《应用数理统计》习题解答

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i

clnL(ln

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较有效

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iii

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0.766,

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为(24.

a

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f(x

x|y)

e

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x))(x,1

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0600001(d)[]30002(d)E[]0.950.3(d)E[]80003dimax决策。）R[]0.751500[,d]0.758000应买

）)((x(P(x)0.750.950.250.30.7875()E[）)((x(P(x)0.750.950.250.30.7875()E[d|x]La)()L()P|x11111210.750.25571.2,0.750.95()E[d|x]La)(x)L(,)(|x)0.250.2125()E[)x](1L)()L,)(xx)571.2*0.78751412*0.2125(((d)1500750500应请np37.

）)(P()111110.750.950.250.30.7875'()E[(dx]11La1P2a(x10.750.25571.2,0.750.95'()E[(dx]LaP(xL2aP(x12122220.250.2125()E[(d)]1La)P)L,)(xx)571.2*0.78751412*0.2125112112(d)1500750500应请B(10,p)(1)EXx

L(p,

xxii

xi(1)10L(p)

ilnp)ln(1p)]iiiii

ii

ii

(xp)(1p)pLn(x)(1p)又，所以效估量(p)

np(1

,R界为

1p(1p)，I(nInp

()n(p)p所以p为p的有效估计量，2pp2

n）hy|x,n

,)f(1

,xy)f(x

,x|)Cii(1)10i

,

x

)nx为参数nx10n贝塔分布的.|(nx1,10所以(|

nx(nxnn《应用数理统计》参答案1．

B1，pi

B(10,i(|iiii(1|p0.2)ii0.20.830.20.8740.2[0.624190.03299]0.408

0.8

pii0.5230.50.540.50.989260.623050.3662

0.5

2.

(a(

1.5)(1.5)0.93370.066819(xa(

)1.5)(1.5)3.

H：500,H:1,

x(为拒绝域sn而

，所以机器工作正.4.

222H：:1222

1000U

|2.51.962

拒绝H，即认为元件不合.05.

:x0.5099,

x10.20.5099n201.7291，值显7.

H：0.04%,H:

2(0.037%)2

8.5561

(n=16.9198.

所以不否定即认为0H：0.005,:0.005

ns2

(0.005)

(=15.5079.

所以拒绝即为导线不合.0:212与

12

nnor

12

<F1

n10.

n2n(122(59,39)1.74F.0H：H:1212S12Sn(n271=1=Sn(812

=0.5611.

F(7,6)F(7,6)所以接受H两产品精度无显著差0

5mHa,H:a01225m=未n121T

S212

1

(为拒绝域又

2

0.93962.2622，0.975接受H,即量无显著差012.

1

1.2

2.1247.91.22

n=27.48813.

即劣0H：a,H:a0121=知12

x

n

9

0.975

(9)2.2622，14.

接受H,即认前后均无显著.0H：(),H:F(x)F(xF(x)为泊松分布0010ivi估计值inpi

2i

ppi0

,ppe10

,n

i

inp

7v2inpiii15.

<m12.5921-0.95接受H即认为分布为泊松分布.0(),HF(xFx),F(为泊松分布0i,估计为i0.805i

i

i

0.805

pe

0.805

n

i

v2iippii

i

(1-0.95能认为分布为泊松布016.

np010(),:(x)F(x),P{}010np010

110n

miii

110

5.12517.

<((9)1-,布.0H：F(H:F(x)(),P{X}k10

p.p未估值

x0n100

0.01p010

0.368,p0.368,n

i

v2inp

i

i

v2inp

i

18.

<(m1-接受H,能认布二项布0F()(yH:F(x,y)(x)y),r拒绝域为{n

i

ni..k

1-

(r1)}n

0.0479

(2)c

和患者14737210健康12322和受H,认为慢性气炎与吸烟量无关.0

ATiijATiij《应用数理统计》参答案1.

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