STAT 514 - HW 10 Solution Key

# STAT 514 - HW 10 Solution Key - 3.“ 83 The CLT tells us...

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Unformatted text preview: 3.“ 83' The CLT tells us that Z = (Exihnp)/,|np(1—p) i's epproﬁmateiy 11(0, 1). For a. test that rejects H0 when > c, we need» ﬁnd c and n to satisfy c—n(.49) _ c-n(.51) m P(Z > n(‘49)(.51))-— .01 and P(Z > 1105904”) —- .99. We thus want c-—n(.49) c-n(.51) Ifn(.49)(".5"'1) = 2'33 9”“ lin"('.s1)(.49')'= "2‘33 Solving these equations gives n =' 13,567 and c = 6,783.5. 8'15 From the NCYmaLmPeerson Lemma. the UMP test rejects He if q" “13 (2"?)“3/Kmi2/(2ﬁ) (“o nexp 1 y 2 1 1 1> k f 3%" = m ’ ‘ —- or some k a0. {(11003 (2! Drawn/2:21} [(20%) I {2 3§(o% After some algebra, this is equivalent to rejecting if . W-c sin 1...}... 2x12) ( 0%.)”- This is the UMP teat ofaize a, where (3 == P,O(ZX]-2 > c). To determine c to obtain a speciﬁed a, use the fact that ~ x121, Thus a = I’d-0(ZXiz/a'g > c/crﬁ): P(x§ > clears), so we must have c/ag = xg'm which mm c =_.. 9% xg'a‘ 8.1“. a.) Size =: P(reject HG ] H0 is true) 1 => Type I error -..: 1. Power = P(reject H0 1 HA is true) 1 =9 Type 11 error :2 0. 1:) Size = P(reject 30 l H0 is true) 0 a": Type I error :3 9. 0 :Typellerrorzl. H Power = P(reject Ha I HA is true) 1| 7‘ (“H 7 or more successes _.. 1 a, 3') P'va’luc = P{ out of 10 Bernoulli trialslo — E = 15%“? + (13°)@)8(%)2 + (19°)G)9(%)1 + i3)(%)1°(%)° = “1375- 5) P-Valae=P{X_>.3EA=1}=1-—P(x<31A;1) —12 —111 410 =1-{Em2‘rL ET+LTJT*”‘0393' .c). P-velue: Hzxizszaxms} .—. 1—P(Y < 913A: 3), (Yziiaxi-vi’oiesonan) "‘3 III a I al—e [y-fﬁ-FET'I‘E-F +Tf+ﬂ]~ 0038 ‘25" 1'1 Thepdfonis c __ III9 r<yle)=-;~y‘1/”) 1e v , y>0. By the Neymeaneerson Lernma, the UMP test will reject if 1 “1/2ey—yI/2 a, ﬂy I 2) k_ W "W )> To see the form of this rejection region, we compute ted/wowéy—3/2ey“¥1’2( «em-1:). which is negative for y < 1 and positive for y > 1. Thus f(y I 2)/f(y I 1) is decreasing for y g 1 and increasing for y a 1. Hence, rejecting for {(y I 2)/f(y I 1) > k is equivalent to rejecting for y 5 Co or y 2 c1. To obtain a size or test, the constant co and c1 must satisfy moo wcl and ﬂcol2)“f(c1l2) W‘W' a:P(Y\$¢8I9=1)+P(YZCII9=1)zl-we +8 Solving theee two equations numerically, for a r...“ .10, yields on = .976546 and c} = 3.637798. The Type 11 error probability is 'l0-32. _w a) b) -‘5 C15- ‘0 1/2 c 1 w w w ml y 1/2e y dyad, y c02.699834. P(c9<Y<cII9:2)=J. First calculate the MLEs under 131: p2 m p. We have X .X - 11(pl1) =1: 113%? 3-"93116 "211 " 2:233 Taking logs and differentiating yield the following equations for the MLES: )m“x1"‘2""""n-1 we...“ 9 _ _, 3" MP- Site 610 L x a: wipi~-—-———§mﬁ~_~_—i—w39, :2 , .,n—1, x 1 1-29" Ei=3p, . _ x + _ - . _ a with solutions 13 3 %, pi = g,l#3,...,n*1, and pH = (n: «- )3.“ ind/m. Except for I: the ﬁrst and second calls, we have expected = observed, since both are equai to xi. For the ﬁrst two terms, expected = mi) = (xi+x2)/ 2 and we get . 2 2 x + X + 2(observed-expocted)2=(x1— 12x2) +(‘2” 123(2) J‘z“"2l2 expected J‘1‘”‘2 xﬁxz ' "ﬂxz ' T T The hypothesis about conditional probabilities is giVen by He: P(changeIinitial agree) = P(changeIinitial disagree) or, in terms of the parameters H - ﬁlm—m F2 0‘ p1+P3 92+P4' This is the same as p1p4 = p2p3, which is not the same as p1 = p2. ...
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