Chapter 20:
Additional Topics in Sampling
20.1 Answers should refer to each of the steps outlined in Figure 20.1
20.2 Answers should refer to each of the steps outlined in Figure 20.1
20.3 Answers should refer to each of the steps outlined in Figure 20.1
20.4 Answers should refer to each of the steps outlined in Figure 20.1
20.5 Answers should refer to each of the steps outlined in Figure 20.1
20.6 Answers should deal with issues such as (a) the identification of the correct
population, (b) selection (nonresponse) bias, (c) response bias
20.7 Answers should deal with issues such as (a) the identification of the correct
population, (b) selection (nonresponse) bias, (c) response bias
20.8 Answers should deal with issues such as (a) the identification of the correct
population, (b) selection (nonresponse) bias, (c) response bias
20.9 While the recall method will lower the number of nonresponses, a bias may
be built in since this technique will reduce the participation of individuals
who are absent on Thursday evenings (e.g., night students, second shift
workers, shopping mall employees, etc.)
20.10 Within Minitab, go to Calc  Make Patterned Data… in order to generate
a simple set of numbers of size ‘n’ or ‘N’. Enter first value as 1, last value
as the total number of stocks traded on the New York Stock Exchange.
Go to Calc  Random Data  Sample from Columns… in order to
generate a simple random sample of size ‘n’ from the number of all stocks.
“Sample ____ rows from column(s):” For Exercise 20.10 enter 20 as
the number of rows to sample from. The results will be the observation
numbers in the list to include in the sample.
500
Statistics for Business & Economics, 6th edition
20.11 Within Minitab, go to Calc  Make Patterned Data… in order to generate
a simple set of numbers of size ‘n’ or ‘N’. Enter first value as 1, last value
as the total number of houses advertised for sale in your city. Go to Calc
 Random Data  Sample from Columns… in order to generate a
simple random sample of size ‘n’ from the number of all houses advertised
for sale in your city. “Sample ____ rows from column(s):” For Exercise
20.11 enter 15 as the number of rows to sample from. The results will
be the observation numbers in the list to include in the sample.
20.12 Within Minitab, go to Calc  Make Patterned Data… in order to generate
a simple set of numbers of size ‘n’or ‘N’. Enter first value as 1, last value
as 12,723. Go to Calc  Random Data  Sample from Columns… in
order to generate a simple random sample of size ‘n’. “Sample ____ rows
from column(s):” Enter 100 as the number of rows to sample from. The
results will be the observation numbers in the list to include in the sample.
20.13 Within Minitab, go to Calc  Make Patterned Data… in order to generate
a simple set of numbers of size ‘n’ or ‘N’. Enter first value as 1, last value
as 984 which is the total number of pages in the text. Go to Calc 
Random Data  Sample from Columns… in order to generate a simple
random sample of size ‘n’. “Sample ____ rows from column(s):” Enter
50 as the number of rows to sample from. The results will be the
observation numbers in the list to include in the sample.
20.14 x  9.7, s  6.2 , ˆ x 
9.7  1.96 (.7519)
( s)2 N  n
(6.2)2 139

= .7519
n
N
50 189
(8.2262, 11.1738)
20.15 a. x  127.43
s 2 N  n (43.27) 2 760
2
ˆ

b.  x 
= 28.9216
n N
60 820
c. 127.43  1.645 ( 28.9216 )
(118.5834, 136.2766)
d. [137.43 – 117.43]/2 = 10 = z / 2 28.9216 , solving for z: 1.86
tabled value of .9686 yields a confidence level of 93.72% or an  of
.0628
(5.32)2 85
= .6936
40 125
7.28  2.58 (.6936) (5.4904, 9.0696)
20.16 ˆ x 
Chapter 20: Additional Topics in Sampling
20.17 a. false: as n increases, the confidence interval becomes narrower for a
given N and s2
b. true
c. true: the finite population correction factor is larger to account for the
fact that a smaller proportion of the population is represented as N
increases relative to n.
d. true
2
20.18 ˆ x 
( s) 2 N  n s 2  n  2  1 1 
 1    s   
n
N
n  N
n N 
20.19 99% confidence interval:
Nx  Z / 2 Nˆ x  N   Nx  Z / 2 Nˆ x
where, Nx  (189)(9.7)  1833.30
s2
(6.2) 2
N ( N  n) 
189(189  50) = 142.1167
n
50
1833.30 2.58(142.1167)
1466.6390 < N  < 2199.9610
Nˆ x 
s 2 N  n (43.27) 2 760

n N
60 820
Nx  Z / 2 Nˆ x  N   Nx  Z / 2 Nˆ x , where
Nx  (820)(127.43)  104, 492.6
20.20 95% confidence interval: using ˆ x 2 
s2
(43.27)2
N ( N  n) 
820(820  60) = 4,409.8619
n
60
104,492.6 1.96(4409.8619)
95,849.2706 < N  < 113,135.9294
Nˆ x 
20.21 90% confidence interval:
Nx  Z / 2 Nˆ x  N   Nx  Z / 2 Nˆ x , where Nx  (125)(7.28)  910
s2
(5.32)2
N ( N  n) 
125(125  40) = 86.7054
n
40
910 1.645(86.7054)
767.3696 < N  < 1,052.6304
Nˆ x 
501
502
Statistics for Business & Economics, 6th edition
20.22 x = 143/35 = 4.0857
90% confidence interval:
Nx  Z / 2 Nˆ x  N   Nx  Z / 2 Nˆ x , where
Nx  (120)(4.0857)  490.2857
s2
(3.1)2
N ( N  n) 
120(120  35) = 52.9210
n
35
490.2857 1.645(52.9210)
403.2307 < N  < 577.3407
Nˆ x 
20.23
p̂ = x/n = 39/400 = .0975
 pˆ  [ pˆ (1  pˆ ) /(n  1)][( N  n) / N ]
 [(.0975)(.9025) /(399)][(1395  400) /1395] = .0125
95% confidence interval: .0975  1.96(.0125): .073 up to .1220
20.24
p̂ = 56/100 = .56
 pˆ  [ pˆ (1  pˆ ) /(n  1)][( N  n) / N ]
 [(.56)(.44) / 99][(420 100) / 420] = .0435
90% confidence interval: .56  1.645(.0435): .4884 up to .6316
20.25
p̂ = 37/120 = .3083
 pˆ  [ pˆ (1  pˆ ) /(n  1)][( N  n) / N ]
 [(.3083)(.6917) /(119)][(257 120) / 257] = .0309
95% confidence interval: .3083  1.96(.0309): .2477 up to .3689
20.26
p̂ = 31/80 = .3875
 pˆ  [ pˆ (1  pˆ ) /(n  1)][( N  n) / N ]
 [(.3875)(.6125) /(79)][(420  80) / 420] = .0493
90% confidence interval: .3875  1.645(.0493): .3064 up to .4686
128.688 < Np < 196.812 or between 129 and 197 students intend to take
the final.
20.27 a. xst 
820(290)  540(352)  440(427)
1 k
= 342.089
N j xj =

1800
N j 1
Chapter 20: Additional Topics in Sampling
503
s 2 N  n (47) 2 820  120

 15.714
n N
120
820
(61) 2 540  90
(93)2 440  90

 34.454 , ˆ 2 x3 
 76.443
90
540
90
440
(820)2 (15.714)  (540) 2 (34.454)  (440) 2 (76.443)

 10.93
(1800)2
b. ˆ 2 x1 
ˆ 2 x
2
ˆ 2 x
st
c. 95% confidence interval: 342.089  1.96 10.93
335.609 up to 348.569
s 2 N  n (12.3)2 208  50

 2.2984
n N
50
208
90% confidence interval: 43.3  1.645 2.2984 : 40.806 up to 45.794
152(27.6)  127(39.2)  208(43.3)
1 k
b. xst   N j x j =
= 37.3306
487
N j 1
20.28 a. x3  43.3 , ˆ 2 x3 
(7.1)2 152  40
(9.9) 2 127  40
 .9286 , ˆ 2 x2 
 1.6785
40
152
40
127
(152)2 (.9286)  (127) 2 (1.6785)  (208) 2 (2.2984)

 .6239
(487)2
c. ˆ 2 x1 
ˆ 2 x
st
90% confidence interval: 37.3306  1.645 .6239 : 36.0313 up to
38.6299
95% confidence interval: 37.3306  1.96 .6239 : 35.7825 up to 38.8787
s 2 N  n (.8) 2 240  40

 .0133
n N
40
240
90% confidence interval: 2.5 1.645 .0133 : 2.3102 up to 2.6897
2.5(240)  3.6(190)  3.9(350)  2.8(280)
1 k
b. xst   N j x j =
= 3.2387
1060
N j 1
20.29 a. ˆ 2 x1 
c. ˆ
2
x1
ˆ 2 x
3
(.8)2 240  40
(.9)2 190  40
2

 .0133 , ˆ x2 
 .0160
40
240
40
190
(1.2)2 350  40
(.7)2 280  40

 .0319 , ˆ 2 x4 
 .0105
40
350
40
280
ˆ 2 x 
st
(240)2 (.0133)  (190) 2 (.0160)  (350) 2 (.0319)  (280)2 (.0105)
 .0054
(1060)2
90% confidence interval: 3.2387  1.645 .0054 : 3.1177 up to 3.3596
95% confidence interval: 3.2387  1.96 .0054 : 3.0947 up to 3.3827
504
Statistics for Business & Economics, 6th edition
s 2 N  n (1.04)2 632  50

 .0199 ; 3.12  1.96 .0199 :
n N
50
632
2.8435 up to 3.3965
(.86)2 529  50
2
ˆ
b.  x2 
 .0134 ; 3.37  1.96 .0134 : 3.1431 up to
50
529
3.5969
(632)2 (.0199)  (529) 2 (.0134)
c. ˆ 2 xst 
 .0087 ; xst = 3.2339
(1161)2
20.30 a. ˆ 2 x1 
3.2339  1.96 .0087 : 3.0513 up to 3.4166
20.31 a. 90% confidence interval:
Nx  Z / 2 Nˆ x  N   Nx  Z / 2 Nˆ x , where
Nx  (208)(43.3)  9006.4
s2
(12.3) 2
Nˆ x 
N ( N  n) 
208(208  50) = 315.3409
n
50
9006.4 1.645(315.3409)
8487.6642 < N  < 9525.1358
b. from Exercise 20-28: xst  37.3306, ˆ 2 xst  .6239
Nxst  (487)(37.3306) = 18,180.0022
N 2ˆ 2 xst  (152)2 .9286  (127)21.6785  (208)2 2.2984  147,964.8785
Nˆ xst  147964.8785  384.6620
90% confidence interval: 18,180.0022  1.645(384.6620):
17,547.2332 < N  < 18,812.7716
20.32 a. Nxst = 237(120) + 198(150) + 131(180) = 81,720
b. ˆ
2
x1
ˆ 2 x
2
ˆ 2 x
st
s 2 N  n 932 120  40


 144.15 ,
n N
40 120
642 150  45
47 2 180  50

 63.7156 , ˆ 2 x3 
 31.9078
45 150
50 180
(120)2 (144.15)  (150)2 (63.71556)  (180)2 (31.9078)

 22.4354
(450)2
95% confidence interval: 181.6(450)  1.96 22.4354 (450) :
77,542.3153 < N  < 85,897.6847
20.33 a. pˆ st  [364
10
8
 1031 ] /1395  .1638
40
60
Chapter 20: Additional Topics in Sampling
pˆ1 (1  pˆ1 ) N1  n1 .25(.75) 364  40

 .0043
n1  1
N1
40  1
364
.1333(.8667) 1031  60
ˆ 2pˆ 2 
 .0018
60  1
1031
(364)2 (.0043)  (1031) 2 (.0018)
ˆ 2pˆ st 
 .0013
(1395)2
b. ˆ 2pˆ1 
95% confidence interval: .1638  1.96 .0013 : .0931 up to .2345
6
14
 50 ] /150  .3467
25
25
ˆ
ˆ
p
(1

p
)
N
.24(.76) 100  25
1
1  n1

 .0057
b. ˆ 2pˆ1  1
n1  1
N1
25  1 100
.56(.44) 50  25
ˆ 2pˆ 2 
 .0051 ,
25  1
50
(100)2 (.0057)  (50) 2 (.0051)
ˆ 2pˆ st 
 .0031
(150)2
20.34 a. pˆ st  [100
90% confidence interval: .3467  1.645 .0031 : .2550 up to .4383
95% confidence interval: .3467  1.96 .0031 : .2375 up to .4559
31
29
34
 127  208 ] / 487  .7214
40
40
50
p
(1

p
)
N

n
.775(.225)
152  40
1
1
1

 .0033
b. ˆ 2pˆ1  1
n1  1
N1
40  1
152
.725(.275) 127  40
.68(.32) 208  50
ˆ 2pˆ 2 
 .0035 , ˆ 2pˆ 3 
 .0034
40  1
127
50  1
208
(152)2 (.0033)  (127) 2 (.0035)  (208) 2 (.0034)
ˆ 2pˆ st 
 .0012
(487)2
20.35 a. pˆ st  [152
90% confidence interval: .7214  1.645 .0012 : .6649 up to .7779
95% confidence interval: .7214  1.96 .0012 : .6541 up to .7887
208
130  55.52 = 56 observations
487


208(12.3)
b. n3  
 130  67.95 = 68 observations
152(7.1)  127(9.9)  208(12.3) 
20.36 a. n3 
20.37 a. n1 
240
160  36.23 = 37 observations
1060
505
506
Statistics for Business & Economics, 6th edition


240(.8)
b. n1  
 160  31.38 = 32
 240(.8)  190(.9)  350(1.2)  280(.7) 
observations
632
100  54.43 = 55 observations
1161


632(1.04)
b. n1  
 100  59.09 = 60 observations
 632(1.04)  529(.86) 
20.38 a. n1 
120
135  36 observations
450


120(93)
b. n1  
 135  51.56 = 52 observations
120(93)  150(64)  180(47) 
20.39 a. n1 
1031
100  73.91 = 74 observations
1395


1031(219.9)
b. n2  
 100  87.71 = 88 observations
 364(87.3)  1031(219.9) 
20.40 a. n2 
2000
812(20000) 2
 1020.4 , n 
20.41  x 
 261.0038 = 262
1.96
811(1020.4) 2  (20000) 2
observations
20.42 How large n?
N 2
(400)(10, 000) 2
n

= 57.988, take 58
2
2
2
( N  1) x   2 (399)(1215.8055)  (10, 000)
observations
.05
 .0194
2.575
.25 N
(.25)320
n

 216.18 = 217 observations
2
( N  1) px  .25 319(.0194) 2  .25
20.43  pˆ 
20.44  pˆ 
.04
417(.25)
 .0243 , n 
 210.33 = 211 observations
1.645
416(.0243) 2  .25
Chapter 20: Additional Topics in Sampling
507
25
 15.1976
1.645
 200(400)2  86250000
20.45 Proportional allocation:  x 
N 
2
j
j
 500(150)2 
86250000
 224.05 = 225 observations
1400(15.1976) 2  86250000 /1400
Optimal allocation:  N j j  500(150)   200(400)  325000
n
n
(325000)2 /1400
 195.98 = 196 observations
1400(15.1976) 2  86250000 /1400
500
 255.1020
1.96
 1150(4000)2  2120(6000)2  930(800000)2  15424 107
20.46 Proportional allocation:  x 
N 
2
j
j
15424 107
 497.47 = 498 observations
4200(255.1020) 2  15424 107 / 4200
Optimal allocation:
 N j j  1150(4000)  2120(6000)  930(8000)  24760000
n
n
(24760000) 2 / 4200
 470.78 = 471 observations
4200(255.1020) 2  15424 107 / 4200
20.47 a. xc 
28(29.6)  35(18.4) 
343
 42(24.1)
 24.979
b.
ˆ
2
xc
65  10
(28)2 (29.6  24.979) 2   (42) 2 (24.1  24.979) 2

 1.655
65(10)(34.3)2
10  1
90% confidence interval: 24.979  1.645 1.655 : 22.8627 up to 27.0953
20.48 a. xc 
69(83)  75(64) 
497
 71(98)
 91.6761
b.
(52  8)
(69) 2 (83  91.67605634) 2    (71) 2 (98  91.67605634) 2
8 1
52(8)(61.125) 2
= 66.409
99% confidence interval: 91.6761  2.58 66.4090
70.6920 up to 112.6602
ˆ 2XC 
20.49 a. pˆ c 
12   26
 .4636
343
508
Statistics for Business & Economics, 6th edition
b.
2
2
 12

 26

(28)   .4636    (42) 2   .4636 
65  10
 28

 42
  .0012

2
65(10)(34.3)
10  1
2
ˆ 2pˆ
c
90% confidence interval: .4636  1.645 .0012 : .4069 up to .5203
20.50 a. pˆ c 
24   34
 .4507
497
b.
2
ˆ 2pˆ
c
2
 24

 34

(69) 2   .4507    (71) 2   .4507 
52  8
 69

 71
  .0013

2
52(8)(62.125)
8 1
95% confidence interval: .4507  1.96 .0013 : .38 up to .5214
20.51
128  131  172
 .2310
1866
2
2
2



2  128
2  131
2  172
(611) 
 .231  (521) 
 .231  (734) 
 .231
50  3
611
521
734





  .0001

2
50(3)(622)
3 1
pˆ c 
ˆ 2pˆ
c
90% confidence interval: .231  1.645 .0001 : .2146 up to .2489
5000
720(37600)2
 3039.5 , n 
 126.34 = 127
1.645
719(3039)2  (37600)2
observations.
Additional sample observations needed are 127-20 = 107
20.52  x 
5
 3.0395
1.645
(3200(40)  800(58)) 2 / 4000
n
 195.49 = 196
4000(3.0395)2  (3200(40) 2  800(58) 2 ) / 4000
observations. Additional sample observations needed: 196-60 = 136
20.53  x 
20
 10.2
1.96
(100(105)  180(162)  200(183)) 2 / 480
n
 159.35 = 160
480(10.2)2  (100(105) 2  180(162) 2  200(183) 2 ) / 480
observations. Additional sample observations needed: 160-30 = 130
20.54  x 
Chapter 20: Additional Topics in Sampling
20.55- 20.57 Discussion questions – various answers
747
(11.44) 2 90  10
 74.7 , s = 11.44, ˆ 2 x 
 11.633
10
10
90
90% confidence interval: 74.7  1.645 11.633 : 69.089 up to 80.311
b. The interval would be wider; the z-score would increase to 1.96
20.58 a. x 
(149.92) 2 272  50
20.59 a. ˆ x 
 366.888
50
272
99% confidence interval: 492.36  2.58 366.888 :
442.9139 up to 541.7501
b. 95% confidence interval: 492.36  1.96 366.888 :
454.8175 up to 529.9025
c. The 90% interval is narrower; the z-score would decline to 1.645
2
38
.623(.377) 100  61
 .623 , ˆ 2pˆ 
 .0015
61
61
100
90% confidence interval: .623  1.645 .0015 : .559 up to .687
b. If the sample information is not randomly selected, the resulting
conclusions may be biased
20.60 a. pˆ 
20.61
36
.6(.4) 148  60
 .6 , ˆ pˆ 
 .0024
60
60
148
95% confidence interval: .6  1.96 .0024 : .504 up to .696
pˆ 
20.62 a. xst  11.5845
ˆ 2 x  .7321 , ˆ 2 x  1.9053 , ˆ 2 x  1.7508
(352)2 (.7321)  (287) 2 (1.9053)  (331) 2 (1.7508)
ˆ 2 x 
 .4671
(970)2
1
2
3
st
99% confidence interval for managers in subdivision 1:
9.2  2.575 .7321 : 6.997 up to 11.403
b. 99% confidence interval for all managers:
11.5845  2.575 .4671 : 9.8247 up to 13.3444
509
510
Statistics for Business & Economics, 6th edition
120(1.6)  180(.74) 325.2

 1.084
300
300
s 2 N  n (.98)2 120  20
ˆ 2 x1 

 .0400
n N
20
120
(.56)2 180  20
ˆ 2 x2 
 .0139 ,
20
180
(120)2 (.04)  (180)2 (.0139)
ˆ 2 xst 
 .0114
(300)2
95% confidence interval for mean number of errors per page in the
book: 1.084  1.96 .0114 : .8747 up to 1.2933
20.63 a. xst 
b. Nxst  Z / 2 Nˆ xst  N   Nxst  Z / 2 Nˆ xst
where, Nxst  (120)(1.6)  (180)(.74)  325.2
N 2ˆ 2 xst  (120)2 .0400  (180)2 .0139  1,029.3408
Nˆ xst  1029.3408  32.08334 , 325.2 2.58(32.08334)
99% confidence interval for the total number of errors in the book:
242.4279 up to 407.9721 or from 243 total errors up to 408 total
errors.
20.64
pˆ1 
pˆ pˆ st
9
15
 .45, pˆ 2 
 .75
20
20
pˆ (1  pˆ1 ) N1  n1 .45(.55) 120  20
 .63, ˆ 2pˆ1  1

 .0109
n1  1
N1
20  1 120
ˆ 2pˆ 
2
pˆ1 (1  pˆ1 ) N1  n1 .75(.25) 180  20

 .0088
n1  1
N1
20  1 180
(120)2 (.0109)  (180) 2 (.0088)
 .0049
st
(300)2
90% confidence interval is:
.63  1.645 .0049 or 0.5147 up to 0.7453
ˆ 2pˆ 
352
80  29.03 = 30 observations
970


352(4.9)
b. n1  
 80  22.7 = 23 observations
 352(4.9)  287(6.4)  331(7.6) 
20.65 a. n1 
120
40  16 observations
300


120(.98)
b. n1  
 40  21.54 = 22 observations
120(.98)  180(.56) 
20.66 a. n1 
Chapter 20: Additional Topics in Sampling
20.67 Refer to section 20.4 – Stratified Random Sampling
2000
328(12000) 2
 1215.8 , n 
 75.28 = 76
1.645
327(1215.8) 2  (12000) 2
observations
20.68  x 
20.69  pˆ 
.06
527(.25)
 .0306 , n 
 177.43 = 178 observations
1.96
526(.0306) 2  .25
20.70 Various answers. Answers should include a discussion of the potential for
stratification of the population. Because different counties utilize different
ballots and ballot techniques, a stratification by county may prove
reasonable. The method used by the county could also be utilized in the
stratification – e.g., butterfly ballots and hand-punched vs. electronic
ballots.
511