北师大统计学基础习题答案5

Exercise 5

1. P201: 4.59.Let Y1?Y2?Y3?Y4?Y5be the order statistics of a random sample of size n from a distribution with p.d.f. f(x)?e?x,0?x??,zero elsewhere..Show that Z1?Y2and Z2?Y4?Y2are independent. Solution: Sincef(x)?e?x,0?x??, so F(x)?1?e?x,0?x??

g2,4(y2,y4)?5![F(y2)][F(y4)?F(y2)][F(y4)]f(y2)f(y4) 1!1!1!?120(1?e?y2)(e?y2?e?y4)(e?y4)e?y2e?y4 SinceY2?Z1，Y4?Z1?Z2，J?10?1. 11g(z1,z2)?120(1?e?z1)(e?z1?eg1(z1)?g(y2)??0?z?z21)e?z?z21e?z1e?z?z21?120e?4z1(1?e?z1)(1?e?z2)e?2z25![F(y2)][1?F(y2)]3[F(y4)]'f(y2)?20e?4z1(1?e?z1) 1!3!?0g2(z2)??g(z1,z2)dz1??120e?4z1(1?e?z1)(1?e?z2)e?2z2dz1

?120(1?e?z2)e?2z2e?4z1?e?5z1?1(|0?|0)?120?(1?e?z2)e?2z2?6(1?e?z2)e?2z2 ?4?520Thus, g(z1,z2)?g1(z1)g2(z2).So Z1andZ2are independent.

P274: 6.15. LetXbe the mean of a random sample of size n from a

distribution that isN(?,9). Find n such thatPr(X?1???X?1)?0.90, approximately.

9n(X??)Solution: Since X~N(?,),~N(0,1).

n3Pr(X?1???X?1)?Pr(|X??|?1)?Pr(|n(X??)nn|?)?2?()?1?0.90333 Thus ?(nn)=0.95, ?1.645 and n?24.35. 33Because n must be an integer, so n=24 or 25.

P275: 6.18. LetX1,X2,?Xn,Xn?1 be a random sample of size 9 from a

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distribution that isN(?,?2).

(a) If?is known, find the length of a 95 percent confidence interval for if this interval is based on the random variable.

(b) If?is unknown, find the expected value of the length of a 95 percent confidence interval for if this interval is based on the random variable8(X??)/S. (c) Compare these two answers. Solution:

(a) Since X??Xi?19i9~N(?,?29),thus

9(X??)?~N(0,1).

Pr(|

9(X??)?|<1.96)=0.95?Pr(?1.96?3(X??)??1.96)?0.95

?Pr(X?1.96?1.96????X?)?0.95. 331.96?1.96?,X?)and 33The 95 percent confidence interval for?is (X?the length of it is

98??1.31? 75(b) Since ? is unknown, then T?X??S/n?1?8(X??)~t(8)

S Pr(?b?8(X??)?b)?0.95?2Pr(T?b)?1?0.95?PrT(?b)?0.975

S2.306?82.306?8 From TABLE IV of Appendix B, we know b=2.306, thus

Pr(X????X?)?0.95. The 95 percent confidence 2.306?84.6128interval for ? is (X?

2.306?8,X?) with the length L?S

E(L)?

4.6124.612?4.612?E(S)?E((9S2/?2)2)?8833872x2971??0?132xxedx 4?(4)212x1?438?(4)2

4.61?2??0?14.61?21922?y2xedx?2yedy?2?()44?02?(4)2?(4)23838?4.61?292

Then E(L)?1.49?

(c) From (a) we know the answer is1.31?, from (b) we know the answer is

1.49?. The two methods yield results that are in substantial agreement, which shows the length of the confidence interval for ? is almost the same with the parameter ? known or unknown.

P279: 6.30.Let two independent random samples, each of size 10, from two normal distributionsN(?1,?2)and N(?2,?2)yieldx?4.8,

2s12?8.64,y?5.6,s2?7.88. Find a 95 percent confidence interval for.

Solution: Let X1,X2,?X10andY1,Y2,?Y10denote, respectively, independent random samples from the two distributions N(?1,?2)and

N(?2,?), thenX?2?Xi?110i10~N(?1,?210)andY??Yi?110i10~N(?2,?210).

X?Y~N(?1??2,?252).(10S12?10S2)/?2~?2(18)

T?(X?Y)?(?1??2)10(S?S)11(?)1810102122?(X?Y)?(?1??2)(S?S)92122~t(18).

Pr(?b?T?b)?0.95?b?2.101and

bb22(S12?S2)??1??2?(X?Y)?(S12?S2))?0.95 33So the random interval

2.1012.10122S12?S2,(X?Y)?S12?S2). is((X?Y)?33Pr((X?Y)?2Let x?4.8,s12?8.64,y?5.6,s2?7.88, we get a 95 percent confidence

interval (-3.6,2.0).

2. Use Splus or R software to compute the mean, standard deviation, skewness and kurtosis of the following dataset.

-0.4292, 0.0064, 0.1181, -0.6282, 2.2010, -1.7623, 0.0921,

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1.8742, 1.4538, 0.3575, -1.5848, 0.4993, 0.7762, -0.2638, -1.1003, -2.2480, 0.5419, -0.4018, -0.3562, -0.5872. Solution: > x<-c(-0.4292,0.0064,0.1181,-0.6282,2.2010,-1.7623,0.0921,1.8742,1.4538,0.3575,-1.5848,0.4993,0.7762,-0.2638,-1.1003,-2.2480,0.5419,-0.4018,-0.3562,-0.5872) > a=mean(x) > a

[1] -0.072065 > b=sd(x) > b

[1] 1.143067

> c=mean(((x-a)/b)^3) > c

[1] 0.1234913

> d=mean(((x-a)/b)^4)-3 > d

[1] -0.5280221

So the mean of the dataset is -0.072065, the standard deviation is 1.143067, the skewness is 0.1234913 and the kurtosis is -0.5280221.

3*. P203: 4.73. Let Y1?Y2???Ynbe the order statistics of a random sample of size n from the exponential distribution with p.d.f.

f(x)?e?x,0?x??,zero elsewhere..

(a) Show thatZ1?nY1,Z2?(n?1)(Y2?Y1),Z3?(n?2)(Y3?Y2),…

Zn?Yn?Yn?1, are independent and that eachZihas the exponential distribution.

Solution: Since g(y1,y2,?yn)?n!e?y1e?y2?e?yn, and

Y1?ZZ1ZZZZ,Y2?1?2,?,Yn?1?2??n?1?Zn and the Jacobian is nnn?1nn?124

10???n110??nn?11J???????

n!111?0nn?12111?1nn?12g(z1,z2,?zn)?|J|g(y1,y2,?yn)?egk(zk)????0??0?z1ne?(z1z2?)nn?1?e?(z1z2????zn)nn?1?e?z1e?z2?e?zn??0??g(z1,z2,?zn)dz1?dzk?1dzk?1?dzn?e?zk

0?So g(z1,z2,?zn)?g1(z1)g2(z2)?gn(zn) and gk(zk)?e?zk,0?zk?? Which shows that z1,z2,?znare independent and that each zi has the exponential distribution.

(b) Demonstrate that all linear functions ofY1,Y2,?,Yn, such as?aiYi, can

1nbe expressed as linear functions of independent random variables. Solution:

a1Y1?a2Y2???anYn?b1nY1?b2(n?1)(Y2?Y1)???bk(n?k?1)(Yk?Yk?1)???bn(Yn?Yn?1)?b1Z1?b2Z2???bnZn. Then b1n?b2(n?1)?a1,b2(n?1)n?b3(n?2)?a2, ?,bk(n?k?1)?bk?1(n?k)?ak,?,bn?1(2)?bn(1)?an?1,bn(1)?an. Thenb1?na1?a2??ana?a3??ana???an,b2?2,?,bk?k,?bn?an

nn?1n?k?1nSo

?aiYi??biZi,bi?i?1i?1ai??an.Since Z1,Z2,?,Znare independent, then

n?i?1b1Z1,b2Z2,?,bnZnare independent. So

?aYcan be expressed as linear

iii?1nfunctions of independent random variables.

P275: 6.19.Let X1,X2,?Xn,Xn?1be a random sample of size n+1, n>1, from a distribution that isN(?,?). Let X??Xi/n and

21n5

S??(Xi?X)2/n. Find the constant c so that the statistic

21ncX?Xn?1?has a t-distribution. If n=8, determine k such

thatPr(X?kS???X?kS)?0.80. The observed interval is(x?ks,x?ks) often called an 80 percent prediction interval forX9. Solution: X~N(?,?2/n)，Xn?1~N(?,?2)，X?Xn?1~N(0,X?Xn?1X?Xn?1n?1?n

n?12?), n~N(0,1).nS2?2~?2(n?1),n?1?nnS2?2(n?1)?X?Xn?1n?1n??~t(n?1).nn?1STherefore c?n?1. n?1Pr(X?kS?x9?X?kS)?Pr(X?X97X?X97?k)?Pr(?k)?0.80 S9S97k?1.415,k?1.60. So k=1.60. 9

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