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confidence bands for the Exercise yocardial infarc 13.8 In a paper in American Statistician, Hauck [1983] derived logistic re
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Answer #1

a) Let's run the following code in R to graph the probability of MI for smokers and non-smokers separately:

Code:

X2 = seq(from=0,to=550,by=1)
p1 = -2.2791 + 0.7682*1 + 0.001952*(X2-100)
p2 = -2.2791 + 0.7682*0 + 0.001952*(X2-100)
py.s = exp(p1)/(1+exp(p1))
py.ns = exp(p2)/(1+exp(p2))
plot(x=X2,y=py.s)
plot(x=X2,y=py.ns)


Output:

For Smokers:

py.s 0.15 0.20 0.25 0.30 0.35 8

Non smokers:

髦3 CN 100 20N) 300 400 500 0 X2

b) We have to assume the sample size as we don't know the value. Let's take N = 100

Coefficient Standard Error t p-value
Smoking 0.7682 0.3137 2.448836468 0.016
Serum 0.001952 0.001608 1.213930348 0.229

Hence, the smoking variable is significant here and not the serum triglyceride variable.

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