Solutions:
1.
a.
H0:   600
Manager’s claim.
Ha:  > 600
3.
7.
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:  = 32
Specified filling weight
Ha:   32
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.
a.
H0:   8000
Ha:  > 8000
9.
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.
z
b.
x  0
/ n

19.4  20
2 / 50
 2.12
Area = .4830
p-value = .5000 - .4830 = .0170
c.
p-value  .05, reject H0
d.
Reject H0 if z  -1.645
-2.12  -1.645, reject H0
9-1
Chapter 9
Reject H0 if z  1.645
13.
a.
z
x  0
/ n

52.5  50
8 / 60
 2.42
2.42  1.645, reject H0
b.
z
x  0
/ n

51  50
8 / 60
 .97
.97 < 1.645, do not reject H0
c.
z
x  0
/ n

51.8  50
8 / 60
 1.74
1.74  1.645, reject H0
17. a.
H0:   44.75
Ha:   44.75
b.
z
x   0 43.95  44.75

 1.54
/ n
5.48 / 111
p-value = 2(.5000 - .4382) = .1236
c.
p-value > .05, do not reject H0. We cannot conclude that the mean length of a work week has
changed.
d.
Reject H0 if z  -1.96 or z  1.96
z = -1.54; cannot reject H0
19.
H0:   45064
Ha:  > 45064
z
x  0
/ n

46196  45064
4562.96 / 75
 2.15
Area = .4842
p-value = .5000 - .4842 = .0158
p-value  .05, reject H0. Conclude that there has been an increase in the mean monthly earnings of
service industry workers.
9-2
Hypothesis Testing
21. a.
H0:   15
Ha:  > 15
b.
z
x 
/ n

17  15
4 / 35
 2.96
c.
p-value = .5000 - .4985 = .0015
d.
p-value  .01; reject H0; the premium rate should be charged.
23. a.
b.
t
x  0
s/ n

14  12
4.32 / 25
 2.31
Degrees of freedom = n – 1 = 24
Using t table, p-value is between .01 and .025.
Actual p-value = .0147
c.
p-value  .05, reject H0.
c.
With df = 24, t.05 = 1.711
Reject H0 if t  1.711
2.31 > 1.711, reject H0.
25. a.
t
x  0
s/ n

44  45
5.2 / 36
 1.15
Degrees of freedom = n – 1 = 35
Using t table, p-value is between .10 and .20
Actual p-value = .1282
p-value > .01, do not reject H0
b.
t
x  0
s/ n

43  45
4.6 / 36
 2.61
Using t table, p-value is between .005 and .01
Actual p-value = .0066
p-value  .01, reject H0
9-3
Chapter 9
c.
t
x  0
s/ n

46  45
5 / 36
 1.20
Using t table, area in upper tail is between .10 and .20
p-value (lower tail) is between .80 and .90
Actual p-value = .8809
p-value > .01, do not reject H0
29. a.
H0:  = 5600
Ha:   5600
b.
t
x  0
s/ n

5835  5600
520 / 25
 2.26
Degrees of freedom = n – 1 = 24
Using t table, area in tail is between .01 and .025
Actual p-value = .0332
c.
p-value  .05; reject H0. The mean diamond price in New York City differs.
d.
df = 24
t.025 = 2.064
Reject H0 if t < -2.064 or t > 2.064
2.26 > 2.064; reject H0
31.
H0:   4946
Ha:  > 4946
t
x  0
s/ n

5310  4946
1249 / 64
 2.33
Degrees of freedom = n - 1 = 63
Using t table, p-value is between .01 and .025
Actual p-value = .0114
Reject H0; High income households are paying a higher mean water bill.
9-4
Hypothesis Testing
33. a.
H0:   280
Ha:  > 280
b.
286.9 - 280 = 6.9 yards
c.
t
x  0
s/ n

286.9  280
10 / 9
 2.07
Degrees of freedom = n – 1 = 8
Using t table, p-value is between .025 and .05
Actual p-value = .0361
c.
35. a.
b.
p-value  .05; reject H0. The population mean distance for the new driver is greater than the USGA
approved driver.
z
p  p0
p0 (1  p0 )
n

.175  .20
.20(1  .20)
400
 1.25
Area in tail = (.5000 - .3944) = .1056
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
37. a.
H0: p  .40
Ha: p > .40
b.
p
z
189
 .45
425
p  p0
p0 (1  p0 )
n

.45  40
.40(1  .40)
425
 1.88
Area = .4699
p-value = .5000 - .4699 = .0301
c.
p-value  .05; reject H0. Conclude more than 40% access less than seven hours per week.
9-5
Chapter 9
41. a.
H0: p  .75
Ha: p < .75
b.
z
p  p0
p0 (1  p0 )
n

.72  .75
.75(1  .75)
300
 1.20
Area = .3849
p-value = .5000 - .3849 = .1151
c.
43. a.
p-value > .05; do not reject H0. The executive's claim cannot be rejected.
H0: p  .48
Ha: p  .48
b.
p
z
360
 .45
800
p  p0
p0 (1  p0 )
n

.45  .48
.48(1  .48)
800
 1.70
Area = .4554
p-value = 2(.5000 - .4554) = .0892
c.
p-value > .05; do not reject H0. There is no reason to conclude the proportion has changed.
9-6
Hypothesis Testing
Reject H0 if z  -1.96 or if z  1.96
47.
x 

n

10
200
 .71
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
 = .5000 - .3051 = .1949
b.
 = 22.5
z
21.39  22.5
10 / 200
 1.57
 = .5000 - .4418 = .0582
c.
 = 21
z
21.39  21
10 / 200
 .55
 = .5000 +.2088 = .7088
9-7
Chapter 9
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
 = .5000 -.4251 = .0749
c.
Power = 1 - .0749 = .9251
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
9-8
Hypothesis Testing
1.00
.75
.50
Power
.25
.00
15.6
15.8
16.0 16.2
Possible Values of u
( z  z )2  2
n
59.
At 0 = 25,
 = .02.
z.02 = 2.05
At a = 24,
 = .20.
z.20 = .84
( 0   a ) 2

(1.96  1.645)2 (10)2
 325
(20  22)2
55.
 =3
n
( z  z )2  2
( 0   a ) 2

(2.05  .84)2 (3)2
 75.2 Use 76
(25  24)2
9-9
16.4