{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Mth562-HW10-11

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

This preview shows pages 1–12. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}