ECON/MGMT 201: Applied Statistics
Solution Final Exam Practice Problems
1. You wish to test whether the reliability of your company’s product differs from the reliability of a
competitor’s product. You have collected samples and found the following:
Your Product Competitor’s Product
Sample Size
1200
1000
Number of Failures
83
57
Assume a 5% level of confidence. What is your conclusion? What is the 95% confidence interval?
Answer:
H0: p1-p2 = 0 ; Ha: p1-p2  0
 = 5% ; test stat: z because 1200>30 and 1000>30  CLT
p1 =83/1200 = 0.0692 ; p 2 =57/1000 = 0.057
p1 1  p1  p2 1 p2 


n1
n2
0.0692  0.057
z
 1.174
0.0104
s p1  p2 
0.06921  0.0692 0.0571  0.057 

 0.0104
1200
1000
rejection rule: z > 1.96 or z<-1.96  fail to reject the null. There is insufficient evidence to
conclude that the reliabilities are different.
2. You have been asked to advise a consumer electronics company concerning its marketing strategy.
The company is considering advertisements in several different magazines. The per-month cost of
advertising is $12,000 for Time, $16,000 for Newsweek, and $20,000 for Reader’s Digest. You have
collected historical data on the response to advertisements in those magazines and found the
following:
Time
Newsweek
Reader’s Digest
Month
Revenues
Revenues
Revenues
January 2000
$53,389
$102,376
$12,349
February 2000
$121,689
$124,574
$52,348
March 2000
$15,999
$112,553
$47,439
April 2000
$120,738
$59,668
$109,483
May 2000
$115,292
$45,087
$125,610
June 2000
$24,508
$101,590
$137,062
July 2000
$97,345
$69,587
$75,944
August 2000
$12,250
$28,977
$159,637
September 2000
$129,029
$26,858
$109,519
October 2000
$13,681
$39,778
$111,394
November 2000
$47,491
$28,574
$177,211
December 2000
$54,971
$101,763
$18,492
You are interested in whether advertising is more effective in one of the magazines than the others.
How might you address the problem? Which magazine (if any) is more effective than the others?
You may assume a 5% level of confidence.
Answer: Define the profit to be the revenues less the cost.
sample mean
Time
$67,199
Newsweek
$70,115
Reader’s Digest
$94,707
standard deviation
$46,648
$36,612
$53,479
We only know how to compare two populations at a time, so consider the three possible pairs.
Time vs. Newsweek
H0: T - N = 0
Ha: T - N  0
: 5%
test stat: t (n<30)
s: 41932
s x1  x2 : 17880
Newsweek vs. Reader’s Digest
H0: N - R = 0
Ha: N - R  0
: 5%
test stat: t (n<30)
s: 45828
s x1  x2 : 19541
Time vs. Reader’s Digest
H0: T - R = 0
Ha: T - R  0
: 5%
test stat: t (n<30)
s: 50180
s x1  x2 : 21397
t: -0.163
t: -1.26
t: -1.29
None of the differences are significant, so it does not appear that advertising in one magazine is
significantly more effective than in the others.
3. Historically, one professor has failed 6% of the students in an introductory class while another
professor has failed 9% in the same class. 312 have taken the class from the first professor while 220
have taken it from the second professor. Is one professor more difficult (in terms of achieving a
passing grade) than the other?
Answer:
H0: p1-p2 = 0 ; Ha: p1-p2  0
 = 5% ; test stat: z because 312>30 and 220>30  CLT
p1 1  p1  p2 1  p 2 


n1
n2
0.06  0.09
z
  1.28
0.0235
s p1  p2 
0.061  0.06 0.091  0.09

 0.0235
312
220
rejection rule: z > 1.96 or z<-1.96  fail to reject the null. There is insufficient evidence to
conclude that the professors differ in difficulty.
4. New graduates from W&L earn an average of $40,000 per year with a standard deviation of $10,000.
New graduates from Stanford earn an average of $45,000 per year with a standard deviation of
$12,000. The samples include 100 people from each university. Do Stanford graduates earn more
than W&L graduates on average?
Answer:
H0: S-W  0 ; Ha: S-W > 0
 = 5% ; test stat: z because 100>30 and 100>30  CLT
s x1  x2 
z
s12 s 22
10000 2 12000 2



 1562
n1 n2
100
100
45000  40000
 3.2 , which is significant at the 5% (and even the 1% level). We
1562
therefore reject the null (i.e., Stanford grads earn more).
5. Suppose the average life expectancy is 74 years. A sample of 40 vegetarians found that the average
age at death was 78 with a standard deviation of 15. Do vegetarians live longer than average? What
is the 95% confidence limit? What is the 80% confidence limit? Sketch the power curve for the test
by plotting at least three points.
Answer:
H0:   74 ; Ha:  > 74
=5% (initially)
test stat: z, because 40>30  CLT
sx 
15
40
 2.37
We reject the null if x > 74+1.6452.37 = 77.9.
95% confidence limit = 78 – 1.6452.37 = 74.1 (i.e., we are 95% sure that the population mean is
greater than 74.1.
80% confidence limit = 78 – 0.842.37 = 76.0 (i.e., we are 80% sure that the population mean is
greater than 76.0.
Assuming the 95% confidence level and arbitrarily choosing =75, =80, and =85…..
77.9  75
 1.22   = 0.8888 and power = 0.1112
2.37
77.9  80
  0.89   = 0.5-0.3133 = 0.1867 and power = 0.8133
  Px  77.9   80 ; z 
2.37
77.9  85
  2.99   = 0.5-0.4986 = 0.0014 and power = 0.9986
  Px  77.9   85 ; z 
2.37
  Px  77.9   75 ; z 
We can then plot the power vs.  to get the power curve.
power
1
74
80
population mean
Note that we could do the power curve for the 80% confidence level (or any other level for that
matter).
6. On average, a company spends $32,500 per year to maintain its buildings. This includes, heating and
electricity, so the figure can be quite volatile. The historical standard deviation is $8,000 based on a
sample size of six. A recent news article suggested that companies of this type spend $25,000 per
year on average. Should the company be alarmed?
Answer:
H0:   25000 ; Ha:  > 25000
Suppose =5%
test stat: t, because 6<30
8000
sx 
 3266
6
32500  25000
t
 2.30
3266
The t table shows a t-stat of 2.015 for a 5% tail and 5 degrees of freedom. We therefore
reject the null. The company should be concerned.
7. A company wants to test the reliability of a new component. Assuming that the proportion of defects
is likely to be about 0.04, what sample size should be chosen so that the margin of error (given a 95%
confidence level) is 0.0005?
Answer:
We want z  s p  0.0005 . s p 
1.96 
0.041  0.04
and z=1.96 (assuming a two-tailed test), so
n
0.041  0.04
 0.0005  n=590,070.
n
8. BRIEFLY discuss the similarities between hypothesis testing and detecting outliers.
Answer: The techniques used to identify outliers are quite similar to those used to test
hypotheses. The basic intuition behind hypothesis testing is that we examine a sample and then
ask whether the null hypothesis would be an outlier given the distribution implied by the
sample.
9. You are interested in evaluating a product for possible sale in your store. The company has
sent 100 randomly-chosen samples for you to test. You plan to go through with the deal as
long as you believe no more than 5% of the products you subsequently purchase will be
defective. You believe a 90% confidence level is appropriate for the test. What test statistic
is appropriate? Specify the null and alternative hypotheses for the test. Sketch the power
curve for the test.
Answer: test stat = z because n=100>30 (CLT). H0:   5% ; Ha:  > 5%.
=10% corresponds to z=1.28. s p 
0.05  0.95
 0.02179 , so we reject if
100
p  0.05  1.28  0.2179  0.0779 . Let  = {6%, 7%, 8%}.
0.0779  0.06
 0.82  area=0.2939  =0.794 and power = 0.206
0.02179
0.0779  0.07
 0.36  area=0.1406  =0.641 and power = 0.359
=7%: z 
0.02179
0.0779  0.08
 0  0.10  area=0.0398  =0.460 and power = 0.540.
=8%: z 
0.02179
=6%: z 
power
(
1
0.5
2%
4%
6%
8%
10%

10. Two professors teach the same course. The professors give the same final exam to both groups of
students. Last term, professor A’s students averaged 75 on the final with a standard deviation of 12.
Meanwhile, professor B’s students averaged 80 on the final with a standard deviation of 10. There
were 16 students in each class. Assuming that teaching quality can be measured by student
performance, is professor B a better teacher than professor A? Use a 10% level of significance in
addressing the issue. Answer the question by calculating the p-value. Be sure to briefly state the
logic behind the rejection decision.
Answer: H0: B - A  0 ; Ha: B - A > 0. =10%, test stat=t because 16 < 30.
s
15  12 2  15  10 2
5
1 1
1.28 .
 11.045 ; s x A  xB  11.045 

 3.905 . t 
3.905
30
16 16
From the t table, we see that the p-value is slightly greater than 10%, so we do not reject the
null. There is insufficient evidence to conclude that B is a better teacher.
Answer the question by establishing a 90% confidence limit. Be sure to briefly state the logic behind
the rejection decision.
Answer: 90% confidence limit = 0 + 1.313.905 = 5.116. Since 5 < 5.116, we do not reject the
null. Note that 1.31 comes from the t table with a 10% tail and 30 degrees of freedom.
Answer the question by establishing a cutoff for the z-score or t-score (whichever is appropriate). Be
sure to briefly state the logic behind the rejection decision.
Answer: The rejection rule is t>1.31. Since 1.28<1.31, we do not reject the null.
11. The city council is planning a community-wide picnic. The council wants to estimate the proportion
of residents who will attend the picnic. To do so, the council will randomly select residents and make
phone calls. What is the margin of error if the council phones 30 people?
Answer: Recall that when we do not know  or have a reasonable estimate of it, we use p=0.5 to
get the most conservative estimate of the standard error of the sample proportion. Assume
=5%. s p 
0.5  1  0.5
 0.0913 . Margin of error = z  s p = 1.960.0913 = 0.179.
30
If the desired margin of error is 4% at the 95% confidence level, what sample size should the council
use?
Answer: 0.04  1.96
601 people.
0.5  1  0.5
. Solving for n gives 600.25. So, the council should call
n