I have the model
am.glm = glm(formula=am ~ hp + I(mpg^2), data=mtcars, family=binomial)
which gives
> summary(am.glm)
Call:
glm(formula = am ~ hp + I(mpg^2), family = binomial, data = mtcars)
Deviance Residuals:
Min 1Q Median 3Q Max
-1.5871 -0.5376 -0.1128 0.1101 1.6937
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -18.71428 8.45330 -2.214 0.0268 *
hp 0.04689 0.02367 1.981 0.0476 *
I(mpg^2) 0.02811 0.01273 2.207 0.0273 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 43.230 on 31 degrees of freedom
Residual deviance: 20.385 on 29 degrees of freedom
AIC: 26.385
Number of Fisher Scoring iterations: 7
Given a value of hp
I would like to find the values of mpg
that would lead to a 50% probability of am
.
I haven't managed to find anything that can be used to output such predictions. I have managed to code something using
#Coefficients
glm.intercept<-as.numeric(coef(am.glm)[1])
glm.hp.beta<-as.numeric(coef(am.glm)[2])
glm.mpg.sq.beta<-as.numeric(coef(am.glm)[3])
glm.hp.mpg.beta<-as.numeric(coef(am.glm)[4])
#Constants
prob=0.9
c<-log(prob/(1-prob))
hp=120
polyroot(c((glm.hp.beta*hp)+glm.intercept-c, glm.hp.mpg.beta*hp,glm.mpg.sq.beta))
Is there a more elegant solution? Perhaps a predict
function equivalent?