大学课件商务与经济统计习题答案(第8版,中文版)
Chapter 21
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.
Chapter 21
Solutions:
1. a. x = 215 is an estimate of the population mean.
b. 2.7386
x
s=
c. 215 ± 2(2.7386) or 209.5228 to 220.4772
2. a. Estimate of population total = N x = 400(75) = 30,000
b. Estimate of Standard Error =
x
Ns
400320
x
Ns=
c. 30,000 ± 2(320) or 29,360 to 30,640
3. a. p= .30 is an estimate of the population proportion
b. .0437
p
s=
c. .30 ± 2(.0437) or .2126 to .3874
4. B = 15
2
22
(70)4900
72.9830
67.1389
(15)(70)
4450
n===
+
A sample size of 73 will provide an approximate 95% confidence interval of width 30.
5. a. x= 149,670 and s = 73,420
10,040.83
x
s==
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
x
s= N
x
s= 771(10,040.83) = 7,741,479.93
Sample Survey
approximate 95% confidence interval
115,395,770 ± 2(7,741,479.93)
or
$99,912,810.14 to $130,878,729.86
c.
p
= 18/50 = 0.36 and .0663p s =
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.
222
(31.3)979.69
336.00512.9157(2.5)(31.3)
4724
n ===+
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: 2x = 103
Stratum 3: 3x = 210
b. Stratum 1 1x = 138
1 6.3640x s =
138 ± 2(6.3640)
or
125.272 to 150.728
Stratum 2
2x = 103
Chapter 21
24.2817
x
s==
103 ± 2(4.2817)
or
94.4366 to 111.5634 Stratum 3
3
x= 210
38.6603
x
s==
210 ± 2(8.6603)
or
192.6794 to 227.3206
c.
200250100
138103210
550550550
st
x
??????
=++
? ? ?
??????
= 50.1818 + 46.8182 + 38.1818
= 135.1818
st
x
s=
3.4092
=
approximate 95% confidence interval
135.1818 ± 2(3.4092)
or
128.3634 to 142.0002
8. a. Stratum 1:
11
N x = 200(138) = 27,600
Stratum 2:
22
N x = 250(103) = 25,750
Stratum 3:
33
N x = 100(210) = 21,000
b.
st
N x = 27,600 + 25,750 + 21,000 = 74,350
Note: the sum of the estimate for each stratum total equals
st
N x
c.
st
N x= 550(3.4092) = 1875.06 (see 7c)
Sample Survey
approximate 95% confidence interval
74,350 ± 2(1875.06)
or
70,599.88 to 78,100.12
9. a. Stratum 1
1p = .50
1.1088p s ==
50 ± 2(.1088)
or
.2824 to .7176
Stratum 2
2p = .78
2.0722p s
.78 ± 2(.0722)
or
.6356 to .9244
Stratum 3
3p = .21
3.0720p s =
.21 ± 2(.0720)
or .066 to .354
b. 200250100
(.50)(.78)(.21).5745550550550
st p =
++=
c.
st p s =
.0530==
d.
approximate 95% confidence interval
Chapter 21
.5745 ± 2(.0530)
or
.4685 to .6805
10. a.
[]
2
2
2
2
22300(150)600(75)500(100)(140,000)92.8359196,000,00015,125,000
(20)(1400)300(150)600(75)500(100)2
n ++=
==+????+++
?????
Rounding up we choose a total sample of 93.
1300(150)9330140,000n ??== ???
2600(75)9330140,000n ??== ???
3500(100)9333140,000n ??== ???
b. With B = 10, the first term in the denominator in the formula for n changes.
22
2
2(140,000)(140,000)305.653049,000,00015,125,000(10)
(1400)15,125,000
4n ===+??+ ???
Rounding up, we see that a sample size of 306 is needed to provide this level of precision.
1300(150)30698140,000n ??
== ???
2600(75)30698140,000n ??
== ???
3500(100)306109140,000n ??
== ???
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.
22
2
(140,000)(140,000)274.606056,250,00015,125,000(15,000)15,125,0004
n ===++
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.
Sample Survey
1300(150)27598140,000n ??
== ???
2600(75)27588140,000n ??
== ???
3500(100)27598140,000n ??
== ???
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. 1x = 29.5333
2x = 64.775
3x = 45.2125 4x = 53.0300 b. Indianapolis
29.5332
± 29.533 ± 10.9086(.9177)
or
19.5222 to 39.5438
Louisville
64.7752
± 64.775 ± 17.7248(.9068)
or
48.7022 to 80.8478
St. Louis
45.21252
± 45.2125 ± (13.7238) (.9487)
or
32.1927 to 58.2323
Memphis
53.03002
±53.0300 ± 18.7719(.9258)
or
35.6510 to 70.4090
Chapter 21
c.
381455803705.426923362338233823310st p ????????????????=+++= ??? ??? ??? ???????????????????
d.
11111115(1)
66()38(32)33.777815p p N N n n ????
???-????-==-
22222253(1)
88()45(37)55.747817p p N N n n ????
???-????-==-
33333335(1)
88()80(72)192.857117p p N N n n ????
???-????-==-
44444455(1)
1010()70(60)116.666719
p p N N n n ????
???-????-==-
.0857st p s ==
approximate 95% confidence interval
.4269 ± 2(.0857)
or
.2555 to .5983
12. a. St. Louis total = 11N x = 80 (45.2125) = 3617
In dollars: $3,617,000
b. Indianapolis total = 11N x = 38 (29.5333) = 1122.2654 In dollars: $1,122,265
c.
3845807029.533364.77545.212553.030048.7821233233233233st x ????????=+++= ? ? ? ?????????
22
11111(13.3603)()38(32)36,175.5176
s N N n n -==
22
22222(25.0666)()45(37)130,772.18
s N N n n -==
Sample Survey
22
33333(19.4084)()80(72)271,213.918
s N N n n -==
22
44444(29.6810)()70(60)370,003.9410
s N N n n -==
st x s =
3.8583=
=
approximate 95% confidence interval
2st st x x s ±
48.7821 ± 2(3.8583)
or
41.0655 to 56.4987
In dollars: $41,066 to $56,499
d.
approximate 95% confidence interval
2st st x Nx Ns ±
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
13.
[]
2
2
22
22250(80)38(150)35(45)(11,275)27.33943,404,0251,245,875
(30)(123)50(80)38(150)35(45)4n ++=
==+????+++ ???
??
Rounding up we see that a sample size of 28 is necessary to obtain the desired precision.
150(80)281011,275n ??== ???
238(150)281411,275n ??
== ???
335(45)28411,275n ??
== ???
Chapter 21
b. []
[]
2
2
2
2
2222
50(100)38(100)35(100)123(100)333,404,025123(100)(30)(123)50(100)38(100)35(100)4
n ++=
=
=+????+++ ???
??
150(100)331312,300n ??== ???
238(100)331012,300n ??
== ???
335(100)33912,300n ??
== ???
This is the same as proportional allocation . Note that for each stratum
h h N n n N ??= ???
14. a. 750
1550
i c i x x M ∑=
==∑ c X M x = = 300(15) = 4500
15
.3050
i c i a p M ∑=
==∑ b. 2()i c i x x M ∑- = [ 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
1.4708c x s == c x X s Ms == 300(1.4708) = 441.24
2()i c i a p M ∑- = [ 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
.0484c p s c. approximate 95% confidence Interval for Population Mean: 15 ± 2(1.4708) or
12.0584 to 17.9416
Sample Survey
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
80130
c x =
= c X M x = = 600(80) = 48,000
13
.10130
c p =
= b. 2()i c i x x M ∑- = [ 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
7.6861c x s = approximate 95% confidence Interval for Population Mean: 80 ± 2(7.6861) or
64.6278 to 95.3722
c. X s = 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
2()i c i a p M ∑- = [ 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
Chapter 21
.0211c p s approximate 95% confidence
Interval for Population Proportion: .10 ± 2(.0211)
or
.0578 to .1422
16. a. 2000
4050
c x =
= Estimate of mean age of mechanical engineers: 40 years
b. 35
.7050
c p =
= Estimate of proportion attending local university: .70
c. 2()i c i x x M ∑- = [ 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
2.0683c x s ==
approximate 95% confidence Interval for Mean age:
40 ± 2(2.0683) or
35.8634 to 44.1366
d. 2()i c i a p M ∑- = [ 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
.0883c p s =
approximate 95% confidence
Interval for Proportion Attending Local University:
Sample Survey
.70 ± 2(.0883) or
.5234 to .8766
17. a. 17(37)35(32)57(44)11,240
36.9737173557304
c x ++???+=
==++???+
Estimate of mean age: 36.9737 years
b. Proportion of College Graduates: 128 / 304 = .4211
Proportion of Males: 112 / 304 = .3684
c. 2()i c i x x M ∑- = [ 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
2.2394c x s ==
approximate 95% confidence Interval for Mean Age of Agents: 36.9737 ± 2(2.2394) or
32.4949 to 41.4525
d. 2()i c i a p M ∑- = [ 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
.0386c p s = approximate 95% confidence
Interval for Proportion of Agents that are College Graduates:
.4211 ± 2(.0386) or
.3439 to .4983
e. 2()i c i a p M ∑- = [ 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
Chapter 21
.0270c p s =
approximate 95% confidence
Interval for Proportion of Agents that are Male: .3684 ± 2(.0270) or
.3144 to .4224 18. a. p = 0.19
0.0206p s =
=
Approximate 95% Confidence Interval: 0.19 ± 2(0.0206) or
0.1488 to 0.2312 b. p = 0.31
0.0243p s =
=
Approximate 95% Confidence Interval: 0.31 ± 2(0.0243) or
0.2615 to 0.3585 c. p = 0.17
0.0197p s =
=
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
0.0262p s =
Sample Survey
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
s==
0.0222
p
b. p= .31
s==
0.0206
p
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. 204.9390
s==
x
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
s = 3000 (204.9390) = 614,817
x
Chapter 21
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
.0304
p
s
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.
21. a. p= 1/3
.0840
p
s=
Approximate 95% Confidence Interval:
.3333 ± 2(.0840)
or
.1653 to .5013
b. 2X = 760 (19 / 45) = 320.8889
c. p= 19 / 45 = .4222
.0722
p
s
Approximate 95% Confidence Interval:
.4222 ± 2(.0722)
or
.2778 to .5666
d.
38010760192607
.3717 140030140045140025
st
p
????????????
=++=
??? ??? ???
????????????
Sample Survey
(1)(1/3)(2/3)
()380(350)
129h h h h h h p p N N n n ??-∑-=??-?? (19/45)(26/45)(7/25)(18/25)
760(715)
260(235)
4424
++
= 1019.1571 + 3012.7901 + 513.2400 = 4545.1892
.0482st p s == Approximate 95% Confidence Interval: .3717 ± 2(.0482) or
.2753 to .4681
22. a. X = 380 (9 / 30) + 760 (12 / 45) + 260 (11 / 25) = 431.0667 Estimate approximately 431 deaths due to beating.
b. 38097601226011.3079140030140045140025st p ????????????
=++= ??? ??? ???????????????
[](1)()1
h h h h h h p p N N n n -∑--
= (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
.0452st p s == Approximate 95% Confidence Interval: .3079 ± 2(.0452) or
.2175 to .3983
c.
380217603426015.7116140030140045140025st p ????????????=++= ??? ??? ???????????????
[](1)()
1
h h h h h h p p N N n n -∑--
Chapter 21
= (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
.0443st p s = Approximate 95% Confidence Interval: .7116 ± 2(.0443) or
.6230 to .8002
d. X = 1400 (.7116) = 996.24
Estimate of total number of black victims ≈ 996
23. a. []
2
2
2222223000(80)600(150)250(220)100(700)50(3000)(20)(4000)3000(80)600(150)250(220)100(700)50(3000)4
n ++++=
??+++++ ???
366,025,000,000
170.73651,600,000,000543,800,000
=
=+
Rounding up, we need a sample size of 171 for the desired precision.
13000(80)17168605,000n ??
== ???
2600(150)17125605,000n ??
== ???
3250(220)17116605,000n ??
== ???
4100(700)17120605,000n ??
== ???
550(3000)17142605,000n ??
== ???
24. a. 14(61)7(74)96(78)23(69)71(73)29(84)18,066
75.27514796237129240
c x +++++=
==+++++
Estimate of mean age is approximately 75 years old.
Sample Survey
b.
122308102284
.35
14796237129240
c
p
+++++
===
+++++
2
()
i c i
a p M
∑-= [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
.0760
c
p
s=
Approximate 95% Confidence Interval:
.35 ± 2(.0760)
or
.198 to .502
X= 4800 (.35) = 1680
Estimate of total number of Disabled Persons is 1680.