{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Dr. Hackney STA Solutions pg 137

# Dr. Hackney STA Solutions pg 137 - 8-16Solutions Manual for...

This preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: 8-16Solutions Manual for Statistical InferenceWe now choose the constantcto achieve sizeα, and wereject if|¯x-θ|> tn-1,α/2ps2/n.c. Again, see Chapter 5 of Lehmann (1986).8.39 a. From Exercise 4.45c,Wi=Xi-Yi∼n(μW,σ2W), whereμX-μY=μWandσ2X+σ2Y-ρσXσY=σ2W. TheWis are independent because the pairs (Xi,Yi) are.b. The hypotheses are equivalent toH:μW= 0 vsH1:μW6= 0, and, from Exercise 8.38, ifwe rejectHwhen|¯W|> tn-1,α/2pS2W/n, this is the LRT (based onW1,...,Wn) of sizeα. (Note that ifρ >0, VarWican be small and the test will have good power.)8.41 a.λ(x,y) =supHL(μX,μY,σ2|x,y)supL(μX,μY,σ2|x,y)=L(ˆμ,ˆσ2|x,y)L(ˆμX,ˆμY,ˆσ21|x,y).UnderH, theXis andYis are one sample of sizem+nfrom a n(μ,σ2) population, whereμ=μX=μY. So the restricted MLEs areˆμ=∑iXi+∑iYin+m=n¯x+n¯yn+mandˆσ2=∑i(Xi-ˆμ)2+∑i(Yi-ˆμ)2n+m.To obtain the unrestricted MLEs, ˆμx, ˆμy, ˆσ2, useL(μX,μY,σ2|x,y) = (2πσ2)-(n+m)/2e-[Σi(xi-μX)2+Σi(yi-μY)2]/2σ2....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online