应用抽样技术2答案

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,应用抽样技术答案,第二章 抽样技术基本概念,2.7,（,1,）抽样分布：,3 3.67 4.33 5 5.67 6.33 7,1/10 1/10 2/10 2/10 2/10 1/10 1/10,（,2,）期望为,5,，方差为,4/3,（,3,）抽样标准误,1.155,（,4,）抽样极限误差,2.263,（,5,）置信区间（,3.407,，,7.933,）,第三章 简单随机抽样,3.3,为调查某中学学生的每月购书支出水平，在全校名学生中，用不放回简单随机抽样的方法抽得一个的样本。对每个抽中的学生调查其上个月的购书支出金额,y,i,（如表,1,所示）。,(1),在,95%,的置信度下估计该校学生该月平均购书支出额；,(2),试估计该校学生该月购书支出超出,70,元的人数；,(3),如果要求相对误差限不超过,10%,，以,95%,的置信度估计该校学生该月购书支出超出,70,元的人数比例，样本量至少应为多少。,样本,序号,支出额（元）,样本,序号,支出额（元）,样本,序号,支出额（元）,1,2,3,4,5,6,7,8,9,10,85,62,42,15,50,39,83,65,32,46,11,12,13,14,15,16,17,18,19,20,20,75,34,41,58,63,95,120,19,57,21,22,23,24,25,26,27,28,29,30,49,45,95,36,25,45,128,45,29,84,表,1 30,名学生某月购书支出金额的样本数据,3.3,解：,(1),依据题意和表,1,的数据，有：,因此，对该校学生某月的人均购书支出额的估计为,56.07,（元），由于置信度,95%,对应的,t=1.96,所以，可以以,95%,的把握说该学生该月的人均购书支出额大约在,56.071.965.115,，即元之间。,，,(2),易知，,N=1750,，,n=30,，,的,95%,的置信区间为,:,的,95%,的置信区间为,:,(159,，,776),(3)N=1750,，,n=30,，,n,1,=8, t=1.96, p=0.267,q=1-0.267=0.733,由此可计算得：,计算结果说明，至少应抽取一个样本量为,659,的简单随机样本，才能满足,95%,置信度条件下相对误差不超过,10%,的精度要求。,n = n0/1+(n01)/N,= 1054.64/1+1053.64/1750=658.2942 = 659,3.5,要调查甲乙两种疾病的发病率，从历史资料得知，甲种疾病的发病率为,8,，乙种疾病的发病率为,5,，求：,(1),要得到相同的标准差,0.05,，采用简单随机抽样各需要多大的样本量？,(2),要得到相同的变异系数,0.05,，又各需要多大的样本量？,3.5,解：已知,P1= 0.08, Q1= 1-P1 = 0.92;,P2= 0.05, Q2 = 1 P2 = 0.95;,V(p) = 0.05*0.05,，,(1),由,得：,由,得：,(2),第四章 分层抽样,4.3,解：,（,1,） ，,（,2,）按比例分配,n=186,，,n,1,=57,，,n,2,=92,，,n,3,=37,（,3,）,Neyman,分配,n=175,，,n,1,=33,，,n,2,=99,，,n,3,=43,4.5,，置信区间（,60.63,，,90.95,）元。,4.6,解 已知,W,1,=0.2,，,W,2,=0.3,，,W,3,=0.5,，,P,1,=0.1,，,P,2,=0.2,，,P,3,=0.4,P=,h,W,h,P,h,=0.28,，,Q=1P=0.72,n=100,的简单随机抽样估计方差：,V(P,srs,) (1f )/100PQ 0.28*0.72/100,= 0.002016,按比例分配的分层抽样的估计方差：,V(P,prop,) ,h,W,h,2,(1f,h,)/n,h, P,h,Q,h, n,-1,h,W,h,P,h,Q,h,= n,-1,0.2*0.1*0.9+0.3*0.2*0.8+0.5*0.4*0.6,= 0.186 n,-1,故,n 92.26 93,4.8,解 已知,W,1,=0.7,，,W,2,=0.3,，,p,1,=1/43,，,p,2,=2/57,（,1,）简单随机抽样,P,srs,=(1+2)/100=0.03,V(P)=PQ/(n-1)=0.03*0.97/99=0.0002937,（,2,）事后分层,P,pst,=,h,W,h,p,h,=0.7*1/43+0.3*2/57=0.0268,V(P,pst,) =,h,W,h,2,(1f,h,)/(n,h,1)p,h,q,h,=0.7,2,*1/42(1/43)(42/43)+0.3,2,*1/56(2/57)(55/57),=0.00031942,第五章 比率估计与回归估计,5.2 N,2000, n,36, 1,0.95, t,1.96,f,= n/N,0.018,，,0.000015359,，,0.00392,置信区间为,40.93%,42.47%,。,第五章 比率估计与回归估计,5.3,当 时用第一种方法，,当 时用第二种,方法，当 时两种方法都可使用。这是因为：,， ，,若 则,0,0,5.4,解,：,V(Y,R,)(1f)/nY,2,C,Y,2,+C,X,2,2rC,Y,C,X,V(Y,srs,)=(1f)/nS,Y,2,=(1f)/n C,Y,2,Y,2,故,V(Y,R,)/V(Y,srs,) = 12rC,X,/C,Y,C,X,2,/C,Y,2,2,/1.063,2,= 1-0.397076,= 0.602924,5.5,证明：由（,5.6,）得：,5.6,解,(1),简单估计,:,总产量,:,Y,srs,=(N/n),i=1,n,Y,i,=(140/10)1400+1120+480,=176400(,斤,),v(Y,srs,)=N,2,(1f)/nS,Y,2,=140,2,(110/140)/10*194911.1,= 354738222,se(Y,srs,)= 18834.496,5.6,解,(2),比率估计,:,R =,i=1,n,Y,i,/ ,i=1,n,X,i,= 12600/29.7,= 424.2424,Y,R,= XR = 460*424.2424,= 195151.5(,斤,),v(Y,R,)=N,2,(1f)/n *,i=1,n,(y,i,RX,i,),2,/(n-1),=140,2,(110/140)/90*124363.5,= 25149054,se(Y,srs,)= 5014.883,面积,/,亩,产量,/,斤,3,1400,2.5,1120,4.2,1710,3.6,1500,1.8,720,5.2,1980,3.2,1310,2.4,1080,2.6,1300,1.2,480,29.7,12600,5.6,解,(3),回归估计,:,回归系数,b = S,xy,/S,xx,2,= 370.5965,y,lr,=xb(xX)=1260370.5965*(2.97460/140)=1377.089,Y,lr,=Ny,lr,=192792.47(,斤,),v(Y,lr,)=N,2,(1f)/n *,i=1,n,y,i,yb(x,i,x),2,/(n-2),=140,2,(110/140)/80*89480.59,= 20356834,se(Y,lr,)= 4511.855,5.7,解：,故估计量 虽然与 一样都是 的无偏估计，,但方差不小于 的方差，,当,时,，,故 不优于 。,0.223,9,0.251,4,0.154,8,0.057,3,0.048,7,0.102,2,0.067,6,0.098,1,第六章 不等概率抽样,6.1,假设对某个总体，事先给定每个单位的与规模成比例的比值,Z,i,，如下表，试用代码法抽出一个,n=3,的,PPS,样本。,表,1,总体单位规模比值,6.1,解：令,则可以得到下表，从,1,1000,中产生,n=3,个随机数，设为,108,，,597,，,754,，则第二、第六和第七个单位入样。,i,M,i,累计,M,i,代码,1,2,3,4,5,6,7,8,98,102,57,251,67,48,154,223,98,200,257,508,575,623,777,1000,198,99200,201257,258508,509575,576623,627777,7781000,M,0,=1000,281,954,1 085,1 629,215,798,920,1 834,5,6,7,8,1 353,639,650,608,1 238,746,512,594,1,2,3,4,子公司序号,子公司,序号,6.3,欲估计某大型企业年度总利润，已知该企业有,8,个子公司，下表是各子公司上年利润,X,i,和当年利润,Y,i,的数据，以,M,i,作为单位,X,i,大小 的度量,对子公司进行,PPS,抽样，设,n=3,试与简单随机抽样作精度比较。,表,2,某企业各子公司上年与当年利润（单位：万元）,对子公司进行抽样，根据教材（,6.7,）式：,显然对 抽样，估计量的精度有显著的提高,。,如果对子公司进行简单随机抽样，同样样本量时 的简单估计方差为：,抽样的设计效应是：,6.4,解,(1) PPS,的样本抽样方法可采用代码法或拉希里法,.,(2),若在时间长度,2,、,8,、,1,、,7h,中打入电话数量分别为,8,、,29,、,5,、,28,，则客户打入电话的总数：,Y,HH,=(35/4)8/2+29/8+5/1+28/7=145.46875,(3),估计量的方差估计,v(Y,HH,)=n(n1),-1,i=1,n,(y,i,/z,i,Y,HH,),2,=35,2,/(4*3)(8/24.15625),2,+(29/84.15625),2,+(5/14.15625),2,+(28/74.15625),2,=106.4697,6.5,设总体,N=3,z,i,=1/2,1/3,1/6,，,Y,i,=10,8,5,采取的,n=2,的,PS,抽样，求,i,ij,(,i,j,=1,2,3),。,解：,(1),所有可能样本为：（,10,，,8,），（,10,，,5,），（,8,，,10,），（,8,，,5,），（,5,，,10,），（,5,，,8,），其概率分别为：,所以：,6.6,解,(1),简单随机抽样简单估计,Y=2+3+6+8+11+14=44,S,2,=(N1),-1,i=1,N,(Y,i,Y),2,=(2*322),2,+(3*322),2,+(6*322),2,+(8*322),2,+(11*322),2,+(14*322),2,/(5*9),= 322/15 = 21.4667,总值估计的方差估计,V(Y,srs,) = N,2,(1f)/nS,2,= 36(12/6)/2322/15,=1288/5 = 257.6,6.6,解,(2),简单随机抽样比率估计,X=1+2+4+7+9+13=36,，,Y=2+3+6+8+11+14=44,，,R=44/36=11/9,，,f=2/6=1/3,总值估计的方差估计,V(Y,R,) N,2,(1f)/n,i=1,N,(Y,i,RX,i,),2,/(N1),= 36(12/6)/10(21*11/9),2,+(32*11/9),2,+(64*11/9),2,+(87*11/9),2,+(119*11/9),2,+(1413*11/9),2,= (12/5)*(488/81),= 14.46,6.6,解,(3) PPS,抽样汉森,赫维茨估计,X=1+2+4+7+9+13=36,，,Y=2+3+6+8+11+14=44,，,取,Z,i,=X,i,/X, (i=1,2,6),总值估计的方差估计,V(Y,HH,) = (1/n),i=1,N,Z,i,(Y,i,/Z,i,Y),2,= (1/nX),i=1,N,X,i,(XY,i,/X,i,Y),2,= (1/72)1*(36*2/144),2,+2*(36*3/244),2,+4*(36*6/444),2,+7*(36*8/744),2,+9*(36*11/944),2,+13*(36*14/1344),2,= 24.96,第七章 整群抽样,7.1,（略）,7.3,解： 不是 的无偏估计，此因类似于,有,因为对群进行简单随机抽样，故,， ，从而 ，若取,则,7.2,样本,耐用时数,1,1036,1075,1125,995,1088,1065,1023,988,1002,994,2,1047,1126,1183,1058,1142,1098,945,968,1036,987,3,1046,1153,1087,984,1224,998,1032,976,1103,958,4,1153,1078,1039,1006,1214,1076,986,994,1048,1126,5,1216,1094,1096,1035,1004,1053,1004,1122,1080,1152,6,964,1136,1185,1021,1007,948,1024,975,1083,994,7,1113,1093,1005,1088,997,1034,985,997,1005,1120,8,1047,1097,1136,989,1073,1102,976,984,1004,1082,样,本,耐用时数,均值,标准差,1,1036,1075,1125,995,1088,1065,1023,988,1002,994,1039.1,47.09907,2,1047,1126,1183,1058,1142,1098,945,968,1036,987,1059,78.46443,3,1046,1153,1087,984,1224,998,1032,976,1103,958,1056.1,85.65493,4,1153,1078,1039,1006,1214,1076,986,994,1048,1126,1072,73.9324,5,1216,1094,1096,1035,1004,1053,1004,1122,1080,1152,1085.6,66.5736,6,964,1136,1185,1021,1007,948,1024,975,1083,994,1033.7,77.44539,7,1113,1093,1005,1088,997,1034,985,997,1005,1120,1043.7,53.65952,8,1047,1097,1136,989,1073,1102,976,984,1004,1082,1049,57.28098,y = (1/80),ij,y,ij,= 1054.78,s,b,2,= (10/7),i,(y,i,y),2,= 3017.65,V(y) = (1f)/(aM)s,b,2,= (18/2000)/(8*10)*3017.65,= 37.5697,Se(y) = 6.1294,(1),以每盒灯泡为群实施整群抽样,y = (1/80),ij,y,ij,= 1054.78,s,2,= (1/79),ij,(y,ij,y),2,= 4628.667,V(y) = (1f)/(aM)s,2,= (180/20000)/(8*10)*4628.667,= 57.6269,Se(y) = 7.5912,(2),以从,20000,个灯泡中按简单随机抽样,y = (1/80),ij,y,ij,= 1054.78,S,w,2,= (1/a) ,i,s,i,2,= 1/(a(M1),ij,(y,ij,y,i,),2,= 4721.0056,r = (s,b,2,s,w,2,)/s,b,2,+(M1)s,w,2, = -0.04723,Deff = V(y)/V(y) = 1+(M1)r,= 0.6694,7.4,对,7.2,题群内相关系数进行估计,7.5,解：由于农户是调查单位，故以村为抽样单位的抽样是整群抽样，村即是群。对于村既有生猪存栏数，也有户数，因此在村大小不等的整群抽样下，既可使用简单估计量估计生猪存栏数，也可以户数为辅助指标构造比率估计和回归估计来估计生猪存栏数,。,（,1,）简单估计量,（,2,）以户数为辅助变量的比率估计量,314.452,， ,98880,，,365.718,， ,133750,0.934,（,3,）以户数为辅助变量的回归估计量,108000,0.803,（,100000,200475,）,112015,显然以户数为辅助变量构造回归估计量效果最好。此因各村生猪存栏数与村的规模（户数）有高度相关性，,r,0.934,，故采用回归估计量精度最高。,企业,已婚女职工人数,/,人,M,i,平均理想婚龄,/,岁,y,i,1,495,24.1,2,1020,22.8,3,844,25.5,4,1518,24.6,5,635,25.8,6,394,23.7,7,2346,24.5,7.6,7.6,(1),按简单随机抽样抽取,简单估计量估计,y = (1/7) ,i,M,i,y,i,= 25321.1571,M = 35680/35 = 1019.4286,Y = y / M = 24.8386,v(y) = (1f)/(a(a1)M,2,) ,i,(y,i,y),2,= (17/35)/(42*1019.4286,2,)*1711911436,= 31.3768,Se(y) = 5.6015,7.6,(2),按简单随机抽样抽取,采用比率估计量估计,Y,R,= ,i,y,i,/ ,i,M,i,= 177248.1/7252 = 24.4413,v(y) = (1f)/(a(a1)m,2,) ,i,(y,i,Ym,i,),2,= (17/35)/(42*1019.4286,2,)*4536349.45,= 0.0831445,Se(y) = 0.2883,7.6,(3),按,PPS,抽样抽取,抽样概率与企业女职工人数成比例,Y,HH,= ,i,y,i,/ a = 24.4286,v(Y) = 1/(a(a1) ,i,(y,i,y),2,= (1/42)*6.3542857,= 0.15129,se(Y) = 0.38896,7.7,证明,分别以 记整群抽样、简单随机抽样的估计量：,7.8,市县编号,社会从业人员数,/,万人,m,i,第三产业从业人员数,/,万人,t,i,18,37.60,7.00,32,41.30,7.39,43,34.40,6.30,65,28.90,4.97,87,57.60,11.23,p,R,= ,i,t,i,/,i,m,i,= 36.89/199.8 = 0.1846,7.8,v(p,R,) = (1f)/(a(a1)m,2,) ,i,(t,i,p,R,m,i,),2,= (15/110)/(20*39.96,2,)*0.549388,= 0.00001642,se(p,R,) = 0.004052,