45. a. H0: p = .39
Ha: p .39
.385p=
0
00
.385 .39 .18
(1 ) .39(1 .39)
300
pp
zpp
n
−−
= = = −
−−
b.
0: .30Hp
0: .30Hp
.399p=
0
.399 .30 3.74
pp
−−
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.
46. a. H0:
= 16
Ha:
16
016.32 16 2.19
x
−−
c.
015.82 16 1.23
/ .8/ 30
x
zn
−−
= = = −
Because z < 0, p-value is two times the lower tail area
Using normal table with z = -1.23: p-value = 2(.1093) = .2186
Using Excel: p-value = 2*NORM.S.DIST(-1.23,TRUE) = . 2187
47. a. H0:
= 900
Ha:
900
b.
.025
xz n
180
935 1.96 200
d.
0935 900 2.75
/ 180/ 200
x
zn
−−
= = =
Because z > 0, p-value is two times the upper tail area
48. a. H0:
4
Ha:
> 4
b.
04.5 4 2.58
/ 1.5/ 60
x
zn
−−
= = =
p-value is the upper tail area at z =2.58
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
055,144 53,900 2.392
/ 5200 / 100
x
tsn
−−
= = =
Ha:
25
024.0476 25.0 1.05
/ 5.8849 / 42
x
tsn
−−
= = = −
Degrees of freedom = n – 1 = 41
Because t < 0, p-value is two times the lower tail area
Ha:
> 520
b. Sample mean: 637.94
Sample standard deviation: 148.4694
0637.94 520 5.62
/ 148.4694 / 50
x
tsn
−−
= = =
Degrees of freedom = n – 1 = 49
d. Using the critical value approach we would:
Reject H0 if t
.05
t
= 1.677
Since t = 5.62 > 1.677, we reject H0.
52. H0:
125,000
Ha:
> 125,000
0130,000 125,000 2.26
/ 12,500/ 32
x
−−
Degrees of freedom = 32 – 1 = 31
p-value is upper-tail area
Using t table: p–value is between .01 and .025
Using Excel: p-value = 1-T.DIST(2.26,31,TRUE) = .0155
p-value .05; reject H0. Conclude that the mean cost is greater than $125,000 per lot.
53. H0:
= 86
Ha:
86
80x=
20s=
080 86 1.90
x
−−
54. a. H0: p .80
Ha: p .80
455 .84
542
p==
0
.84 .80 2.33
pp
−−
b. H0: p .75
Ha: p .75
423 .78
542
p==
0
00
.78 .75 1.61
(1 ) .75(1 .75)
542
pp
zpp
n
−−
= = =
−−
55. a. H0: p ≤ .39
Ha: p > .39
b.
0
00
.41 .39 .87
(1 ) .39(1 .39)
450
pp
zpp
n
−−
= = =
−−
56. a. H0: p .30
Ha: p > .30
b.
136 .34
400
p==
(34%)
c.
0
00
.34 .30 1.75
(1 ) .30(1 .30)
400
pp
zpp
n
−−
= = =
−−
57. a. H0: p .079
Ha: p > .079
b.
0
00
.108 .079 2.15
(1 ) (.079)(1 .079)
400
pp
zpp
n
−−
= = =
−−
Upper tail p-value is the area to the right of the test statistic
Using normal table with z = 2.15: p-value = 1-.9842 = .0158
58. H0: p .90
Ha: p < .90
49 .8448
58
p==
0
00
.8448 .90 1.40
(1 ) .90(1 .90)
58
pp
n
−−
−−
p-value is the lower-tail area
Using normal table with z = -1.40: p-value =.0808
Using Excel: p-value = NORM.S.DIST(-1.40,TRUE) = . 0808
p-value > .05; do not reject H0. Claim of at least 90% cannot be rejected.
59. a. The point estimate of the proportion of people aged 65–69 working is
180 .30
600
p==
b. H0: p .27
Ha: p > .27
0
.30 .27 1.66
pp
−−