12
12.1 Chapter a x t b x t c x t
/ 2s / 2s / 2s
/ n = 510
2.064(125/ 25 ) = 510
51.60; LCL = 458.40, UCL = 561.60
/ n = 510 / n = 510
2.009(125/ 50 ) = 510 1.984(125/ 100 ) = 510
35.51; LCL = 474.49, UCL = 545.51 24.80; LCL = 485.20, UCL = 534.80
d. The interval narrows.
12.2 a x t b x t c x t
/ 2s / 2s
/ 2s
/ n = 1,500
1.984(300/ 100 ) = 1,500
59.52; LCL = 1,440.48, UCL = 1,559.52
/ n = 1,500 / n = 1,500
1.984(200/ 100 ) = 1,500 1.984(100/ 100 ) = 1,500
39.68; LCL = 1,460.32, UCL = 1,539.68 19.84; LCL = 1,480.16, UCL = 1,519.84
d. The interval narrows.
12.3 a x t b x t a x t
/ 2s / 2s
/ 2s
/ n = 700
1.645(100/ 400 ) = 700
8.23; LCL = 691.77, UCL = 708.23
/ n = 700 / n = 700
1.96(100/ 400 ) = 700 2.576(100/ 400 ) = 700
9.80; LCL = 690.20, UCL = 709.80 12.88; LCL = 687.12, UCL = 712.88
d. The interval widens.
12.4 a x t b x t c x t
/ 2s / 2s
/ 2s
/ n = 10
1.984(1/ 100 ) = 10
.20; LCL = 9.80, UCL = 10.20
/ n = 10 / n = 10
1.984(4/ 100 ) = 10 1.984(10/ 100 ) = 10
.79; LCL = 9.21, UCL = 10.79 1.98; LCL = 8.02, UCL = 11.98
d The interval widens.
12.5 a x t b x t c x t
/ 2s / 2s
/ 2s
/ n = 120
2.009(15/ 51 ) = 120
4.22; LCL = 115.78, UCL = 124.22
/ n = 120 / n = 120
1.676(15/ 51 ) = 120 1.299(15/ 51 ) = 120
3.52; LCL = 116.48, UCL = 123.52 2.73; LCL = 117.27, UCL = 122.73
d The interval narrows.
12.6 a x t b x t c x t
/ 2s / 2s
/ 2s
/ n = 63
1.990(8/ 81 ) = 63
1.77; LCL = 61.23, UCL = 64.77
/ n = 63 / n = 63
2.000(8/ 64 ) = 63 2.030(8/ 36 ) = 63
2.00; LCL = 61.00, UCL = 65.00 2.71; LCL = 60.29, UCL = 65.71
237
d The interval widens.
12.7
H0 :
= 20 > 20
t
,n 1
H1 :
a Rejection region: t
t x s/ n 23 20
t .05 ,9 = 1.833
1 .05 , p-value = .1597. There is not enough evidence to infer that the population
9 / 10
mean is greater than 20. b Rejection region: t
t x s/ n 23 20
t
,n 1
t .05 , 29 = 1.699
1 .83 , p-value = .0391. There is enough evidence to infer that the population mean is
9 / 30
greater than 20. c Rejection region: t
t x s/ n 23 20
t
,n 1
t .05 , 49
1.676
2 .36 , p-value = .0112. There is enough evidence to infer that the population mean is
9 / 50
greater than 20. d As the sample size increases the test statistic increases [and the p-value decreases].
12.8
H0 :
H1 :
= 180 180
t
180 200
/ 2, n 1
Rejection region: t at
x s/ n 175 22 /
t .025 ,199
1 .972 or t
t
/ 2, n 1
t .025 ,199 = 1.972
3 .21, p-value = .0015. There is enough evidence to infer that the population
mean is not equal to 180. b t
x s/ n 175 45 / 180 200 1 .57 , p-value = .1177. There is not enough evidence to infer that the
population mean is not equal to 180. c t
x s/ n 175 60 / 180 200 1 .18 , p-value = .2400. There is not enough evidence to infer that the
population mean is not equal to 180. d. As the s increases, the test statistic increases and the p-value increases.
12.9 Rejection region: t
t
,n 1
t .05 ,99
1 .660
238
at
x s/ n
145
150
1 .00 , p-value = .1599. There is not enough evidence to infer that the population
50 / 100
mean is less than 150. bt
x s/ n 140 150 2 .00 , p-value = .0241. There is enough evidence to infer that the population
50 / 100
mean is less than 150. ct
x s/ n 135 150 3 .00 , p-value = .0017. There is enough evidence to infer that the population
50 / 100
mean is less than 150 d The test statistics decreases and the p-value decreases.
12.10
H0 : H0 :
= 50 50
t
/ 2, n 1
a Rejection region: t
t x s/ n 52 15 / 50 25
t .05 , 24
1 .711 or t
t
/ 2,n 1
t .05 , 24
1 .711
.67 , p-value = .5113. There is not enough evidence to infer that the population
mean is not equal to 50. b Rejection region: t
t x s/ n 52 50
t
/ 2,n 1
t .05 ,14
1 .761 or t
t
/ 2, n 1
t .05 ,14
1 .761
.52 , p-value = .6136. There is not enough evidence to infer that the population mean
15 / 15
is not equal to 50. c Rejection region: t
t x s/ n 52 50
t
/ 2, n 1
t .05 , 4
2 .132 or t
t
/ 2, n 1
t .05 , 4
2 .132
.30 , p-value = .7804. There is not enough evidence to infer that the population mean
15 / 5
is not equal to 50. d The test statistic decreases and the p-value increases.
12.11 Rejection region: t at
x s/ n 585 600
t
,n 1
t .10 , 49
1.299
2 .36 , p-value = .0112. There is enough evidence to infer that the population
45 / 50
mean is less than 600. bt
x s/ n 590 600 1 .57 , p-value = .0613. There is enough evidence to infer that the population
45 / 50
mean is less than 600.
239
ct
x s/ n
595
600
.79 , p-value = .2179. There is not enough evidence to infer that the population
45 / 50
mean is less than 600. d The test statistic increases and the p-value increases.
12.12 Rejection region: t at
x s/ n 106 100
t
,n 1
t .01,99
2 .364
1 .71, p-value = .0448. There is not enough evidence to infer that the population
35 / 100
mean is greater than 100. bt
x s/ n 106 100 2 .40 , p-value = .0091. There is enough evidence to infer that the population
25 / 100
mean is greater than 100. ct
x s/ n 106 100 4 .00 , p-value = .0001. There is enough evidence to infer that the population mean
15 / 100
is greater than 100 d The test statistic increases and the p-value decreases.
12.13 a x t b x z
/2
/ 2s
/ n = 40
2.365(10/ 8 ) = 40
8.36; LCL = 31.64, UCL = 48.36
/ n = 40
1.96(10/ 8 ) = 40
6.93; LCL = 33.07, UCL = 46.93
/2
c The student t distribution is more widely dispersed than the standard normal; thus, z
t
/2.
is smaller than
12.14 a x t b x z
/2
/ 2s
/ n = 175
2.132(30/ 5 ) = 175
28.60; LCL = 146.40, UCL = 203.60
/ n = 175
1.645(30/ 5 ) = 175
22.07; LCL = 152.93, UCL = 197.07
/2
c The student t distribution is more widely dispersed than the standard normal; thus, z
t
/2.
is smaller than
12.15 a x t 16,017.59 b x
z
/2
/ 2s
/ n = 15,500
1.645(9,950/ 1,000 ) = 15,500
517.59; LCL = 14,982.41, UCL =
/ n = 15,500
1.645(9,950/ 1,000 ) = 15,500
517.59; LCL = 14,982.41, UCL =
16,017.59 c With n = 1,000 the student t distribution with 999 degrees of freedom is almost identical to the standard normal distribution.
240
12.16 a x t b x z
/2
/ 2s
/ n = 350
2.576(100/ 500 ) = 350
11.52; LCL = 338.48, UCL = 361.52
/ n = 350
2.575(100/ 500 ) = 350
11.52; LCL = 338.48, UCL = 361.52
c With n = 500 the student t distribution with 999 degrees of freedom is almost identical to the standard normal distribution.
12.17
H0 : H0 :
= 70 > 70
t
,n 1
a Rejection region: t
t x s/ n 74 .5 70 9 / 11
t .05 ,10 = 1.812
1 .66 , p-value = .0641. There is not enough evidence to infer that the population
mean is greater than 70. b Rejection region: z
z x / n 74 .5 70 9 / 11
z
z .05 = 1.645
1 .66 , p-value = P(Z > 1.66) = .5 P(0 <Z< 1.66) = .5 .4515 = .0485. There is
enough evidence to infer that the population mean is greater than 70. c The Student t distribution is more dispersed than the standard normal.
12.18
H0 : H0 :
= 110 < 110
t
,n 1
a Rejection region: t
t x s/ n 103 110 17 / 10
t .10 ,9 = 1.383
1 .30 , p-value = .1126. There is not enough evidence to infer that the population
mean is less than 110. b Rejection region: z
z x / n 103 110 17 / 10
z
z .10 = 1.28
1 .30 , p-value = P(Z <1.30) = .5 P(0 <Z< 1.30) = .5 .4032 = .0968. There is
enough evidence to infer that the population mean is less than 110. c The Student t distribution is more dispersed than the standard normal.
12.19
H0 : H0 :
= 15 < 15
t
,n 1
a Rejection region: t
t .05 ,1499 = 1.645
241
t
x s/ n
14 15 25 / 1,500
1 .55 , p-value = .0608. There is not enough evidence to infer that the population
mean is less than 15. b Rejection region: z
z x / n 14 15 25 / 1,500
z
z .05 = 1.645
1 .55 , p-value = P(Z < 1.55) = .5 P(0 <Z < 1.55) = .5 .4394 = .0606. There
is not enough evidence to infer that the population mean is less than 15. c With n = 1,500 the student t distribution with 1,499 degrees of freedom is almost identical to the standard normal distribution.
12.20 a Rejection region: t
t x s/ n 405 400
t
,n 1
t .05 ,999 = 1.645
1 .58 , p-value = .0569. There is not enough evidence to infer that the population
100 / 1,000
mean is less than 15. b Rejection region: z
t x s/ n 405 400
z
z .05 = 1.645
1 .58 , p-value = P(0 < Z < 1.58) = .5 .4429 = .0571. There is not enough
100 / 1,000
evidence to infer that the population mean is less than 15. c With n = 1,000 the student t distribution with 999 degrees of freedom is almost identical to the standard normal distribution.
12.21
H0 : H0 :
=6 <6
t
,n 1
a Rejection region: t
t x s/ n 5 .69 6
t .05 ,11 = 1.796
.68 , p-value = .2554. There is not enough evidence to support the courier's
1 .58 / 12
advertisement.
12.22 x t
/ 2s
/ n = 24,051
2.145(17,386/ 15 ) = 24,051
9,628; LCL = 14,422, UCL = 33,680
12.23
H0 : H0 :
= 20 > 20
t
,n 1
Rejection region: t
t .05 ,19
1 .729
242
t
x s/ n
20 .85 6 .76 /
20 20
.56 , p-value = .2902. There is not enough evidence to support the doctor's claim.
12.24
H0 : H0 :
=8 <8
t
,n 1
Rejection region: t
t x s/ n
t .01,17
2 .567
7 .91 8 .085 / 18
4.49, p-value = .0002. There is enough evidence to conclude that the average
container is mislabeled.
12.25 x t
/ 2s
/ n = 18.13
2.145(9.75/ 15 ) = 18.13
5.40; LCL = 12.73, UCL =23.53
12.26 x t
/ 2s
/ n = 26.67
1.796(16.52/ 12 ) = 26.67
8.56; LCL = 18.11, UCL =35.23
12.27 x t
/ 2s
/ n = 17.70
2.262(9.08/ 10 ) = 17.70
6.49; LCL = 11.21, UCL =24.19
12.28
H0 : H0 :
= 10 < 10
t
,n 1
Rejection region: t
t x s/ n 7 .10 10 3 .75 / 10
t .10 ,9
1 .383
2 .45 , p-value = .0185. There is enough evidence to infer that the mean
proportion of returns is less than 10%.
12.29 x t
/ 2s
/ n = 7.15
1.972(1.65/ 200 ) = 7.15
.23; LCL = 6.92, UCL = 7.38
12.30 x t
/ 2s
/ n = 4.66
2.576(2.37/ 240 ) = 4.66
.39; LCL = 4.27, UCL = 5.05
Total number: LCL = 100 million (4.27) = 427 million, UCL = 100 million (5.05) = 505 million
12.31
H0 :
= 60 60
t
/ 2,n 1
H0 :
Rejection region: t
t .025 ,161
1.975 or t
t
/ 2,n 1
1 .975
243
t
x s/ n
63 .70
60
2.49, p-value = .0140. There is enough evidence to infer that the mean time
18 .94 / 162
differs from 60 minutes.
12.32
H0 : H0 :
= 45 > 45
t
,n 1
Rejection region: t
t x s/ n
t .05 ,143
1.656
53 .78
45
26.01, p-value = 0. There is enough evidence to infer that the mean time
4 .05 / 144
exceeds 45 hours.
12.33 x t
/ 2s
/ n = 59.04
1.980(20.62/ 122 ) = 59.04
3.70; LCL = 55.34, UCL = 62.74
12.34 x t
/ 2s
/ n = 2.67
1.973(2.50/ 188 ) = 2.67
.36; LCL = 2.31, UCL = 3.03
12.35 a x t
/ 2s
/ n = 62.79
2.052(5.32/ 28 ) = 62.79
2.06; LCL = 60.73, UCL = 64.85
b Prices are required to be normally distributed. The histogram (not shown) is bell shaped.
12.36 x t
/ 2s
/ n = 29.14
2.009(4.62/ 49 ) = 29.14
1.33; LCL = 27.81, UCL = 30.47
12.37 x t
/ 2s
/ n = 13.94
1.960(2.16/ 212 ) = 13.94
.29; LCL = 13.65, UCL = 14.23
Package of 10: LCL = 13.65(10) = 136.5 days, UCL = 14.23(10) = 142.3 days.
12.38
H0 :
H0 :
= 15 > 15
t
,n 1
Rejection region: t
t x s/ n
t .05 ,115
1 .658
15 .27
15
.51, p-value = .3061. There is not enough evidence to infer that the mean
5 .72 / 116
number of commercials is greater than 15.
12.39 x t
/ 2s
/ n = 3.44
1.960(3.33/ 471 ) = 3.44
.30; LCL = 3.14, UCL = 3.74
Total: LCL = 270,509,000(3.14) = 849,398,260, UCL = 270,509,000(3.74) = 1,011,703,660
244
12.40
H0 :
H0 :
= 85 > 85
t
,n 1
Rejection region: t
t x s/ n
t .05 ,84
1.664
89 .27
85
2.28, p-value = .0127. There is enough evidence to infer that an e-grocery will
17 .30 / 85
be successful.
12.41
H0 : H0 :
=2 >2
t
,n 1
Rejection region: t
t x s/ n
t .01,99
2.364
2 .10
2
1.32, p-value = .0956. There is not enough evidence to infer that the recycling
.76 / 100
plant will be profitable.
12.42
H0 : H1 :
2
= 300 300
2 2 1 / 2,n 1 2 .975 , 99
2
a Rejection region:
2
74 .2219 or
2
2 / 2, n 1
2 .025 , 99
129 .561
( n 1)s 2
2
=
(100
1)( 220 ) 300
= 72.60, p-value = .0427. There is enough evidence to infer that the
population variance differs from 300. b Rejection region:
2
2 2 1 / 2,n 1 2 .975 , 49
32 .3574 or
2
2 / 2, n 1
2 .025 , 49
71 .4202
( n 1)s 2
2
=
(50 1)( 220 ) 300
= 35.93, p-value = .1643. There is not enough evidence to infer that the
population variance differs from 300. c Decreasing the sample size decreases the test statistic and increases the p-value of the test.
12.43
H0 :
H1 :
2
= 100 < 100
2 2 1 ,n 1 2 .99 , 49
2
a Rejection region:
2
29 .7067
( n 1)s 2
2
=
(50 1)(80 ) 100
= 39.20, p-value = .1596. There is not enough evidence to infer that the
population variance is less than 100. b Rejection region:
2 2 1 ,n 1 2 .99 , 99
70 .0648
245
2
( n 1)s 2
2
=
(100
1)(80 ) 100
= 79.20, p-value = .0714. There is not enough evidence to infer that the
population variance is less than 100. c Increasing the sample size increases the test statistic and decreases the p-value.
12.44 a LCL =
( n 1)s 2
2 / 2,n 1
=
( n 1)s 2
2 .05 ,14
=
(15 1)(12 ) 23 .6848
= 7.09
UCL =
( n 1)s 2
2 1 / 2,n 1
=
( n 1)s 2 (15 1)(12 )
2 .95 ,14
6 .57063
= 25.57
b LCL =
( n 1)s 2
2 / 2,n 1
=
( n 1)s 2
2 .05 , 29
=
(30 1)(12 ) 42 .5569
(30 1)(12 ) 17 .7083
= 8.18
UCL =
( n 1)s 2
2 1 / 2,n 1
=
( n 1)s 2
2 .95 , 29
=
= 19.65
c Increasing the sample size narrows the interval.
12.45 LCL =
( n 1)s 2
2 / 2,n 1
=
( n 1)s 2
2 .05 , 7
=
(8 1)(. 00093 ) 14 .0671
= .0005,
UCL =
( n 1)s 2
2 1 / 2,n 1
=
( n 1)s 2
2 .95 , 7
=
(8 1)(. 00093 ) 2 .16735
= .0030
12.46
H0 :
H1 :
2
= 250 < 250
2 2 1 ,n 1 2 .90 , 9
2
Rejection region:
2
4.16816
( n 1)s 2
2
=
(10 1)( 210 .22 ) 250
7 .57 , p-value = .4218. There is not enough evidence to infer that the
population variance has decreased.
12.47
H0 : H1 :
2
= 23 23
2 2 1 / 2, n 1 2 .95 , 7
2
Rejection region:
2 .16735 or
2
2 / 2, n 1
2 .05 , 7
14 .0671
246
2
( n 1)s 2
2
=
(8 1)(16 .50 ) 23
5 .02 , p-value = .6854. There is enough evidence to infer that the
population variance has changed.
12.48 LCL =
( n 1)s 2
2 / 2,n 1
=
( n 1)s 2
2 .025 , 9
=
(10 1)(15 .43 ) 19 .0228
7 .30
UCL =
( n 1)s 2
2 1 / 2,n 1
=
( n 1)s 2
2 .975 , 9
=
(10 1)(15 .43 ) 2 .70039
51 .43
12.49 a H 0 :
H1 :
2
= 250 250
2 2 1 / 2, n 1 2 .975 , 24
2
Rejection region:
2
12.4011 or
2
2 / 2, n 1
2 .025 , 24
39.3641
( n 1)s 2
2
=
( 25 1)( 270 .58 ) 250
= 25.9760, p-value = .7088. There is not enough evidence to infer that
the population variance is not equal to 250. b Demand is required to be normally distributed. c The histogram is approximately bell shaped.
12.50
H0 : H1 :
2
= 18 18
2 2 ,n 1 2 .10 , 244
2
Rejection region:
2
272.704
( n 1)s 2
2
=
( 245
1)( 22 .56 ) 18
= 305.81; p-value = .0044. There is enough evidence to infer that the
population variance is greater than 18.
12.51 LCL =
( n 1)s 2
2 / 2,n 1
=
( n 1)s 2
2 .05 ,89
=
(90 1)( 4 .72 ) 113 .145
3 .75
UCL =
( n 1)s 2
2 1 / 2,n 1
=
( n 1)s 2
2 .95 ,89
=
(90 1)( 4 .72 ) 69 .1260
6 .16
12.52
H0 : H1 :
2
= 200 200
2
247
Rejection region:
2
2
2 1
,n 1
2 .95 , 99
77.9295
( n 1)s 2
2
=
(100
1)(174 .47 ) 200
= 86.36; p-value = .1863. There is not enough evidence to infer that the
population variance is less than 200. Replace the bulbs as they burn out.
12.53 LCL =
( n 1)s 2
2 / 2,n 1
=
( n 1)s 2
2 .025 , 24
=
( 25 1)(19 .68 ) 39 .3641
12 .00
UCL =
( n 1)s 2
2 1 / 2,n 1
=
( n 1)s 2
2 .975 , 24
=
( 25 1)(19 .68 ) 12 .4011
38 .09
^ 12.54 a p ^ b p z
/2
z
/2
^ p (1
^ p ) / n = .48
1.96 .48 (1 .48 ) / 500 = .48 .0692
.0438
^ p (1
^ p ) / n = .48
1.96 .48 (1 .48 ) / 200 = .48 1.96 .48 (1 .48 ) / 1000 = .48
^ c p
z
/2
^ p (1
^ p ) / n = .48
.0310
d The interval narrows.
^ 12.55 a p
z
/2
^ p (1
^ p ) / n = .50
1.96 .50 (1 .50 ) / 400 = .50 .0461 .0294
.0490
^ b p
^ c p
z
z
/2
^ p (1
^ p (1
^ p ) / n = .33
^ p ) / n = .10
1.96 .33 (1 .33 ) / 400 = .33 1.96 .10 (1 .10 ) / 400 = .10
/2
d The interval narrows.
12.56
H 0 : p = .60
H 1 : p > .60
a z
^ p p (1 ^ p p (1
p p) / n p p) / n
=
.63 .60 .60 (1 .60 ) / 100 .63 .60 .60 (1 .60 ) / 200
= .61, p-value = P(Z > .61) = .5 .2291 =.2709
b z
=
= .87, p-value = P(Z > .87) = .5 .3078 = .1922
c z
^ p p (1
p p) / n
=
.63 .60 .60 (1 .60 ) / 400
= 1.22, p-value = P(Z > 1.22) = .5 .3888 = .1112
d The p-value decreases.
^ p p (1
12.57 a z
p p) / n
=
.73 .70 .70 (1 .70 ) / 100
= .65, p-value = P(Z > .65) = .5 .2422 =.2578
248
b z
^ p p (1 ^ p p (1
p p) / n p p) / n
=
.72
.70
= .44, p-value = P(Z > .44) = .5 .1700 =.3300
.70 (1 .70 ) / 100 .71 .70 .70 (1 .70 ) / 100
c z
=
= .22, p-value = P(Z > .22) = .5 .0871 =.4129
d. The z statistic decreases as does the p-value.
12.58 n =
z
/2
^ p (1 B
^ p)
2
2
=
1 .645 .5(1 .5) .03
= 752
12.59 a.5
.03
b Yes, because the sample size was chosen to produce this interval.
^ 12.60 a p
z
/2
^ p (1
^ p ) / n = .75
1.645 .75 (1 .75 ) / 752 = .75
.0260
b The interval is narrower. c Yes, because the interval estimate is better than specified.
12.61 n =
z
/2
^ p (1 B
^ p)
2
2
=
1 .645 .75 (1 .75 ) .03
= 564
12.62 a.75
.03
b Yes, because the sample size was chosen to produce this interval.
^ 12.63 a p
z
/2
^ p (1
^ p ) / n = .92
1.645 .92 (1 .92 ) / 564 = .92
.0188
b The interval is narrower. c Yes, because the interval estimate is better than specified.
^ 12.64 a p
z
/2
^ p (1
^ p ) / n = .5
1.645 .5 (1 .5 ) / 564 = .5
.0346
b The interval is wider. c No because the interval estimate is wider (worse) than specified.
^ 12.65 p = 259/373 = .69
^ p z
/2
^ p (1
^ p ) / n = .69
1.96 .69 (1 .69 ) / 373 = .69
.0469; LCL = .6431, UCL = .7369
249
12.66
H 0 : p = .25
H 1 : p < .25
^ p = 41/200 = .205
^ p p (1 p p) / n .205 .25 1 .47 , p-value = P(Z < 1.47) = .5 .4292 =.0708. There is
z
=
.25 (1 .25 ) / 200
enough evidence to support the officer's belief.
^ 12.67 p = 204/314 = .65
^ p z
/2
^ p (1
^ p ) / n = .65
1.96 .65 (1 .65 ) / 314 = .65
.0528; LCL = .5972, = UCL .7028
12.68
H 0 : p = .92 H 1 : p > .92
^ p = 153/165 = .927
^ p p (1 p p) / n .927 .92 .33 , p-value = P(Z > .33) = .5 .1293 =.3707. There is not enough
z
=
.92 (1 .92 ) / 165
evidence to conclude that the airline's on-time performance has improved.
^ 12.69 p = 97/344 = .28
^ p z
/2
^ p (1
^ p ) / n = .28
1.96 .28 (1 .28 ) / 344 = .28
.0474; LCL = .2326, UCL = .3274
^ 12.70 p = 68/400 = .17
^ p z
/2
^ p (1
^ p ) / n = .17
1.96 .17 (1 .17 ) / 400 = .17
.0368; LCL = .1332, UCL = .2068
12.71 LCL = .1332(1,000,000)(3.00) = $399,600, UCL = .2068(1,000,000)(3.00) = $620,400
p 12.72 ~
x n
2 4
1 2 200 4
.0147
~ p
z
~ (1 ~ ) p p
/2
n
4
= .0147
1 .96
.0147 (1 .0147 ) 200 4
= .0147
.0165; LCL = 0 (increased from .0018),
UCL = .0312
250
p 12.73 ~
x n
2 4
3 2 374 4
.0132
~ p
z
~ (1 ~ ) p p
/2
n
4
= .0132 1 .645
.0132 (1 .0132 ) 374 4
= .0132
.0097; LCL = .0035, UCL = .0229
p 12.74 ~
x n
2 4
1 2 385 4
.0077
~ p
z
~ (1 ~ ) p p
/2
n
4
= .0077
2 .575
.0077 (1 .0077 ) 385 4
= .0077
.0114; LCL = 0 (increased from .0037),
UCL = .0191
^ 12.75 p
z
/2
^ p (1
^ p ) / n = .1056
1.96 .1056 (1 .1056 ) / 521 = .1056
.0264; LCL = .0792, UCL
= .1320
12.76 LCL = 75,000(.0792) =5,940, UCL = 75,000(.1320) = 9,900
^ 12.77 p
z
/2
^ p (1
^ p ) / n = .7584
1.96 .7584 (1 .7584 ) / 567 = .7584
.0352; LCL = .7232,
UCL = .7936
12.78
H 0 : p = .90 H 1 : p < .90
z
^ p p (1
p p) / n
=
.8644
.90
= 1.58, p-value =P(Z < 1.58) = .5 .4429 = .0571. There is not
.90 (1 .90 ) / 177
enough evidence to infer that the satisfaction rate is less than 90%.
^ 12.79 p
z
/2
^ p (1
^ p ) / n = .2333
1.96 .2333 (1 .2333 ) / 120 = .2333
.0757; LCL = .1576,
UCL = .3090
12.80
H 0 : p = .80 H 1 : p > .80
z
^ p p (1
p p) / n
=
.8225
.80
= 1.13, p-value = P(Z > 1.13) = .5 .3708 = .1292. There is not
.80 (1 .80 ) / 400
enough evidence to infer that the claim is true.
251
^ 12.81 p
z
/2
^ p (1
^ p ) / n = .7840
1.96 .7840 (1 .7840 ) / 426 = .7840
.0391; LCL = .7449,
UCL = .8231
12.82
H 0 : p = .50 H 1 : p > .50
z
^ p p (1
p p) / n
=
.57
.50
= 1.40, p-value = P(Z > 1.40) = .5 .4192 =.0808. There is enough
.50 (1 .50 ) / 100
evidence to conclude that more than 50% of all business students would rate the book as excellent.
12.83 Codes 1, 2, and 3 have been recoded to 5.
H 0 : p = .90
H 1 : p > .90
z ^ p p (1 p p) / n
=
.96
.90
= 2.00, p-value = P(Z > 2.00) = .5 .4772 =.0228. There is enough
.90 (1 .90 ) / 100
evidence to conclude that more than 90% of all business students would rate the book as at least adequate.
^ 12.84 p
z
/2
^ p (1
^ p ) / n = .0827
1.645 .0827 (1 .0827 ) / 387 = .0827
.0230; LCL = .0597,
UCL = .1057
12.85
H 0 : p = .12
H 1 : p > .12
z ^ p p (1 p p) / n
=
.1625
.12
2 .62 , p-value = P(Z > 2.62) = .5 .4956 = .0044. There is enough
.12 (1 .12 ) / 400
evidence to infer that the proposed newspaper will be financially viable.
^ 12.86 p
z
/2
^ p (1
^ p ) / n = .1914
1.645 .1914 (1 .1914 ) / 810 = .1914
.0227; LCL = .1687,
UCL = .2141 Number: LCL = 270 million (.1687) = 45.55 million, UCL = 270 million (.2141) = 57.81 million
^ 12.87 p
z
/2
^ p (1
^ p ) / n = .2031
1.96 .2031 (1 .2031 ) / 650 = .2031
.0309; LCL = .1722,
UCL = .2340 Number: LCL = 5 million (.1722) = .861 million, UCL = 5 million (.2340) = 1.17 million
252
^ 12.88 p
z
/2
^ p (1
^ p ) / n = .0975
1.96 .0975 (1 .0975 ) / 2000 = .0975
.0130; LCL = .0845,
UCL = .1105 Number: LCL = 100 million (.0845) = 8.45 million, UCL = 100 million (.1105) = 11.05 million
12.89 Codes 3 and 4 were changed to 5
^ p z
/2
^ p (1
^ p ) / n = .7305
1.96 .7305 (1 .7305 ) / 475 = .7305
.0399; LCL = .6906,
UCL = .7704; Market segment size: LCL = 19,108,000 (.6906) = 13,195,985, UCL = 19,108,000 (.7704) = 14,720,803
12.90 Code 2 was changed to 3.
^ p z
/2
^ p (1
^ p ) / n = .5313
1.96 .5313 (1 .5313 ) / 320 = .5313
.0547; LCL = .4766,
UCL = .5860; Market segment size: LCL = 15,517,000 (.4766) = 7,395,402 , UCL = 15,517,000 (.5860) = 9,092,962
^ 12.91 a p
z
/2
^ p (1
^ p ) / n = .2919
1.96 .2919 (1 .2919 ) / 1836 = .2919
.0208; LCL = .2711,
UCL = .3127 b LCL = 107,194,000 (.2711) = 29,060,293, UCL = 107,194,000 (.3127) = 33,519,564
^ 12.92 p
z
/2
^ p (1
^ p ) / n = .1077
1.96 .1077 (1 .1077 ) / 325 = .1077
.0337; LCL = .0740,
UCL = .1414; Market segment size: LCL = 35.6 million(.0740) = 2.634 million, UCL = 35.6 million(.1414) = 5.034 million
^ 12.93 p
z
/2
^ p (1
^ p ) / n = .1748
1.645 .1748 (1 .1748 ) / 412 = .1748
.0308; LCL = .1440,
UCL = .2056; Number: LCL = 187 million(.1440) = 26.928 million, UCL = 187 million(.2056) = 38.447 million
^ 12.94 p
z
/2
^ p (1
^ p ) / n = .1500
1.96 .1500 (1 .1500 ) / 340 = .1500
.0380; LCL = .1120,
UCL = .1880; Number: LCL = 187 million(.1120) = 20.944 million, UCL = 187 million(.1880) = 35.156 million
^ ^ 12.95 p = 4/80 = .05; p
z
/2
^ p (1
^ p ) / n = .05
1.96 .05 (1 .05 ) / 80 = .05
.0478; LCL = .0022,
UCL = .0978; Number: LCL = 2,453(.0022) = 5.4 (rounded to 5), UCL = 2,453(.0978) = 239.9 (rounded to 240)
253
^ ^ 12.96 p = 29/559 = .0519; p
z
/2
^ p (1
^ p ) / n = .0519
2.575 .0519 (1 .0519 ) / 559 = .0519
.0242;
LCL = .0277, UCL = .0761; Number: LCL = 118,653(.0277) = 3,287, UCL = 118,653(.0761) = 9,029
12.97 x t
/ 2s
/ n = 229.18
1.96(67.36/ 500 ) = 229.18
5.92; LCL = 223.26, UCL = 235.10
Total value: 73,544(223.26) = $16,419,433, UCL = 73,544(235.10) = $17,290,194
^ ^ 12.98 p = 14/125 = .112; p z
^ ^ p (1 p )
/2
N
n
n
N 1
= .112 1 .645
.112 (1 .112 ) 125
2, 490 2, 490
125 1
= .112
.0452; LCL = .0668, UCL = .1572; Number: LCL = 2,490(.0668) = 166,
UCL = 2,490(.1572) = 391
^ ^ 12.99 p = 5/85 = .0588; p z
^ ^ p (1 p )
/2
N
n
n
N 1
= .0588 1 .645
.0588 (1 .0588 ) 85
1,864 1,864
85 1
= .0588
.0410;
LCL = .0178, UCL = .0998; Number: LCL = 1,864(.0410) = 76, UCL = 1,864(.0998) = 186
12.100 x t
s
/2
N
n
n
N 1
= 313 .47 1 .984
55 .53 100
1, 431 100 1, 431 1
= 313.47
10.63; LCL = 302.84,
UCL = 324.10; Total: LCL = 1,431(302.84) = $433,364, UCL = 1,431(324.10) = $463,787
^ ^ 12.101 p = 18/200 = .09; p z
^ ^ p (1 p )
/2
N
n
n
N 1
= .09 1 .96
.09 (1 .09 ) 200
3,745
200
3,745 1
= .09
.0386;
LCL = .0514, UCL = .1286; Number: LCL = 3,745(.0514) = 192, UCL = 3,745(.1286) = 482
12.102 x t
s
/2
N
n
n
N 1
= 12 ,940 1 .653
4,139 188
2,684 2,684
100 1
= 12,940
490; LCL = 12,450,
UCL = 13,430; Total: LCL = 2,684(12,450) = $33,415,800, UCL = 2,684(13,430 = $36,046,120
^ ^ 12.103 p = 38/317 = .1199; p
z
/2
^ p (1
^ p ) / n = .1199
1.96 .1199 (1 .1199 ) / 317 = .1199
.0358;
LCL = .0841, UCL = .1557; Number: LCL = 102,412(.0841) = 8,613, UCL = 102,412(.1557) = 15,946
12.104 a
H0 :
H1 :
= 30 > 30
254
1 2 3 4 5 6 7 8 9 10 11 12
A t-Test: Mean
B
C
D
Mean Standard Deviation Hypothesized Mean df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail
Costs 31.95 7.19 30 124 3.04 0.0015 1.6572 0.0030 1.9793
t = 3.04, p-value = .0015; there is enough evidence to infer that the candidate is correct.
1 2 3 4 5 6 7
A B t-Estimate: Mean
C
D
Mean Standard Deviation LCL UCL
Costs 31.95 7.19 30.68 33.23
b LCL = 30.68, UCL = 33.23 c The costs are required to be normally distributed.
12.105
H0 :
= 60 < 60
H1 :
1 2 3 4 5 6 7 8 9 10 11 12
A t-Test: Mean
B
C
D
Mean Standard Deviation Hypothesized Mean df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail
Times 57.79 6.58 60 23 -1.64 0.0569 1.7139 0.1138 2.0687
t = 1.64, p-value = .0569. There is not enough evidence to conclude that the supplier's assertion is correct.
255
12.106
H0 : H1 :
2
= 17 > 17
2
1 2 3 4 5 6 7 8 9 10 11 12 13
2
A Chi Squared Test: Variance
B
C
D
Sample Variance Hypothesized Variance df chi-squared Stat P (CHI<=chi) one-tail chi-squared Critical one tail P (CHI<=chi) two-tail chi-squared Critical two tail
Left-tail Right-tail Left-tail Right-tail
Times 27.47 17 19 30.71 0.0435 10.1170 30.1435 0.0869 8.9065 32.8523
= 30.71, p-value = .0435. There is enough evidence to infer that problems are likely.
12.107
1 2 3 4 5 6 A z-Estimate: Proportion Sample Proportion Observations LCL UCL B Resolution 0.358 215 0.304 0.412
LCL = .304, UCL = .412
12.108 a
1 2 3 4 5 6 7 A B t-Estimate: Mean C D
Mean Standard Deviation LCL UCL
Marks 71.88 10.03 69.03 74.73
LCL = 69.03, UCL = 74.73 b
H0 :
= 68 > 68
H1 :
256
1 2 3 4 5 6 7 8 9 10 11 12
A t-Test: Mean
B
C
D
Mean Standard Deviation Hypothesized Mean df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail
Marks 71.88 10.03 68 49 2.74 0.0043 1.6766 0.0086 2.0096
t = 2.74, p-value = .0043; there is enough evidence to infer that students with a calculus background would perform better in statistics than students with no calculus?
12.109
1 2 3 4 5 6 7
A B t-Estimate: Mean
C
D
Mean Standard Deviation LCL UCL
Points 117.54 50.24 108.19 126.89
LCL = 108.19, UCL = 126.89
12.110
A 1 z-Estimate: Proportion 2 3 Sample Proportion 4 Observations 5 LCL 6 UCL B Insurance 0.632 250 0.582 0.682
LCL = .582, UCL = .682
12.111 a
1 2 3 4 5 6 7
A B t-Estimate: Mean
C
D
Mean Standard Deviation LCL UCL
Times 6.91 0.23 6.84 6.98
257
LCL = 6.84, UCL = 6.98 b The histogram is bell shaped. c
H0 :
=7 < 7
H1 :
1 2 3 4 5 6 7 8 9 10 11 12
A t-Test: Mean
B
C
D
Mean Standard Deviation Hypothesized Mean df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail
Times 6.91 0.23 7 74 -3.48 0.0004 1.2931 0.0008 1.6657
t = 3.48, p-value = .0004; there is enough evidence to infer that postal workers are spending less than seven hours doing their jobs.
12.112
1 2 3 4 5 6 7
A B t-Estimate: Mean
C
D
Time Mean Standard Deviation LCL UCL 6.35 2.16 6.05 6.65
LCL = 6.05, UCL = 6.65
12.113
A B 1 t-Estimate: Mean 2 3 4 Mean 5 Standard Deviation 6 LCL 7 UCL C D
Times 5.79 2.86 5.11 6.47
LCL = 5.11, UCL = 6.47
258
12.114
A 1 z-Estimate: Proportion 2 3 Sample Proportion 4 Observations 5 LCL 6 UCL B Tourist 0.667 72 0.558 0.776
LCL = .558, UCL = .776
12.115
H0 : H1 :
2
=4 >4
2
1 2 3 4 5 6 7 8 9 10 11 12 13
2
A Chi Squared Test: Variance
B
C
D
Sample Variance Hypothesized Variance df chi-squared Stat P (CHI<=chi) one-tail chi-squared Critical one tail P (CHI<=chi) two-tail chi-squared Critical two tail
Left-tail Right-tail Left-tail Right-tail
Lengths 6.52 4 99 161.25 0.0001 77.0463 123.2252 0.0002 73.3611 128.4220
= 161.25, p-value = .0001; there is enough evidence to conclude that the number of springs requiring
reworking is unacceptably large.
12.116
H 0 : p = .90 H 1 : p < .90
1 2 3 4 5 6 7 8 9 10 11
A B z-Test: Proportion
C
D
Sample Proportion Observations Hypothesized Proportion z Stat P(Z<=z) one-tail z Critical one-tail P(Z<=z) two-tail z Critical two-tail
Springs 0.86 100 0.9 -1.33 0.0912 1.2816 0.1824 1.6449
259
z = 1.33, p-value = .0912; there is enough evidence to infer that less than 90% of the springs are the correct length.
12.117
A B 1 t-Estimate: Mean 2 3 4 Mean 5 Standard Deviation 6 LCL 7 UCL C D
Service 1.10 0.98 0.94 1.26
LCL = .94, UCL = 1.26
12.118 a H 0 :
H1 :
= 9.8 < 9.8
1 2 3 4 5 6 7 8 9 10 11 12
A t-Test: Mean
B
C
D
Time Mean Standard Deviation Hypothesized Mean df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail 9.16 2.64 9.8 149 -2.97 0.0018 1.2873 0.0036 1.6551
t = 2.97, p-value = .0018; there is enough evidence to infer that enclosure of preaddressed envelopes improves the average speed of payments?
b
H0 : H1 :
2
= 10.24 (3.22) < 10.24
2
260
1 2 3 4 5 6 7 8 9 10 11 12 13
2
A Chi Squared Test: Variance
B
C
D
Time Sample Variance Hypothesized Variance df chi-squared Stat P (CHI<=chi) one-tail chi-squared Critical one tail P (CHI<=chi) two-tail chi-squared Critical two tail 6.98 10.24 149 101.58 0.0011 127.3493 171.5069 0.0021 121.7870 178.4854
Left-tail Right-tail Left-tail Right-tail
= 101.58, p-value = .0011; there is enough evidence to infer that the variability in payment speeds
decreases when a preaddressed envelope is sent.
12.119 n =
z
/2
^ p (1 B
^ p)
2
2
=
2 .575 .5(1 .5) .02
= 4144
12.120
1 2 3 4 5 6 A z-Estimate: Proportion Sample Proportion Observations LCL UCL B Concert 0.1533 600 0.1245 0.1822
Proportion: LCL = .1245, UCL = .1822 Total: LCL = 400,000(.1245) = 49,800 UCL = 400,000(.1822) = 72,880
12.121 Number of cars:
H0 :
= 125 < 125
H1 :
261
1 2 3 4 5 6 7 8 9 10 11 12
A t-Test: Mean
B
C
D
Mean Standard Deviation Hypothesized Mean df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail
Cars 125.80 3.90 125 4 0.46 0.3351 3.7469 0.6702 4.6041
t = .46, p-value = .3351; there is not enough evidence to infer that the employee is stealing by lying about the number of cars.
Amount of time
H0 :
= 3.5 > 3.5
H1 :
1 2 3 4 5 6 7 8 9 10 11 12
A t-Test: Mean
B
C
D
Time Mean Standard Deviation Hypothesized Mean df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail 3.61 0.40 3.5 628 7.00 0 2.3323 0 2.5837
t = 7.00, p-value = 0; there is enough evidence to infer that the employee is stealing by lying about the amount of time.
262
Case 12.1 95% confidence interval estimate of mean weekly consumption per student:
1 2 3 4 5 6 7
A B t-Estimate: Mean
C
D
Mean Standard Deviation LCL UCL
Cans 1.316 1.115 1.218 1.414
A 1 2 3 4 5 6 7 8 9 10 Case 12.1 Mean number of cans/ student / week Number of cans sold annually Gross revenue Less 35% university take Cost to produce cans Net profit Current profit
B
C
1.218 2,436,000 $1,827,000 $1,187,550 $487,200 $500,350 $484,000
A 1 2 3 4 5 6 7 8 9 10 Case 12.1 Mean number of cans/ student / week Number of cans sold annually Gross revenue Less 35% university take Cost to produce cans Net profit Current profit
B
C
1.414 2,828,000 $2,121,000 $1,378,650 $565,600 $613,050 $484,000
Estimated Mean Number of Cans per Student LCL = 1.218 UCL = 1.414 Revenue $1,187,550 1,378,650 Cost $487,200 565,600 Profit $500,350 613,050 Current Profit $484,000 484,000 Net $ 16,350 129,050
Pepsi should sign the exclusivity agreement.
263
Case 12.2 Estimated Mean Number of Cans per Student LCL = 1.218 UCL = 1.414 Revenue $1,187,550 1,378,650 Cost $487,200 565,600 Profit $500,350 613,050 Current Profit $855,910 1,071,290 Net $355,560 458,240
Coke would not sign the exclusivity agreement. Coke is expected to lose from the exclusivity agreement because they currently have a much larger share of the market and would not gain by paying for exclusivity.
Case 12.3 Exclude "missing" licenses:
1 2 3 4 5 6 A z-Estimate: Proportion Sample Proportion Observations LCL UCL B Insured 0.0300 233 0.0081 0.0520
Estimated number of uninsured drivers: LCL = 4,505,665 .0081 = 36,496 UCL = 4,505,665 .0520 = 234,295
Include "missing" licenses with uninsured:
1 2 3 4 5 6 A z-Estimate: Proportion Sample Proportion Observations LCL UCL B Insured 0.0924 249 0.0564 0.1283
Estimated number of uninsured drivers: LCL = 4,505,665 .0564 = 254,145 UCL = 4,505,665 .1283 = 578,227
It is quite likely that the "missing" licenses are uninsured.
264
Case 12.4
a 95% confidence interval estimate of the mean medical costs for each of the four age categories:
1 2 3 4 5 6 7 A B t-Estimate: Mean C D
Mean Standard Deviation LCL UCL
Age:45-64 1808 826 1643 1972
1 2 3 4 5 6 7
A B t-Estimate: Mean
C
D
Mean Standard Deviation LCL UCL
Age:65-74 4494 1820 4381 4607
1 2 3 4 5 6 7
A B t-Estimate: Mean
C
D
Mean Standard Deviation LCL UCL
Age:75-84 8074 3186 7876 8272
1 2 3 4 5 6 7
A B t-Estimate: Mean
C
D
Mean Standard Deviation LCL UCL
Age:85+ 15957 6207 15572 16342
265
b Estimated annual costs for 2003 Estimate of mean Age category Number (1,000s) 45 to 64 65 to 74 75 to 84 85 and over Total 7,932 2,187 1,423 450 11,992 LCL 1,784 4,381 7,876 UCL 1,877 4,607 8,272 Estimate of total (1,000s) LCL 14,150,688 9,581,247 11,207,548 7,007,400 41,946,883 UCL 14,888,364 10,075,509 11,771,056 7,353,900 44,088,829
15,572 16,342
Estimated annual costs for 2006 Estimate of mean Age category Number (1,000s) 45 to 64 65 to 74 75 to 84 85 and over Total 8,678 2,253 1,486 563 12,980 LCL 1,784 4,381 7,876 UCL 1,877 4,607 8,272 Estimate of total (1,000s) LCL 15,481,552 9,870,393 11,703,737 8,767,036 45,822,717 UCL 16,288,606 10,379,571 12,292,192 9,200,546 48,160,915
15,572 16,342
Estimated annual costs for 20011 Estimate of mean Age category Number (1,000s) 45 to 64 65 to 74 75 to 84 85 and over Total 9,649 2,609 1,546 692 14,496 LCL 1,784 4,381 7,876 UCL 1,877 4,607 8,272 Estimate of total (1,000s) LCL 17,213,816 11,430,029 12,176,296 10,775,824 51,595,965 UCL 18,111,173 12,019,663 12,788,512 11,308,664 54,228,012
15,572 16,342
Estimated annual costs for 2016 Estimate of mean Age category Number (1,000s) 45 to 64 65 to 74 75 to 84 85 and over Total 9,883 3,273 1,645 784 15,585 LCL 1,784 4,381 7,876 UCL 1,877 4,607 8,272 Estimate of total (1,000s) LCL 17,631,272 14,339,013 12,956,020 12,208,448 57,134,753 UCL 18,550,391 15,078,711 13,607,440 12,812,128 60,048,670
15,572 16,342
266
Estimated annual costs for 2021 Estimate of mean Age category Number (1,000s) 45 to 64 65 to 74 75 to 84 85 and over Total 9,840 3,886 1,938 846 16,510 LCL 1,784 4,381 7,876 UCL 1,877 4,607 8,272 Estimate of total (1,000s) LCL 17,554,560 17,024,566 15,263,688 13,173,912 63,016,726 UCL 18,469,680 17,902,802 16,031,136 13,825,332 66,228,950
15,572 16,342
Estimated annual costs for 2026 Estimate of mean Age category Number (1,000s) 45 to 64 65 to 74 75 to 84 85 and over Total 9,636 4,364 2,459 930 17,389 LCL 1,784 4,381 7,876 UCL 1,877 4,607 8,272 Estimate of total (1,000s) LCL 17,190,624 19,118,684 19,367,084 14,481,960 70,158,352 UCL 18,068,772 20,104,948 20,340,848 15,198,060 73,730,628
15,572 16,342
267
268