Download Chapter 8

Chapter 8: Estimation
Chapter 8: Estimation
8-1
a. An unbiased point estimator of the population mean is the sample mean:
 X i  520.95
X
n
Results for: Sugar.xls
Descriptive Statistics: Weights
Variable
Weights
N
100
Mean
520.95
Median
518.75
TrMean
520.52
Variable
Weights
Minimum
504.70
Maximum
544.80
Q1
513.80
Q3
527.28
StDev
9.45
b. Evidence of non-normality?
Normal Probability Plot
.999
.99
Probability
.95
.80
.50
.20
.05
.01
.001
505
515
525
535
545
Weights
Average: 520.948
StDev: 9.45150
N: 100
Anderson-Darling Normality Test
A-Squared: 1.989
P-Value: 0.000
The distribution shows evidence of non-normality
c. The unbiased point estimate of the variance of the sample mean:
s 2 (9.45)2
Var ( X )  
 .8930
n
100
SE Mean
0.95
113
114
8-2
Instructor’s Solutions Manual for Statistics for Business & Economics, 5 th Edition
a. Evidence of non-normality?
Normal Probability Plot
.999
.99
Probability
.95
.80
.50
.20
.05
.01
.001
85
95
105
115
125
HousePrice8_
Average: 101.375
StDev: 14.2020
N: 8
Anderson-Darling Normality Test
A-Squared: 0.194
P-Value: 0.837
No evidence of non-normality.
b. The minimum variance unbiased point estimator of the population mean is the
 X i  101.375
sample mean: X 
n
c. The unbiased point estimate of the variance of the sample mean:
2
xi  nx 2 83627  8(101.375)2

2
sx

 201.6964
n 1
7
2
 x2
sx
201.6964
ˆ (X ) 
Var ( X ) 
; Var

 25.2121
n
n
8
x 3
d. px    .375
n 8
Chapter 8: Estimation
a. Check each variable for normal distribution:
Normal Probability Plot
.999
.99
Probability
.95
.80
.50
.20
.05
.01
.001
45
50
55
Meals
Average: 50.1
StDev: 2.46842
N: 30
Anderson-Darling Normality Test
A-Squared: 0.413
P-Value: 0.318
Normal Probability Plot
.999
.99
.95
Probability
8-3
.80
.50
.20
.05
.01
.001
15
20
25
Attendance
Average: 21.24
StDev: 2.50466
N: 25
Anderson-Darling Normality Test
A-Squared: 0.377
P-Value: 0.383
115
116
Instructor’s Solutions Manual for Statistics for Business & Economics, 5 th Edition
Normal Probability Plot
.999
.99
Probability
.95
.80
.50
.20
.05
.01
.001
9
10
11
12
13
14
15
16
Ages
Average: 12.24
StDev: 1.96384
N: 25
Anderson-Darling Normality Test
A-Squared: 0.569
P-Value: 0.126
No evidence of non-normality in Meals, Attendance or Ages
b. Unbiased estimates of population mean and variance:
Descriptive Statistics: Meals, Attendance, Ages
Variable
Meals
Attendan
Ages
N
30
25
25
Mean
50.100
21.240
12.240
Median
50.000
21.000
12.000
TrMean
50.192
21.348
12.217
Variable
Meals
Attendan
Ages
Minimum
45.000
15.000
9.000
Maximum
55.000
25.000
16.000
Q1
48.000
19.500
10.000
Q3
52.000
23.500
14.000
StDev
2.468
2.505
1.964
Variable
Unbiased estimate of mean Unbiased estimate of variance (s2)
Meals
50.100
(2.468)2 = 6.0910
Attendance
21.240
(2.505)2 = 6.2750
Ages
12.240
(1.964)2 = 3.8573
1
1
 
8-4
a. E ( X )  E ( X 1 )  E ( X 2 )    
2
2
2 2
1
3
 3
E (Y )  E ( X 1 )  E ( X 2 )  

4
4
4 4
1
2
 2
E (Z )  E ( X1 )  E ( X 2 )  

3
3
3 3
SE Mean
0.451
0.501
0.393
Chapter 8: Estimation
1
1
12 2
 Var ( X 1 )  Var ( X 2 ) 

n 4
4
2 8
4
2
1
9
5
Var (Y )  Var ( X 1 )  Var ( X 2 ) 
16
16
8
2
1
4
5
Var ( Z )  Var ( X 1 )  Var ( X 2 ) 
9
9
9
is
most
efficient
since
Var ( X )  Var (Y )  Var ( Z )
X
Var (Y ) 5
c. Relative efficiency between Y and X :
  1.25
Var ( X ) 4
Var ( Z ) 10
Relative efficiency between Z and X :
  1.111
Var ( X ) 9
b. Var ( X ) 
2
8-5
Calculate a weighted average of the proportions:
24
32
pmen 
 .40, pwomen 
 .533
60
60
480
370
ptotal 
; pmen 
; pwomen  .5647(.40)  .4353(.533)  .458
480  370
480  370
8-6
a. Evidence of non-normality?
Normal Probability Plot for Leak Rates (
ML Estimates
99
ML Estimates
95
Mean
0.0515
StDev
0.0216428
90
Goodness of Fit
Percent
80
AD*
70
60
50
40
30
0.596
20
10
5
1
0.00
0.05
0.10
Data
No evidence of nonnormality exists.
b. The minimum variance unbiased point estimator of the population mean is
 X i  .0515
the sample mean: X 
n
117
118
Instructor’s Solutions Manual for Statistics for Business & Economics, 5 th Edition
c. The unbiased point estimate of the variance of the sample mean:
s 2 x  (.0216428)2  .0004684
Var ( X ) 
8-7
 2X
n
ˆ (X ) 
; Var
s 2 X .0004684

 .00000937
n
50
a. Evidence of non-normality?
Normal Probability Plot
.999
.99
Probability
.95
.80
.50
.20
.05
.01
.001
3.6
3.7
3.8
3.9
4.0
4.1
Weights
Average: 3.80787
StDev: 0.102407
N: 75
Anderson-Darling Normality Test
A-Squared: 0.511
P-Value: 0.191
No evidence of the data distribution coming from a non-normal population
b. The minimum variance unbiased point estimator of the population mean is the
 X i  3.8079
sample mean: X 
n
Descriptive Statistics: Weights
Variable
Weights
N
75
Mean
3.8079
Median
3.7900
TrMean
3.8054
Variable
Weights
Minimum
3.5700
Maximum
4.1100
Q1
3.7400
Q3
3.8700
StDev
0.1024
SE Mean
0.0118
c. Minimum variance unbiased point estimate of the population variance is the
sample variance s2:
s2 = (.1024)2 = .01049
8-8
a. n  25, x  2.90,   .45, z.025  1.96
 = 2.90  1.96(.45/5) = 2.7236 up to 3.0764
x  z  

n

b. 2.99 – 2.90 = .09 = z / 2 (.45 / 5), z / 2  1
  2[1  Fz (1)]  .3174
100(1-.3174)% = 68.26%
Chapter 8: Estimation
8-9
a. n  16, x  4.07,   .12, z.005  2.58
4.07  2.58(.12/4) = 3.9926 up to 4.1474
b. narrower since the z score for a 95% confidence interval is smaller than the z
score for the 99% confidence interval
c. narrower due to the smaller standard error
d. wider due to the larger standard error
8-10
a. n  9, x  187.9,   32.4, z.10  1.28
187.9  1.28(32.4/3) = 174.076 up to 201.724
b. 210.0 – 187.9 = 22.1 = z / 2 (32.4 / 3), z / 2  2.05
  2[1  Fz (2.05)]  .0404
100(1-.0404)% = 95.96%
8-11
a. 95% confidence interval:
Results for: TOC.xls
One-Sample T: Leak Rates (cc/sec.)
Variable
Leak Rates (
N
50
Mean
0.05150
StDev
0.02186
SE Mean
0.00309
95.0% CI
( 0.04529, 0.05771)
SE Mean
0.00309
98.0% CI
( 0.04406, 0.05894)
b. 98% confidence interval:
One-Sample T: Leak Rates (cc/sec.)
Variable
Leak Rates (
8-12
N
50
Mean
0.05150
StDev
0.02186
a.
Results for: Sugar.xls
Descriptive Statistics: Weights
Variable
Weights
N
100
Mean
520.95
Median
518.75
TrMean
520.52
Variable
Weights
Minimum
504.70
Maximum
544.80
Q1
513.80
Q3
527.28
StDev
9.45
SE Mean
0.95
90% confidence interval:
Results for: Sugar.xls
One-Sample T: Weights
Variable
Weights
N
100
Mean
520.948
StDev
9.451
SE Mean
0.945
90.0% CI
( 519.379, 522.517)
b. narrower since a smaller value of z will be used in generating the 80%
confidence interval.
8-13
n  457, x  3.59, s  1.045
3.69 – 3.59 = .1 = z / 2 (1.045/ 457), z / 2  2.05
  2[1  Fz (2.05)]  .0404
100(1-.0404)% = 95.96%
119
120
8-14
Instructor’s Solutions Manual for Statistics for Business & Economics, 5 th Edition
n  174, x  6.06, s  1.43
6.16 – 6.06 = .1 = z / 2 (1.43/ 174), z / 2  .922
  2[1  Fz (.92)]  .3576
100(1-.3576)% = 64.24%
8-15 n  9, x  157.82, s  38.89, t8,.025  2.306
margin for error:  2.306(38.89/3) =  29.8934
8-16 n  7, x  74.7143, s  6.3957, t6,.025  2.447
margin for error:  2.447(6.3957/ 7 ) =  5.9152
8-17
a. n  10, x  16.37, s  5.3757, t9,.005  3.25
16.37  3.25(5.3757/ 10 ) = 10.8452 up to 21.895
b. narrower since the t-score will be smaller for a 90% confidence interval than
for a 99% confidence interval
8-18
n  25, x  42, 740, s  4, 780, t24,.05  1.711
42,740  1.711(4780/5) = 41,104.28 up to 44,375.72
8-19
n  9, x  16.222, s  4.790, t8,.10  1.86
We must assume a normally distributed population
16.222  1.86(4.790/3) = 13.252 up to 19.192
8-20
a. unbiased point estimate of proportion:
Tally for Discrete Variables: Adequate Variety
Adequate Variety
1
2
N=
Count CumCnt
135
135
221
356
356
Percent CumPct
37.92 37.92
62.08 100.00
x 135

 .3792
n 356
b. 90% confidence interval:
n  356, p  135 / 356  .3792, z.05  1.645
p
p(1  p)
= .3792  (1.645) .3792(.6208) / 356 =
n
.3369 up to .4215
p  z / 2
Chapter 8: Estimation
8-21
n  189,
p  57 /189  .3016, z.025  1.96
p(1  p)
= .3016  (1.96) .6984(.3016) /189 =
n
.2362 up to .3670
p  z / 2
8-22
p(1  p)
n
p(1  p)
.8790(.121)
.90 - .8790 = .021,

 .0133
n
600
.021 = z / 2 (.0133), z / 2  1.58
m arg in for error  z
  2[1  Fz (1.58)]  .1142
100(1-.1142)% = 88.58%
8-23
n  320,
p  80 / 320  .25, z.025  1.96
p(1  p)
= .25  (1.96) .25(.75) / 320 =
n
95% confidence interval: .2026 up to .2974
p  z / 2
8-24
n  95,
p  67 / 95  .7053, z.005  2.58
p(1  p)
= .7053  (2.58) .7053(.2947) / 95 =
n
99% confidence interval: .5846 up to .8260
Using PHStat, the result is:
Sample Proportion
0.705263158
Z Value
-2.57583451
Standard Error of the
Proportion
0.046776854
Interval Half Width
0.120489435
Confidence Interval
Interval Lower Limit
0.584773723
Interval Upper Limit
0.825752593
p  z / 2
8-25
n  320,
p  240 / 320  .75, z.025  1.96
p(1  p)
= .75  (1.96) .75(.25) / 320 =
n
95% confidence interval: .7026 up to .7974
p  z / 2
121
122
8-26
Instructor’s Solutions Manual for Statistics for Business & Economics, 5 th Edition
p(1  p)
n
p(1  p)
.495(.505)
.545-.445 = .100, p = .495,

 .0355
n
198
.05 = z / 2 (.0355), z / 2  1.41
m arg in for error  z
  2[1  Fz (1.41)]  .0793
100(1-.1586)% = 84.14%
8-27
n  420,
p  223/ 420  .5310, z.025  1.96
p(1  p)
= .5310  (1.96) .5310(.4690) / 420 =
n
95% confidence interval: .4833 up to .5787
The margin for error is .0477
p  z / 2
8-28
n  246,
p  40 / 246  .1626, z.01  2.326
p(1  p)
= .1626  (2.326) .1626(.8374) / 246 =
n
98% confidence interval: .1079 up to .2173
Using PHStat, the result is:
Sample Proportion
0.1626
Z Value
-2.3263
Standard Error of the
Proportion
0.0235
Interval Half Width
0.0547
Confidence Interval
Interval Lower Limit
0.1079
Interval Upper Limit
0.2173
p  z / 2
8-29
a. n  246,
p  232 / 246  .9431, z.01  2.326
p(1  p)
= .9431  (2.326) .9431(.0569) / 246 =
n
98% confidence interval: .9087 up to .9775
b. n  246, p  10 / 246  .0407, z.01  2.326
P  z / 2
p(1  p)
= .0407  (2.326) .0407(.9593) / 246 =
n
98% confidence interval: .0114 up to .0699
p  z / 2
Chapter 8: Estimation
8-30
n  50, s 2  (.000478)2 ,  2 49,.025  70.222,  2 49,.975  31.555
(n  1) s 2
 2 n1, / 2
2 
(n  1)s 2
 2 n1,1 / 2
=
49(.000478)2
49(.000478)2
2
=
 
70.222
31.555
1.59E-7 <  2 < 3.55E-7
8-31
n  10, s 2  28.898,  29,.05  16.92,  29,.95  3.33
(n  1) s 2
 2 n1, / 2
2 
(n  1)s 2
 2 n1,1 / 2
=
9(28.898)
9(28.898)
2 
=
16.92
3.33
15.3713 up to 78.1027
8-32
n  20, s 2  6.62,  219,.025  32.85,  219,.975  8.91
(n  1) s 2
 2 n1, / 2
2 
(n  1)s 2
 2 n1,1 / 2
=
19(6.62)
19(6.62)
2 
=
32.85
8.91
3.8289 up to 14.1167
Bounds = 5.1439
Similar values are found with PHStat,
Degrees of Freedom
19
Sum of Squares
125.78
Single Tail Area
0.025
Lower Chi-Square Value
8.9065
Upper Chi-Square Value
32.852
Results
Interval Lower Limit for Variance
3.8285
Interval Upper Limit for Variance
14.122
Assumption: Population from which sample was drawn has an approximate normal
distribution.
8-33
n  18, s 2  108.16,  217,.05  27.59,  217,.95  8.67
(n  1) s 2
 2 n1, / 2
2 
(n  1)s 2
 2 n1,1 / 2
=
17(108.16)
17(108.16)
2 
=
27.59
8.67
66.6444 up to 212.0784
Assume that the population is normally distributed
123
124
8-34
Instructor’s Solutions Manual for Statistics for Business & Economics, 5 th Edition
a. n  15, s 2  (2.36)2 ,  214,.05  26.12,  214,.95  5.63
14(5.5696)
14(5.5696)
2 
=
26.12
5.63
2.9852 up to 13.8498
From PHStat, similar results.
Interval Lower Limit for Variance
2.9854
Interval Upper Limit for Variance
13.853
Assumption: Population from which sample was drawn has an approximate normal
distribution.
b. wider since the chi-square statistic for a 99% confidence interval is larger than
for a 95% confidence interval
8-35
n  9, s 2  .7875,  28,.05  15.51,  28,.95  2.73
8(.7875)
8(.7875)
2 
=
15.51
2.73
.4062 up to 2.3077
8-36
Let X = Without Passive Solar; Y = With Passive Solar; di  xi  yi
n  10,
d
i
 373, d  37.3,
d
2
i
 16,719, t9,.05  1.833
sd  [16, 719  (10)(37.3) 2 ] / 9  17.6575
37.3  1.833(17.6575) / 10
27.0649   x   y  47.5351
8-37
di  xi  yi , xi  after course
n  6,
d
i
 45, d  7.5,
d
sd  [2, 099  (6)(7.5) 2 ] / 5  18.77
7.5  1.476(18.77) / 6
3.8103   x   y  18.8103
2
i
 2, 099, t5,.10  1.476
Chapter 8: Estimation
2
8-38
2
sy
sx

=
nx n y
95% confidence interval: ( X  Y )  t( v , / 2)
2
 s 2   s y 2  
 x   

 nx   n y  
where v 
=
2 2
2 2


 sx 
sy
 /( n y  1)

 /(nx  1)  

 nx 
 ny 
2
 2.532   8.612  



6

  9 

= 9.940
v
2
2
 2.532 
 8.612 

 /(6  1)  
 /(9  1)
 6 
 9 
2
(76.12 – 74.61)  t10,.025
2
sy
sx

=
nx n y
(2.53)2 (8.61)2
=

6
9
-5.286 up to 8.306
1.51  2.228
8-39
Descriptive Statistics: Machine 1, Machine 2
Variable
Machine
Machine
N
100
100
Mean
520.95
513.75
Median
518.75
514.05
TrMean
520.52
513.91
Variable
Machine
Machine
Minimum
504.70
496.50
Maximum
544.80
527.00
Q1
513.80
510.33
Q3
527.28
517.68
StDev
9.45
5.49
SE Mean
0.95
0.55
95% confidence level: assuming normal populations and similar variances
(520.95  513.75)  (1.96)
(9.45)2 (5.49)2

100
100
5.0579 up to 9.3421
8-40
nx  138, x  36,558, sx  11,624, z.05  1.645
ny  266, y  37, 499, s y  16,521
(36,558  37, 499)  (1.645)
-3,270.41 up to 1,388.41
(11, 624)2 (16,521) 2

138
266
125
126
8-41
Instructor’s Solutions Manual for Statistics for Business & Economics, 5 th Edition
nx  190, x  .517, sx  .148, z.005  2.58
ny  417, y  .489, s y  .159
(.148)2 (.159) 2
(.517  .489)  (2.58)

190
417
-.0062 up to .0622
8-42
nx  9, x  9.78, s 2 x  17.64, t17,.05  1.74
ny  10, y  15.1, s 2 y  27.01
(9.78  15.10)  (1.74)
8(17.64)  9(27.01) 19
17
90
-9.1207 up to –1.5193
8-43
nx  12, x  135, 000, sx  56, 000, t25,.025  2.06
ny  15,
y  408, 000, s y  43, 000
(435, 000  408, 000)  (2.06)
11(56, 000) 2  14(43, 000) 2 27
25
180
-12,209.98 up to 66,209.98
8-44
nx  21, x  72.1, sx  11.3, t37,.10  1.303
ny  18,
y  73.8, s y  10.6
20(11.3)2  17(10.6)2 39
(72.1  73.8)  (1.303)
37
378
-6.2971 up to 2.8971
8-45
nx  120,
px 
( px  p y )  z / 2
x 85

 .7083, ny  163,
n 120
px (1  px ) p y (1  p y )

=
nx
ny
(.7083  .4785)  (2.326)
.2298  .132657
.0971 up to .3625
py 
y 78

 .4785, z.01  2.33
n 163
(.7083)(.2917) (.4785)(.5215)
=

120
163
Chapter 8: Estimation
8-46
Results for: Library.xls
Tabulated Statistics: Class, Adequate Variety
Rows: Class
1
Columns: Adequate
2
All
1
73
50.69
54.07
20.56
71
49.31
32.27
20.00
144
100.00
40.56
40.56
2
26
25.49
19.26
7.32
76
74.51
34.55
21.41
102
100.00
28.73
28.73
3
19
28.79
14.07
5.35
47
71.21
21.36
13.24
66
100.00
18.59
18.59
4
17
39.53
12.59
4.79
26
60.47
11.82
7.32
43
100.00
12.11
12.11
135
38.03
100.00
38.03
220
61.97
100.00
61.97
355
100.00
100.00
100.00
All
Cell Contents -Count
% of Row
% of Col
% of Tbl
pseniors  17 / 43  .3953,
p freshmen  73 /144  .5069
(.3953)(.6047) (.5069)(.4931)
=

43
144
-.1116  .1405 = -.2521 up to .0289
(.3953  .5069)  (1.645)
8-47
p freshmen  80 /138  .5797,
(.5797  .7396)  (1.96)
psophs  71/ 96  .7396
(.5797)(.4203) (.7396)(.2604)
=

138
96
-.1599  .1204
-.2803 up to -.0395
8-48
nx  100,
px  .61, ny  100,
.1  (.61  .54)  .03  z / 2
  2[1  Fz (.43)]  .6672
p y  .54
(.61)(.39) (.54)(.46)

, z / 2  .43
100
100
100(1-.6672)% = 33.28% confidence level
127
128
8-49
Instructor’s Solutions Manual for Statistics for Business & Economics, 5 th Edition
nx  510,
px  .6275, n y  332,
(.6275  .6024)  (1.645)
p y  .6024, z.05  1.645
(.6275)(.3725) (.6024)(.3976)

510
332
.0251  .0565
-.0314 up to .0816
8-50
a. z.05  1.645, B  .04
.25( z / 2 ) 2
(.25)(1.645) 2
=
n
 422.8 , take n = 423
B2
(.04) 2
(.25)(1.96)2
b.
 600.25 , take n = 601
(.04)2
(.25)(2.33)2
c.
 542.89 , take n = 543
(.05)2
8-51
z.005  2.58, B  .05
n
8-52
z.05  1.645, B  .03
n
8-53
.25( z / 2 ) 2
(.25)(2.58) 2
=
 665.64 , take n = 666
B2
(.05) 2
.25( z / 2 ) 2
(.25)(1.645) 2
=
 751.7 , take n = 752
B2
(.03) 2
a. n  10, x  257, s  37.2, t9,.05  1.833
 = 257  1.833(37.2/ 10 ) = 235.4318 up to 278.5628
x  t  s

n

assume that the population is normally distributed
b. 85 and 98% confidence intervals:
Using Minitab:
One-Sample T: NewPrescriptions8_53
Variable
NewPrescript
N
10
Mean
257.0
StDev
37.2
One-Sample T: NewPrescriptions8_53
Variable
NewPrescript
N
10
Mean
257.0
StDev
37.2
SE Mean
11.8
(
85.0% CI
238.5,
275.5)
SE Mean
11.8
(
98.0% CI
223.8,
290.2)
c. The sample size would need to quadruple in order to cut in half the bounds of
the confidence interval
Chapter 8: Estimation
8-54
n  16, x  150, s  12, t15,.025  2.131
 = 150  2.131(12/4) = 143.607 up to 156.393
x  t  s

n

Using PHStat,
Standard Error of the
Mean
3
Degrees of Freedom
15
t Value
2.131450856
Interval Half Width
6.394352567
Confidence Interval
Interval Lower Limit
143.61
Interval Upper Limit
156.39
It is recommended that he stock 157 gallons.
8-55
n  50, x  30, s  4.2, z.05  1.645
= 30  1.645(4.2/ 50 ) = 29.0229 up to 30.9771
8-56
Results from Minitab:
Descriptive Statistics: Passengers8_56
Variable
Passenge
Variable
Passenge
N
50
Minimum
86.00
Mean
136.22
Maximum
180.00
One-Sample T: Passengers8_56
Variable
Passengers8_
N
50
Mean
136.22
Median
141.00
Q1
118.50
StDev
24.44
TrMean
136.75
Q3
152.00
SE Mean
3.46
Using results from PHStat,
Data
Sample Standard
Deviation
Sample Mean
Sample Size
Confidence Level
Standard Error of the Mean
Degrees of Freedom
t Value
Interval Half Width
Confidence Interval
Interval Lower Limit
Interval Upper Limit
(
StDev
24.44
95.0% CI
129.27, 143.17)
SE Mean
3.46
24.43925414
136.22
50
95%
3.456232466
49
2.009574018
6.945554964
129.27
143.17
129
130
8-57
Instructor’s Solutions Manual for Statistics for Business & Economics, 5 th Edition
a. 88% confidence level for proportion of defects due to incorrect label
Tally for Discrete Variables: defect
defect
dent
Incorrect Label
Missing label
Wrong color
N=
Count CumCnt
13
13
8
21
14
35
13
48
48
p  8 / 48  .1667,
Percent CumPct
27.08 27.08
16.67 43.75
29.17 72.92
27.08 100.00
pz
p(1  p)
(.1667)(.8333)
=
 .1667  1.56
n
48
.0828 up to .2506
b. 92% confidence interval for proportion of defects due to missing label
p(1  p)
(.2917)(.7083)
=
p  14 / 48  .2917, p  z
 .2917  1.75
n
48
.1769 up to .4065
8-58
a. nx  225,
px  .6222, ny  210,
p y  .5714, z.05  1.645
px  p y  .6222  .5714  .0508
b. Minitab results:
Test and CI for Two Proportions
Sample
X
N Sample p
1
140
225 0.622222
2
120
210 0.571429
Estimate for p(1) - p(2): 0.0507937
95% CI for p(1) - p(2): (-0.0413643, 0.142952)
8-59
Assume both populations are distributed normally with equal variances
nx  40, x  340, sx  20, t.05  1.671
n y  50, y  285, s y  30
(nx  1) sx  (n y  1) s y
1 1
where s p 

nx  n y  2
nx ny
2
( X  Y )  tnx  ny 2, / 2 s p
sp 
(40  1)202  (50  1)302
= 29.63007
40  30  2
(340  285)  (1.671)(29.63007)
44.49695 up to 65.50305
1
1

40 50
2
Chapter 8: Estimation
8-60
Assume both populations are distributed normally with equal variances
nx  15, x  470, sx  5, t25,.05  1.708
ny  12,
y  460, s y  7
(nx  1) sx  (n y  1) s y
2
(470  460)  (1.708)
nx  n y  2
2
1 1

nx n y
(15  1)52  (12  1)7 2 1 1
(470  460)  (1.708)

15  12  2
15 12
10  (1.708)(5.9632)(.3873)
10  3.9447 = 6.055 up to 13.945
Since both endpoints of the confidence interval are positive, this provides
evidence that the new machine provides a larger mean filling weight than the old
8-61
90% confidence level:
z.05  1.645, B  .025
.25( z / 2 ) 2
(.25)(1.645) 2
=
 1082.4 , take n = 1,083
B2
(.025) 2
85% confidence level:
z.075  1.44, B  .025
n
n
.25( z / 2 ) 2
(.25)(1.44) 2
=
 829.4 , take n = 830
B2
(.025) 2
8-62 a. The minimum variance unbiased point estimator of the population mean is the
 X i  27  3.375. The unbiased point estimate of the
sample mean: X 
n
8
2
2
 xi  nx  94.62  8(3.375)2  .4993
variance: s 2 x 
n 1
7
x 3
b. px    .375
n 8
8-63
98% confidence interval for student pair:
Results for: Student Pair.xls
Paired T-Test and CI: COURSE, NO COURSE
Paired T for COURSE - NO COURSE
N
Mean
COURSE
6
70.67
NO COURSE
6
66.17
Difference
6
4.50
StDev
16.03
14.19
4.14
SE Mean
6.55
5.79
1.69
98% CI for mean difference: (-1.18, 10.18)
131