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CH7
7.3
For each of the following change, indicate whether a confidence interval for  will
become longer or shorter:
a. An increase in the level of confidence
b. An increase in the sample size.
c. A decrease in the level of confidence.
d. A decrease in the sample size.
7.11
In an article in Accounting and Business Research, Carslaw and Kaplan investigate
factors that influence “audit delay” for firms in New Zealand. Audit delay, which is
defined to be the length of time (in days) from a company’s financial year-end to the
date of the auditor’s report, has been found to affect the market reaction to the report.
This is because late reports seem to often be associated with lower returns and early
reports seem to often be associated with higher returns.
Carslaw and Kaplan investigated audit delay for two kinds of public
companies—owner-controlled and manager-controlled companies. Here a company is
considered to be owner controlled if 30 percent or more of the common stock is
controlled by a single outside investor ( an investor not part of the management group
or board of directors). Otherwise, a company is considered manager controlled. It was
felt that the type of control influences audit delay. To quote Carslaw and Kaplan:
Large external investors, having an acute need for timely information, may be
expected to pressure the company and auditor to start and to complete the audit as
rapidly as practicable.
a. Suppose that a random sample of 100 public owner-controlled companies in New
Zealand is found to give a mean audit delay of x  82.6 days with a standard
deviation of s  32.83 days. Calculate a 95 percent confidence interval for the
population mean audit delay for all public owner-controlled companies in New
Zealand.
b. Suppose that a random sample of 100 public manager-controlled companies in New
Zealand is found to give a mean audit delay of x  93 days with a standard deviation
of s  37.18 days. Calculate a 95 percent confidence interval for the population mean
audit delay for all public manager-controlled companies in New Zealand.
c. Use the confidence intervals you computed in parts a and b to compare the mean
audit delay for all public owner-controlled companies versus that of all public
manager-controlled companies. How do the means compare? Explain.
7.21
In its October 7, 1991, issue, Fortune magazine reported on the rapid rise of fees and
expenses charged by various types of mutual funds. As stated in the article:
a. Suppose the average annual expense for a random sample of 12 stock funds is 1.63
percent with a standard deviation of 0.31 percent. Calculate a 95 percent confidence
interval for the mean annual expense charged by all stock funds. Assume that stock
fund annual expenses are approximately normally distributed.
b. Suppose the average annual expense for a random sample of 12 municipal bond
funds is 0.89 percent with a standard deviation of 0.23 percent. Calculate a 95 percent
confidence interval for the mean annual expense charged by all municipal bond funds.
Assume that municipal bond fund expenses are approximately normally distributed.
c. Use the 95 percent confidence intervals you computed in parts a and b to compare
the average annual expense for stock funds with that for municipal bond funds. How
do the averages compare? Explain.
7.29
Referring to Exercise7.11a(page 267), regard the sample of 100 public owner
controlled companies for which s  32.83 as a preliminary sample. How large a
random sample of public owner-controlled companies is needed to make us
a. 95 percent confident that x , the sample mean audit delay, is within four days of
 , the true mean audit delay?
b. 99 percent confident that x is within four days of  ?
7.47
In the book Cases in Finance, Nunnally and Plath present a case in which the estimated
percentage of uncollectible accounts varies with the age of the account. Here the age of an
unpaid account is the number of days elapsed since the invoice date.
Suppose an accountant believes the percentage of accounts that will be uncollectible
increases as the ages of the accounts increase. To test this theory, the accountant randomly
selects 500 accounts with ages between 31 and 60 days from the accounts receivable ledger
dated one year ago. The accountant also randomly selects 500 accounts with ages between
61 and 90 days from the accounts receivable ledger dated one year ago.
a. If 10 of the 500 accounts with ages between 31 and 60 days were eventually classified
as uncollectible, find a point estimate of and a 95 percent confidence interval for the
proportion of all accounts with ages between 31 and 50 days that will be uncollectible.
b. If 27 of the 500 accounts with ages between 61 and 90 days were eventually classified
as uncollectible, find a point estimate of and a 95 percent confidence interval for the
proportion of all accounts with ages between 61 and 90 days that will be uncollectible.
c. Based on these intervals, is there strong evidence that the percentage of accounts aged
between 61 and 90 days that will be uncollectible is higher than the percentage of
accounts aged between 31 and 60 days that will be uncollectible? Explain.
CH8
8.3
Define each of the following:
a. Type I error b. Type II error
c. 
d. 
8.15
Explain what a p-value is, and explain how it is used to test a hypothesis..
8.31
THE BANK CUSTOMER WAITING TIME CASE
Letting  be the mean waiting time under the new system, we found in Exercise
8.8(page 304) that we should test H 0 :   6 versus H a :   6 in order to attempt
to provide evidence that  is less than six minutes. The random sample of 100
waiting times yields a sample mean of x  5.46 minutes and a sample standard
deviation of s  2.475 . Moreover, Figure 8.6 gives the MINITAB output obtained
when we use the waiting time data to test H 0 :   6 versus H a :   6 .
On this output the label “SE Mean,” which stands for “the standard error of the
mean,” denotes the quantity s
n , and the label “Z” denotes the calculated test
statistic.
a. Use rejection points to test H 0 vs H a at each of   0.10,0.05,0.01 and 0.0001
b. Calculate the p-value and verify that it equals 0.0146(0.015 rounded), as shown on
the MINITAB output. Use the p-value to test H 0 vs H a at each of
  0.10,0.05,0.01 and 0.0001 .
c. How much evidence is there that the new system has reduced the mean waiting
time to below six minutes?
Figure 8.6
MINITAB Output of the Test of H 0 :   6 versus H a :   6 in the
Bank Customer Waiting Time Case
Test of mu = 6.000
vs mu < 6.000.
The assumed sigma = 2.47
Variable
N
Mean
StDev
SE Mean
Z
P
WaitTime
100
5.460
2.475
0.247
-2.18 0.015
8.89
Suppose we test H 0 : p  0.3 vesus H a : p  0.3 and that a random sample of n =
100 gives a sample proportion pˆ  0.20
a. Test H 0 vs H a at the 0.01 level of significance by using a rejection point. What
do you conclude?
b. Find the p-value for this test
c. Use the p-value to test H 0 vs H a by setting  equal to 0.10, 0.05, 0.001.
What do you conclude at each value of  ?
8.69
A consumer electronics firm has developed a new type of remote control button that is
designed to operate longer before becoming intermittent. A random sample of 35 of
the new buttons is selected and each is tested in continuous operation until becoming
intermittent. The resulting lifetimes are found to have a sample mean of
x  1,241.2 hours and a sample standard deviation of s  110.8
a. Independent tests reveal that the mean lifetime (in continuous operation) of the
best remote control button on the market is 1,200 hours. Letting  be the mean
lifetime of the population of all remote control buttons that will or could
potentially be produced, set up the null and alternative hypotheses needed to
attempt to provide evidence that the new button’s mean lifetime exceeds the mean
lifetime of the best remote button currently on the market.
b. Using the previously given sample results, test the hypotheses you set up in part a
by setting  equal to 0.10, 0.05, 0.01, and 0.001. What do you conclude for each
value of  ?
c. Suppose that x  1,241.2 and s  110.8 had been obtained by testing a sample of
100 buttons. Test the hypotheses you set up in part a by setting  equal to 0.10,
0.05, 0.01, and 0.001. Which sample (the sample of 35 or the sample of 100) gives
a more statistically significant result? That is , which sample provides stronger
evidence that H a is true?
d. If we define practical importance to mean that  exceeds 1,200 by an amount
that would be clearly noticeable to most consumers, do you think that the result
has practical importance?
Explain why the samples of 35 and 100 both indicate the same degree of practical
importance.
e. Suppose that further research and development effort improves the new remote
control button and that a random sample of 35 buttons gives x  1,524.6 hours
and s  102.8 hours. Test your hypotheses of part a by setting  equal to 0.10,
0.05, 0.01, and 0.001.
(1) Do we have a highly statistically significant result? Explain.
(2) Do you think we have a practically important result? Explain.
CH9
9.7
In an article in Accounting and Business Research, Carslaw and Kaplan(1991) study
the effect of control (owner versus manager control) on audit delay (the length of time
from a company’s financial year-end to the date of the auditor’s report) for public
companies in New Zealand. Suppose a random sample of 100 public owner-controlled
companies in New Zealand gives a mean audit delay of x1  82.6 days with a
standard deviation of s1  32.83 days, while a random sample of 100 public
manager-controlled companies in New Zealand gives a mean audit delay of
x2  93 days with a standard deviation of s2  37.18 days. Assuming the samples are
independent:
a. Let 1 be the mean audit delay for all public owner-controlled companies in New
Zealand, and let  2 be the mean audit delay for all public manager-controlled
companies in New Zealand. Calculate a 95 percent confidence interval for
1   2 . Based on this interval, can we be 95 percent confident that the mean
audit delay for all public owner-controlled companies in New Zealand is less than
that for all public manager-controlled companies in New Zealand? If so, by how
much?
b. Consider testing the null hypothesis H 0 : 1   2  0 versus H a : 1   2  0 .
Interpret (in writing) the meaning (in practical terms) of each of H 0 and H a .
c. Use a rejection point to test the null hypothesis H 0 : 1   2  0 versus
H a : 1   2  0 at the 0.05 level of significance. Based on this test, what do you
conclude about how 1 and  2 compare? Write your conclusion in practical
terms.
d. Find the p-value for testing H 0 : 1   2  0 versus H a : 1   2  0 . Use the
p-value to test H 0 and H a by setting  equal to 0.10, 0.05, 0.025, 0.01, and
0.001. How much evidence is there that 1 is less than  2 ?
9.21
In the book Business Research Methods, Donald R. Cooper and C. William Emory
(1995) discuss a manager who wishes to compare the effectiveness of two methods
for training new salespeople. The authors describe the situation as follows:
The company selects 22 sales trainees who are randomly divided into two
experimental groups—one receives type A and the other type B training. The
salespeople are then assigned and managed without regard to the training they have
received. At the year’s end, the manager reviews the performances of salespeople in
these groups and finds the following results:
A Group
B Group
Average Weekly Sales
x1  $1,500
x2  $1,300
Standard Deviation
s1  225
s2  251
a. Set up the null and alternative hypotheses needed to attempt to establish that type
A training results in higher mean weekly sales than does type B training.
b. Because different sales trainees are assigned to the two experimental groups, it is
reasonable to believe that the two samples are independent. Assuming that the
normality assumption holds, test the hypotheses you set up in part a at levels of
significance 0.10, 0.05, 0.01, and 0.001. How much evidence is there that type A
training produces results that are superior to those of type B?
c. Calculate a 95 percent confidence interval for the difference between the mean
weekly sales obtained when type A training is used and the mean weekly sales
obtained when type B training is used. Interpret this interval.
9.31
On its website, the Statesman Journal newspaper (Salem, Oregon, 1999) reports
mortgage loan interest rates for 30-year and 15-year fixed-rate mortgage loans for a
number of Willamette Valley lending institutions. Of interest is whether there is any
systematic difference between 30-yaer rates and 15-year rates (expressed as annual
percentage rate or APR) and, if there is , what is the size of that difference. Table 9.3
displays mortgage loan rates and the difference between 30-year and 15-year rates for
nine randomly selected lending institutions. Assuming that the population of paired
differences is normally distributed:
a. Set up the null and alternative hypotheses needed to determine whether there is a
difference between mean 30-year rates and mean 15-year rates.
b. Figure 9.9 gives the MINITAB output for testing the hypotheses that you set up in
part a. Use the output and rejection points to test these hypotheses by setting 
equal to 0.10, 0.05, 0.01, and 0.001. How much evidence is there that mean
mortgage loan rates for 30-year and 15-year terms differ?
c. Figure 9.9 gives the p-value for testing the hypotheses that you set up in part a.
Use p-value to test these hypotheses by setting  equal to 0.10, 0.05, 0.01, and
0.001. How much evidence is there that mean mortgage loan rates for 30-year and
15-year terms differ?
d. Calculate a 95 percent confidence interval for the difference between mean
mortgage loan rates for 30-year rates versus 15-year rates. Interpret this interval.
Lending Institution
30-Year
15-Year
Difference
American Mortgage N.W. Inc.
6.715
6.599
0.116
City and Country Mortgage
6.648
6.367
0.281
Commercial Bank
6.740
6.550
0.190
Landmark Mortgage Co.
6.597
6.362
0.235
Liberty Mortgage,
6.425
6.162
0.263
MaPS Credit Union
6.880
6.583
0.297
Mortgage Brokers, Inc.
6.900
6.800
0.100
Mortgage First Corp.
6.675
6.394
0.281
Silver Eagle Mortgage
6.790
6.540
0.250
Inc.
Figure9.9 MINITAB Output of a Paired Difference t Test of the Mortgage Loan Rate Data
in Table 9.3
Paired T-Test and CI: 30-Year, 15-Year
Paired T for 30-Year - 15-Year
N
Mean
StDev
SE Mean
30-Year
9
6.7078
0.1463
0.0488
15-Year
9
6.4841
0.1840
0.0613
Difference
9
0.2237
0.0727
0.0242
95% CI for mean difference: (0.1677, 0.2796)
T-Test of mean difference = 0 (vs not = 0): T-Value = 9.22 P-Value = 0.000
9.47
Consider the display panel situation in Exercise 9.38, and let  A ,  B , and C represent
the mean times to stabilize the emergency condition when using display panels A, B,
and C, respectively. Figure 9.15 give the MINITAB output of a one-way ANOVA of
the display panel data in Table 9.9 (page 372)
a. Test the null hypothesis that  A ,  B , and C are equal by setting   0.05 . On the
basis of this test, can we conclude that display panels A, B, and C have different
effects on the mean time to stabilize the emergency condition?
b. Consider the pairwise differences  A   B ,  A   C ,  B   C . Find a point
estimate of and a Tukey simultaneous 95 percent confidence interval for each
pairwise difference. Interpret the results by describing the effects of changing
from using each display panel to using each of the other panels. Which display
panel minimizes the time required to stabilize the emergency condition?
c. Find an individual 95 percent confidence interval for each pairwise difference in
part b. Interpret the results.
Display panel study data ( time, in seconds, required to stabilize air traffic emergency condition)
Display Panel
A
B
C
21
24
40
27
21
36
24
18
35
26
19
32
Figure 9.15 MINITAB Output of a One-Way ANOVA of the Display Panel Study Data In Table 9.9
One-way ANOVA: Time versus Display
Tukey's pairwise comparisons
Analysis of Variance for Time
Critical value = 3.95
Source
DF
SS
MS
F
Display
2
500.17
250.08
Error
9
74.75
8.31
Total
11
574.92
P
Intervals for (column level mean) –(row level mean)
A
30.11 0.000
B
B
-1.692
9.692
C
-16.942
-20.942
-5.558
-9.558
Individual 95% CIs For Mean
Based on Pooled StDev
Level
N
Mean
StDev
A
4
24.500
2.646
B
4
20.500
2.646
C
4
35.750
3.304
--+---------+---------+---------+---(-----*----)
(----*-----)
(-----*----)
--+---------+---------+---------+---Pooled StDev =
2.882
18.0
24.0
30.0
36.0