I've created an example table (just in order to create a function) with:

ex<-data.frame(b=c(rep('A',50),rep('B',30), rep('C',20)), 

Then I run a logistic regression model (MASS library)

modT<-glm(clase~1, family=binomial, data = ex)
modT<-stepAIC(modT, scope = fm, family=binomial, data =ex, k = 4)

As you can see coefficients are not significant, but I've created the class using them. So I don't understand why this is happening.

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  • $\begingroup$ First, you should actually draw the outcome rather than round it up (not least to remove the numerical errors), e.g. as ex$clase <- rbinom(nrow(ex), prob=1/(1+exp(-lineal)), size=1). $\endgroup$ Commented Apr 12, 2016 at 2:59
  • $\begingroup$ Second, you would not be surprised to fail to recover the original model and coefficients if you had defined your data as ex<-data.frame(b=c(rep('A',5),rep('B',3), rep('C',2)), fl=round(runif(10,0,1),0),r=runif(10,0,0.5)). Maybe ask yourself why that is. $\endgroup$ Commented Apr 12, 2016 at 3:01

1 Answer 1


You have three problems, you are not randomizing when creating the dependent variable, and you are using the r variable in the GLM stepwise algorithm (and not when creating the dependent variable). In order to randomize, you need to create a variable that will be 1,0 with probability = ex$clase (according to your terminology). You are also running a model with intercept, and there is none defined in the simulation.

If you run this, you will see that the pvalues are almost 0. Some parameter combinations will yield a flatter likelihood, which will end up showing in greater pvalues ( with these ones you will get almost 0 for all of them) - since the stderrors will correspond to the inverse of the information matrix (the curvature of the likelihood)

ex <- data.frame(b=c(rep('A',50),rep('B',30), rep('C',20)), 
fl = round(runif(1000,0,1),0),r=runif(1000,0,1))
ex2 <- cbind(ex,model.matrix(~b-1,ex))
ex2$prob <- 1/(1+exp(-2*ex2$fl + ex2$bA + ex2$bB + ex2$bC))

ex2$clase <- ifelse((ex2$r>ex2$prob),1,0)

fm <- as.formula(clase~fl-1 + bA + bB + bC )
modT <- glm(clase~fl-1 + bA + bB + bC , family=binomial, data = ex2)
modT <- stepAIC(modT, scope = fm , family=binomial, data =ex, k = 4)
  • $\begingroup$ +1 for pointing out the absence of a random draw in the dgp. On the other hand the absence of an intercept in the dgp seems unproblematic; it should just be estimated to be zero... $\endgroup$ Commented Apr 12, 2016 at 3:03
  • $\begingroup$ yep, I solved it with sapply(ex$prob, function (p){rbinom(1,1,p)}) $\endgroup$
    – GabyLP
    Commented Apr 12, 2016 at 3:06

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