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Economics for Management Studies

MGEB12H3

Daga

Winter

Description

ECMB12H3 Lec30
Quantitative Methods in Economics II
Division of Management
University of Toronto at Scarborough
January – May 2005
Dr. Yu
Test 1
Date: Tuesday, February 2, 2005
Time allowed: Two (2) hours
Aidsallowed: Calculator and one aid sheet (two 8.5”x11” pages) prepared by
student.
Notes:
• This test consists of 20 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.
• Statistical tables are provided separately. Do not hand in the tables. Only hand in
your test papers.
• Show your work in each question in Part II.
• This test is worth 30% of your course grade.
Print Last Name: Solution
Given Name(s):
Student Number:
Do not write on the space below, for markers only
Page Question Total Marks
2-5 1-15 45
5 16 10
6 17 15
7 18 10
8 19 10
9 20 10
Total 100
1 Part I. Multiple Choice. 3 points in each question. No part mark.
Circle only one answer. If there are more than one correct answer, circle the best one.
1. The null hypothesis H 0:µ = 10 is tested against H a µ ≠10 . Suppose H 0s
rejected at the significance levelα =0.05. Which one of the following statements
is true?
(a) H 0ust be rejected at α = 0.01 level.
(b) H 0ust be accepted at α =0.01 level.
√ (c) The p-value must be less than 0.05.
(d) The 95% confidence interval must contain the value 10.
(e) The power of the test is always equal to 0.95.
2. In a hypothesis test H : p = 0. versus H : p > 0.2 , where p the percentage of
0 a
voters in favour of a policy. A large random sample shows that 25% of voters in
the sample are in favour. If the p-value is calculated as 0.05, the number of voters
in this sample is closest to
64(a) √ (b) 174 (c) 246 (d) 363 (e) 500
3. The results of a recent poll on the preference of shoppers regarding two products
are shown below.
ShoppersShoFppverring
Product Surveye Pdrhiuct
A 805060
B 906012
The 95% confidence interval estimate for the difference between the percentages
of shoppers favoring the products is closest to
√ (a) -0.024 to 0.064 (b) 0.6 to 0.7 (c) 0.024 to 0.7
(d) 0.02 to 0.3 (e) -0.04 to 0.013
4 Let X 1X 2e a random sample of size 2 from a normal distribution with mean
7 and variance 8. The probability P − 2 < X 1 X <22 is closest to
(a) 0.0688 (b) 0.1915 (c) 0.25 (d) 0.32 √ (e) 0.383
2 5. Let X be the mean of a random sample of size 4, from a normal distribution
with mean 10 and variance 9. Let Y be the mean of a random sample of size 4,
from a normal distribution with mean 3 and variance 4. Suppose X and Y are
independent. The probability P X > 2Y )is closest to
(a) 0.175 (b) 0.25 (c) 0.45 (d) 0.75 √ (e) 0.945
6. A normal population has an unknown mean µ and unknown variance σ 2 . A
random sample X ,X ,...,X selected from the population gives a sample
1 2 6
variance of 24. Denote the sample mean by X . To test the null hypothesis
H :µ = 8 versus H : µ < 8 using α =0.05, the null hypothesis will be rejected
0 a
if
(a) X ≤ 4.(b) X ≥ 4.71 √ (c) X ≤ 3.97
(d) X ≥ 3.97 (e) none of these
7. The appropriate null and alternative hypotheses for the test are:
(a) 36H : µ = $ , 480: µ < $ √ (b) H : µ = $480, H480: µ < $
0 1 0 1
(c) 48H 0 µ = $ , 481: µ > $ (d) H 0:0 = $ , 361: µ = $
(e) 48H 0 µ < $ , 481: µ ≥ $
Questions 8-9. Use the following information.
Some people who bought X-Game gaming systems complained about having
received defective systems. The industry standard for such systems has been
ninety-eight percent non-defective systems. To test if the percentage of defective
systems produced by X-Game has exceeded the industry standard, a random
sample of 400 units is selected.
8. Assume a significance level of 0.05, we will conclude that the percentage of
defective systems produced by X-Game has exceeded the industry standard if the
number of defective items in the sample exceeds
(a) 7 (b)0 √ (c) 13 (d) 18 (e) 21
9. In this sample of 400 units, 10 units were defective. Thep-value is closest to
0.1(a)7 √ (b) 0.2389 (c) 0.2611 (d) 0.3143 (e) 0.3791
3 Questions 10 to 13. Use the information in following information.
Suppose you are interested in conducting the statistical test ofH 10µ =
0
against 1a0:µ > where µ is the mean of a population. A random sample of
100 observations is selected from this population and the sample mean is denoted
by X . Assume that you know from previous experiments that the population
standard deviation is 80.
10. Suppose the decision rule is to rejectH if X > 108, then the probability of
0
making a Type I error is closest to
(a) 0.05 (b) 0.10 √ (c) 0.1587 (d) 0.2340 (e) 0.3413
11. Suppose the decision rule iα =0.05. Which one of the following statements is
true?
√ (a) Reject 0 ifX > 113.16 (b) Reject 0 if X < 113.16
(c) Accept H0if X > 86.64 (d) Accept H0if X < 86.64
(e) None of the above is true.
12. Suppose the sample mean is X =110. The p-value of the test is closest to
(a) 0.05 √ (b) 0.1056 (c) 0.2112 (d) 0.3944 (e) 0.45
13. For α =0.05, the power of the testµa=110 is closest to
0.1(a)4 √ (b) 0.3446 (c) 0.395 (d) 0.6554 (e) 0.95
Questions 14-15. Use the following information.
Salary information regarding male and female employees of a large company is
shown below.
e l a m e F e l a M
646 Sa mizele
Sample Mean Salary (in $1,000) 44 41
Populaaiian1e282
14. The 95% confidence interval for the difference between the means of the two
populations is
(a) 0 to 6.92 (b) -2 to 2 (c) -1.96 to 1.96
√ (d) -0.92 to 6.92 (e) –2.12 to 5.23
4 15. We will conclude that average salary of males is significantly greater than females
if the significance level is closest to
√ (a) 0.0668 (b) 0.0334 (c) 1.336 (d) 1.96 (e) 1.645
Part 2. Show your work in all questions.
16. 10arks)
Suppaotse X1,X2,...,n form a random sample from a normal distribution for
2
which the mean µ is unknown and the variance isσ =1 . It is desired to test
H 0:µ ≤ 0 against H a µ >0 . Suppose the probability of a type II error is 0.05
when µ =1. If the sample size is 16, find the significance level.
(10 marks)
Solution: LetX be the sample mean and c be the critical value.
teTto 0 H 0:µ ≤ versus H a µ > 0 ; the decision rule is to rejectH 0f
X ≥ c and to accept H if X < c .
0
β =P(type II error)=P(accept 0 when H i0 false)= P X < c when µ > 0 ).
Given that whenµ =1, β =0.05.
=H0.nce P X < c when µ =1 = P X −1 < c − = P Z]< 4(c− 1 .
1/ 16 1/ 16
From the standard normal table, 4(c-1) = -1.645, and c = 0.58875
α =significance level=P(reject0H when H 0s true)=P X ≥ c when µ = 0 )
X − 0 c −0
= P ≥ = P Z ≥ 4c = P Z ≥2.355 = 0.0091
1/ 16 1/ 16
5 17. (15ints)
When people are not under any stress, their systolic blood pressures have an
average of 120 mmHg with a standard deviation of 12 mmHg. As a class research
project, Rosana wants to see whether the stress of final exams increases the blood
pressures of first year students. She takes a random sample of 36 first year
students during the final exam period. She finds that the average systolic blood
pressure in the sample is 125.2 mmHg.
(a) Test if the stress of final exams increases the blood pressures on the first year
students. State the null and alternative hypotheses, the decision rule, the test

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