STAT 431
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Stat 431 ASSIGNMENT 3 SOLUTIONS
- Exam (elaborations) • 7 pages • 2022
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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 
. ...
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Stat 431 Assignment 3 solutions|very helpful
- Exam (elaborations) • 12 pages • 2022
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- CA$11.81
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Stat 431 Assignment 3 solutions 
1. [ 12 marks] 
(a) We start by fitting the interaction model 
logit(πi) = β0 + β1xi1 + β2xi2 + β3xi1xi2 
The Wald-test for H0 : β3 (coefficient of interaction term) gives a p−value << 0.05, 
hence we can not drop the interaction term. Since we can not further simplify this 
model, it is the best logit model we can fit. [2] 
cells<-("", header=T) 
cells$resp<-cbind(cells$Cell, 200-cells$Cell) 
fit1<-glm(resp~tnf+ifn+tnf*ifn, family=binomi...
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