Chapter 22
Sample Survey
Learning Objectives
1.
Learn what a sample survey is and how it differs from an experiment as a method of collecting data.
2.
Know about the methods of data collection for a survey.
3.
Know the difference between sampling and nonsampling error.
4.
Learn about four sample designs: (1) simple random sampling, (2) stratified simple random
sampling, (3) cluster sampling, and (4) systematic sampling.
5.
Lean how to estimate a population mean, a population total, and a population proportion using the
above sample designs.
6.
Understand the relationship between sample size and precision.
7.
Learn how to choose the appropriate sample size using stratified and simple random sampling.
8.
Learn how to allocate the total sample to the various strata using stratified simple random sampling.
22 1
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Chapter 22
Solutions:
1.
2.
a.
x = 215 is an estimate of the population mean.
b.
sx
c.
215 2(2.7386) or 209.5228 to 220.4772
a.
Estimate of population total = N x = 400(75) = 30,000
b.
Estimate of Standard Error = Ns x
20
50
800 50
2.7386
800
8 400 80
Nsx 400
320
400
80
3.
c.
30,000 2(320) or 29,360 to 30,640
a.
p = .30 is an estimate of the population proportion
b.
1000 100 (.3)(.7)
sp
.0437
1000 99
c.
.30 2(.0437) or .2126 to .3874
4.
B = 15
n
(70) 2
4900
72.9830
2
2
(15)
(70)
67.1389
4
450
A sample size of 73 will provide an approximate 95% confidence interval of width 30.
5.
a.
x = 149,670 and s = 73,420
sx
771 50 73, 420
10, 040.83
771 50
approximate 95% confidence interval
149,670 2(10,040.83)
or
$129,588.34 to $169,751.66
b.
�
X = N x = 771(149,670) = 115,395,570
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Sample Survey
sx$ = N sx = 771(10,040.83) = 7,741,479.93
approximate 95% confidence interval
115,395,570 2(7,741,479.93)
or
$99,912,610.14 to $130,878,529.86
c.
771 50 (.36)(.64)
p = 18/50 = 0.36 and s p
.0663
49
771
approximate 95% confidence interval
0.36 2(0.0663)
or
0.2274 to 0.4926
This is a rather large interval; sample sizes must be rather large to obtain tight confidence intervals
on a population proportion.
6.
B = 5000/2 = 2500 Use the value of s for the previous year in the formula to determine the
necessary sample size.
n
(31.3) 2
979.69
336.0051
(2.5)2 (31.3) 2 2.9157
4
724
A sample size of 337 will provide an approximate 95% confidence interval of width no larger than
$5000.
7.
a.
Stratum 1: = 138
Stratum 2: x2 = 103
Stratum 3: x3 = 210
b.
Stratum 1
x1 = 138
30 200 20
sx1
6.3640
200
20
138 2(6.3640)
or
125.272 to 150.728
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Chapter 22
Stratum 2
x2 = 103
25 250 30
sx2
4.2817
250
30
103 2(4.2817)
or
94.4366 to 111.5634
Stratum 3
x3 = 210
50 100 25
sx3
8.6603
100
25
210 2(8.6603)
or
192.6794 to 227.3206
c.
200
250
100
xst
138 550 103 550 210
550
= 50.1818 + 46.8182 + 38.1818
= 135.1818
1
sxst
2
(550)
(30) 2
(25) 2
(50) 2
200(180)
250(220)
100(75)
20
30
25
1
2
(550)
3,515,833.3 3.4092
approximate 95% confidence interval
135.1818 2(3.4092)
or
128.3634 to 142.0002
8.
a.
Stratum 1: N1 x1 = 200(138) = 27,600
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Sample Survey
Stratum 2: N 2 x2 = 250(103) = 25,750
Stratum 3: N 3 x3 = 100(210) = 21,000
b.
N xst = 27,600 + 25,750 + 21,000 = 74,350
Note: the sum of the estimate for each stratum total equals N xst
c.
sxst = 550(3.4092) = 1875.06 (see 7c)
approximate 95% confidence interval
74,350 2(1875.06)
or
70,599.88 to 78,100.12
9.
a.
Stratum 1
p1 = .50
200 20 (.50)(.50)
s p1
.1088
19
200
.50 2(.1088)
or
.2824 to .7176
Stratum 2
p2 = .78
250 30 (.78)(.22)
s p2
.0722
29
250
.78 2(.0722)
or
.6356 to .9244
Stratum 3
p3 = .21
100 25 (.21)(.79)
s p3
.0720
24
100
.21 2(.0720)
22 5
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Chapter 22
or
.066 to .354
200
250
100
(.50)
(.78)
(.21) .5745
550
550
550
b.
pst
c.
1
s pst
2
(550)
(.5)(.5)
(.78)(.22)
(.21)(.79)
250(220)
100(75)
200(180)
19
29
24
1
2
(550)
(473.6842 325.4483 51.8438) .0530
d.
approximate 95% confidence interval
.5745 2(.0530)
or
.4685 to .6805
10. a.
n
300(150) 600(75) 500(100)
2
(20) 2
2
2
2
(1400) 2
300(150) 600(75) 500(100)
2
(140, 000) 2
92.8359
196, 000, 000 15,125, 000
Rounding up we choose a total sample of 93.
300(150)
n1 93
30
140, 000
600(75)
n2 93
30
140, 000
500(100)
n3 93
33
140, 000
b.
With B = 10, the first term in the denominator in the formula for n changes.
n
(140, 000) 2
(140, 000) 2
305.6530
2
49, 000, 000 15,125, 000
2 (10)
(1400)
15,125, 000
4
Rounding up, we see that a sample size of 306 is needed to provide this level of precision.
300(150)
n1 306
98
140, 000
600(75)
n2 306
98
140, 000
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Sample Survey
500(100)
n3 306
109
140,000
Due to rounding, the total of the allocations to each strata only add to 305. Note that even though
the sample size is larger, the proportion allocated to each stratum has not changed.
c.
n
(140, 000) 2
(140, 000) 2
274.6060
(15, 000) 2
56, 250, 000 15,125, 000
15,125, 000
4
Rounding up, we see that a sample size of 275 will provide the desired level of precision.
The allocations to the strata are in the same proportion as for parts a and b.
300(150)
n1 275
98
140, 000
600(75)
n2 275
88
140, 000
500(100)
n3 275
98
140, 000
Again, due to rounding, the stratum allocations do not add to the total sample size. Another item
could be sampled from, say, stratum 3 if desired.
11. a.
b.
x1 = 29.5333
x2 = 64.775
x3 = 45.2125
x4 = 53.0300
Indianapolis
13.3603 38 6
29.533 2
38
6
29.533 10.9086(.9177)
or
19.5222 to 39.5438
Louisville
25.0666 45 8
64.775 2
45
8
64.775 17.7248(.9068)
or
48.7022 to 80.8478
St. Louis
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Chapter 22
19.4084 80 8
45.2125 2
80
8
45.2125 (13.7238) (.9487)
or
32.1927 to 58.2323
Memphis
29.6810 70 10
53.0300 2
70
10
53.0300 18.7719(.9258)
or
35.6510 to 70.4090
c.
38 1 45 5 80 3 70 5
pst
.4269
233 6 233 8 233 8 233 10
d.
1 5
6 6
p (1 p1 )
N1 ( N1 n1 ) 1
38(32) 33.7778
n1 1
5
5 3
8 8
p (1 p2 )
N 2 ( N 2 n2 ) 2
45(37) 55.7478
n2 1
7
3 5
8 8
p (1 p3 )
N 3 ( N 3 n3 ) 3
80(72) 192.8571
n3 1
7
5 5
10 10
p (1 p4 )
N 4 ( N 4 n4 ) 4
70(60) 116.6667
n4 1
9
1
1
s pst
33.7778 55.7478 192.8571 116.6667
(399.0494) .0857
2
(233) 2
(233)
approximate 95% confidence interval
.4269 2(.0857)
or
.2555 to .5983
12. a.
St. Louis total = N1 x1 = 80 (45.2125) = 3617
In dollars: $3,617,000
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Sample Survey
b.
Indianapolis total = N1 x1 = 38 (29.5333) = 1122.2654
In dollars: $1,122,265
c.
38
45
80
70
xst
29.5333
64.775
45.2125
53.0300 48.7821
233
233
233
233
N1 ( N1 n1 )
s12
(13.3603) 2
38(32)
36,175.517
n1
6
N 2 ( N 2 n2 )
s22
(25.0666) 2
45(37)
130, 772.1
n2
8
N 3 ( N 3 n3 )
s32
(19.4084) 2
80(72)
271, 213.91
n3
8
N 4 ( N 4 n4 )
s42
(29.6810) 2
70(60)
370, 003.94
n4
10
1
sxst
36,175.517 130, 772.1 271, 213.91 370, 003.94
2
(233)
1
(808,165.47) 3.8583
(233)2
approximate 95% confidence interval
xst 2sxst
48.7821 2(3.8583)
or
41.0655 to 56.4987
In dollars: $41,066 to $56,499
d.
approximate 95% confidence interval
Nxst 2 Nsxst
233(48.7821) 2(233)(3.8583)
11,366.229 1797.9678
or
9,568.2612 to 13,164.197
In dollars: $9,568,261 to $13,164,197
22 9
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Chapter 22
n
13.
50(80) 38(150) 35(45)
2
(30)2
2
2
2
(123) 2
50(80) 38(150) 35(45)
4
(11, 275) 2
27.3394
3, 404, 025 1, 245,875
Rounding up we see that a sample size of 28 is necessary to obtain the desired precision.
50(80)
n1 28
10
11, 275
38(150)
n2 28
14
11, 275
35(45)
n3 28
4
11, 275
b.
n
50(100) 38(100) 35(100)
2
(30) 2
2
2
2
(123)
50(100) 38(100) 35(100)
4
2
123(100)
2
3, 404, 025 123(100) 2
33
50(100)
n1 33
13
12,300
38(100)
n2 33
10
12,300
35(100)
n3 33
9
12,300
This is the same as proportional allocation . Note that for each stratum
N
nh n h
N
14. a.
xc
xi
750
15
Mi
50
�
X M xc = 300(15) = 4500
pc
b.
ai
15
.30
M i 50
( xi xc M i ) 2
= [ 95 15 (7) ]2 + [ 325 15 (18) ]2 + [ 190 15 (15) ]2 + [ 140 15 (10)]2
= (10)2 + (55)2 + (35)2 + (10)2
= 4450
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Sample Survey
25 4
sxc
2
(25)(4)(12)
4450
1.4708
3
s X� Ms xc = 300(1.4708) = 441.24
(ai pc M i ) 2
= [ 1 .3 (7) ]2 + [ 6 .3 (18) ]2 + [ 6 .3 (15) ]2 + [2 .3 (10) ]2
= (1.1)2 + (.6)2 + (1.5)2 + (1)2
= 4.82
25 4 4.82
s pc
.0484
2
(25)(4)(12)
3
c.
approximate 95% confidence
Interval for Population Mean:
15 2(1.4708)
or
12.0584 to 17.9416
d.
approximate 95% confidence
Interval for Population Total:
4500 2(441.24)
or
3617.52 to 5382.48
e.
approximate 95% confidence
Interval for Population Proportion:
.30 2(.0484)
or
.2032 to .3968
15. a.
10,400
xc
80
130
�
X M xc = 600(80) = 48,000
13
pc
.10
130
b.
( xi xc M i ) 2
= [ 3500 80 (35) ]2 + [ 965 80 (15) ]2 + [ 960 80 (12) ]2
+ [ 2070 80 (23) ]2 + [ 1100 80 (20) ]2 + [ 1805 80 (25) ]2
= (700)2 + (235)2 + (0)2 + (230)2 + (500)2 + (195)2
= 886,150
22 11
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Chapter 22
30 6
sxc
2
(30)(6)(20)
886,150
7.6861
5
approximate 95% confidence
Interval for Population Mean:
c.
80 2(7.6861)
or
64.6278 to 95.3722
s �X = 600(7.6861) = 4611.66
approximate 95% confidence
Interval for Population Total:
48,000 2(4611.66)
or
38,776.68 to 57,223.32
d.
(ai pc M i ) 2
= [ 3 .1 (35) ]2 + [ 0 .1 (15) ]2 + [ 1 .1 (12) ]2 + [4 .1 (23) ]2
+ [ 3 .1 (20) ]2 + [ 2 .1 (25) ]2
= (.5)2 + (1.5)2 + (.2)2 + (1.7)2 + (1)2 + (.5)2
= 6.68
6.68
30 6
s pc
.0211
2
(30)(6)(20) 5
approximate 95% confidence
Interval for Population Proportion:
.10 2(.0211)
or
.0578 to .1422
16. a.
xc
2000
40
50
Estimate of mean age of mechanical engineers: 40 years
b.
pc
35
.70
50
Estimate of proportion attending local university: .70
c.
( xi xc M i ) 2
= [ 520 40 (12) ]2 + ∙ ∙ ∙ + [ 462 40 (13) ]2
= (40)2 + (7)2 + (10)2 + (11)2 + (30)2 + (9)2 + (22)2 + (8)2 + (23)2
+ (58)2
= 7292
22 12
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Sample Survey
120 10
sxc
2
(120)(10)(50 /12)
7292
2.0683
9
approximate 95% confidence
Interval for Mean age:
40 2(2.0683)
or
35.8634 to 44.1366
d.
(ai pc M i ) 2
= [ 8 .7 (12) ]2 + ∙ ∙ ∙ + [ 12 .7 (13) ]2
= (.4)2 + (.7)2 + (.4)2 + (.3)2 + (1.2)2 + (.1)2 + (1.4)2 + (.3)2
+ (.7)2 + (2.9)2
= 13.3
13.3
120 10
s pc
.0883
2
(120)(10)(50
/12)
9
approximate 95% confidence
Interval for Proportion Attending Local University:
.70 2(.0883)
or
.5234 to .8766
17. a.
17(37) 35(32) 57(44) 11, 240
xc
36.9737
17 35 57
304
Estimate of mean age: 36.9737 years
b.
Proportion of College Graduates: 128 / 304 = .4211
Proportion of Males: 112 / 304 = .3684
c.
( xi xc M i ) 2
= [ 17 (37) (36.9737) (17) ]2 + ∙ ∙ ∙ + [ 57 (44) (36.9737) (44) ]2
= (.4471)2 + (174.0795)2 + (25.3162)2 + (460.2642)2 + (173.1309)2
+ (180.3156)2 + (94.7376)2 + (400.4991)2
= 474,650.68
150 8
sxc
2
(150)(8)(40)
474, 650.68
2.2394
7
approximate 95% confidence
Interval for Mean Age of Agents:
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Chapter 22
36.9737 2(2.2394)
or
32.4949 to 41.4525
d.
(ai pc M i ) 2
= [ 3 .4211 (17) ]2 + ∙ ∙ ∙ + [ 25 .4211 (57) ]2
= (4.1587)2 + (.7385)2 + (2.9486)2 + (10.2074)2 + (.1073)2 + (3.0532)2
+ (.2128)2 + (.9973)2
= 141.0989
150 8
s pc
2
(150)(8)(40)
141.0989
.0386
7
approximate 95% confidence
Interval for Proportion of Agents that are College Graduates:
.4211 2(.0386)
or
.3439 to .4983
e.
(ai pc M i ) 2
= [ 4 .3684 (17) ]2 + ∙ ∙ ∙ + [ 26 .3684 (57) ]2
= (2.2628)2 + (.8940)2 + (2.5784)2 + (3.6856)2 + (3.8412)2 + (1.5792)2
+ (.6832)2 + (5.0012)2
= 68.8787
150 8
s pc
2
(150)(8)(40)
68.8787
.0270
7
approximate 95% confidence
Interval for Proportion of Agents that are Male:
.3684 2(.0270)
or
.3144 to .4224
18. a.
p = 0.19
sp
(0.19)(0.81)
0.0206
363
Approximate 95% Confidence Interval:
0.19 2(0.0206)
or
0.1488 to 0.2312
b.
p = 0.31
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Sample Survey
sp
(0.31)(0.69)
0.0243
363
Approximate 95% Confidence Interval:
0.31 2(0.0243)
or
0.2615 to 0.3585
c.
p = 0.17
sp
(0.17)(0.83)
0.0197
373
Approximate 95% Confidence Interval:
0.17 2(0.0197)
or
0.1306 to 0.2094
d.
The largest standard error is when p = .50.
At p = .50, we get
sp
(0.5)(0.5)
0.0262
363
Multiplying by 2, we get a bound of B = 2(.0262) = 0.0525
For a sample of 363, then, they know that in the worst case ( p = 0.50), the bound will be
approximately 5%.
e. If the poll was conducted by calling people at home during the day the sample results would only be
representative of adults not working outside the home. It is likely that the Louis Harris organization
took precautions against this and other possible sources of bias.
19. a.
Assume (N n) / N 1
p = .55
sp
b.
(0.55)(0.45)
0.0222
504
p = .31
sp
(0.31)(0.69)
0.0206
504
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Chapter 22
c.
The estimate of the standard error in part (a) is larger because p is closer to .50.
d.
Approximate 95% Confidence interval:
.55 2(.0222)
or
.5056 to .5944
e.
Approximate 95% Confidence interval:
.31 2(.0206)
.2688 to .3512
20. a.
sx
3000 200 3000
204.9390
3000
200
Approximate 95% Confidence Interval for Mean Annual Salary:
23,200 2(204.9390)
or
$22,790 to $23,610
b.
N x = 3000 (23,200) = 69,600,000
sx$ = 3000 (204.9390) = 614,817
Approximate 95% Confidence Interval for Population Total Salary:
69,600,000 2(614,817)
or
$68,370,366 to $70,829,634
c.
p = .73
3000 200 (.73)(.27)
sp
.0304
3000 199
Approximate 95% Confidence Interval for Proportion that are Generally Satisfied:
.73 2(.0304)
or
.6692 to .7908
d. If management administered the questionnaire and anonymity was not guaranteed we would expect
a definite upward bias in the percent reporting they were “generally satisfied” with their job. A
procedure for guaranteeing anonymity should reduce the bias.
22 16
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Sample Survey
21. a.
p = 1/3
380 30 (1/ 3)(2 / 3)
sp
.0840
29
380
Approximate 95% Confidence Interval:
.3333 2(.0840)
or
.1653 to .5013
b.
�
X 2 = 760 (19 / 45) = 320.8889
c.
p = 19 / 45 = .4222
760 45 (19 / 45)(26 / 45)
sp
.0722
44
760
Approximate 95% Confidence Interval:
.4222 2(.0722)
or
.2778 to .5666
d.
380 10 760 19 260 7
pst
.3717
1400 30 1400 45 1400 25
p (1 ph )
(1/ 3)(2 / 3)
N h ( N h nh ) h
380(350)
29
nh 1
760(715)
(19 / 45)(26 / 45)
(7 / 25)(18 / 25)
260(235)
44
24
= 1019.1571 + 3012.7901 + 513.2400 = 4545.1892
1
s pst
4545.1892 .0482
2
(1400)
Approximate 95% Confidence Interval:
.3717 2(.0482)
or
.2753 to .4681
22 17
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Chapter 22
22. a.
�
X = 380 (9 / 30) + 760 (12 / 45) + 260 (11 / 25) = 431.0667
Estimate approximately 431 deaths due to beating.
b.
380 9 760 12 260 11
pst
.3079
1400 30 1400 45 1400 25
N h ( N h nh )
ph (1
ph )
nh 1
= (380) (380 30) (9 / 30) (21 / 30) / 29 + (760) (760 45) (12 / 45) (33 / 45) / 44 +
(260) (260 25)(11 / 25) (14 / 25) / 24
= 4005.5079
1
s pst
4005.5079 .0452
2
(1400)
Approximate 95% Confidence Interval:
.3079 2(.0452)
or
.2175 to .3983
c.
380 21 760 34 260 15
pst
.7116
1400 30 1400 45 1400 25
N h ( N h nh )
ph (1
ph )
nh 1
= (380) (380 30) (21 / 30) (9 / 30) / 29 + (760) (760 45) (34 / 45) (11 / 45) / 44 +
(260) (260 25) (15 / 25) (10 / 25) / 24
= 3855.0417
1
s pst
3855.0417 .0443
2
(1400)
Approximate 95% Confidence Interval:
.7116 2(.0443)
or
.6230 to .8002
d.
�
X = 1400 (.7116) = 996.24
Estimate of total number of victims is 996.
22 18
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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Sample Survey
23. a.
n
3000(80) 600(150) 250(220) 100(700) 50(3000)
2
(20) 2
2
2
2
2
2
(4000) 2
3000(80) 600(150) 250(220) 100(700) 50(3000)
4
366, 025, 000, 000
170.7365
1, 600,000, 000 543,800, 000
Rounding up, we need a sample size of 171 for the desired precision.
b.
3000(80)
n1 171
68
605, 000
600(150)
n2 171
25
605, 000
250(220)
n3 171
16
605,000
100(700)
n4 171
20
605, 000
50(3000)
n5 171
42
605, 000
24. a.
14(61) 7(74) 96(78) 23(69) 71(73) 29(84) 18,066
xc
75.275
14 7 96 23 71 29
240
Estimate of mean age is approximately 75 years old.
b.
12 2 30 8 10 22
84
pc
.35
14 7 96 23 71 29 240
(ai pc M i ) 2
= [12 .35 (14) ]2 + [ 2 .35 (7) ]2 + [30 .35 (96) ] 2
+ [ 8 .35 (23) ]2 + [ 10 .35 (71) ]2 + [ 22 .35 (29) ]2
= (7.1)2 + (.45)2 + (3.6)2 + (.05)2 + (14.85)2 + (11.85)2
= 424.52
100 6
424.52
s pc
.0760
2
(100)(6)(48) 5
Approximate 95% Confidence Interval:
.35 2(.0760)
or
.198 to .502
22 19
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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 22
�
X = 4800 (.35) = 1680
Estimate of total number of Disabled Persons is 1680.
22 20
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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.