Chapter 9
Hypothesis Tests
Learning Objectives
1.
Learn how to formulate and test hypotheses about a population mean and/or a population proportion.
2.
Understand the types of errors possible when conducting a hypothesis test.
3.
Be able to determine the probability of making these errors in hypothesis tests.
4.
Know how to compute and interpret p-values.
5.
Be able to use critical values to draw hypothesis testing conclusions.
6.
Be able to determine the size of a simple random sample necessary to keep the probability of
hypothesis testing errors within acceptable limits.
7.
Know the definition of the following terms:
null hypothesis
alternative hypothesis
Type I error
Type II error
one-tailed test
two-tailed test
p-value
level of significance
critical value
power curve
9-1
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Chapter 9
Solutions:
1.
2.
3.
4.
5.
a.
H0:   600
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.
a.
H0:   14
Ha:  > 14
Manager’s claim.
Research hypothesis
b.
There is no statistical evidence that the new bonus plan increases sales volume.
c.
The research hypothesis that  > 14 is supported. We can conclude that the new bonus plan
increases the mean sales volume.
a.
H0:  = 32
Ha:   32
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.
a.
H0:   220
Ha:  < 220
Specified filling weight
Overfilling or underfilling exists
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.
a.
Conclude that the population mean monthly cost of electricity in the Chicago neighborhood is
greater than $104 and hence higher than in the comparable neighborhood in Cincinnati.
b. The Type I error is rejecting H0 when it is true. This error occurs if the researcher concludes that the
population mean monthly cost of electricity is greater than $104 in the Chicago neighborhood when
the population mean cost is actually less than or equal to $104.
6.
c.
The Type II error is accepting H0 when it is false. This error occurs if the researcher concludes that
the population mean monthly cost for the Chicago neighborhood is less than or equal to $104 when it
is not.
a.
H0:   1
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.
The label claim or assumption.
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Hypothesis Tests
7.
8.
9.
a.
H0:   8000
Ha:  > 8000
Research hypothesis to see if the plan increases average sales.
b.
Claiming  > 8000 when the plan does not increase sales. A mistake could be implementing the
plan when it does not help.
c.
Concluding   8000 when the plan really would increase sales. This could lead to not
implementing a plan that would increase sales.
a.
H0:   220
Ha:  < 220
b.
Claiming  < 220 when the new method does not lower costs. A mistake could be implementing
the method when it does not help.
c.
Concluding   220 when the method really would lower costs. This could lead to not
implementing a method that would lower costs.
a.
z=
b.
Lower tail p-value is the area to the left of the test statistic
x − 0
/ n
=
19.4 − 20
2 / 50
= −2.12
Using normal table with z = -2.12: p-value =.0170
c.
p-value  .05, reject H0
d.
Reject H0 if z  -1.645
-2.12  -1.645, reject H0
10. a.
b.
z=
x − 0
/ n
=
26.4 − 25
6 / 40
= 1.48
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
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.
b.
z=
x − 0
/ n
=
14.15 − 15
3 / 50
= −2.00
Because z < 0, p-value is two times the lower tail area
Using normal table with z = -2.00: p-value = 2(.0228) = .0456
c.
p-value  .05, reject H0
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Chapter 9
d.
Reject H0 if z  -1.96 or z  1.96
-2.00  -1.96, reject H0
12. a.
z=
x − 0
/ n
=
78.5 − 80
12 / 100
= −1.25
Lower tail p-value is the area to the left of the test statistic
Using normal table with z = -1.25: p-value =.1056
p-value > .01, do not reject H0
b.
z=
x − 0
/ n
=
77 − 80
12 / 100
= −2.50
Lower tail p-value is the area to the left of the test statistic
Using normal table with z = -2.50: p-value =.0062
p-value  .01, reject H0
c.
z=
x − 0
/ n
=
75.5 − 80
12 / 100
= −3.75
Lower tail p-value is the area to the left of the test statistic
Using normal table with z = -3.75: p-value ≈ 0
p-value  .01, reject H0
d.
z=
x − 0
/ n
=
81 − 80
12 / 100
= .83
Lower tail p-value is the area to the left of the test statistic
Using normal table with z = .83: p-value =.7967
p-value > .01, do not reject H0
Reject H0 if z  1.645
13.
a.
b.
z=
x − 0
/ n
=
52.5 − 50
8 / 60
= 2.42
2.42  1.645, reject H0
x − 0 51 − 50
z=
=
= .97
 / n 8 / 60
.97 < 1.645, do not reject H0
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Hypothesis Tests
c.
z=
x − 0
/ n
=
51.8 − 50
= 1.74
8 / 60
1.74  1.645, reject H0
14. a.
z=
x − 0
/ n
=
23 − 22
10 / 75
= .87
Because z > 0, p-value is two times the upper tail area
Using normal table with z = .87: p-value = 2(1 - .8078) = .3844
p-value > .01, do not reject H0
b.
z=
x − 0 25.1 − 22
=
= 2.68
 / n 10 / 75
Because z > 0, p-value is two times the upper tail area
Using normal table with z = 2.68: p-value = 2(1 - .9963) = .0074
p-value  .01, reject H0
c.
z=
x − 0
/ n
=
20 − 22
10 / 75
= −1.73
Because z < 0, p-value is two times the lower tail area
Using normal table with z = -1.73: p-value = 2(.0418) = .0836
p-value > .01, do not reject H0
15. a.
b.
H0:   
Ha:  < 1056
z=
x − 0
/ n
=
910 − 1056
1600 / 400
= −1.83
Lower tail p-value is the area to the left of the test statistic
Using normal table with z = -1.83: p-value =.0336
c.
p-value  .05, reject H0. Conclude the mean refund of “last minute” filers is less than $1056.
d.
Reject H0 if z  -1.645
-1.83  -1.645, reject H0
9-5
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Chapter 9
16. a.
b.
H0:   3173
Ha:  > 3173
z=
x − 0
/ n
=
3325 − 3173
1000 / 180
= 2.04
p-value = 1.0000 - .9793 = .0207
c.
17. a.
b.
p-value < .05. Reject H0. The current population mean credit card balance for undergraduate
students has increased compared to the previous all-time high of $3173 reported in the original
report.
H0:  = 24.57
Ha:   24.57
z=
x − 0
/ n
=
23.89 − 24.57
2.4 / 30
= −1.55
Because z < 0, p-value is two times the lower tail area
Using normal table with z = -1.55: p-value = 2(.0606) = .1212
c.
p-value > .05, do not reject H0. We cannot conclude that the population mean hourly wage for
manufacturing workers differs significantly from the population mean of $24.57 for the goodsproducing industries.
d.
Reject H0 if z  -1.96 or z  1.96
z = -1.55; cannot reject H0. The conclusion is the same as in part (c).
18. a.
b.
H0:  = 192
Ha:   192
z=
x − 0
/ n
=
182 − 192
55 / 150
= −2.23
Because z < 0, p-value is two times the lower tail area
Using normal table with z = – 2.23: p-value = 2(.0129) = .0258
c.
p-value = .0258   = .05
Reject H0 and conclude that the mean number of restaurant meals eaten by young millennials has
changed in 2012.
19.
H0:  ≥ 12
Ha:  < 12
z=
x −
0 = 10 − 12 = −1.77
 n 8 50
9-6
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Hypothesis Tests
p-value is the area in the lower tail
Using normal table with z = -1.77: p-value = .0384
p-value  .05, reject H0. Conclude that the actual mean waiting time is significantly less than the
claim of 12 minutes made by the taxpayer advocate.
20. a.
H0:   838
Ha:  < 838
x − 0
z=
c.
Lower tail p-value is area to left of the test statistic.
/ n
=
745 − 838
b.
300
60
= −2.40
Using normal table with z = -2.40: p-value = .0082.
d.
21. a.
p-value  .01; reject H 0 . Conclude that the annual expenditure per person on prescription drugs is
less in the Midwest than in the Northeast.
H0:   15
Ha:  > 15
x −
z=
c.
Upper tail p-value is the area to the right of the test statistic
/ n
=
17 − 15
b.
4 / 35
= 2.96
Using normal table with z = 2.96: p-value = 1.0000 - .9985 = .0015
d.
22. a.
b.
p-value  .01; reject H0; the premium rate should be charged.
H0:  = 8
Ha:   8
z=
x −
8.4 − 8.0
=
= 1.37
 / n 3.2 / 120
Because z > 0, p-value is two times the upper tail area
Using normal table with z = 1.37: p-value = 2(1 - .9147) = .1706
c.
d.
p-value > .05; do not reject H0. Cannot conclude that the population mean waiting time differs from
8 minutes.
x  z.025 ( / n )
8.4 ± 1.96 (3.2 / 120)
8.4 ± .57
(7.83 to 8.97)
Yes;  = 8 is in the interval. Do not reject H0.
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Chapter 9
23. a.
b.
t=
x − 0
s/ n
=
14 − 12
4.32 / 25
= 2.31
Degrees of freedom = n – 1 = 24
Upper tail p-value is the area to the right of the test statistic
Using t table: p-value is between .01 and .025
Exact p-value corresponding to t = 2.31 is .0149
c.
p-value  .05, reject H0.
d.
With df = 24, t.05 = 1.711
Reject H0 if t  1.711
2.31 > 1.711, reject H0.
24. a.
b.
t=
x − 0
s/ n
=
17 − 18
4.5 / 48
= −1.54
Degrees of freedom = n – 1 = 47
Because t < 0, p-value is two times the lower tail area
Using t table: area in lower tail is between .05 and .10; therefore, p-value is between .10 and .20.
Exact p-value corresponding to t = -1.54 is .1303
c.
p-value > .05, do not reject H0.
d.
With df = 47, t.025 = 2.012
Reject H0 if t  -2.012 or t  2.012
t = -1.54; do not reject H0
25. a.
t=
x − 0
s/ n
=
44 − 45
5.2 / 36
= −1.15
Degrees of freedom = n – 1 = 35
Lower tail p-value is the area to the left of the test statistic
Using t table: p-value is between .10 and .20
Exact p-value corresponding to t = -1.15 is .1290
p-value > .01, do not reject H0
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Hypothesis Tests
b.
t=
x − 0
s/ n
=
43 − 45
4.6 / 36
= −2.61
Lower tail p-value is the area to the left of the test statistic
Using t table: p-value is between .005 and .01
Exact p-value corresponding to t = -2.61 is .0066
p-value  .01, reject H0
c.
t=
x − 0
s/ n
=
46 − 45
5 / 36
= 1.20
Lower tail p-value is the area to the left of the test statistic
Using t table: p-value is between .80 and .90
Exact p-value corresponding to t = 1.20 is .8809
p-value > .01, do not reject H0
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.
Exact p-value corresponding to t = 2.10 is .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.
Exact p-value corresponding to t = -2.57 is .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.
9-9
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Chapter 9
Exact p-value corresponding to t = 1.54 is .1285
p-value > .05, do not reject H0
27. a.
b.
H0:   13.04
Ha:  < 13.04
t=
x − 0
=
s/ n
12.75 − 13.04
2 / 100
= −1.45
Degrees of freedom = n – 1 = 99
p-value is the lower tail area at the test statistic
Using t table: p-value is between .05 and .10
Exact p-value corresponding to t = -1.45 is .0751
c.
p-value > .05; do not reject H0. We cannot conclude that the cost of a restaurant meal is significantly
cheaper than a comparable meal fixed at home.
d.
df = 99
t.05 = -1.66
Reject H0 if t  -1.66
-1.45 > -1.66; do not reject H0
28. a.
b.
H0:   9
Ha:  < 9
t=
x − 0
s/ n
=
7.27 − 9
6.38 / 85
= −2.50
Degrees of freedom = n – 1 = 84
Lower tail p-value is P(t ≤ -2.50)
Using t table: p-value is between .005 and .01
Exact p-value corresponding to t = -2.50 is .0072
c.
29. a.
b.
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.
H0:  = 90,000
Ha:   90,000
t=
x − 0
s/ n
=
85, 272 − 90,000.00
11,039.23 / 25
= −2.14
Degrees of freedom = n – 1 = 24
Because t < 0, p-value is two times the lower tail area
9 - 10
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Hypothesis Tests
Using t table: area in lower tail is between .01 and .025; therefore, p-value is between .02 and .05.
Exact p-value corresponding to t = -2.14 is .0427
c.
p-value  .05; reject H0. The mean annual administrator salary in Ohio differs significantly from the
national mean annual salary.
d.
df = 24
t.025 = 2.064
Reject H0 if t < -2.064 or t > 2.064
-2.14 < -2.064; reject H0. The conclusion is the same as in part (c).
30. a.
b.
H0:  = 6.4
Ha:   6.4
Using Excel or Minitab, we find x = 7.0 and s = 2.4276
t=
x − 0
s/ n
=
7.0 − 6.4
2.4276 / 40
= 1.56
df = n - 1 = 39
Because t > 0, p-value is two times the upper tail area at t = 1.56
Using t table: area in upper tail is between .05 and .10; therefore, p-value is between .10 and .20.
Exact p-value corresponding to t = 1.56 is .1268
c.
Most researchers would choose  = .10 or less. If you chose  = .10 or less, you cannot reject H0.
You are unable to conclude that the population mean number of hours married men with children in
your area spend in child care differs from the mean reported by Time.
H0:   423
Ha:  > 423
31.
t=
x − 0
s/ n
=
460.4 − 423.0
101.9 / 36
= 2.20
Degrees of freedom = n - 1 = 35
Upper tail p-value is the area to the right of the test statistic
Using t table: p-value is between .01 and .025.
Exact p-value corresponding to t = 2.02 is .0173
Because p-value = .0173 < α, reject H0; Atlanta customers have a higher annual rate of consumption
of Coca Cola beverages.
32. a.
b.
H0:  = 10,192
Ha:   10,192
t=
x − 0
s/ n
=
9750 − 10,192
1400 / 50
= −2.23
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Chapter 9
Degrees of freedom = n – 1 = 49
Because t < 0, p-value is two times the lower tail area
Using t table: area in lower tail is between .01 and .025; therefore, p-value is between .02 and .05.
Exact p-value corresponding to t = -2.23 is .0304
c.
33. a.
p-value  .05; reject H0. The population mean price at this dealership differs from the national mean
price $10,192.
H0:   1503
Ha:  < 1503
b.
$1440 - $1503 = -$63
c.
t=
x − 0
s/ n
=
1440 − 1503
165 / 25
= −1.91
Degrees of freedom = n – 1 = 24
p-value is lower-tail area
Using t table: p-value is between .025 and .05
Exact p-value corresponding to t = -1.91 is .0341
d.
34. a.
p-value  .05; reject H0. The population mean automobile premium is lower in Pennsylvania than the
national mean.
H0:  = 2
Ha:   2
b.
x=
c.
s=
d.
t=
xi 22
=
= 2.2
n
10
 ( xi − x )
n −1
x − 0
s/ n
=
2
= .516
2.2 − 2
.516 / 10
= 1.22
Degrees of freedom = n - 1 = 9
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.
Exact p-value corresponding to t = 1.22 is .2535
e.
p-value > .05; do not reject H0. No reason to change from the 2 hours for cost estimating purposes.
9 - 12
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Hypothesis Tests
35. a.
b.
z=
p − p0
p0 (1 − p0 )
n
=
.175 − .20
.20(1 − .20)
400
= −1.25
Because z < 0, p-value is two times the lower tail area
Using normal table with z = -1.25: p-value = 2(.1056) = .2112
c.
p-value > .05; do not reject H0
d.
z.025 = 1.96
Reject H0 if z  -1.96 or z  1.96
z = − 1.25; do not reject H0
36. a.
z=
p − p0
p0 (1 − p0 )
n
=
.68 − .75
.75(1 − .75)
300
= −2.80
Lower tail p-value is the area to the left of the test statistic
Using normal table with z = -2.80: p-value =.0026
p-value  .05; Reject H0
b.
z=
.72 − .75
.75(1 − .75)
300
= −1.20
Lower tail p-value is the area to the left of the test statistic
Using normal table with z = -1.20: p-value =.1151
c.
p-value > .05; Do not reject H0
.70 − .75
z=
= −2.00
.75(1 − .75)
300
Lower tail p-value is the area to the left of the test statistic
Using normal table with z = -2.00: p-value =.0228
p-value  .05; Reject H0
d.
z=
.77 − .75
.75(1 − .75)
300
= .80
Lower tail p-value is the area to the left of the test statistic
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Chapter 9
Using normal table with z = .80: p-value =.7881
p-value > .05; Do not reject H0
37. a.
b.
H0: p  .113
Ha: p > .113
p=
z=
52
= .13
400
p − p0
p0 (1 − p0 )
n
=
.13 − .113
.113(1 − .113)
400
= 1.07
Using normal table with z = 1.07: p-value = 1.0000 - .8577 = .1423
c.
38. a.
b.
p-value > .05; do not reject H0. We cannot conclude that there has been an increase in union
membership.
H0: p = .64
Ha: p  .64
p=
z=
52
= .52
100
p − p0
p0 (1 − p0 )
n
=
.52 − .64
.64(1 − .64)
100
= −2.50
Because z < 0, p-value is two times the lower tail area
Using normal table with z = -2.50: p-value = 2(.0062) = .0124
c.
p-value  .05; reject H0. Proportion differs from the reported .64.
d.
Yes. Since p = .52, it indicates that fewer than 64% of the shoppers believe the supermarket brand
is as good as the name brand.
39. a.
b.
H0: p = .577
Ha: p  .577
Proportion of stay at home residents in Arkansas p =
z=
p − p0
p0 (1 − p0 )
n
=
.6167 − .577
.577(1 − .577)
120
74
= .6167
120
= .88
Because z > 0, p-value is two times the upper tail area
Using normal table with z = .88: p-value = 2(.1894) = .3788
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© 2017 Cengage Learning® . May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Hypothesis Tests
c.
Do not reject H0. Do not conclude that the proportion of stay at home residents in Arkansas is
higher than the overall proportion of .577.
89
Stay at home residents in Virginia p =
= .494
180
z=
.494 − .577
.577(1 − .577)
180
= −2.25
Because z < 0, p-value is two times the lower tail area
Using the normal table with z = -2.25: p-value = 2(.0122) = .0244
Reject H0. Conclude that the proportion of stay at home residents in Virginia does differ significantly
from the overall proportion of .577.
d.
40. a.
From the results obtained in parts (b) and (c) we would expect the number of stay at home residents
to be higher in Arkansas than Virginia. The sample results show that 61.7% of the current residents
of Arkansas were born there while only 49.4 % of the current residents of Virginia were born there.
Perhaps the proximity to Washington D. C. causes a number of people to move there from other
parts of the country.
Sample proportion: p = .35
Number planning to provide holiday gifts: np = 60(.35) = 21
b.
H0: p  .46
Ha: p < .46
z=
p − p0
p0 (1 − p0 )
n
=
.35 − .46
.46(1 − .46)
60
= −1.71
p-value is area in lower tail
Using normal table with z = -1.71: p-value = .0436
c.
41. a.
b.
Using a .05 level of significance, we can conclude that the proportion of business owners providing
gifts has decreased from last year to this year. The smallest level of significance for which we could
draw this conclusion is .0436; this corresponds to the p-value = .0436. This is why the p-value is
often called the observed level of significance.
H0: p  .53
Ha: p < .53
z=
p − p0
p0 (1 − p0 )
n
=
.46 − .53
.53(1 − .53)
300
= −2.43
p-value is the lower-tail area at the test statistic
Using normal table with z = ˗2.43: p-value =.0075
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© 2017 Cengage Learning® . May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 9
c.
42. a.
b.
p-value   = .01 ; reject H0. We conclude that there has been a statistically significant decline in
the proportion of American families owning stocks or stock funds over the ten-year period.
p = 12/80 = .15
p(1 − p )
.15(.85)
=
= .0399
n
80
p  z.025
p(1 − p)
n
.15  1.96 (.0399)
.15  .0782 or .0718 to .2282
c.
H0: p = .06
Ha: p  .06
p = .15
z=
p − p0
p0 (1 − p0 )
n
=
.15 − .06
.06(.94)
80
= 3.38
p-value ≈ 0
We conclude that the return rate for the Houston store is different than the U.S. national return rate.
43. a.
b.
H0: p ≤ .10
Ha: p > .10
There are 13 “Yes” responses in the Eagle data set.
p=
c.
z=
13
= .13
100
p − p0
p0 (1 − p0 )
n
=
.13 − .10
.10(1 − .10)
100
= 1.00
Upper tail p-value is the area to the right of the test statistic
Using normal table with z = 1.00: p-value = 1 - .8413 = .1587
p-value > .05; do not reject H0.
On the basis of the test results, Eagle should not go national. But, since p > .13, it may be worth
expanding the sample size for a larger test.
44. a.
H0: p  .50
Ha: p > .50
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© 2017 Cengage Learning® . May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Hypothesis Tests
b.
Using Excel we find that 92 of the 150 physicians in the sample have been sued.
So, p =
z=
92
= .6133
150
p − p0
p0 (1 − p0 )
n
=
.6133 − .50
(.50)(.50)
150
= 2.78
p-value is the area in the upper tail at z = 2.78
Using normal table with z = 2.78: p-value = 1 – .9973 = .0027
c.
45. a.
Since p-value = .0027  .01, we reject H0 and conclude that the proportion of physicians over the age
of 55 who have been sued at least once is greater than .50.
H0: p = .39
Ha: p  .39
p = .385
z=
p − p0
p0 (1 − p0 )
n
=
.385 − .39
.39(1 − .39)
300
= −.18
Because z < 0, p-value is 2 times the lower tail area at z = -.18
Using normal table with z = -.18: p-value = 2(.4286) = .8572
p − value  .05; do not reject H0. We cannot conclude that bullish sentiment differs significantly
from its long-term average of .39.
b.
H 0 : p  .30
H 0 : p  .30
p = .399
z=
p − p0
p0 (1 − p0 )
n
=
.399 − .30
.30(1 − .30)
300
= 3.74
Because z > 0, p-value is the upper tail area at z = 3.74
Using normal table with z = 3.74: p-value ≈ 0
p − value   = .01; reject H0. We conclude that bearish sentiment is significantly greater than its
long-term average of .30.
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© 2017 Cengage Learning® . May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 9
c.
It would be dangerous to generalize these results. The target population for this study is members of
the American Association of Individual Investors. We might be willing to extend these results to the
target population of all individual investors. But, should not extend the results to a target population
of all investors.
x =
46.

n
=
5
120
= .46
c = 10 - 1.645 (5 / 120 ) = 9.25
Reject H0 if x  9.25
a.
When  = 9,
z=
9.25 − 9
5 / 120
= .55
P(Reject H0) = (1.0000 - .7088) = .2912
b.
Type II error
c.
When  = 8,
z=
9.25 − 8
5 / 120
= 2.74
 = (1.0000 - .9969) = .0031
47.
Reject H0 if z  -1.96 or if z  1.96
x =

n
=
10
200
= .71
9 - 18
© 2017 Cengage Learning® . May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Hypothesis Tests
c1 = 20 - 1.96 (10 / 200 ) = 18.61
c2 = 20 + 1.96 (10 / 200 ) = 21.39
a.
 = 18
z=
18.61 − 18
10 / 200
= .86
 = 1.0000 - .8051 = .1949
b.
 = 22.5
z=
21.39 − 22.5
10 / 200
= −1.57
 = 1.0000 - .9418 = .0582
c.
 = 21
z=
21.39 − 21
10 / 200
= .55
 = .7088
48. a.
H0:   15
Ha:  > 15
Concluding   15 when this is not true. Fowle would not charge the premium rate even though
the rate should be charged.
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© 2017 Cengage Learning® . May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 9
b.
Reject H0 if z  2.33
z=
x − 0
/ n
=
x − 15
4 / 35
= 2.33
Solve for x = 16.58
Decision Rule:
Accept H0 if x < 16.58
Reject H0 if x  16.58
For  = 17,
z=
16.58 − 17
= −.62
4 / 35
 = .2676
c.
For  = 18,
z=
16.58 − 18
= −2.10
4 / 35
 = .0179
49. a.
H0:   25
Ha:  < 25
Reject H0 if z  -2.05
z=
x − 0
/ n
=
x − 25
3 / 30
= −2.05
Solve for x = 23.88
Decision Rule:
Accept H0 if x > 23.88
Reject H0 if x  23.88
b.
For  = 23,
z=
23.88 − 23
3 / 30
= 1.61
 = 1.0000 -.9463 = .0537
9 - 20
© 2017 Cengage Learning® . May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Hypothesis Tests
c.
For  = 24,
z=
23.88 − 24
= −.22
3 / 30
 = 1.0000 - .4129 = .5871
d.
50. a.
b.
The Type II error cannot be made in this case. Note that when  = 25.5, H0 is true. The Type II
error can only be made when H0 is false.
Accepting H0 and concluding the mean average age was 28 years when it was not.
Reject H0 if z  -1.96 or if z  1.96
z=
x − 0
/ n
=
x − 28
6 / 100
Solving for x , we find
at
at
z = -1.96,
z = +1.96,
x = 26.82
x = 29.18
Decision Rule:
Accept H0 if 26.82 < x < 29.18
Reject H0 if x  26.82 or if x  29.18
At  = 26,
z=
26.82 − 26
6 / 100
= 1.37
 = 1.0000 - .9147 = .0853
At  = 27,
z=
26.82 − 27
6 / 100
= −.30
 = 1.0000 - .3821 = .6179
At  = 29,
z=
29.18 − 29
6 / 100
= .30
 = .6179
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© 2017 Cengage Learning® . May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 9
At  = 30,
z=
29.18 − 30
6 / 100
= −1.37
 = .0853
c.
Power = 1 - 
at  = 26,
Power = 1 - .0853 = .9147
When  = 26, there is a .9147 probability that the test will correctly reject the null hypothesis that
 = 28.
51. a.
b.
Accepting H0 and letting the process continue to run when actually over - filling or under - filling
exists.
Decision Rule: Reject H0 if z  -1.96 or if z  1.96 indicates
Accept H0 if 15.71 < x < 16.29
Reject H0 if x  15.71 or if x  16.29
For  = 16.5
z=
16.29 − 16.5
.8 / 30
= −1.44
 = .0749
c.
Power = 1 - .0749 = .9251
9 - 22
© 2017 Cengage Learning® . May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Hypothesis Tests
d. The power curve shows the probability of rejecting H0 for various possible values of . In particular,
it shows the probability of stopping and adjusting the machine under a variety of underfilling and
overfilling situations. The general shape of the power curve for this case is
1.00
.75
.50
Power
.25
.00
15.6
15.8
16.0 16.2
16.4
Possible Values of u
c = 0 + z.01
52.

= 15 + 2.33
n
At  =  z =
16.32 − 17
4 / 50
4
50
= 16.32
= −1.20
 = .1151
At  =  z =
16.32 − 18
4 / 50
= −2.97
 = .0015
Increasing the sample size reduces the probability of making a Type II error.
53. a.
b.
Accept   100 when it is false.
Critical value for test:
c = 0 + z.05

n
At  = 120 z =
= 100 + 1.645
119.51 − 120
75 / 40
75
40
= 119.51
= −.04
 = .4840
c.
At  = 130 z =
119.51 − 130
75 / 40
= −.88
 = .1894
d.
Critical value for test:
9 - 23
© 2017 Cengage Learning® . May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 9
c = 0 + z.05

n
= 100 + 1.645
At  =  z =
113.79 − 120
75 / 80
75
80
= 113.79
= −.74
 = .2296
At  =  z =
113.79 − 130
75 / 80
= −1.93
 = .0268
Increasing the sample size from 40 to 80 reduces the probability of making a Type II error.
( z + z ) 2  2
54.
n=
55.
n=
56.
At 0 = 3,
( 0 −  a )
2
( z + z ) 2  2
( 0 −  a ) 2
=
(1.645 + 1.28)2 (5)2
= 214
(10 − 9)2
=
(1.96 + 1.645) 2 (10) 2
= 325
(20 − 22) 2
 = .01.
z.01 = 2.33
At a = 2.9375,  = .10.
z.10 = 1.28
 = .18
n=
57.
( z + z  ) 2  2
( 0 −  a ) 2
=
(2.33 + 1.28) 2 (.18) 2
= 108.09 Use 109
(3 − 2.9375) 2
At 0 = 400,
 = .02.
z.02 = 2.05
At a = 385,
 = .10.
z.10 = 1.28
 = 30
n=
58.
( z + z ) 2  2
( 0 −  a )
2
=
(2.05 + 1.28) 2 (30) 2
= 44.4 Use 45
(400 − 385) 2
At 0 = 28,
 = .05. Note however for this two - tailed test, z / 2 = z.025 = 1.96
At a = 29,
 = .15.
z.15 = 1.04
 =6
n=
( z / 2 + z ) 2  2
( 0 −  a ) 2
=
(1.96 + 1.04) 2 (6) 2
= 324
(28 − 29) 2
9 - 24
© 2017 Cengage Learning® . May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Hypothesis Tests
59.
At 0 = 25,
 = .02.
z.02 = 2.05
At a = 24,
 = .20.
z.20 = .84
 =3
n=
60. a.
( z + z ) 2  2
( 0 −  a ) 2
=
(2.05 + .84) 2 (3) 2
= 75.2 Use 76
(25 − 24) 2
H0:  = 16
Ha:   16
b.
z=
x − 0
/ n
=
16.32 − 16
.8 / 30
= 2.19
Because z > 0, p-value is two times the upper tail area
Using normal table with z = 2.19: p-value = 2(.0143) = .0286
p-value  .05; reject H0. Readjust production line.
c.
z=
x − 0
/ n
=
15.82 − 16
.8 / 30
= −1.23
Because z < 0, p-value is two times the lower tail area
Using normal table with z = -1.23: p-value = 2(.1093) = .2186
p-value > .05; do not reject H0. Continue the production line.
d.
Reject H0 if z  -1.96 or z  1.96
For x = 16.32, z = 2.19; reject H0
For x = 15.82, z = -1.23; do not reject H0
Yes, same conclusion.
61. a.
H0:  = 900
Ha:   900
b.
x  z.025

n
935  1.96
180
200
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© 2017 Cengage Learning® . May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 9
935  25
c.
d.
(910 to 960)
Reject H0 because  = 900 is not in the interval.
x − 0
935 − 900
z=
=
= 2.75
 / n 180 / 200
Because z > 0, p-value is two times the upper tail area
Using normal table with z = 2.75: p-value = 2(.0030) = .0060
62. a.
b.
H0:  ≤ 4
Ha:  > 4
z=
x − 0
/ n
=
4.5 − 4
1.5 / 60
= 2.58
p-value is the upper tail area at z =2.58
Using normal table with z = 2.58: p-value = 1.0000 - .9951 = .0049
c.
63.
p-value  .01, reject H0. Conclude that the mean daily background television that children
from low-income families are exposed to is greater than four hours.
The hypothesis test that, if rejected, will allow us to conclude that the mean starting salary for
business graduates in 2013 is significantly higher than the mean starting salary for business
graduates in 2012 is as follows.
H0:  ≤ 53,900
Ha:  > 53,900
t=
x − 0
s/ n
=
55,144 − 53,900
5200 / 100
= 2.392
Degrees of freedom = n – 1 = 100 – 1 = 99
p-value is upper-tail area
Using t table: p-value is less than .01
Exact p-value corresponding to t = 2.392 is .0093
p-value  .01; reject H0. The mean starting salary for business majors has increased in 2013.
64.
H0:  ≤ 30.8
Ha:  > 30.8
t=
x − 0
s/ n
=
32.7234 − 30.8
12.1041/ 47
= 1.0894
Degrees of freedom = n – 1 = 46
Using t table, area in the upper tail is between .20 and .10 and p-value is between .10 and .20.
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© 2017 Cengage Learning® . May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Hypothesis Tests
Exact p-value is .1408
Because p-value > α = .05, do not reject H0. There is no evidence to conclude that the mean age of
recently wed British men exceeds the mean age for those wed in 2013.
65. a.
b.
H0:  ≤ 520
Ha:  > 520
Sample mean: 637.94
Sample standard deviation: 148.4694
t=
x − 0
s/ n
=
637.94 − 520
148.4694 / 50
= 5.62
Degrees of freedom = n – 1 = 49
p-value is the area in the upper tail
Using t table: p-value is < .005
Exact p-value corresponding to t = 5.62  0
c.
We can conclude that the mean weekly pay for all women is higher than that for women with only a
high school degree.
d.
Using the critical value approach we would:
Reject H0 if t  t.05 = 1.677
Since t = 5.62 > 1.677, we reject H0.
66.
H0:   125,000
Ha:  > 125,000
t=
x − 0
s/ n
=
130,000 − 125,000
12,500 / 32
= 2.26
Degrees of freedom = 32 – 1 = 31
Upper tail p-value is the area to the right of the test statistic
Using t table: p-value is between .01 and .025
Exact p-value corresponding to t = 2.26 is .0155
p-value  .05; reject H0. Conclude that the mean cost is greater than $125,000 per lot.
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Chapter 9
H0:  = 86
Ha:   86
67.
x = 80
s = 20
t=
x − 0
s/ n
=
80 − 86
20 / 40
= −1.90
Degrees of freedom = 40 - 1 = 39
Because t < 0, p-value is two times the lower tail area
Using t table: area in lower tail is between .025 and .05; therefore, p-value is between .05 and .10.
Exact p-value corresponding to t = -1.90 is .0648
p-value > .05; do not reject H0.
There is not a statistically significant difference between the population mean for the nearby county
and the population mean of 86 days for Hamilton county.
68. a.
H0: p  .80
Ha: p  .80
p=
z=
455
= .84
542
p − p0
p0 (1 − p0 )
n
=
.84 − .80
.80(1 − .80)
542
= 2.33
p-value is the area in the upper tail
Using normal table with z = 2.33: p-value = 1.0000 - .9901 = .0099
p-value  .05; reject H0. We conclude that over 80% of airline travelers feel that use of the full body
scanners will improve airline security.
b.
H0: p  .75
Ha: p  .75
p=
z=
423
= .78
542
p − p0
p0 (1 − p0 )
n
=
.78 − .75
.75(1 − .75)
542
= 1.61
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© 2017 Cengage Learning® . May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Hypothesis Tests
p-value is the area in the upper tail
Using normal table with z = 1.61: p-value = 1.0000 - .9463 = .0537
p-value > .01; we cannot reject H0. Thus, we cannot conclude that over 75% of airline travelers
approve of using full body scanners. Mandatory use of full body scanners is not
recommended.
Author’s note: The TSA is also considering making the use of full body scanners optional. Travelers
would be given a choice of a full body scan or a pat down search.
69. a.
H0: p = .6667
Ha: p  .6667
b.
p=
c.
z=
355
= .6502
546
p − p0
p0 (1 − p0 )
n
=
.6502 − .6667
.6667(1 − .6667)
546
= −.82
Because z < 0, p-value is two times the lower tail area
Using normal table with z = -.82: p-value = 2(.2061) = .4122
p-value > .05; do not reject H0; Cannot conclude that the population proportion differs from 2/3.
70. a.
H0: p  .30
Ha: p > .30
b.
p=
c.
z=
136
= .34 (34%)
400
p − p0
p0 (1 − p0 )
n
=
.34 − .30
.30(1 − .30)
400
= 1.75
p-value is the upper-tail area
Using normal table with z = 1.75: p-value = 1.0000 - .9599 = .0401
d.
71. a.
b.
p-value  .05; reject H0. Conclude that more than 30% of the millennials either live at home with
their parents or are otherwise dependent on their parents.
H0: p  .079
Ha: p > .079
z=
p − p0
=
.108 − .079
= 2.15
p0 (1 − p0 )
(.079)(1 − .079)
400
n
Upper tail p-value is the area to the right of the test statistic
9 - 29
© 2017 Cengage Learning® . May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 9
Using normal table with z = 2.15: p-value = 1-.9842 = .0158
Since p-value = .0158   = .05 , reject H 0 . We can conclude that the unemployment rate for
18 to 34 year olds is significantly higher than the national unemployment rate of 7.9% for all adults.
c.
You can tell the campaign manager that the observed level of significance is .0158 and that this
means the results would have been significant even with  as small as .0158. Most reasonable
people would reject the null hypotheses and conclude that the proportion of adults aged 18 to 34 who
were unemployed was higher than the national unemployment rate for all adults.
H0: p  .90
Ha: p < .90
72.
p=
z=
49
= .8448
58
p − p0
p0 (1 − p0 )
n
=
.8448 − .90
.90(1 − .90)
58
= −1.40
Lower tail p-value is the area to the left of the test statistic
Using normal table with z = -1.40: p-value =.0808
p-value > .05; do not reject H0. Claim of at least 90% cannot be rejected.
73. a.
The point estimate of the proportion of people aged 65-69 working is
p=
180
= .30
600
b.
H0: p  .27
Ha: p > .27
c.
z=
p − p0
p0 (1 − p0 )
n
=
.30 − .27
.27(1 − .27)
600
= 1.66
p-value is the upper tail area at z = 1.66
Using normal table with z = 1.66: p-value = 1 - .9515 = .0485
p-value  .05; reject H0.
We conclude that the proportion of people aged 65-69 has increased since 2007.
74. a.
H0:   72
Ha:  > 72
Reject H0 if z  1.645
9 - 30
© 2017 Cengage Learning® . May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Hypothesis Tests
z=
x − 0
/ n
=
x − 72
20 / 30
= 1.645
Solve for x = 78
Decision Rule:
Accept H0 if x < 78
Reject H0 if x  78
b.
For  = 80
z=
78 − 80
20 / 30
= −.55
 = .2912
c.
For  = 75,
z=
78 − 75
20 / 30
= .82
 = .7939
d.
For  = 70, H0 is true. In this case the Type II error cannot be made.
e.
Power = 1 - 
1.0
.8
P
o
w
e
r
.6
.4
.2
72
75.
76
78
80
74
Possible Values of 
Ho False
82
84
H0:   15,000
Ha:  < 15,000
At 0 = 15,000,  = .02.
z.02 = 2.05
At a = 14,000,  = .05.
z.10 = 1.645
9 - 31
© 2017 Cengage Learning® . May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 9
n=
( z + z )2  2
( 0 −  a )
2
=
(2.05 + 1.645)2 (4, 000)2
= 218.5
(15, 000 − 14, 000) 2
Use 219
H0:  = 120
Ha:   120
76.
At 0 = 120,
 = .05. With a two - tailed test, z / 2 = z.025 = 1.96
At a = 117,
 = .02.
n=
b.
( z / 2 + z ) 2  2
( 0 −  a ) 2
=
z.02 = 2.05
(1.96 + 2.05) 2 (5) 2
= 44.7 Use 45
(120 − 117) 2
Example calculation for  = 118.
Reject H0 if z  -1.96 or if z  1.96
z=
x − 0
/ n
=
x − 120
5 / 45
Solve for x .
At z = -1.96, x = 118.54
At z = +1.96, x = 121.46
Decision Rule:
Accept H0 if 118.54 < x < 121.46
Reject H0 if x  118.54 or if x  121.46
For  = 118,
z=
118.54 − 118
5 / 45
= .72
 = .2358
Other Results:
If  is
117
118
119
121
122
123
z
2.07
.72
-.62
+.62
+.72
-2.07

.0192
.2358
.7291
.7291
.2358
.0192
9 - 32
© 2017 Cengage Learning® . May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.