應用抽樣技術答案

.第二章2.1判斷題：（1）錯；（2）錯；（3）對；（4）錯；（5）錯；（6）錯；（7）錯；（8）錯；（9）對；（10）對；（11）錯；（12）錯；（13）錯。2.3選擇題：（1）b；（2）b；（3）d；（4）c；（5）c。2.7（1）抽樣分布：??33.674.3355.676.337??Pr??1/101/102/102/102/101/101/10（2）期望為5，方差為4/3（3）抽樣標準誤=√4/3=1.155（4）抽樣極限誤差=1.96*1.155=2.263（5）置信區(qū)間=(5.67-2.263,5.67+2.263)=（3.407,7.933）。若區(qū)間兩端只考慮抽樣分布的可能性取值，則可得該抽樣分布作為離散分布的置信區(qū)間為[3,7]第三章3.1 判斷題是否為等概率抽樣：（1）是；（2）否；（3）是；（4）否。3.21Yi5.5（1）YN21(YiY)26.25NS21(YiY)28.33N1（2）樣本：（2,5）（2,6）（2,9）（5,6）（5,9） (6,9)Ey13.545.55.577.55.56Es21(4.5824.50.584.5)8.3363.3..(1)yi1682yi21182661f130/17500.03276n30y1682/3056.067s21n1nny21182663050.067^2826.2712n1i1n1i1301vy1fs20.03276826.27127.07nse(y)vy5.203t se(y) 1.965.203 10.19895%置信度下置信區(qū)間為（56.067-10.198,56.067+10.198）=（45.869,66.265）.因此，對該校學生某月的人均購書支出額的估計為56.07（元），由于置信度95%對應的t1.96，所以，可以以95%的把握說該學生該月的人均購書支出額大約在45.87~66.27元之間。(2)易知N=1750,n=30，??=8，t=1.961n181f130/1750p0.267n10.03389n30301pqp(1p)0.267(10.267)0.1957(1f)pq11v(p)10.033890.9570.081440.0167n2n230P的95%的置信區(qū)間為：pt(1f)pq10.2671.960.081140.0167(0.0907,0.4433)n12n則N1的估計值為?Np467.25467，其95%的置信區(qū)間為：N11750(0.0907,0.4433)(159,776)(3)n0t2q1.962(10.267)1054.642p0.120.267..可得最少的樣本量：nn01054.64658.2948659。1n0111054.641N17503.4(1)N200000n70n161f110.01449n1n1701該地區(qū)擁有私人汽車的比例估計：pn160.086n70q1p10.0860.914pq0.0860.9140.0786因而比例估計的標準差：se(p)(1f)pq0.014490.0860.9140.0338n1(2)t=1.965%t2pq1.9620.0786120.79121n020.052由于N很大，最少的樣本量nn0n0121。n011N3.5已知P10.08Q10.92P20.05Q20.95V0.052CV0.05要得到相同的標準差0.05，則所需的樣本量由n0PQ得：V(P)n010.080.9229.4430n020.050.95190.0520.052要得到相同的變異系數(shù)0.05，則所需的樣本量由n0Q得：2CV(P)Pn010.924600n020.957600。0.0520.080.0520.05第四章..4.3根據(jù)題中所給的數(shù)據(jù)，可以得到以下相關結果hnhNhWhfhyhWhyh2sh1102560.30330.039111.23.397294.40002104200.49760.023825.512.6896302.50003101680.19910.059526.05.1754848.8889合計308441.000021.2621321.2621（元）購買彩票的平均支出：ystWhyhh132方差估計量：v(yst)Wh2(1fh)sh11.3104h1nh標準差：()()11.31043.3631seystvyst(2)t=1.9610%Y22V0.121.26211.1768t1.963Whsh2h1348.1395按比例分配時：3Whsh2348.1395n0h1295.84V1.1768所需樣本量為nn0295.84219.05220n0295.8411N844各層樣本量；n1W1n0.303322066.443167n2W2n0.4976220109.0083109n3W3n0.199122043.603344按尼曼分配時：W1s10.303394.40002.9470W2s20.4976302.50008.6550..W3s30.1991848.88895.79953Whsh2.94708.65505.799517.4016h1所需樣本量為n(Whsh)217.40162190.54191V(Whsh2)/N1.1768348.1395/844各層樣本量；n1nW1s11912.947032.26803217.4016Whshn2nW2s21918.655094.76729517.4016Whshn3nW3s31915.799563.50106417.4016Whsh4.4hWhnh在家人數(shù)nhiphqhWh*phwh2phqhnh-110.1830270.90.10.1620.00010120.2130280.9333330.0666670.1969.46E-0530.1430270.90.10.1266.08E-0540.0930260.8666670.1333330.0783.23E-0550.1630280.9333330.0666670.1493335.49E-0560.2230290.9666670.0333330.2126675.38E-05合計0.9240.000397??????=∑?????=0.924()∑????≈v2(1-??)??0.000397??????=?????-1√??(??????)≈0.019924?2=0.000026(2)當按比例分配時，v=()????=∑??????≈2662??=??=2658???001+????=??????1≈478??2≈558??3≈373??4≈239??5≈425??6≈585按尼曼分配時..2(∑???√????)n= ≈2561??+∑????????????√??????=???∑??√????? ? ???1≈536 ??2≈519 ??3≈416 ??4≈303 ??5≈396 ??6≈3914.5依題意，可算得樣本量n=200，并有如下表數(shù)據(jù)12345678910TotalWhyh7.2983.6413.9744.25611.4469.16414.115.0963.1683.6475.792Whsh2904.05355.944739.65526.901325.85918.842132.48522.24170.37295.7511892.07故可得平均支出的分層估計：yWhyh75.79其方差估計：v(y)1Whsh211892.0759.46n200標準差：se(y)v(y)59.46047.7195%的置信區(qū)間為：(75.791.967.71)(60.68,90.91)。4.6W1 0.2,W2 0.3,W3 0.5 P1 0.1,P2 0.2, P3 0.4PWhPh0.28樣本量為100的簡單隨機抽樣估計方差：V1fS21NPQ1PQ0.002016nnN1n按比例分配分層抽樣的樣本量為：n0WhPhQh0.0180.0480.120.186V0.00201692.260.002016nn0934.6另解 已知W1=0.2，W2=0.3，W3=0.5，..P1=0.1，P2=0.2，P3=0.4P=ΣhWhPh=0.28，Q=1—P=0.72n=100的簡單隨機抽樣估計方差：V(Psrs) ≈[(1 —f ’)/100]PQ ≈=0.002016按比例分配的分層抽樣的估計方差：2V(Pprop) ≈ΣhWh[(1—fh)/nh]PhQh-1≈nΣhWhPhQh-1故 n≈92.26 ≈934.8(1) 由題設，n 100 n1 3所以，對于差錯率的簡單估計： p n1 3 0.03n 100q1p0.97pq0.030.970.0291因而上述估計的標準差：(1f)pqpq0.02910.0171se(p)n1n11001(2)由題設取層權，WA0.7,WB0.3每層的差錯率估計分別為：pAn1A1n1B20.03509nA430.02326,pB57nB差錯率的事后分層估計：ppstWApAWBpB0.0268sA2nApAqA0.02326sB2nB1pBqB0.03446nA1nB方差：v(ppst)1f(WAsA2WBsB2)1[(1WA)sA2(1WB)sB2]0.000269nn2標準誤：()(ppst)0.0002690.0164seppstv..第五章5.1記總產(chǎn)量為Y，總播種面積為XX=77000N=200n=101f110/2000.095n10yyi142.2xxi2706y14.22x270.622307.222818328xiyi43330.6yixisy2n1(yi2ny2)31.6818sx21(xi2nx2)9564.931n1sxy1(yixinyx)539.0311n1?y142.20.05255故有Rx2706總產(chǎn)量的比率估計：??0.05255770004046.34YRRX其方差：?2(1f)(yi?)22(1f)?2?NRxiN225484.173v(YR)nn1n(syRsx2Rsyx)總產(chǎn)量的簡單估計：?20014.222844YsrsNy?N2(1f)(yiy)2N2(1f)2120390.8v(Ysrs)nn1nsy..比率估計的方差較小，因此效果較好。5.3當CX時，用第一種方法，即用y進行估計；當CX時，用第二種方法，2CYX2CY即用y進行估計；當CX時，兩種方法都可使用。x2CY1f21f22y1f21f22VynSYnCYYVXnX2SYnRCYy?1f222VV(R)nR(CYCX2CYCX)x當CX時，Vyy1fR2CX(2CYCX)02CYXxn當CX時，Vyy1fR2CX(2CYCX)02CYXxn當CX時，Vyy1fR2CX(2CYCX)02CYXxn5.4?2222RSYSXCX2CXV(YR)V(yR)SYRSX?212V(y)SYCYCYV(Ysrs)由題設，cx/cy1.054/1.0630.9915?cX2cX12r10.9915220.6960.99150.6029v(YR)?cYcYv(Ysrs)所以與簡單估計量比，比率估計量的精度提高了1-0.6029=39.71%.5.6記總產(chǎn)量為Y，總播種面積為XX=460N=1401f110/140n=10n0.0928610yyi12600xxi29.7y1260x2.97..s2y 194911.111 sx2 1.3468 sxy 499.111易得總產(chǎn)量的簡單估計：?Ny140112176400（斤）Ysrsvy1fsy20.09286194911.11118098.89n?Nvy14018098.89140134.532118834.5se(Ysrs)比率估計?y12600424.2424Rx29.7??yX12600460195151.5（斤）YRRXx29.7?N2(1f)222?(sy?25149053.77v(YR)nRsx2Rsyx)??5014.88se(