Chapter 21Sample SurveyLearning Objectives1. 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 theabove 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 21Solutions:1. a. x = 215 is an estimate of the population mean.b. 2.7386xs=c. 215 ± 2(2.7386) or 209.5228 to 220.47722. a. Estimate of population total = N x = 400(75) = 30,000b. Estimate of Standard Error =xNs400320xNs=c. 30,000 ± 2(320) or 29,360 to 30,6403. a. p= .30 is an estimate of the population proportionb. .0437ps=c. .30 ± 2(.0437) or .2126 to .38744. B = 15222(70)490072.983067.1389(15)(70)4450n===+A sample size of 73 will provide an approximate 95% confidence interval of width 30.5. a. x= 149,670 and s = 73,42010,040.83xs==approximate 95% confidence interval149,670 ± 2(10,040.83)or$129,588.34 to $169,751.66b. X= N x= 771(149,670) = 115,395,570xs= Nxs= 771(10,040.83) = 7,741,479.93Sample Surveyapproximate 95% confidence interval115,395,770 ± 2(7,741,479.93)or$99,912,810.14 to $130,878,729.86c.p= 18/50 = 0.36 and .0663p s =approximate 95% confidence interval0.36 ± 2(0.0663)or0.2274 to 0.4926This 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)979.69336.00512.9157(2.5)(31.3)4724n ===+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 = 103Stratum 3: 3x = 210b. Stratum 1 1x = 1381 6.3640x s =138 ± 2(6.3640)or125.272 to 150.728Stratum 22x = 103Chapter 2124.2817xs==103 ± 2(4.2817)or94.4366 to 111.5634 Stratum 33x= 21038.6603xs==210 ± 2(8.6603)or192.6794 to 227.3206c.200250100138103210550550550stx⎛⎫⎛⎫⎛⎫=++⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭= 50.1818 + 46.8182 + 38.1818= 135.1818stxs=3.4092=approximate 95% confidence interval135.1818 ± 2(3.4092)or128.3634 to 142.00028. a. Stratum 1:11N x = 200(138) = 27,600Stratum 2:22N x = 250(103) = 25,750Stratum 3:33N x = 100(210) = 21,000b.stN x = 27,600 + 25,750 + 21,000 = 74,350Note: the sum of the estimate for each stratum total equalsstN xc.stN x= 550(3.4092) = 1875.06 (see 7c)Sample Surveyapproximate 95% confidence interval74,350 ± 2(1875.06)or70,599.88 to 78,100.129. a. Stratum 11p = .501.1088p s ==50 ± 2(.1088)or.2824 to .7176Stratum 22p = .782.0722p s.78 ± 2(.0722)or.6356 to .9244Stratum 33p = .213.0720p s =.21 ± 2(.0720)or .066 to .354b. 200250100(.50)(.78)(.21).5745550550550st p =++=c.st p s =.0530==d.approximate 95% confidence intervalChapter 21.5745 ± 2(.0530)or.4685 to .680510. a.[]222222300(150)600(75)500(100)(140,000)92.8359196,000,00015,125,000(20)(1400)300(150)600(75)500(100)2n ++===+⎛⎫⎡⎤+++⎪⎣⎦⎝⎭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.2222(140,000)(140,000)305.653049,000,00015,125,000(10)(1400)15,125,0004n ===+⎛⎫+ ⎪⎝⎭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 thoughthe sample size is larger, the proportion allocated to each stratum has not changed.222(140,000)(140,000)274.606056,250,00015,125,000(15,000)15,125,0004n ===++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 Survey1300(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 = 29.53332x = 64.7753x = 45.2125 4x = 53.0300 b. Indianapolis29.5332± 29.533 ± 10.9086(.9177)or19.5222 to 39.5438Louisville64.7752± 64.775 ± 17.7248(.9068)or48.7022 to 80.8478St. Louis45.21252± 45.2125 ± (13.7238) (.9487)or32.1927 to 58.2323Memphis53.03002±53.0300 ± 18.7719(.9258)or35.6510 to 70.4090Chapter 21c.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)116.666719p p N N n n ⎛⎫⎛⎫⎪⎪-⎝⎭⎝⎭-==-.0857st p s ==approximate 95% confidence interval.4269 ± 2(.0857)or.2555 to .598312. a. St. Louis total = 11N x = 80 (45.2125) = 3617In dollars: $3,617,000b. Indianapolis total = 11N x = 38 (29.5333) = 1122.2654 In dollars: $1,122,265c.3845807029.533364.77545.212553.030048.7821233233233233st x ⎛⎫⎛⎫⎛⎫⎛⎫=+++= ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭2211111(13.3603)()38(32)36,175.5176s N N n n -==2222222(25.0666)()45(37)130,772.18s N N n n -==Sample Survey2233333(19.4084)()80(72)271,213.918s N N n n -==2244444(29.6810)()70(60)370,003.9410s N N n n -==st x s =3.8583==approximate 95% confidence interval2st st x x s ±48.7821 ± 2(3.8583)or41.0655 to 56.4987In dollars: $41,066 to $56,499d.approximate 95% confidence interval2st st x Nx Ns ±233(48.7821) ± 2(233)(3.8583)11,366.229 ± 1797.9678or9,568.2612 to 13,164.197In dollars: $9,568,261 to $13,164,19713.[]222222250(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 21b. [][]2222222250(100)38(100)35(100)123(100)333,404,025123(100)(30)(123)50(100)38(100)35(100)4n ++===+⎛⎫⎡⎤+++ ⎪⎣⎦⎝⎭150(100)331312,300n ⎛⎫== ⎪⎝⎭238(100)331012,300n ⎛⎫== ⎪⎝⎭335(100)33912,300n ⎛⎫== ⎪⎝⎭This is the same as proportional allocation . Note that for each stratumh h N n n N ⎛⎫= ⎪⎝⎭14. a. 7501550i c i x x M ∑===∑ c X M x = = 300(15) = 450015.3050i 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 = 44501.4708c x s == c x X s Ms == 300(1.4708) = 441.242()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) or12.0584 to 17.9416d. approximate 95% confidenceInterval for Population Total:4500 ± 2(441.24)or3617.52 to 5382.48e. approximate 95% confidenceInterval for Population Proportion:.30 ± 2(.0484)or.2032 to .396815. a.10,400 80130c x ==c X M x = = 600(80) = 48,00013.10130c 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= 886,1507.6861c x s =approximate 95% confidenceInterval for Population Mean:80 ± 2(7.6861)or64.6278 to 95.3722c. X s = 600(7.6861) = 4611.66approximate 95% confidenceInterval for Population Total:48,000 ± 2(4611.66)or38,776.68 to 57,223.322()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 = 6.68.0211c p sapproximate 95% confidenceInterval for Population Proportion:.10 ± 2(.0211)or.0578 to .142216. a.2000 4050c x ==Estimate of mean age of mechanical engineers: 40 yearsb. 35.7050c p ==Estimate of proportion attending local university: .70c. 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= 72922.0683c x s ==approximate 95% confidenceInterval for Mean age:40 ± 2(2.0683)or35.8634 to 44.1366d. 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:.70 ± 2(.0883)or.5234 to .876617. a.17(37)35(32)57(44)11,24036.9737173557304c x ++⋅⋅⋅+===++⋅⋅⋅+Estimate of mean age: 36.9737 yearsb. Proportion of College Graduates: 128 / 304 = .4211Proportion of Males: 112 / 304 = .3684c. 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= 474,650.682.2394c x s ==approximate 95% confidenceInterval for Mean Age of Agents:36.9737 ± 2(2.2394)or32.4949 to 41.4525d. 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% confidenceInterval for Proportion of Agents that are College Graduates:.4211 ± 2(.0386)or.3439 to .4983e. 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= 68.8787.0270c p s =approximate 95% confidenceInterval for Proportion of Agents that are Male:.3684 ± 2(.0270)or.3144 to .422418. a. p = 0.190.0206p s ==Approximate 95% Confidence Interval:0.19 ± 2(0.0206)or0.1488 to 0.2312b. p = 0.310.0243p s ==Approximate 95% Confidence Interval:0.31 ± 2(0.0243)or0.2615 to 0.3585c. p = 0.170.0197p s ==Approximate 95% Confidence Interval:0.17 ± 2(0.0197)or0.1306 to 0.2094d. The largest standard error is when p = .50.At p = .50, we get0.0262p s =Multiplying by 2, we get a bound of B = 2(.0262) = 0.0525For a sample of 363, then, they know that in the worst case (p= 0.50), the bound will beapproximately 5%.e. If the poll was conducted by calling people at home during the day the sample results would only berepresentative 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≈ 1p= .550.0222s==pb. p= .31s==0.0206pc. 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 .5944e. Approximate 95% Confidence interval:.31 ± 2(.0206).2688 to .351220. a. 204.9390s==xApproximate 95% Confidence Interval for Mean Annual Salary:23,200 ± 2(204.9390)or$22,790 to $23,610b. N x = 3000 (23,200) = 69,600,000s = 3000 (204.9390) = 614,817xApproximate 95% Confidence Interval for Population Total Salary:69,600,000 ± 2(614,817)or$68,370,366 to $70,829,634c. p= .73.0304psApproximate 95% Confidence Interval for Proportion that are Generally Satisfied:.73 ± 2(.0304)or.6692 to .7908d. If management administered the questionnaire and anonymity was not guaranteed we would expect adefinite upward bias in the percent reporting they were “generally satisfied” with their job. Aprocedure for guaranteeing anonymity should reduce the bias.21. a. p= 1/3.0840ps=Approximate 95% Confidence Interval:.3333 ± 2(.0840)or.1653 to .5013b. 2X = 760 (19 / 45) = 320.8889c. p= 19 / 45 = .4222.0722psApproximate 95% Confidence Interval:.4222 ± 2(.0722)or.2778 to .5666d.38010760192607.3717 140030140045140025stp⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫=++=⎪⎪ ⎪⎪ ⎪⎪⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭(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 .468122. a. X = 380 (9 / 30) + 760 (12 / 45) + 260 (11 / 25) = 431.0667Estimate approximately 431 deaths due to beating.b. 38097601226011.3079140030140045140025st p ⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫=++= ⎪⎪ ⎪⎪ ⎪⎪⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭[](1)()1h 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 .3983c. 380217603426015.7116140030140045140025st p ⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫=++= ⎪⎪ ⎪⎪ ⎪⎪⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭[](1)()1h h h h h h p p N N n n -∑--= (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 .8002d. X = 1400 (.7116) = 996.24Estimate of total number of black victims ≈ 99623. a. []222222223000(80)600(150)250(220)100(700)50(3000)(20)(4000)3000(80)600(150)250(220)100(700)50(3000)4n ++++=⎛⎫+++++ ⎪⎝⎭366,025,000,000170.73651,600,000,000543,800,000==+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)18,06675.27514796237129240c x +++++===+++++Estimate of mean age is approximately 75 years old.b.122308102284.3514796237129240cp+++++===+++++2()i c ia 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= 424.52.0760cps=Approximate 95% Confidence Interval:.35 ± 2(.0760)or.198 to .502X= 4800 (.35) = 1680Estimate of total number of Disabled Persons is 1680.。