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For a binomial model, with probit link function:

model = glm(response~A+B+C, family = binomial("probit"), na.action = na.omit)

where A and B are continuous, C is a binary qualitative variable (sex, M or F). Coefficients of the model:

  (Intercept)          A                B                   C 
-10.326460234   0.008913341   0.253518215        -0.291546432 

Given C = sexF, where sexM is eliminated from the model matrix (I think this is to eliminate aliasing problems?), can the fitted probability for a combination with C = sexM (coefficient not provided) be obtained using the difference between the intercept and C coefficients (-10.03491) and applying a formula such as:

$$-10.326 +0.0089x + 0.254x -0.292x= probit(P) $$

I'm sorry if my explanation is unclear, I'm studying GLMs for the first time and trying to get to grips with an exercise Q.

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  • $\begingroup$ It's useful for these type of questions related to exercises or homework question to add the self-study tag to your question $\endgroup$ – vigos Mar 19 at 22:38
  • $\begingroup$ well for a start you would need separate variables for your fitted model and not just use x for a,b and c, more like $x_1,x_2,x_3$ and $x_3$ would be a zero or one depending on if it refers to male or female $\endgroup$ – vigos Mar 19 at 22:46
  • $\begingroup$ Thanks for adding the tag and pointing out a mistake in the formula. I meant to put them as $x_1$,$x_2$,$x_3$. So for this case if C = sexF (1), to determine fitted prob for sexM (C = 0) the formula would be in the form below? $$-10.326 _0.0089x_1 +0.254x_2 - 0.292(0) = probit(P)$$ $\endgroup$ – thatsnotmyname71 Mar 20 at 11:56
  • $\begingroup$ See stats.stackexchange.com/questions/141255/… $\endgroup$ – kjetil b halvorsen Mar 20 at 17:28

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