12.83 a
n/)p
ˆ
1(p
ˆ
zp
ˆ2/ −
= .50
1.96
400/)50.1(50.−
= .50
.0490
n/)p
ˆ
1(p
ˆ
zp
ˆ2/ −
n/)p
ˆ
1(p
ˆ
zp
ˆ2/ −
400/)10.1(10.−
12.84
p:H0
= .60
p:H1
> .60
a
n/)p1(p
pp
ˆ
z−
−
=
=
100/)60.1(60.
60.63.
−
−
= .61, p-value = P(Z > .61) = 1 – .7291 =.2709
n/)p1(p
n/)p1(p
pp
ˆ
z−
−
=
d The p-value decreases.
12.85 a
n/)p1(p
pp
ˆ
z−
−
=
=
100/)70.1(70.
70.73.
−
−
= .65, p-value = P(Z > .65) = 1 – .7422 =.2578
n/)p1(p
pp
ˆ
z−
−
=
100/)70.1(70.
−
−
n/)p1(p
pp
ˆ
z−
−
=
100/)70.1(70.
70.71.
−
−
12.87a .5
.03
b Yes, because the sample size was chosen to produce this interval.
12.89 n =
2
2/
B
)p
ˆ
1(p
ˆ
z
−
=
2
03.
)75.1(75.645.1
−
= 564
12.90a .75
.03
b Yes, because the sample size was chosen to produce this interval.
n/)p
ˆ
1(p
ˆ
zp
ˆ2/ −
12.92 a
n/)p
ˆ
1(p
ˆ
zp
ˆ2/ −
= .5
1.645
564/)5.1(5. −
= .5
.0346
b The interval is wider.
c No because the interval estimate is wider (worse) than specified.
12.93
p
ˆ
= 259/373 = .69
n/)p
ˆ
1(p
ˆ
zp
ˆ2/ −
= .69
1.96
373/)69.1(69.−
= .69
.0469; LCL = .6431, UCL = .7369
p
ˆ
12.95
p
ˆ
= 204/314 = .65
n/)p
ˆ
1(p
ˆ
zp
ˆ2/ −
= .65
1.645
314/)65.1(65.−
= .65
.0443; LCL = .6057, UCL = .6943
p
ˆ
12.97
p
ˆ
= 97/344 = .28
n/)p
ˆ
1(p
ˆ
zp
ˆ2/ −
= .28
1.96
344/)28.1(28.−
= .28
.0474; LCL = .2326, UCL = .3274
p
ˆ
12.99 LCL = .1332(1,000,000)(3.00) = $399,600, UCL = .2068(1,000,000)(3.00) = $620,400
12.100
0147.
4200
21
4n
2x
p
~=
+
+
=
+
+
=
12.101
0132.
4374
23
4n
2x
p
~=
+
+
=
+
+
=
12.102
0077.
4385
21
4n
2x
p
~=
+
+
=
+
+
=
12.103
n/)p
ˆ
1(p
ˆ
zp
ˆ2/ −
= .2680
1.96
974/)2680.1(2680.−
= .2680
.0278; LCL = .2402, UCL
= .2958
12.105
n/)p
ˆ
1(p
ˆ
zp
ˆ2/ −
= .1056
1.96
521/)1056.1(1056. −
= .1056
.0264; LCL = .0792, UCL = .1320
12.106 LCL = 75,000(.0792) =5,940, UCL = 75,000(.1320) = 9,900
n/)p
ˆ
1(p
ˆ
zp
ˆ2/ −
12.108
p:H0
= .90
p:H1
< .90
12.109
n/)p
ˆ
1(p
ˆ
zp
ˆ2/ −
= .2333
1.96
120/)2333.1(2333. −
= .2333
.0757; LCL = .1576,
UCL = .3090
n/)p
ˆ
1(p
ˆ
zp
ˆ2/ −
n/)p
ˆ
1(p
ˆ
zp
ˆ2/ −
12.112
p:H0
= .50
p:H1
> .50
12.113 Codes 1, 2, and 3 have been recoded to 5.
p:H0
n/)p1(p
pp
ˆ
z−
−
=
=
100/)90.1(90.
90.96.
−
−
= 2.00, p-value = P(Z > 2.00) = 1 – .9772 =.0228. There is enough evidence
to conclude that more than 90% of all business students would rate the book as at least adequate.
12.115 H0: p = .2155
H1: p ≠ .2155
n/)p1(p
pp
ˆ
z−
−
=
12.116a.
n/)p
ˆ
1(p
ˆ
zp
ˆ2/ −
= .2031
1.96
650/)2031.1(2031. −
= .2031
.0309; LCL = .1722,
12.117
2.118a
Proportion: LCL = .4438, UCL = .4807. Number: LCL = 104,094,270, UCL = 112,755,571
b
Proportion: LCL = .0429, UCL = .0626. Number: LCL = 10,051,434, UCL = 14,677,058
12.119 H0: p = .10
H1: p > .10
12.120
12.121
12.122 p = proportion of Republican supporters among Democrats and Republicans only
H0: p = .5
H1: p < .5
12.123
12.124 p = proportion of conservatives among liberals and conservatives only
H0: p = .5
H1: p > .5
12.125
12.126
12.127 p = proportion of moderates among moderates and conservatives only
H0: p = .5
H1: p > .5
12.1218
12.129
12.130
12.131
12.132 H0: p = .33
H1: p < .33
12.133 p = proportion of Americans who support the Democrats among Democrats and Republicans only
H0: p = .5
H1: p > .5
z = 7.26, p-value = 0. There is sufficient evidence to infer that there are more Democrats than Republicans.
12.134 p = proportion of Americans who are conservatives among liberals and conservatives only
12.135
12.136 Codes 3 and 4 were changed to 5
12.137 Code 2 was changed to 3.
12.138a.
n/)p
ˆ
1(p
ˆ
zp
ˆ2/ −
= .2919
1.96
1836/)2919.1(2919.−
= .2919
.0208; LCL = .2711,
UCL = .3127
b LCL = 120,728,000 (.2711) = 32,729,361, UCL = 120,728,000 (.3127) = 37,751,646
n/)p
ˆ
1(p
ˆ
zp
ˆ2/ −
n/)p
ˆ
1(p
ˆ
zp
ˆ2/ −
n/)p
ˆ
1(p
ˆ
zp
ˆ2/ −
UCL = .1880; Number: LCL = 203,882,000(.1120) = 22,834,784, UCL = 203,882,000(.1880) = 38,329,816
12.142
12.143
12.144 a
:H0
= 30
:H1
> 30
b
12.145a
b
11.46
2
0:H
= 17
2
1:H
> 17
12.147
12.148
12.149
12.150
12.151
12.152a
b
:H0
= 68
:H1
> 68
12.153
12.154 H0: p = .75
H1: p > .75