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Unformatted text preview: STAT 409 Fall 2009
Homework #14
(due Wednesday, December 9, by 4:30 p.m.)
1. When correctly adjusted, a machine that makes widgets operates with a 5% defective b) rate. However, there is a 10% chance that a disgruntled employee kicks the machine,
in which case the defective rate jumps up to 30%. Suppose that a widget made by this machine is selected at random and is found to be
defective. What is the probability that the machine had been kicked? P(D) = 0.90 x 0.05 + 0.10 x 0.30 = 0.075. 0.10x 0.30 0.030 A random sample of 20 widgets was examined, 4 widgets out of these 20 are found to
be defective. What is the probability that the machine had been kicked? P(X=4K') = [240](0.05)4(0.95)16 = 0.0133,
P(X=4IK) = [240)(0.30)4(0.70)16 = 0.1304. 0.10x 0.1304 2. 7.21 of the observations of a random sample from a Poisson distribution 7.21 Let Y be the sum
be a gamma one with parameters at and 13. with mean 6. Let the prior p.d.f. of 6 (3) Find the posterior p.d.f. of 6, given Y = y. (b) If the loss function is [w()=) — 612, ﬁnd the Bayesian point estimate w(y). (c) Show that this w(y) is a weighted average of the maximum likelihood estimate y/Il
and the prior mean m8, with respective weights of 11/01 +1/f3) and (1/ﬁ)/(n +1/ﬂ) (0K3 X\ X1 H X“ we. 1.1.05 POC350H __.7 1, Z X; Cs ?o§“>sw ._ t i“ “We bxan 8f“)
TV(’A\ \HBWA w at” “L r e, ‘6‘
UA+J~ *\ v? +“\ :7 Cs Gama/tax Witt“
NW 4.: ﬁ'ho’» NW (5; t
4;; Ma (‘03 cl .LJHB : 3‘“ n +OL '56
h w 9*h “+35 (5 “+46 3. Consider a random sample X1, X2, , X” from a distribution with p.d.f. f(x;?\.)=7txk_1, O<x<l, zero elsewhere, 7V>0. A a) Find the maximum likelihood estimator of X, )L. b) Let 7x. have a prior p.d.f. which is gamma with parameters 0L and 9. Find the
conditional mean of 7» given X1=x1, X2 =x2, ,Xn :26". Show that it is A a weighted average of the maximum likelihood estimator k and the prior mean 0L9.
[Hint f(x;?») = ix“ = iiexh‘xq
h, __ n n FA'\
m Lm —: n a 14‘: a (mi J” ”9/ "" ":54
(‘33 ﬁtﬂx @(okwtda) : Min—q \e 9x ntﬁaﬂm‘
vme W"
P (AhPOL'l —’A(lé~~=_22~rr\)
“ v . . Q,
:9 \KS (1ka bu‘M,‘
NQM ,L :_ n+9.) NW 6: k 4. Let X 1 , X 2, , X n be a random sample from the distribution with probability density function
3
f(x)=39xze_ex x>0 9>O. a) Find a sufﬁcient statistic Y = u(X1, X2, ,Xn) for 9. f(x1,x2,...xn;6) =f(x1;9)f(x2;6) ...f(xn;9) _ez.n x.3 n
_ n n F11 2
l: n
By Factorization Theorem, Y = Z X is a sufﬁcient statistic for 6.
i=1 OR f(x;?») = exp{—Gx3+ln0+ln3+21nx}. : K(x)=x3. n
:> Y = Z X; is a sufﬁcient statistic for )L.
i=1 A b) Obtain the maximum likelihood estimator of 9, 9 . n —Gxg
L(6)=H 39x36
i=1 n n
lnL(6)= nln9+z 1n(3xl2)—GZ 363 . _ 1
1:1 1:1 I n n "
(lnL(6))=6—in3=0 :> 6: n
1:1 b) Let X1,X2,.. density function . , X n be a random sample from the distribution with probability f(x)=39xze"9x3 x>0 9>0. 12
Let W = Z X? . Find the probability distribution of W.
i=1 3 FX(x) = 1 — e—ex , x>0. Let V=X3.
Fv(v)=P(vsv)=P(stl/3)=1—e“9", v>0.
V has an Exponential distribution with mean n n
:> W = 2V2. = has aGamma distribution with 0t =11 = 12
i=1 i 1 ll .1.
9. and B: usual 9 = Use part (a) to suggest a conﬁdence interval for 9 With ( l — Ct) 100 % conﬁdence level. n
2 W/B = 2 9 ZX has a chisquare distribution with r = 2 OL = 2 n d.f.
i=1 2 2
X _ (n) X (n)
:> 1 05/2 a/2 ) : 1_a
n_ n
22X? 22x?
i=1 i=1
2 2
X _ (n) X (n)
A ( l ~0t) 100 % conﬁdence interval for 9: —IHL, 0::2 6. a) 7.24 7.24 Consider a random sample X1. X2 , . , . , X” from a distribution with, p.d.f. ft); £9} = ﬂags—8"), 0 : x <: 00.
Let {9 have a prior p.d.fi which is gamma with a z 4 and the usual :9 = _1_ /4. Find the
conditional mean off}, given X1 = x1, X2 x2, , X” :1» =x is b) Show that the conditional mean of 9 given X1 = x1 , X2 = x2, , X n n A a weighted average of the maximum likelihood estimator 9 and the prior mean. (What are the weights?)
(M HR LH 4H3 y“ 1
"\T(e)mC(aLa,+o\\o\ :: 9 q/ i Deena1
(Ht) ‘3‘
n+LK ~\ ~9i‘iiéx33
: b o u g Q, Let X1, X2 , , X 12 be a random sample of size n = 12 from the distribution with
probability density function f(x).——39x26‘ex3 x>0 G>0. Find a uniformly most powerful rejection region of size OL = 0.05 for testing H0:9=3 vs. H1:9<3. Let 9<3
n —3x3
( ) Higxie L H x x ...x 1:1
7» a , a = 0’ 19 2: 9 n =
(XI xz x”) L(H1;X1,xz,..,xn) n 2 ~6xl3
H 39x1. 6
i=1
= {ijnexp (6—3)§1x3
\6 i=1 1
n
7»(x1,x2,..,xn)sk C} (9—3)ZX?Sk1 n
RejectHO if ac.
i=1 1 n
2X? has a Gamma distribution with OL = n = 12 and B = usual 9 = i=1 n n
Then 2— =’ 2 9 has a X2 ( 2 OL = 24 degrees of freedom) distribution.
i=1 i=1 n
0.05 = OL = P(RejectH0H0istrue) = P(2X§ 20 19:3)
i=1 = P(6%X§’ 26c[9=3) = P(x2(24)26c).
i=1 :> 60 = X30504) = 36.42. :> 0:6.07. 12
RejectHO if 26.07.
i=1 ...
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 Spring '11
 Stepanov

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