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Unformatted text preview: 541 a.) Xi ~ exponentialﬁ). px = 1, ValX z 1. From the CLT, in is approximately r(l, 1 [{1} 55.55 So in «1 in  '
i/‘W andP(1/~m%< <x)—»P(ng).
b (1 I ~ .._ d __ _ 1 «2/2
) EP(Z S X) — EFZ(X) — {2(3) ... We .
M???“ %(,§S‘i‘“'+“) (Wagwmm) ‘ «_~ ﬁFW(xﬁ+n) = fw(xﬁ + n) «a = rleJmﬁ‘l‘ n)n“1e‘<“'ﬁ+m)ﬁ _ Therefore, (1/1100) (xii+n)n1e“(m+nkﬁs 3;..3'1‘2/2 1r
yields n! as nu+1l2€nw321n as n—rao. Substituting 1: == 0 " 5.43 (a) Let X ~ n(0,0'2). For Va > 0, limPQY— yl<g)=1imPQ\/Z(Ynwy)f<\/Ea)w—.PQX<OO)=1, LIZ—”mu [Iv“N70 Mam 5.44 (6083/2116 (3LT, ﬂu; w p)w‘L>n(0,p(1—p))
(b) Let g(x)= x(1—x) then g‘(x)=1—2x. For p ¢1/2,i.e., g'(p) i G by the Delta Method, J; (510’; )— g( p))—d> 11(0, 0‘2 [g‘ (p)]2 ), then «5020 — 1;)“ p(1  p))—~"‘i>n(0,(i *pr 19(1” p)) f5,24 a) ForU ~uniform(0,1),_EU :1/2, Veri=1/12. Then 355i 2 3—1/2
x— “Elm6 1211— 5: «IE(—1~m) is in the form T((ﬁwEU)/U) with n = 12, so X is approximately n(0,1) by the Central
Limit Theorem. b) The approximation does not have the same range as Z ~ n(0, 1) where ¢o < Z < +¢o, since *6<X<'8. . 2 '
c) . EX=E(:E21U:;6)= 22EUi 6‘— —(§:1%)n6:66=0.
i“1i=i i= 'VarX—TV 22Ui =6) Vat 2U—12VHU1=1
i=1  i=1 EX3 2 0 since X is symmetric about 0. (In fact, all add moments of X are 0.) Thus, the ﬁrst three moments of X all agree with the first three moments of a. staminad
normal. The fourth moment is not easy to get, one way to do it' is to get the mgf of X. Sinee EetU 2: (e t1)/t, Eet( 2111; 6)_¢_6;(§f_‘1)12=(et/213't/2)12 Computing the feurth derivative and evaluating it at t .—.. 0 gives us 3x4. This is a lengthy
calculation, but is made easier with a symbolic nianipnlation computer package. The
answer is EX4 2329/10, slightly smalles than E Z4 a 3, Where Z ~ n(0,1). Problem from notes
Contruct a 95% conﬁdence interval for p. Justify your method eompietely.
X; N independent Bernouiﬁﬁp)‘ . 13 = 3;; Xi 1. B th . WWE—zv} “id
y e CLT we know that W N(O,1) 2. By the WLLN we know that 33 —+p p
3. Deﬁne a function g(m)m \/$(i «w :73) .Then 91(55) is a. )continuous function. We apply theorem 5.54a11dgetg(p)——>Pg(p)= 1/p(pi—);Mp\/ (1—39
4.1381338 another continuous function Mas) 23/1113.) We apply theorem 5.5.4 again and get h(\/p(1wp))—>Ph(M)= V___:(:: 21. 5‘ ﬂow) 2 \fﬁfﬁ—p) $41 13) x/ﬂﬁ p) d ‘
«ea—:5) v’pe—mx man 5) By SlutSkyS woo—p) N(0,1) So the 95% conﬁdence interval for p is: 13 :i: 1.96 x 23 1—29)
n ...
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 Spring '07
 SENTURK,DAMLA
 Statistics

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