# 2014-02 Test1S.pdf

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University of Toronto Scarborough

Economics for Management Studies

MGEB12H3

Daga

Winter

Description

MGEB12H3S – L01, L30, L60
Quantitative Methods in Economics II
Test 1
Friday February 7, 2014
7 –9 pm
Victor Yu
Last Name (Print) __Solution____________________
First/Other Names
Student Number
Time allowed: Two (2) hours
Aids allowed: Any Calculator
One aid sheet (two 8.5”x11” pages) prepared by student.
This test consists of 22 questions in 10 pages including this cover page.
It is the student’s responsibility to hand in all pages of this test. Any missing
page will get zero mark.
Show your work in each question in Part II.
This test is worth 25% of your course grade.
Turn to the last page, enter your name and student number again.
The University of Toronto's Code of Behaviour on Academic Matters applies to all University of
Toronto Scarborough students. The Code prohibits all forms of academic dishonesty including, but
not limited to, cheating, plagiarism, and the use of unauthorized aids. Students violating the Code
may be subject to penalties up to and including suspension or expulsion from the University.
Management, 1265 Military Trail, Toronto, ON, M1C 1A4, Canada 1
www.utsc.utoronto.ca/mgmt Part I. Multiple Choice. 3 marks in each question. No part mark.
Circle only one answer. If there are more than one correct answer, circle the best one.
1. In a given hypothesis test, the null hypothesis is rejected at the 0.1 and 0.05 level of
significance, but is accepted at the 0.01 level. The most accurate statement about the p-value
for this test is:
(a) p-value = 0.01 √(b) 0.01 < p-value < 0.05 (c) 0.05 < p-value < 0.10
(d) p-value = 0.10 (e) p-value > 0.10
2. A random sample X ,X ,..., X is selected from a normal population with mean . Suppose
1 2 10
a 95% confidence interval for is X 3.39 , where X is the sample mean. To test
H 0 30 versus H a 30 at the 0.01 level of significance, the null hypothesisH 0 will
be accepted when
(a) 24.735 X 35.265 (b) 23.25 X 35.75 (c)28.65 X 31.35
√(d) 25.125 X 34.875 (e) 26.875 X 33.125
3. Are Japanese managers more motivated than American managers? A randomly selected group
of each were administered an Attitudes Test (AT), where higher AT scores indicate more
motivation. The AT scores from this sample are summarized as follows:
American Japanese
Sample Size 200 100
Mean AT score 65 70
Standard Deviation 20 10
To determine if the average AT score of American managers are significantly lower than the
average AT score of Japanese managers, the test statistic is closest to
(a) –6.9 (b) –4.75 √(c) –2.89 (d) 2.89 (e) 4.75
4. The mean life of a certain bulb is 1000 hours and its standard deviation is 100 hours. A new
brand of bulb has been introduced to the market. A sample of size n bulbs of the new brand
reveals a sample mean of 1010 and the standard deviation is still 100 hours. What is the
minimum n that would allow you to reject the hypothesis that the mean life of the new brand of
bulb is 1000 at the 5% level?
(a) 180 (b) 200 (c) 205 (d) 300 √(e) 400
5. Just prior to the provincial election, a poll found that 46% of 1200 registered voters said they
would vote for a candidate. To test the null hypothesis that the true proportion is 50%, the value
of the test statistic is closest to
(a) –1.464 (b) –1.96 √ (c) –2.771 (d) –2.78 (e) –3.471
p p0 0.46 0.5
Solution: Z 2.771
p 0 0n 0.50.5/1200
2 6. A random sample of 50 Honda Accords driven at 55 miles per hour revealed an average of 40.5
miles per gallon and a variance of 16 miles per gallon. Honda claims the Accord gets 42 miles
to the gallon on the highway. The test statistic for the null hypothesis that Honda’s claim is true
is closest to
(a) –1.645 (b) –1.84 (c) –2.0 (d) –2.45 √ (e) –2.65
X 40.5 42
Solution: t 0 2.65
S / n 16 / 50
Questions 7–8, your friend is about to toss two 6-sided dice that he claims are fair (equal probability of
each outcome). You suspect that the dice are loaded so that a six is more likely than the other
outcomes. You decide you will reject your friend's claim that the dice are fair if both dice turn
out to be 6.
7. What is the probability that you reject the claim given that it is true?
1 1 1 1 1
√(a) 36 (b) 8 (c)6 (d) 4 (e) 2
Solution: Probability that you reject the claim given that it is true
=Probability that both dice turn out to be 6 given that his claim is true
=Probability that both dice turn out to be 6 given that the both dice are fair
=Probability that both dice turn out to be 6 given that each die has 1/6 probability of
getting 6
= 1/6 * 1/6 = 1/36
8. Now suppose that the claim is false and that the probability of a 6 isfor each die. What is
2
the probability that you reject his claim?
1 1 1 1 1
(a) (b) (c) √ (d) (e)
36 8 6 4 2
Solution:
Probability that you reject his claim given that his claim is false
= Probability that both dice turn out to be 6 given that his claim is false
= Probability that both dice turn out to be 6 given that the probability of a 6 is ½ for each die
= ½ * ½ =1/4
Questions 9–10, a random sample of size 16 is drawn from a normal population having mean and
standard deviation =6 for the purpose of testing H 0 30 versus H a 30 . The
experimenter decides to accept H if the sample mean is between 27 and 33, otherwise reject
0
H0 .
9. The significance level in this test is closest to
(a) 0.01 (b) 0.0195 (c) 0.0275 (d) 0.0316 √ (e) 0.0456
Solution: P X 33 P X 27 P Z 2 P Z 2 0.0456
10. If the actual population mean is 34, the power of the test is closest to
(a) 0.2486 (b) 0.2514 √(c) 0.7486 (d) 0.7514 (e) 0.9544
Solution: P 27 X 33 | 34 P 4.67 Z 0.67 0.2514
3 Questions 11-13, an auto insurance company wishes to estimate p, the percentage of policy holders
who file claims within a year. The company also wishes to test if over 8% of their policy
holders file claims each year, i.e., the hypotheses to be testH0: p = 0.08 andH a p > 0.08.
The company will select a large random sample and let p be the percent of policy
holders in the sample, who file claims within a year.
11. Suppose a 95% confidence interval for the percent of policy-holders filing claims is (0.0853,
0.1147). The null hypothesis H0will be rejected at a significance level
√ (a) 0.0016 (b) 0.01 (c) 0.05 (d) 0.1 (e) 0.1375
12. With a sample of 400, the null hypothesiH 0will be rejected at a significance level 0.05 if
(a) p > 0.08 √(b) p > 0.1023 (c) p > 0.12
(d) p > 0.1522 (e) p > 0.2
13. Suppose the p-value calculated from a sample of 400 policy holders is 0.015. The number of
policy holders who filed a claim in this sample is closest to
(a) 40 √(b) 44 (c) 48 (d) 52 (e) 56
Questions 14–17, a random sample X 1X 2..., X9 is selected from a population that is normally
distributed with mean and standard deviation . Let X be the sample mean and S 2 be the
sample variance. The hypothesis to be tested iH 0 5 verses H a 5 at a significance
level .
14. If is known to be 3 and =0.05, H0 will be rejected when
(a) X 1.645 (b) X 1.86 (c) X 1.96
√(d) X 6.645 (e) X 6.86
15. If is known to be 3 and H 0 is rejected whenX 6.5 , the value of is closest to
(a) 0.01 (b) 0.0275 (c) 0.05 √(d) 0.0668 (e) 0.1
16. If is known to be 3 and the p-value of the test is 0.025, the sample mXanis closest to
(a) 5.306 √(b) 6.96 (c) 7.306 (d) 8.122 (e) 8.96
17. If is unknown but its sample estimate is S = 3, at a significance level=0.05, H 0 will be
rejected when
(a) X 1.645 (b) X 1.86 (c) X 1.96
(d) X 6.645 √(e) X 6.86
4 Part II Show your work in each question.
18. (9 marks)
A special battery must have a life of at least 400 hours. Previous experience indicates that the
standard deviation of the battery life is 30 hours. A hypothesis test H 0 400 versus
H : 400 is to be conducted using a 0.02 level of significance, whee is the average life
1
of this type of battery. If the batteries from a particular production run have an actual life of
385 hours, the production manager would like a sampling procedure that erroneously concludes
that the batch is acceptable only 10% of the time. What sample size is recommended for the
hypothesis test?
(9 marks)
Solution: For =0.02, rejectH if Z < -2.05, accepH if Z -2.05.
0 0
X 400 30
That is, acceptH 0 if 2.05 , or X 400 2.05 .
30/ n n
Given that =P(Type II error at =385)=P(accept H 0 when =385)=0.1
P X 400 2.05 30 when 385 0.1
n
400 2.05 30/ n 385
P 0.1
30/ n
P Z 0.5 n 2.05 0.1
0.5 n 2.05 1.28
2
Solving for n, we have n 1.28 2.05 6.66 44.355 .
0.5
Since n must be an integer, the

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