抽样技术比率估计和回归估计习题答案
第六章比率估计和回归估计作业解答
P117
5.1
解：N =200, X =77000, =385, n =10
2
=270. 6, s x =9564. 93333,
2
=14. 22, s y =31. 6818, s xy =539. 031
ˆ==0. 0525, R
ˆ=R ˆX =0. 0525⨯77000=4042. 5 Y R
2
N (1-f ) 2ˆ22
ˆ) =ˆs ) =5498. 19 v (Y (s y +R s x -2R R xy
n
ˆ) =v (Y ) =74. 15(万斤) se (Y R R
若用简单估计量：
2
N ˆ) =(1-f ) s 2=120390. 84 v (Y y
n
ˆ) =v (Y ) =346. 97(万斤) se (Y
5.2
解
ˆN =2000, n =36, R ==41. 7%，c x =0. 09, c y =0. 085, r =0. 79, z =1. 96
ˆ) =(1-f ) R ˆ2(C 2+C 2-2rC C ) =0. 00001536 v (R y x x y
n
ˆ) =v (R ) =0. 392% se (R
ˆ-z ⋅se (R ˆ), R ˆ+z ⋅se (R ˆ)], [R 即【40. 93%，42. 47%】
5.4
1-f 222
(C Y +C X -2ρC Y C X ) n
1-f 21-f 22
V (srs ) =S Y =C Y
n n V (R ) =
22
V (R ) C Y +C X -2ρC Y C X
==0.6029 2
V (srs ) C Y
5.5
证明：
1-f
n
V (R ) =
∑(Y -RX )
i
i
N
2
N -1
=
N -n 2
S d Nn
令
N -n 2
S d =V ，有 Nn
22
n (NV +S d ) =NS d 2S d
2NS d
n ==2
S NV +S d
1+d
NV
5.6
解：N =140, X =460, =3. 2857, n =10
2
=2. 97, s x =1. 3468,
2
=1260, s y =194911. 1111,
∑xy -
s xy =
1
∑x ∑y =499. 1111 n -1
简单估计量：
ˆ=N =140⨯1260=176400(斤) Y
2
N (1-f ) 2
ˆ) =v (Y s y =
n
1402⨯(1-
10
10)
⨯194911. 1111=354749137. 2
ˆ) =v (Y ) =18834. 79(斤) se (Y R R
比率估计：
ˆ==1260=424. 2424 R
2. 98
ˆ=R ˆX =424. 2424⨯460=195151. 50(斤) Y R
2
N (1-f ) 2ˆ22
ˆˆs ) =25157117. 83 v (Y R ) =(s y +R s x -2R xy
n
ˆ) =v (Y ) =5015. 69(斤) se (Y R R
回归估计：
b =
s xy
2s x
=
499. 1111
=370. 5904
1. 3468
lr =+b (-) =1376. 9954(斤) v (lr ) =
1-f n -12
(s y -bs xy ) =1038. 9636
n n -2
se (lr ) =v (lr ) =32. 2330(斤)
ˆ=N =140⨯1376. 9954=192779. 36(斤) Y lr lr ˆ) =Nse () =4512. 62(斤) se (Y lr lr
5.8
解：（1）由题可知，分层抽样为按比例抽样，有
ˆ==N ∑W =45900(万斤) Y st h h
h
v (st ) =
1-f 2
∑W h s yh =258. 0886 n h
se (st ) =v (st ) =16. 0651(万斤)
ˆ) =Nse () =4819. 54(万斤) se (Y st st
（2）分别比率估计
. 1111ˆ=1=200. 8333=1. 0406 R ˆ=2=121R =1. 061 312
11932114. 1111
ˆ=R ˆX =1. 0406⨯24500=25494. 7(万斤) Y R 111ˆ=R ˆX =1. 0613⨯21200=22499. 56(万斤) Y R 222ˆ=∑Y ˆ=1. 0613⨯21200=47994. 26(万斤) Y Rs Rh
2
N 2h (1-f h ) ˆˆ2s 2-2R ˆs ) v (Y Rs ) =∑(s yh +R xh h xyh
n h
=589048. 22+1576514. 55=2165562. 77
ˆ) =v (Y ) =1471. 59(万斤) se (Y Rs Rs
（3）联合比率估计
st =∑W h h =153(万斤) st =∑W h h =145. 6667(万斤)
ˆ=st =153=1. 0503 R c
st 145. 6667
ˆ=R ˆX =1. 0503⨯(24500+21200) =47998. 71(万斤) Y Rc c
2
N h ˆ) =∑(1-f h ) (s 2+R ˆ2s 2-2R ˆs ) v (Y Rc yh c xh c xyh
n h
=568712. 5337+1569007. 36=2137719. 89
ˆ) =v (Y ) =1462. 09(万斤) se (Y Rc Rc
（4）分别回归估计
b 1=
s xy 1
2
s x 1s xy 2197. 858256111==1. 1350 b 2=2==1. 1065 5132s x 2178. 81111
lr 1=1+b 1(1-1) =213. 5075(万斤) lr 2=2+b 2(2-2) =125. 1683(万斤) lrs =∑W h lrh =160. 5040(万斤)
2W (1-f h ) 2h v () =∑(n -1)(s lrs h yh -b h s xyh ) n h (n h -2)
=6. 7391+2. 3030=9. 0421
se (lrs ) =v (lrs ) =3. 01(万斤)
ˆ==300⨯160. 504=48151. 2(万斤) Y lrs lrs ˆ) =Nse () =300⨯3. 01=903(万斤) se (Y lrs lrs
（5）联合回归估计 2
W ∑h (1-f h ) s yxh h h
b c =≈1. 125322 ∑W h (1-f h ) s xh n h
h
lrc =st +b c (-st ) =153+1. 1253⨯(152. 3333-145. 6667) =160. 5019(万斤)
2
W v (y ) =∑h (1-f h ) (s 2-2b s +b 2s 2)
yh c yxh c xh lrc
h n h
=5. 3979+2. 0362=7. 4341
se (lrc ) =v (lrc ) =2. 7266(万斤)
ˆ=N =300⨯160. 5019=48150. 57(万斤) Y lrc lrc ˆ) =Nse () =300⨯2. 7266=817. 98(万斤) se (Y lrc lrc