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应用抽样技术课后习题答案.ppt_图文

2.71 3 3.67 4.33 5 5.67 6.33 7 1/10 1/10 2/10 2/10 2/10 1/10 1/10 254/3 31.155 42.263 53.4077.933 3.3 yi 1 (1)95% (2)70 (3)10%95% 70 1 30 1 85 11 20 21 49 2 62 12 75 22 45 3 42 13 34 23 95 4 15 14 41 24 36 5 50 15 58 25 25 6 39 16 63 26 45 7 83 17 95 27 128 8 65 18 120 28 45 9 32 19 19 29 29 10 46 20 57 30 84 3.3(1)1 yi 1682, y 1682 30 56.07(), s 2 y (118266 16822 / 30) / 30 798.73 1 f N n 1750 30 0.03276 b2 4ac n n N 30 1750 v( y) 0.03276 798.73 26.168 se( y) v( y) 5.115 56.07 95% t=1.96, 95% 56.071.965.115 50.96--61.19 (2)N=1750n=30 n1 8 t=1.96 p n1 8 0.267 n 30 1 f N n 1750 30 0.03389 n 1 (n 1)N 29 1750 pq p(1 p) 0.267 0.733 0.1957 (1 f ) pq 0.03389 0.1957 0.08144 n 1 1 0.0167 2n P 95%: p (u 1 2 (1 f ) pq 1 ) 0.267 (1.96 0.08144 0.0167) n 1 2n =(0.0907,0.4433) N195%: (159776) (3)N=1750n=30n1=8, t=1.96, p=0.267, q=1-0.267=0.733 n0 t2q r2 p 1.962 0.733 0.01 0.267 1054.64 n = n0/[1+(n0--1)/N] = 1054.64/[1+1053.64/1750]=658.2942 = 659 659 95%10% 3.5 85 (1)0.05 (2)0.05 3.5 P1= 0.08, Q1= 1-P1 = 0.92; 0.95; (1) n0 PQ V ( p) n01 0.08 0.92 0.052 30 P2= 0.05, Q2 = 1 P2 = V(p) = 0.05*0.05 n02 0.05 0.95 0.052 19 Q (2) n0 Cv2 ( p)P n01 0.92 0.052 0.08 4600 n02 0.95 0.052 0.05 7600 4.3 1 yst 20.0 7 s( yst ) 3.08 2 n=186n1=57n2=92n3=37 3Neyman n=175n1=33n2=99n3=43 4.5 yst 75.7 9 60.6390.95 4.6 W1=0.2W2=0.3W3=0.5 P1=0.1P2=0.2P3=0.4 P=hWhPh=0.28Q=1--P=0.72 n=100 V(Psrs) [(1--f ')/100]PQ 0.28*0.72/100 = 0.002016 V(Pprop) hWh2 [(1--fh)/nh] Ph Qh n-1hWh Ph Qh = n-1[0.2*0.1*0.9+0.3*0.2*0.8+0.5*0.4*0.6] = 0.186 n-1 n 92.26 93 4.8 W1=0.7W2=0.3p1=1/43p2=2/57 1 Psrs=(1+2)/100=0.03 V(P)=PQ/(n-1)=0.03*0.97/99=0.0002937 2 Ppst=hWhph=0.7*1/43+0.3*2/57=0.0268 V(Ppst) =hWh2[(1--fh)/(nh--1)]phqh =0.72*[1/42](1/43)(42/43)+0.32*[1/56](2/57)(55/57) =0.00031942 5.2 N2000, n36, 1 0.95, t1.96, f = n/N0.018 v(R^) 0.000015359 se(R^) 0.00392 [40.93%,42.47%] 5.3 V(y) 1 f n 2CSCY2XY21CCnXYf Y2CY2V(Xy)1nXf2 Y 2CY22CCXY1nf R2CY2 V ( y) x V (R^) CX 2CY 1 n f V R2 (y X (CY2 ) V ( CX2 2CY CX ) y) 1 f xn R 2CX (2CY CX )0 CX 2CY V( y )V(y) 1 f X xn R 2C X (2CY CX ) 0 CX 2CY 0 V ( y ) V ( y ) 1 f X xn R2CX (2CY CX ) 5.4 V(YR)[(1--f)/n]Y2[CY2+CX2--2rCYCX] V(Ysrs)=[(1--f)/n]SY2