# 409Pract4ans - 1. Let X1, X2 , , X n be a random sample...

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Unformatted text preview: 1. Let X1, X2 , , X n be a random sample from the distribution with probability density function f(x)=39x2€—ex3 x>0 9>0. a) Find a sufﬁcient statistic Y = u(X1, X2, , X”) for 9. f(x1,x2,...xn;9) =f(x1;9)f(x2;6) ...f(xn;6) _9 .n .3 n = [3’76" 6 2’:le l: 72 By Factorization Theorem, Y = Z Xi3 is a sufﬁcient statistic for 9. i=1 OR f(x;7t) = expl—G-x3+ln9+ln3+21nx}. :> K(x)=x3. n :> Y = Z Xi3 is a sufﬁcient statistic for 9». i=1 A b) Obtain the maximum likelihood estimator of 9, 9 . Fl 71 lnL(9)= n-ln9+Z ln(3xi2 )~9-Z x? ‘ i=1 l=l b) Let X I , X2 , , X n be a random sample from the distribution with probability density function f(x)=36x2e'ex3 x>0 9>0. 11 Let W = Z X; . Find the probability distribution of W. i =1 3 FX(x) = 1 w e‘e’C , x>0. Let v=X3. Fv(v)=p(vsv)=P(xsvl/3)=1—69", v>0. V has anEXponential distribution with mean “usual 6” = n :> W=ZVi = 1:1 I X has a Gamma distribution with 0!, = n 71 =1 and B = “usual 6” = l 9 (“usual X” = 9 Use part (a) to suggest a conﬁdence interval for 9 with ( l — Ct) 100 % conﬁdence level. n 2 W/B = 2 9 2X has a chi-square distribution with r = 2 OL = 2 n d.f. i=1 7’1 :> P(X12_a/2(2n) < 26 < Xi/2(2n)) = l—oc. l: 2 2n 2 2n 3 p<£lj53i2<9<hé§£¥)=1_m 22X? 22X? i=1 i=1 xf_a/2(2n) xix/2W) A ( 1 — or) 100 % conﬁdence interval for G: n n 3 3 c) Suppose n = 5, and x1=0.2, x2=1.2, x3=0.2, x4=0.9, x5=0.3. Use part (b) to construct a 95% conﬁdence interval for 9. n X39300) = 3.247, X3_025(1o) = 20.48. 2 x; =2.5. lﬂ (0.6494, 4.096) 2-2.5 2-2.5 3. Let X 1 , X2 , , X n be a random sample from the distribution with probability density function 3 f(x)=39xze_ex x>0 9>0. Wewishtotest H0:9=3 vs. H129<3. a) If n=5, ﬁndauniforrnly most powerful rejection region of size 0c=0.10. Let 9<3. I’Z —-3x:7’ H[9xl?‘e 1] L H - '=1 )L(xl9x2r":xl’l) = ( Oaxl’xza 3x”) = l L(H1;x1,x2,...,xn) n 2 49x13 H 39xl. 6 i=1 3 ” n 3 = [.6] exp (9—3). xi . lﬂ n 3 k(x1,x2,...,xn)sk <:> (6-3)in skl lﬂ n 3 <3 2x1 26 (since 9<3). i=1 n . . 3 RejectHO if 1221361. 26. 11 Z X has a Gamma distribution with OL = n = 5 and [3 = “usual 8” = 1:1 9 If T has a Gamma(0t, [3 = 1/70 distribution, where 0t is an integer, then 2 T/B = 2 k T has a x 2 ( 2 0L) distribution (a chi—square distribution with 2 0t degrees of freedom n n Then 6; = 2 9 has a X2 ( 2 OL = 10 degrees of freedom) distribution. i=1 i=1 n 0.10 = OL = P(RejectH0|H0istrue) = P(ZX;J 26 [9:3) i=1 5 = P(6ZX§ 26c]9=3) = P(X2(10)26c). i=1 :> 6c = X31000) = 15.99. :> 0:2.665. 5 RejectHO if 2x1? 22.665. i=1 b) Suppose n = 5, and x1=0.2, x2= 1.2, x3=0.2, x4=0.9, x5=0.3. Find the p—Value of the test. n xl-3 = 2.5. has a Gamma distribution With OL = n 1 i=1 5 l: and [3 = “usual 9” = l 9 If T has a Gamma( 0t, [3 = 1A») distribution, Where 0L is an integer, then P(T > t) = P(Y S Ct —— 1 ), Where Y has a Poisson(7tt) distribution. 5 p-Vaiue = P(ZX? Z 2.5[9=3) = P(Poisson(2.5><3)SS—1)= 0.132. i=1 b) Let X1, X2 , , X n be a random sample from the distribution with probability density function 3 f(x)=39x2e‘9x x>0 G>0. We wish to test H0 : 9 = 3 vs. H1 : 9 < 3. Consider the rejection region 5 “RejectHO if 2 x; Z 3”. i=1 Find the signiﬁcance level of this test. If T has a Gamma(oc, B = 1/9») distribution, where 0L is an integer, then P(T > 2‘) = P(Y S Ct — l ), where Y has a PoissonUtl‘) distribution. 5 on = P(RejectHOIHOistrue) = 13(fo 2 3|9=3) i=1 = P(Poisson(3><3)£5—l) = P(Poisson(9)§4) = 0.055. Find the power of this test at 9 = 2 and 0 = l. 5 Power(e=2) = P(RejectHOI0=2) = 13(fo 2 3|9=2) i=1 = P(Poisson(3><2)SS—l) = P(Poisson(6)S4) = 0.285. 5 Power(9=l) = P(RejectH0I9=1) = P(ZXE’Z 3IG=1) i=1 = P(Poisson(3><l)SS—1)= P(Poisson(3)£4) = 0.815. d) (0») Let X I , X2 , , X n be a random sample from the distribution with probability density function f(x)=39xZe—ex3 x>0 9>0. Let the prior p.d.f. of 9 be Gamma(Ot, Find the posteriorp.d.f. of 9, given X1 =361 , X2=x2, ,Xn=xn. Find the conditional mean of 9, given X1 =x1,X2=x2, ,Xn=xn. Show that the conditional mean of 9, given X1 =x1,X2=x2, ,Xn=xn is A a weighted average of the maximum likelihood estimator 9 and the prior mean. (What are the weights?) Use part (a) to construct a ( l — on) 100 % conﬁdence interval for 9, given X1 = x1, X2=x2, ,Xn=xn. Let Ot=3, (i=1. Suppose n=5, and x1=0.2, x2=1.2, x3=0.2, x4=0.9, x5=0.3. Use part (d) to construct a 95 % conﬁdence interval for 9. +-\ *9/ h v—Sx} (QUWLK a}: i d.\ 9 a (5x WBGNZQ ) \’\ n+a~~t ~9LéXE+ﬁd :: a“, a Q :5) Tr(g\&co\+c\3 {3‘ Gamma NW 0L 2 V\+d\ ) l; NQ/w P " h u :23"??? n n+4 (SM “WM3 §XF+35 (C3 Vx+é\ “ n §x3 % A 7: L W V‘ 3 w“ +13 M - Exk Jr (5 2x\ EXB-kl‘g ( 1»Cxaxmmsk(o’fl N 78(14“) . uéxmmw wmm P [Vi—9g (LAMS 2 2C? X? + ﬁg) 9 é 72m+wWQ t \*J\. ’L 7(\'_<}_(lvx +101} 7C:('LV\+LJ\3 . M ,.__7:._.——-—-————-— Cs q (k—ULMooZ, If\ I ) ﬂaw?) zﬂ‘éxf-“a Wécw (K3 zﬂwqu—«e MB. 75:15AM: ((9.0l08 75%0'vs(‘m:?’8‘gw [9.010% 18-8“ \:[o.qgm,q.nool mg m ) mam ...
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## This note was uploaded on 12/15/2011 for the course STAT 409 taught by Professor Stephanov during the Fall '11 term at University of Illinois at Urbana–Champaign.

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409Pract4ans - 1. Let X1, X2 , , X n be a random sample...

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