Quantitative Methods in Economics

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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?

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

· 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?

· 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

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

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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
šp
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

šp
Z .
Test statistic is 0.13
200
š 26
p . Hence
0.1 0.9 / 200
0.13 0.1
0.1 0.9 / 200
0.1
u

špu
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
š
, do not
reject H 0 if p  c
š
. Then D 0.05 = Preject H0 when H0 is true = ¸
· ¹
P¨ © § pš t c when p 0.1
=
·¸¸¸¹
§¨¨¨©

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
šp
, do not reject H 0 if  0.1349
šp
The test statistic is and 0.13
200
š 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

·¸¸¸¹
§¨¨¨©
· ¹
¨ © § š
= 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.
Page 7 of 9
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

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),

· 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 ,

15

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

·¸¸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 .


.
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

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

Last Updated on January 19, 2018