Download Chapter 9 Problems and Solutions

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Chapter 9 Problems and Solutions - Hypothesis Tests
1. The manager of the Danvers-Hilton Resort Hotel stated that the mean guest bill for a
weekend is $600 or less. A member of the hotel’s accounting staff noticed that the total
charges for guest bills have been increasing in recent months. The accountant will use
a sample of future weekend guest bills to test the manager’s claim.
a. Which form of the hypotheses should be used to test the manager’s claim?
Explain.
H0: μ ≥ 600
H0: μ ≤ 600 H0: μ = 600
Ha: μ < 600 Ha: μ > 600 Ha: μ ≠ 600
b.
c.
What conclusion is appropriate when H0 cannot be rejected?
What conclusion is appropriate when H0 can be rejected?
3. A production line operation is designed to fill cartons with laundry detergent to a
mean weight of 32 ounces. A sample of cartons is periodically selected and weighed to
deter- mine whether underfilling or overfilling is occurring. If the sample data lead to a
conclu- sion of underfilling or overfilling, the production line will be shut down and
adjusted to obtain proper filling.
a. Formulate the null and alternative hypotheses that will help in deciding
whether to shut down and adjust the production line.
b. Comment on the conclusion and the decision when H0 cannot be rejected.
c. Comment on the conclusion and the decision when H0 can be rejected.
4. Because of high production-changeover time and costs, a director of manufacturing
must convince management that a proposed manufacturing method reduces costs
before the new method can be implemented. The current production method operates
with a mean cost of $220 per hour. A research study will measure the cost of the new
method over a sample production period.
a. Develop the null and alternative hypotheses most appropriate for this
study.
b. Comment on the conclusion when H0 cannot be rejected.
c. Comment on the conclusion when H0 can be rejected.
5. Nielsen reported that young men in the United States watch 56.2 minutes of
prime-time TV daily (The Wall Street Journal Europe, November 18, 2003). A researcher
believes that young men in Germany spend more time watching prime-time TV. A
sample of German young men will be selected by the researcher and the time they
spend watching TV in one day will be recorded. The sample results will be used to test
the following null and alternative hypotheses.
H0: μ ≤ 56.2 Ha: μ > 56.2
a. What is the Type I error in this situation? What are the consequences of making
this error?
b. What is the Type II error in this situation? What are the consequences of making
this error?
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6. The label on a 3-quart container of orange juice states that the orange juice
contains an average of 1 gram of fat or less. Answer the following questions for a
hypothesis test that could be used to test the claim on the label.
a. Develop the appropriate null and alternative hypotheses.
b. What is the Type I error in this situation? What are the consequences of making
this error?
c. What is the Type II error in this situation? What are the consequences of making
this error?
10.
Consider the following hypothesis test:
H0: μ ≤ 25 Ha: μ > 25
A sample of 40 provided a sample mean of 26.4.
The population standard deviation is 6.
a.
b.
c.
d.
11.
Compute the value of the test statistic.
What is the p-value?
At α = .01, what is your conclusion?
What is the rejection rule using the critical value? What is your conclusion?
Consider the following hypothesis test:
H0: μ = 15
Ha: μ ≠ 15
A sample of 50 provided a sample mean of 14.15.
The population standard deviation is 3.
a.
b.
c.
d.
13.
Compute the value of the test statistic.
What is the p-value?
At α = .05, what is your conclusion?
What is the rejection rule using the critical value? What is your conclusion?
Consider the following hypothesis test:
H0: μ ≤ 50
Ha: μ > 50
A sample of 60 is used and the population standard deviation is 8. Use the critical
value approach to state your conclusion for each of the following sample
results. Use α = .05.
a.
b.
c.
x¯ = 52.5
x¯ = 51
x¯ = 51.8
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17. Wall Street securities firms paid out record year-end bonuses of $125,500 per
employee for 2005 (Fortune, February 6, 2006). Suppose we would like to take a sample
of employ- ees at the Jones & Ryan securities firm to see whether the mean year-end
bonus is different from the reported mean of $125,500 for the population.
a. State the null and alternative hypotheses you would use to test whether the
year-end bonuses paid by Jones & Ryan were different from the population
mean.
b. Suppose a sample of 40 Jones & Ryan employees showed a sample mean
year-end bonus of $118,000. Assume a population standard deviation of u =
$30,000 and com- pute the p-value.
c. With α = .05 as the level of significance, what is your conclusion?
d. Repeat the preceding hypothesis test using the critical value approach.
20. For the United States, the mean monthly Internet bill is $32.79 per household
(CNBC, January 18, 2006). A sample of 50 households in a southern state showed a
sample mean of $30.63. Use a population standard deviation of a = $5.60.
a. Formulate hypotheses for a test to determine whether the sample data
support the con- clusion that the mean monthly Internet bill in the southern state
is less than the national mean of $32.79.
b. What is the value of the test statistic?
c. What is the p-value?
d. At α = .01, what is your conclusion?
26.
Consider the following hypothesis test:
H0: μ = 100 Ha: μ ≠ 100
A sample of 65 is used. Identify the p-value and state your conclusion for each of the
following sample results. Use α = .05.
a.
b.
c.
x¯ = 103 and s = 11.5
x¯ = 96.5 and s = 11.0
x¯ = 102 and s = 10.5
28. A shareholders’ group, in lodging a protest, claimed that the mean tenure for a
chief executive officer (CEO) was at least nine years. A survey of companies reported in
The Wall Street Journal found a sample mean tenure of x̄ = 7.27 years for CEOs with a
standard deviation of s = 6.38 years (The Wall Street Journal, January 2, 2007).
a. Formulate hypotheses that can be used to challenge the validity of the
claim made by the shareholders’ group.
b. Assume 85 companies were included in the sample. What is the p-value
for your hypothesis test?
c. At α = .01, what is your conclusion?
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30. Time Warner Inc.’s CNN has been the longtime ratings leader of cable television
news. Nielsen Media Research indicated that the mean CNN viewing audience was
600,000 viewers per day during 2002 (The Wall Street Journal, March 10, 2003).
Assume that for a sample of 40 days during the first half of 2003, the daily audience
was 612,000 viewers with a sample standard deviation of 65,000 viewers.
a. What are the hypotheses if CNN management would like information on
any change in the CNN viewing audience?
b. What is the p-value?
c. Select your own level of significance. What is your
conclusion?
d. What recommendation would you make to CNN management in this
application?
Chapter 9 Solutions
1. a. H0:   600
Manager’s claim.
Ha:  > 600
b. We are not able to conclude that the manager’s claim is wrong.
c. The manager’s claim can be rejected. We can conclude that  > 600.
3. a. H0:  = 32
Ha:   32
Specified filling weight
Overfilling or underfilling exists
b. There is no evidence that the production line is not operating properly. Allow the
production process to continue.
c. Conclude   32 and that overfilling or underfilling exists. Shut down and adjust
the production line.
4. a. H0:   220
Ha:  < 220
Research hypothesis to see if mean cost is less than $220.
b. We are unable to conclude that the new method reduces costs.
c. Conclude  < 220. Consider implementing the new method based on the
conclusion that it lowers the mean cost per hour.
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5. a. The Type I error is rejecting H0 when it is true. This error occurs if the researcher
concludes that young men in Germany spend more than 56.2 minutes per day
watching prime-time TV when the national average for Germans is not greater than
56.2 minutes.
b. The Type II error is accepting H0 when it is false. This error occurs if the researcher
concludes that the national average for German young men is  56.2 minutes when
in fact it is greater than 56.2 minutes.
6. a. H0:   1
The label claim or assumption.
Ha:  > 1
b. Claiming  > 1 when it is not. This is the error of rejecting the product’s claim
when the claim is true.
c. Concluding   1 when it is not. In this case, we miss the fact that the product is
not meeting its label specification.
10. a.
z
x  0 26.4  25

 1.48
/ n
6 / 40
b. Upper tail p-value is the area to the right of the test statistic
Using normal table with z = 1.48: p-value = 1.0000 - .9306 = .0694
Using Excel: p-value = 1 - NORMSDIST(1.48) = .0694
c. p-value > .01, do not reject H0
d. Reject H0 if z  2.33
1.48 < 2.33, do not reject H0
11. a.
z
x  0
/ n

14.15  15
3/ 50
 2.00
b. Because z < 0, p-value is two times the lower tail area
Using normal table with z = -2.00: p-value = 2(.0228) = .0456
Using Excel: p-value = 2*NORMSDIST(-2.00) = .0456
c. p-value  .05, reject H0
d. Reject H0 if z  -1.96 or z  1.96
-2.00  -1.96, reject H0
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Reject H0 if z  1.645
13.
a.
b.
z
z
z
c.
x  0
/ n
x  0
/ n
x  0
/ n



52.5  50
8 / 60
51  50
8 / 60
b.
.97 < 1.645, do not reject H0
 1.74
17. a. H0:   125,500
z
2.42  1.645, reject H0
 .97
51.8  50
8 / 60
 2.42
1.74  1.645, reject H0
Ha:
  125,500
x  0 118, 000  125,500

 1.58
/ n
30, 000 / 40
Because z < 0, p-value is two times the lower tail area
Using normal table with z = -1.58: p-value = 2(.0571) = .1142
Using Excel: p-value = 2*NORMSDIST(-1.58) = .1141
c. p-value > .05, do not reject H0. We cannot conclude that the year-end bonuses paid
by Jones & Ryan differ significantly from the population mean of $125,500.
d. Reject H0 if z  -1.96 or z  1.96
z = -1.58; cannot reject H0
20. a. H0:   32.79
z
b.
Ha:
 < 32.79
x 
0  30.63  32.79  2.73
 n
5.6 50
c. Lower tail p-value is area to left of the test statistic.
Using normal table with z = -2.73: p-value = .0032.
Using Excel: p-value = NORMSDIST(-2.73) = .0032
H
d. p-value  .01; reject 0 . Conclude that the mean monthly internet bill is less in the
southern state.
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26. a.
t
x  0
s/ n

103  100
11.5 / 65
 2.10
Degrees of freedom = n – 1 = 64
Because t > 0, p-value is two times the upper tail area
Using t table; area in upper tail is between .01 and .025; therefore, p-value is
between .02 and .05.
Using Excel: p-value = TDIST(2.10,64,2) = .0397 p-value  .05, reject H0
b.
t
x  0
s/ n

96.5  100
11/ 65
 2.57
Because t < 0, p-value is two times the lower tail area
Using t table: area in lower tail is between .005 and .01; therefore, p-value is
between .01 and .02.
Using Excel: p-value = TDIST(2.57,64,2) = .0125
p-value  .05, reject H0
c.
t
x  0
s/ n

102  100
10.5 / 65
 1.54
Because t > 0, p-value is two times the upper tail area
Using t table: area in upper tail is between .05 and .10; therefore, p-value is between
.10 and .20.
Using Excel: p-value = TDIST(1.54,64,2) = .1285
p-value > .05, do not reject H0
28. a. H0:   9
b.
t
x  0
s/ n

Ha:
7.27  9
6.38 / 85
 2.50
 < 9
Degrees of freedom = n – 1 = 84
Lower tail p-value is the area to the left of the test statistic
Using t table: p-value is between .005 and .01
Using Excel: p-value = TDIST(2.50,84,1) = .0072
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c. p-value  .01; reject H0. The mean tenure of a CEO is significantly lower than 9
years. The claim of the shareholders group is not valid.
30. a. H0:  = 600
b.
t
x  0
s/ n

Ha:
612  600
65 / 40
 1.17
  600
df = n - 1 = 39
Because t > 0, p-value is two times the upper tail area
Using t table: area in upper tail is between .10 and .20; therefore, p-value is between
.20 and .40.
Using Excel: p-value = TDIST(1.17,39,2) = .2491
c. With  = .10 or less, we cannot reject H0. We are unable to conclude there has been a
change in the mean CNN viewing audience.
d. The sample mean of 612 thousand viewers is encouraging but not conclusive for the
sample of 40 days. Recommend additional viewer audience data. A larger sample
should help clarify the situation for CNN..