Dear Troy, use this commend, your will get IC95% and OR. logistic.model <- glm(formula =y~ x1+x2, family = binomial) summary(logistic.model)
sum.coef<-summary(logistic.model)$coef est<-exp(sum.coef[,1]) upper.ci<-exp(sum.coef[,1]+1.96*sum.coef[,2]) lower.ci<-exp(sum.coef[,1]-1.96*sum.coef[,2]) cbind(est,upper.ci,lower.ci) regards. 2010/8/6 Troy S <troysocks-tw...@yahoo.com> > Dear UseRs, > > I have fitted a logistic regression using glm and want a 95% confidence > interval on a response probability. Can I use > > predict(model, newdata, se.fit=T) > > Will fit +/- 1.96se give me a 95% of the logit? And then > exp(fit +/- 1.96se) / (exp(fit +/- 1.96se) +1) to get the probabilities? > > Troy > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > [[alternative HTML version deleted]] ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
