STAT 431

Stat 431 ASSIGNMENT 3 SOLUTIONS Last document update:   ago

Stat 431 ASSIGNMENT 3 SOLUTIONS 1. (a) Given the tolerance distribution, the probability of response the dose x is π(x) = Zx −∞ exp((u − µ)/δ) δ(1 + exp((u − µ)/δ) 2 du = exp((x − µ)/δ) 1 + exp((x − µ)/δ) ⇒ log π(x) 1 − π(x) = − µ δ + 1 δ x This implies that it is most appropriate to choose a logistic link function. (b) The binary logistic regression model is log π(x) 1 − π(x) = β0 + β1x where β0 = − µ δ and β1 = 1 δ . ...

Stat 431 Assignment 3 solutions|very helpful Last document update:   ago

Stat 431 Assignment 3 solutions 1. [ 12 marks] (a) We start by fitting the interaction model logit(πi) = β0 + β1xi1 + β2xi2 + β3xi1xi2 The Wald-test for H0 : β3 (coefficient of interaction term) gives a p−value << 0.05, hence we can not drop the interaction term. Since we can not further simplify this model, it is the best logit model we can fit. [2] cells<-("", header=T) cells$resp<-cbind(cells$Cell, 200-cells$Cell) fit1<-glm(resp~tnf+ifn+tnf*ifn, family=binomi...