# From the last line below, V...

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Practice Problem Solutions 2 1) From the last line below, V ar (log ˆ θ ) 1 /n . n ( g ( ˆ θ ) - g ( θ )) d -→ N (0 , θ 2 [ g ( θ )] 2 ) g ( θ ) = log θ g ( θ ) = 1 θ [ g ( θ )] 2 = 1 θ 2 n (log ˆ θ - log θ ) d -→ N (0 , 1) 2a) G 2 = 2 . 4 on 4 df indicating a good fit.
2c) The Bayesian estimates compare very well to the classical estimates shown in part a).
model { for(i in 1:N) { y[i] ~ dbin(p[i],n[i]) logit(p[i]) <- alpha + beta * x[i] } alpha ~ dnorm(0.0,1.0E-6) beta ~ dnorm(0.0,1.0E-6) LD50 <- -alpha/beta } list(y=c(0,14,29,42,67,73),n=c(75,75,75,75,75,75),N=6, x=c(-0.553,-0.113,0.059,0.185,0.446,0.753)) list(alpha=0,beta=0)