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Mth562-HW10-11 - MTH/STA 562 Exercise 10.93 The null...

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Unformatted text preview: MTH/STA 562 Exercise 10.93. The null hypothesis specifies S20 2 {a2 : a2 = 03}, While (2 = (20 U 52a 2 {02 : 02 2 03}. In the restricted space (20 under H0 : o2 = 03, the likelihood function is TL 1 1 L (9°) = H W27? exp l3? i=1 1 (3h — M2] = and n lnL(QO)=—Elnag—nlnv27r—i2 (35—102. 2 2001:1 Then d 1 " —1 L o —— r— = d“ n ( ) 03 (y M) Which implies that the maximum—likelihood estimate of u is Q. So L (6°) 2 . 1‘” ‘Wl ' In the general space 9, the likelihood function is 1 1 ” 2 L(9)=-—-;-———~71-exp —— (Eh—M) (‘72) /2(V27T) l 202 i=1 and n lnL(Q)=—gln02~nln\/27r—2T:2- (35—1020 i=1 Then a 1 ” —1nL(9)=— (yz—M)=0 8“ 021:1 6 n 1 1 " 2 filnL(Q)=—§; fl (gm—M) =0 These imply that ii = y and E2 = max {(73, i Z (1% — SO i=1 1:1 L = map [—27% . (92' “@2] ‘~ The likelihood ratio statistic is given as L 82 11/2 1 n 1 n _ __ _____ i _ — 2 _ A —— < 2) exp I: 200 (yz + 282 I 0 0 i=1 L (‘3) <03?” WW“ exp i— 1 TL 202 0 i=1 (3% — M2] that implies that and ll _ 2 n/2 _ 2 [M] exp [JR—£5; + :5] if 32 > 03” n03 208 2 Hence, the rejection region /\ S k is equivalent to (n —~ 1) S2" 00 2 A 2 9 (X2) = n—n/2 (X2)n/2 exp (_X Where X2 = Note that 9 (X2) = 1 if 32 S 03, while 9 (X2) is monotonically decreasing function of X2 if 32 > 03.. Hence, the rejection region /\ S k is equivalent to X2 Z c, Where c is determined so that the test has size a. Exercise 10.94. The hypothesis of interest is H0 : p1 = p2 = p3 = p4 = 1) against the alternative that at least one of these equalities is incorrect. In 9, the likelihood function is L (Q) = (200)}? (1 ‘" POM—m 7%" 1:1 and 4 200 4 4 lnL Q =ln ( ) nilnpi+ 200—ni ln 1-2911. ( ) m 12:1} ) ( ) Then a 200 n —n lnL Q =—Z— —z= api- ( ) 101- 1—p1- implies that the maximum-likelihood estimate of pi is A “1’ = 200 for i = 1,2,3,4. So L (62> = 116??) (gm (1 — 4 4 _ [4 (200)] Em SOD—Em- 1? i=1 (1 — p) i=1 i=1 m 1:1 mi and 4 200 4 4 1 L 9 =1 1- l 800 —— 1 l 1 — n ng<m>+<gn>w+< 20> Then 4 4 E m 800 — d i=1 i=1 —1 L Q = — M— = dp 11 ( 0) p 1 _p 0 Which implies that the maximum-likelihood estimate of p is A_ m p _ 800 " So L (‘30) = lfi (333)] (8’32); (1 — 2:1 Thus, the likelihood ratio statistic is given as 4 4 1:: Since the m are large, it follows from Theorem 10.2 that : -2{(Z"z‘>1n<$%>+(800—-;m)1n<1—;;g> <2#411633)+<200—m>1n(1-2%3)l} has an approximate chi-square distribution with 3 degrees of freedom. Since n1 2 4 76, n2 2 53, n3 = 59, n4 2 48, and Em = 236, —2ln)\ = —2 (52676) = 10.54. i—1 The rejection region , for a = 0.05, will be ——-2 lnA > X3053 = 7.81, and the null hypothesis is rejected. The fraction of voters favoring candidate A is not the same in all four wards. Exercise 10.95. Let Y1, Y2, - - - ,Yn be the random sample drawn from the first population, and let X1, X2, - - - ,Xm be the random sample drawn from the second population. Under the null hypothesis H0, 72 m nfirz &—X2 ;( ) z; ) (n—1)S%+(m—1)S§ 2 T+T=”—“a—2——=X 0 0 0 has a chi-square distribution with n+m— 2 degrees of freedom. If Ha is true7 then 5’12 and 822 will tend to be larger than 03 (since they will be estimates of 02 > 03). Under the null hypothesis H0 : a2 = 03, the likelihood function is n 1 1 2 m 1 1 2] L Q = 6X —— Z- __ , ex _____ xi _ ( 0) gaov27r p[ 203“; #1)] gaofi pi 20(2)( #2) _( + ) 1 Z: — ,u’1)2 Z —‘ M2)2 2 (V 2”) (ail—(“WW exp -- 51—w— + L;—~— ‘70 00 and + 1 Z (3% _ ,ul)2 2(132 — [LI/Q2 lnL(90)=-(n+m)ln\/27r—n mlngg__ i=1—2___+ i=1 2 2 ‘70 00 Then a 1 ” ————lnL Q 2—2 i—~ =0 8M1 ( 0) 03 i:1(y #1) a 1 m E (mi‘M2)=—'0 3M2 00 i=1 which imply that the maximum—likelihood estimates of in and #2 are respectively given by ill = y and fig = E. So “(n+m) 1 Z (yz — y>2 + 2 (ml _ if L (90) = (03) _(n+m)/2 exp ~§ ——~——$=1 08 + __—’=1 03 _(n+m) — n m 1 = (0—3) (+ Wexp (3X2). In the space Q, the likelihood function is n LUZ) = H 1 eXI) [~2%2(yi —M1)2] 12% exp [—55% —M2)2] i=1 0 27F i=1 0 ( + ) 1 Z (yz — M02 2 ($1 - M2)2 _ /—_ _ n m 2 —('n,+m)/2 1:1 i=1 _ ( 2W) (a ) eXp «E 02 + 02 and n E (Eli “ #02 2 (Sci — #2)2 + i=1 n+m 1 i=1 1n 02 —' lnL(Q) = ——(n+m)ln\/T— 2 02 02 Similarly we see that the maximum-likelihood estimates of M1 and #2 are respec- tively given by Z171 = y and fig = 5. Then the likelihood is maximized either at 32 = 301“ at 82 = as. If 32 = 03, then L 520 A = (A) :1, L (9) Thus, for k > 1, L (fig) /\ = g k >a l I I A [clam V 3 + 3 (D N *U r-A-‘x l II—l >< [\D + Nil—4 I’__—'—l § 1 “C 0) HM + 3 | H v [5’37 l—l W4 which is decreasing function of X2. Thus, we reject H0 if X2 is too large. The rejection region is 2 (TL-1)5i+(m-1)S§ _ 2 X _ 2 > Xa;n+m——2 ' 00 Exercise 10.96. (a) In the Space (2, the likelihood function is .1103 ——(n1+n2+n3) _,n 2 _n 2 _n 2 1 n1 : (VZW) (0i) 1/ (‘73) 2/ (032,) 3/ 9XP{—%§ (Uh—Inf} 1i=1 1 "2 2 1 “3 2 .exp {*T‘g i=1 (yi — M2) }9XP{—‘2:§i:1(wi-M3) } and lnL(Q) = —(n1+n2+n3)lnv27r—%lna2—glnag—glnofi 1 "1 2 1 "2 2 1 “3 2 “3;? i=1 (“a—#1) —27§i:1(yz—#2) ‘égggfifli—M) - Then a 1 "1 —1 L Q = —— 2— _ 2mm <> figs: #1) 0 a 1 "2 ——1 L Q = — 2-— = 2mm <> M2) 0 a 1 “3 223 n <> 02m 13> a 7111 1 “1 2 5&1an) = _3E+fiz(mz—Hl) =0 i=1 a n21 1 "2 2 ——1 L Q = ———— v——- 2w = a 7131 1 “3 572W”) = ‘30—§+27g_2<wi—~3>2=0 which imply that the maximum—likelihood estimates of 111, 112, 113, a are simply [11:25: “2:377 Ill/3:711?) n1 n2 n3 01—”— ($2*$)702—n— (yz“yl,U3—n— 1 i=1 2 1:1 3 2:1 So : —(n1+n2+n3) 1 “2 ' exp “27;? 71:1 1 "3 _ 2 (yl y) }eXp{ 263. 2:1 (firm/2 (Ea—"2” (Ga—"3” expi 2 17 0-32 and 032) (m + 712 + 723)] .. —(n1+n2+n3) A _n 2 A _n 2 A -n 2 1 = W) (a?) 1/ (as) 2/ (0;) 3/ em [—5 Under the null hypothesis H0, the likelihood function is "1 1 1 2 “2 1 1 L Q = —— i— - ~-— ( O) 2.11:0 27reXp[ 202 (x #1)] 2110 27rexp[ 202 1 eXP[-—1-(w'—M )2] i=1 a 27r 202 7’ 3 Z _(nl+n2+n3) (0.2) ‘(n1+n2+n3)/2 1 n1 2 n2 2 n3 2 eXp{—fi [ZUW—Ml) +Z(yz—M2) +Z(wz‘—M3) i=1 2:1 z=1 and lnL(SZO) — — (m + 112 +113) lnv27r ~- m + Z2 +713 lna2 1 m 2 n2 2 n3 2 :m—m +2<M> +Z<W‘“3)l" i=1 i=1 i=1 Then a 1 “1 am n ( 0) aggm m) a 1 "2 ——l L Q = — i- = 0 6% n < 0) ago 112) a 1 “3 —1 L o = ~—- i— =0 am n < o) 02;?” 113) a 711 + n2 + 713 1 1 n1 2 "2 EQIHL(90)=——‘2—g+274 ;($1*M1) +';(yi— which imply that the maximum-likelihood estimates of ,ul, #2, p3, and 02 are simply fl1=57 M229, fi3=m n1 —:”U_)2 + —@)2 + —m)2:| i=1 w __i__ n1+n2+n3 q H n13? + 77,233 + 77,3832, m + 7L2 + 713 ' —(n1+n2+n3) 0' L (60> = («a .exp {—2%? —_J7)2 + (yi- —'y‘)2 + ——u7)2:l} _(n1+n2+n3) (82)—(n1+n2+n3)/2 exp [.__ (A2) -(n1 +n2+n3) / 2 I r—I 4m+m+mfl. NJ Thus, the likelihood ratio statistic is given as L (v 271-) ‘(n1+n2+n3) (3%)‘711/2 (3%)—n2/2 (8%)‘113/2 eXp [—% (m + 712 + n3)] A2 111/2 A2 112/2 A2 713/2 (01) (‘72) (U3) (82)(n1+n2+n3)/2 H Hence, the rejection region is given by cam/2 (83)”? (Ear/2 (32)(n1+n2+n3)/2 < k' (b) For large values of n1, ng, and 713, the quantity —2 In A will have an approx— imate chi—square distribution with 2 degrees of freedom, since 7" = 3 and To = 1. Hence, the rejection region, with a = 0.05, is —2ln A > Xaow = 5.99. Exercise 10.97.. (a) In the space 9, the likelihood function is _ —a:,'/0 — 2/9 __ —m —n ____ I __ I L - J2! —61€ 1 ZL! —626 y 2 — 01 62 exp < 61 El $1) exp ( 02 E M) and 1m 1 lLQ=—16—16—— i—«— n H mnl “I12 Zy Then 6 m 1 ———- —6—1—9—%Z:EI$i—0 (9 n 1 n 5th”) — 72—922. yi- Which imply that the maximum-likelihood estimates of (91 and 02 are simply d1 = E and 62 = y, and L = (5)4” (@V” exp (—% exp <—% 2 (gym (m—n e‘(m+n)' Under the null hypothesis H0, the likelihood function is L = fl ge—wi/Q fi ée—yi/g = 6-(m+n) exp l—é 111‘ + i izl ,:1 i=1 i=1 and lnL(QO) = —(m+n)ln0—% ., i=1 Then d m+n 1 m n __ Q :_ __ , . 2 6,9111“ 0) 6 + 62(;x1+;%) 0 which imply that the maximum—likelihood estimate of 19 is A 1 m n m§+ny 9: i i =—‘——« m+n<§x+gy> m+n So A ’1 1 m n 7715+ n3 _(m+n) _ L (90) — 0 (m+”)exp [—3 =< m+n > e ("WW2 Then A A _ L (90) (is?) W") e WW2 z W L (a) (firm (:2)— e-<m+“> (mgigyrm (b) Notice that each X,» has a gamma distribution with oz 2 1 and B = 61. It can be easily shown that if H0 is true, 2? has a gamma distribution with a = 1 and fl = 61, which is equivalent to a chi-square distribution with 2 degrees of m freedom. Also, gZXi = 2—31? is the sum of m independent chi—square random i=1 variables and has a chi-square distribution with 2m degrees of freedom. Similarly, £216 = 2573? has a chi-square distribution With 2n degrees of freedom. Since i—l X’ E and Y’ s are independent, an F statistic can be found. That is, 27mY/2m_Z 237/277, ‘ y has an F distribution with 2m numerator degrees of freedom and 2n denominator degrees of freedom. Write A _ Em?” m 1 _ (m§+ny)(m+n) — mE—l—ng m m£+ny n m+n Zi(m+n) ‘ §(m+n) 1 Note that A is small if F is either too large or too small. Then the rejection region, A g k, is equivalent to F 2 01 or F S 02, Where cl and 02 are chosen so that the test has size a, Exercise 10.100. In the space S), the maximum-likelihood estimates are [[71 = 11—17 11/22 = $27 and n1 Dyn- w + at] = n1+n2 A 1 02 = 7L1 + H2 31:1 i=1 as in Emercz'se 10,, 96. Then n1 72'2 13(6) = H012Wexp[—-,%<yt—yl>2l-H 1 exp[—i<y2i—y2>2] i=1 i=1 a V 2” 202 "1 i=1 i=1 2 (mfmm(aZ>—<“I+“2>/2e—<m+nv2. Under the null hypothesis H0, the likelihood function is 711 1 1 2 "2 1 1 2 L o = —— i— - —— i- ( 0) 0 27r exp i 202 (yl H) i a 27r exp i 202 (y2 I”) ] Toll-Hm) —— n n 2 1 n1 722 = (V27?) (02) ‘ 1+ 2” 'exp {‘Q [E (912' - “>2 + Z (W “ Mi} i=1 i=1 and “1 n +71 1 “2 111L620) = "(711 + "2) 1“ V 27“ 1 2 2 111027—— (yu — m2 + (922' - [if] a 02 , 1:1 10 Then a 1 "1 “2 a—Mlano) = ; [;(y1i_ M)+;(y2i—M)] =0 8 n1+n2 1 1 n1 2 n2 2 The first equality implies that n1 n2 Zyli * “Lu—l- 2922- —n2,u = 0 i=1 1'21 01' n1 n2 2 3/11" + Z 921 _ _ fl 2 1:1 i=1 = n1y1 4' n2y2 n1 + n2 R1 + 712 , While the second one implies that 1 n1 722 A2 __ . __ A 2 _ _ A 2 00 “‘ “1 + “2 :(ylz M) + 121: (3/22 M) ] " i=1 A —n n _n n "1 A “2 A L (90) 2 ( 1+ 2) (83) ( 1+ 2)/2 exp {__2_::% (yli _ m2 +Z(y2i _lu)2:l} i=1 i=1 2 (v%0_“””)ee—Wfimvaermvz Thus, the likelihood ratio statistic is given as L ( 271,) —(n1+n2) (8(2))—(n1+n2)/2 e_(n1+n2)/2 82 (nl+n2)/2 A = = _____V_—__ 2 . (m)-(n1+n2) (32)'-(n1+n2)/2 e—(m+n)/2 ( ) [email protected]> Hence, the rejection region is given by A2 (n1+n2)/2 e 00 In order to Show that this test reduces to the two—sample t test of Section 104.8, we observe that (m +n2)32 = (n1 — l)Sf+(n2—1)S22= (n1+n2—2)S: Where ‘azmune+w—ne 1" n1+n2—2 ” 11 Consider n1 : Zam— i=1 “2 (711 + 7193(2) fl)? + Z (921' — m2 i=1 “1 2 i=1 (“1 “ 1) 312 Jrn1 <31 — n2 (gh- — $02 + n1 (n — m2 + Z (2121- — 32)2 + 722 (32 — m2 i=1 _ _. 2 7111/1-l'n2y2 2 _ ——r + n —-1 8 +72 — n1+n2) (2 )2 2<y2 “1712 (—331 — Q2? ’01 + N2 =(m+m_my+fififlgfli p n1 + n2 " Now /\2/(n1+n2) __ a: __ (721 + “2 — 2) _ 1 _ A2 ""(y—§)2_ nn(—_—)2 1 00 ("1 + “2 _ 2) 322’ + W 1 + __1 3113:7121”, 2— (n—1+n2—2)S—I2J _ 1 _ 1 _ 4M: — 1 T2 ’ l + n1+n2‘—2 + n1+n2—2 Where _ _ T = y1 _ y2 Sp./nil+;1§ has a 73 distribution with 721 + n2 — 2 degrees of freedom under H0. Since we consider only M1 > [12, or equivalently yl — $2 > 0, T = VT? Will be positive. Hence small values of /\ imply large values of T, and a one—tailed t test is implied. 12 “131 + n2y2 n1+n2 )2 ...
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