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August 2012 EXAMINATIONS
Solution – Part I
(1) In a random sample of 600 eligible voters, the probability that less than 38% will be in
favour of this policy is closest to (B)
(2) In a large random sample, the probability that less than 42% are in favour of this policy is
0.67. The sample size is closest to (A)
(3) The 90th percentile of daily sales is closest to
(D)
(4) In the next 4 days, the probability that their average daily sales exceed $600 is closest to
(A)
(5) In the next 4 days, the probability that the daily dales exceed $500 in only one of these
days is closest to (E)
(6) If Line 1 and Line 2 are independent, the probability that Line 1 produces more parts than
Line 2 in any single day is closest to (D)
(7) If Line 1 and Line 2 are independent, what is the probability that the average number of
parts produced by Line 1 is greater than that produced by Line 2 in the next 5 days? (E)
(8) This firm specifies that the estimation of this proportion has a margin of error 0.05 with 90%
confidence. The smallest sample size required is closest to (B)
(9)
The smallest sample size required is closest to
(A)
(10)
A 90% confidence interval for the real proportion is closest to
(11)
What is the standard deviation of your sample?
(12)
Which one of the following statements is true?
(13)
How will it change your OLS estimate for the slope of the regression line, ǃ1? (B)
(14)
How do you interpret the slope estimate for x3?
(15)
Which one of these variables will cause perfect multicollinearity?
(D)
(D)
(B)
(C)
(E)
(16) 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)
(17) . If you do not find out about his systematic mistakes, what consequences will they have
on the results of your tests? (B)
(18)
(19)
What kind of data is it? (D)
What is your set of hypotheses corresponding to your research question?
(A)
(20) What is the P-value of your test for the hypotheses that you identified in question (19)?
(C)
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UNIVERSITY OF TORONTO
Faculty of Arts and Science
August 2012 EXAMINATIONS
ECO220Y1Y
Duration - 3 hours
Examination Aids: Calculator
Solution
Part II: Short Answer Questions [60 points]
(21) [12 points] You are hired as a consultant by the marketing department of Crown Bank
and asked to analyze the data of customer satisfaction survey. 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?” If the response is 9 or 10, the customer is considered “delighted” by Crown
Bank. The department wants to know if customers are more likely to be “delighted” in the
areas with more Crown Bank ATMs. They obtained random samples from two areas that
have the same area, but vary in ATM density (number of ATMs per capita). The following
table shows the result.
Area 1
Area 2
ATM density (per km2)
10
3
Total responses
175
175
Responses with 9 or 10
121
105
(a) [4 points] What is the set of hypotheses that the marketing department wants to test? [A
set of hypotheses]
(Solution)
H0: p1-p2=0
H1: p1-p2>0
(b) [8 points] Conduct the test for the hypothesis you identified in question (a) by the P-value
method. Use the significance level Į=0.05. Write a short report to the marketing
department about the result. For full marks, you should clearly state the test statistic, the
P-value, and the decision. [Analysis, 3 items & 3 or 4 sentences]
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(Solution)
Since‫݌‬
ෝ ଵ ‫݊ כ‬ଵ ൌ ͳʹͳ ൐ ͳͲǡ ሺͳ െ ‫݌‬Ƹଵ ሻ ‫݊ כ‬ଵ ൌ ͷͶ ൐ ͳͲǡ ‫݌‬Ƹ ଶ ‫݊ כ‬ଶ ൌ ͳͲͷ ൐ ͳͲǡ ሺͳ െ ‫݌‬Ƹ ଶ ሻ ‫݊ כ‬ଶ ൌ
͹Ͳ ൐ ͳͲ, it satisfies the success/failure conditions. Thus, we can use normal
approximation for the distribution of difference in population proportions.
Test statistic:
z
( pˆ 1 pˆ 2 )
§1
1 ·
pˆ (1 pˆ ) u ¨¨ ¸¸
© n1 n 2 ¹
1.788 , where pˆ
121 105
175 175
0.646
P-value:
P ( Z ! 1.788) 1 0.9633
0.0367
Decision:
Since the P-value is 0.037, less than 0.05, we reject the null hypothesis. There is
sufficient evidence to suggest that customers are more likely to be “delighted” in the
areas with more Crown Bank ATMs.
(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) [2 points] What set of hypotheses does the ABI insurance wish to test? [A set of
hypotheses]
(Solution)
H0:ȝ=77, H1: ȝ>77
(b) [4 points] 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? [Analysis, 3 items & 2-3 sentences]
Page 3 of 16
(Solution)
Since sample size is n=28, degrees of freedom for t statistic is 27. The critical value for
Į=0.05 for one sided test when degrees of freedom is 27 is 1.703. Thus rejection region
is t ! 1.703 .
x P0
SE ( x )
The test statistic is t
78.6 77
4.48 / 28
78.6 77
0.847
1.890 .
Since t=1.890>1.703, we reject the null hypothesis. There is sufficient 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) [5 points] Suppose the true mean life expectancy of policyholders is 80.18 years.
Obtain the power of the test. [Analysis, one value]
(Solution)
Given Į=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 81.18, t statistic
corresponding to the critical value is
78.442 80.18
t
2.053
0.847
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.
Hypothesis Test D=.05 (H0:P=P0,HA:P>P0)
ࢼ
ࢻ
77
78.4
80.18
Page 4 of 16
(d) [4 points] Obtain the 0.99 confidence interval for the mean life expectancy of the
policyholders and interpret the result. [Analysis, a set of values & 1-2 sentences]
(Solution)
For ɋ=n-1=27, the critical value for Į=0.005 is 2.771
x r 2.771u SE( X )
78.6 r 2.771u 0.847 (76.254,80.946)
With 0.99 confidence, the mean life expectancy of policy holders of ABI Insurance is at
least 76.254 years and at most 80.946 years.
(23) [18 points] A researcher would like to know if productivity of factory workers changes
by better lighting in the room. In order to investigate this question, he collected data from
a factory. He randomly chose 17 workers and sent them to work in room 1. He randomly
chose another set of 17 workers and sent them to work in room 2. Then he set the lighting
of room 1 at the regular level and the lighting of room 2 to be brighter. Other than the
lighting, work conditions in the two rooms were identical. He collected data on daily
productivity of each worker in the two rooms.
The theoretical model to be estimated is as follows:
productivityi=ȕ0+ȕ1room2i+ ȕ2agei + ȕ4experiencei+İ
Where
productivityi=number of production by worker i on that day
room2i=1 if worker i is in room 2, 0 if worker i is in room 1
agei=age of worker i
experiencei= years of experience of worker i at the factory
The regression result is given as follows.
productivityi=-4.57+5.94 room2i+ 1.87 agei + 0.79 experiencei+İ
(9.68) (1.63)
(0.51)
(0.50)
n=34, R2=0.8584
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(a) [4 points] What is the set of hypotheses that the researcher would like to test? [A set of
hypotheses]
(Solution)
H0: ȕ1=0, H1: ȕ10
(b) [4 points] Conduct the test you stated in (a) by the rejection region method. For full
marks, you have to clearly state the rejection region, the test statistic, and the decision.
Based on the result of the test, report and interpret the result of the research.
[Analysis,3 items, 2-3 sentences]
(Solution)
Given significance level ߙ ൌ ͲǤͲͷǡ and degrees of freedom ߥ ൌ ͵Ͷ െ Ͷ ൌ ͵Ͳǡ
corresponding critical value is 2.042. Thus, the rejection region is: t>2.042, t<-2.042.
Based on the regression result, the test statistic is
௕
ఱǤవర
ୀ
భ ሻ భǤలయ
‫ ݐ‬ൌ ௌாሺ௕భ
ൌ ͵Ǥ͸ͶͶ
Since t=3.644> 2.042, it is in the rejection region. Thus we reject the null hypothesis in
favor of the alternative. There is sufficient evidence to suggest that ȕ1 is statistically
significantly zero. This estimate suggests that the lighting in a room changes
productivity of workers.
(c) [4 points] Conduct the test of overall significance for this model by the rejection region
method. Use the significance level Į=0.05.For full marks, you have to clearly state the
rejection region, the test statistic, and the decision.[Analysis, 3 items & 1-2 sentences]
(Solution)
With significance level Į=0.05 and degrees of freedom, ߥଵ ൌ ͵ǡ ߥଶ ൌ ͵Ͳǡ the rejection
region is F>2.92.
The F statistics is obtained as follows.
F
R2 / k
(1 R 2 ) / n k 1
60.62
Since F=60.62>2.92, we reject the null hypothesis in favor of the alternative. There is
enough evidence to suggest that at least one beta is statistically significantly different
from zero.
Page 6 of 16
(d) [6 points] Suppose that the researcher lets each worker choose whether to work in the
room with the brighter lighting or the one with the regular lighting. Then, which
assumptions, if any, of the multiple regression model will be violated? Can the
coefficient estimates obtained from this sample be reliable? Explain. [4-5 sentences]
The exogeneity assumption of x is violated. (i.e. E(xjǆi)=0 for all i and j)
If workers choose their room to work in, it is likely to create endogeneity. For example,
workers who care about producing more, may tend to choose brighter room. It means
workers morale may be lurking variables and positively correlate with both xi and yi. In
this case, the coefficient estimate for room1 is biased upward and is not reliable as an
estimate of the effect of brighter lighting to the productivity.
(24) [15 points] A sales manager is interested in determining if there is a relationship
between college GPA and sales performance among salespeople hired within the last
year. He selected a sample of recently hired salespeople and recorded the number of
units each salesperson sold in the last month. Variables obtained were:
IDi= identification number of salesperson i,
unitssold i=the number of units sold last month by salesperson i,
GPAi= college GPA of salesperson i.
The mean of unitssold i was 22.4 units and the mean of GPAi was 3.09.The table below
shows the regression result.
.reg unitssold GPA
Source |
SS
df
MS
-------------+-----------------------------Model | 123.041121
1 123.041121
Residual | 34.5588792
13 2.65837532
-------------+-----------------------------Total |
157.6
14 11.2571429
Number of obs
F( 1,
13)
Prob > F
R-squared
Adj R-squared
Root MSE
=
=
=
=
=
=
15
46.28
0.0000
0.7807
0.7638
1.6305
-----------------------------------------------------------------------------unitssold |
Coef.
Std. Err.
t
P>|t|
[95% Conf. Interval]
-------------+---------------------------------------------------------------GPA |
7.396965
1.087268
6.80
0.000
5.048066
9.745865
_cons | -.4270344
3.381615
-0.13
0.901
-7.73257
6.878501
-----------------------------------------------------------------------------Mean of GPA=3.09
(a) [4 points]Write down the theoretical model that is being estimated. [An equation]
(Solution)
unitssold i=ǃ0+ǃ1 GPAi+ǆi
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(b) [5 points] Interpret the coefficient estimate for GPAi. [Interpretation, 1 sentence]
(Solution)
An increase in GPA by 1 point is associated with an increase of the number of units sold
last month by a salesperson by 7.4 units on average.
(c) [6 points] Obtain the 0.9 prediction band of sales performance for a salesperson with
GPA of 3.0. [Analysis & a pair of values]
(Solution)
yˆQ r tD / 2 u SE 2 (b1 ) u ( xQ x ) 2 s e2
se2
n
.427 7.39 * 3.0 r 1.771u (1.087) 2 (3 3.09) 2 (1.631) 2
(1.631) 2
15
(18.78,24.75)