B 5 pts suppose that ˆ γ 5ˆ β 1 3 2 ˆ β 2 1 6

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(b) (5 pts) Suppose that ˆγ0= 5,ˆβ1=-3.2,ˆβ2= 1.6, and ˆσ= 0.2. Calculate a point estimate ˆτforτ=E[yi|xi= 2.5].Hint:Expressτin terms ofγ0, β1, β2andσ, and then replace the parameters by theirestimated values.
(c) (5 pts) Suppose that se(ˆβ2) = 0.3 and ifTt(n3), we haveP(|T|<1.98) = 95%.Usethis information in addition to the numerical values from part (b) to find a 95% confidenceinterval forλ= exp(β2)-1.
STAT 331, March 19, 2015, 1:30pm-2:20pmPage 5 of 6
NotationFor vectorsx= (x1, . . . , xn) andy= (y1, . . . , yn),¯x=1nnsummationdisplayi=1¯y=1nnsummationdisplayi=1yiSxx=nsummationdisplayi=1(xi-¯x)2Sxy=nsummationdisplayi=1(yi-¯y)(xi-¯x).Linear Regression ResultsThe multiple linear regression model isy∼ N(Xβ, σ2I)⇐⇒yi|xiind∼ N(∑pj=1xijβj, σ2),i= 1, . . . , n.The loglikelihood for this model is(β, σ|y, X) =-12(ˆβ-β)XX(ˆβ-β) +eeσ2-n2log(σ2),where the MLE ofβisˆβ= (XX)1Xy, ande=y-Xˆβ.ˆβandeare independent, and1σ2eeχ2(np).Log-Normal DistributionIfZ∼ N(µ, σ2), thenY=eZhas a log-Normal distribution denoted log -N(µ, σ2), withE[Y] = exp(µ+12σ2)andvar(Y) = [exp(σ2)-1] exp(2µ+σ2).Matrix FormulasInverse of a 2×2 matrix:bracketleftbiggabcdbracketrightbigg1=1ad-bcbracketleftbiggd-b-cabracketrightbiggIfVn×nis a symmetric, positive definite matrix, then we have:V=LL, whereLis lower triangular and the diagonal entriesLii>0.

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