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Unformatted text preview: 11-6Solutions Manual for Statistical Inference11.14 LetXin(i,2). Then from Exercise 11.11CoviaiciXi,iciviXi=2aiviVariaiciXi=2a2ici,Var(iciviXi)=2civ2i,and the Cauchy-Schwarz inequality givesXaivi.Xa2i/ciXciv2i.Ifai=civithis is an equality, hence the LHS is maximized. The simultaneous statement isequivalent toki=1ai(yi-i)2s2pki=1a2i/nMfor alla1,...,ak,and the LHS is maximized byai=ni(yi-i). This produces theFstatistic.11.15 a. Sincet2=F1,, it follows from Exercise 5.19(b) that fork2P[(k-1)Fk-1,a]P(t2a).So ifa=t2,/2, theFprobability is greater than, and thus the-level cutoff for theFmust be greater thant2,/2.b. The only difference in the intervals is the cutoff point, so the Scheffe intervals are wider.c. Both sets of intervals have nominal level 1-, but since the Scheffe intervals are wider,tests based on them have a smaller rejection region. In fact, the rejection region is containedtests based on them have a smaller rejection region....
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- Spring '12