence interval for the population mean time required

for maintenance employees to complete the computer assisted training.

(5 points)

Solution: Let X be the sample mean and S be the sample standard deviation and n be the sample size.

Since the sample size is large, a confidence interval for the population mean time required for

maintenance employees to complete the computer assisted training can be calculated as

·¸¹

§¨©

r

n

S | X Z | |

D / 2 , where we have used the Z value instead of the t. Given that a 80% confidence interval for the mean μ is 48.3, 54.7 , we have | ||

S | X | —————- Equation (1) |

1.28 ¸ 48.3

· ¹

§¨©

n

1.28 ¸ 54.7

· ¹

§¨©

n

S

X —————- Equation (2)

Adding equations (1) and (2), we get X 51.5 . Substitute the value of X 51.5 into equation (2),

we have 51.5 1.28 ¸ 54.7 n S | , and 2.5 S . Therefore a 90% confidence interval for μ is |

· ¹

§¨©

n

·¸¹

§¨©

r

n

S

X 1.645 =51.5r1.6452.5 51.5r 4.1125

which is 47.3875, 55.6125

(b) Another random sample X1, X2,…, X6 is selected. Sample statistics show that 300

6

¦1i

X i and

6¦1

2 15120

i

X i . Based on this sample, find a 90% confidence interval for the population mean time

required for maintenance employees to complete the computer assisted training.

(5 points)

Solution: 50

6

1 300

n¦1i

X i

n

X ,

1 1 1 1 1 2 2 i X n X n X X n S | |||

24 6 300 15120 2 » | ¦ | ¦ | ¦ i |

15

1 | 1 |

50 2.0152 50 4.03 24 50 2.015 / 2 r r ¸ r r S X t | |

6 | n D |

2

1

2

º ¼

ª«¬

º»»¼

ª««¬

·¸¹

§¨©

n

i

n

i

n

i

ii

¸ ¸ · ¹

§¨¨©

· ¹

§¨©

Page 8 of 9

21. (10 points)

The credit department calculates that it would make a profit if the average balance in its accounts is

more than $1500. An accountant has previous information on the variation of the account balances,

he can assume that the standard deviation of account balances is $175. The accountant wishes to

test if the company makes a profit or not, using a 5% significance level.

(a) The accountant selects a random sample of 100 accounts. Find the probability of making a type II

error if in reality, the average credit balance is $1550.

(4 points)

Solution: H0 : P d 1500 , H1 : P ! 1500

For D =0.05, reject H0 if Z !1.645 and accept H0 if Z d1.645 .

That is to accept H0 if 1.645

175/ 100

1500

d

X

, or X d 1528.7875

For P 1550 , E = P(Type II error)=P(accept H0 when P 1550 ) = PX d 1528.7875 | P 1550

= ¸ ¸

· ¹

§¨¨©

d

175/ 100

1528.7875 1550

/ n

X

P

V

P

= PZ d 1.2121429 = 0.1131

(b) If the probability of making a type II error is 0.1 when the real average credit balance is $1550, what

is the required sample size?

(6 points)

Solution: Let n be the required sample size.

For D =0.05, reject H0 if Z !1.645 and accept H0 if Z d1.645 .

That is to accept H0 if 1.645

175/

1500

d

n

X

, or ¸

· ¹

§¨©

d

n

X 1500 1.645 175 .

Given E =0.1 at P 1550 .

0.1 = P(Type II error)=P(accept H0 when P 1550 ) = ¸ ¸

· ¹

P¨ ¨ © § X d 1500 1.645¨ © §175 n ¸ ¹ · | P 1550

= § d X P P | n n 175/ 1500 1.645175/ | 1550 |

n V / |

·¸¸¹

¨¨©

= ¸ ¸

· ¹

§¨¨©

·¸¸¹

§¨¨©

d

175

P Z 1.645 50 n =

·¸¸¹

§¨¨©

d

3.5

P Z 1.645 n

From the standard normal table, 1.28

3.5

1.645 n , n >3.51.645 8 @2 104.80641.

The answer is n = 105.

Page 9 of 9

22. (15 points)

Insurance companies track life expectancy information to assist in determining the cost of life

insurance policies. Last year the average life expectancy of all policyholders was 77 years. ABI

Insurance wants to determine if their clients now have a longer life expectancy, on average, so they

randomly sample some of their recently paid policies. The insurance company will only change

their premium structure if there is evidence that people who buy their policies are living longer than

before. The sample has 28 observations, a mean of 78.6 years, and a standard deviation of 4.48

years.

(a) What set of hypotheses does the ABI insurance wish to test?

(2 points)

Solution: H0:μ=77, H1: μ>77

(b) Conduct the test for the hypotheses you identified in question (a) by rejection region method. For

full marks, you should clearly state the rejection region, the test statistic, and the decision. Based on

the result, what will the insurance company do to its premium structure?

[Use the hint in question 19]

(4 points)

Solution: Since sample size is n=28, degrees of freedom for t statistic is 27. The critical value for α=0.05

for one sided test when degrees of freedom is 27 is 1.703. Thus rejection region is t !1.703 .

0.847 28 4.48 n S SE X | . The test statistic is 1.890 0.847 78.6 77 ( ) 0 SE X x t P |

.

Since t=1.890>1.703, we reject the null hypothesis. There is enough evidence to suggest that the life

expectancy of policy holders for ABI Insurance increased from 77 years. Thus, the company will

change its premium structure.

(c) Suppose the true mean life expectancy of policyholders is 80.18 years. Obtain the power of the test.

(5 points)

Solution: Given α=0.05, the critical value of the test in original unit is c=77+1.703×0.847=78.442.

Given the mean of the distribution under the alternative is 81, t statistic corresponding to the critical

value is | 2.053 0.847 |

78.442 80.18

t

Thus, the power of the test, the probability of rejecting the wrong null, is given by

P(t>-2.052)=1-0.025=0.975.

(d) Obtain the 0.99 confidence interval for the mean life expectancy of the policyholders and interpret

the result.

(4 points)

Solution: For ν=n-1=27, the critical value for α=0.005 is 2.771

x r 2.771u SE(X ) 78.6 r 2.771u0.847 (76.254,80.946)

With 0.99 confidence, the life expectancy of policy holders of ABI Insurance is at least 76.254 years

and at most 80.946 years.

End of Test 3