Question

As part of a study of the association between smoking and risk of squamous cell carcinoma, a logistic regression model was estimated as:

logit(probability of having squamous cell carcinoma) = -4.84 + 4.6*(SMOKER)

where SMOKER=0 for non-smoker and 1 for smoker. What is the predicted probability of a smoker having squamous cell carcinoma?

Answer 1

0.44

Explanation:

Given the estimated logistic regression model on risk of having squamous cell carcinoma

-4.84 + 4.6*(SMOKER)

SMOKER = 0 (non-smoker) ; 1 (SMOKER)

What is the predicted probability of a smoker having squamous cell carcinoma?

exp(-4.84 + 4.6*(SMOKER)) / 1 + exp(-4.84 + 4.6*(SMOKER))

SMOKER = 1

exp(-4.84 + 4.6) / 1 + exp(-4.84 + 4.6)

exp^(-0.24) / (1 + exp^(-0.24))

0.7866278 / 1.7866278

= 0.4402863

= 0.44