Guazzetti Stefano
2010-Aug-07 05:25 UTC
[R] R: Confidence Intervals for logistic regression
a closer look to the help on predict.glm will reveal that the function accepts a 'type' argument. In you case 'type = response' will give you the results in probabilities (that it seems to be what you are looking for). There also is an example on use of the 'type' argument at the end of the page. Stefano -----Messaggio originale----- Da: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]Per conto di Troy S Inviato: Friday, August 06, 2010 6:31 PM A: Michael Bedward Cc: r-help at r-project.org Oggetto: Re: [R] Confidence Intervals for logistic regression Michael, Thanks for the reply. I believe Aline was sgiving me CI's on coefficients as well. So c(pred$fit + 1.96 * pred$se.fit, pred$fit - 1.96 * pred$se.fit) gives me the CI on the logits if I understand correctly? Maybe the help on predict.glm can be updated. Thanks! On 6 August 2010 01:46, Michael Bedward <michael.bedward at gmail.com> wrote:> Sorry about earlier reply - didn't read your email properly (obviously :) > > You're suggestion was right, so as well as method for Aline below, > another way of doing the same thing is: > > pred <- predict(y.glm, newdata= something, se.fit=TRUE) > ci <- matrix( c(pred$fit + 1.96 * pred$se.fit, pred$fit - 1.96 * > pred$se.fit), ncol=2 ) > > lines( something, plogis( ci[,1] ) ) > lines( something, plogis( ci[,2] ) ) > > > > On 6 August 2010 18:39, aline uwimana <rwanuza at gmail.com> wrote: > > 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-twigs at 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 at 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 at 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 at 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. Rispetta l'ambiente: Se non ti ? necessario, non stampare questa mail. "Le informazioni contenute nel presente messaggio di posta elettronica e in ogni suo allegato sono da considerarsi riservate e il destinatario della email ? l'unico autorizzato ad usarle, copiarle e, sotto la propria responsabilit?, divulgarle. Chiunque riceva questo messaggio per errore senza esserne il destinatario deve immediatamente rinviarlo al mittente cancellando l'originale. Eventuali dati personali e sensibili contenuti nel presente messaggio e/o suoi allegati vanno trattati nel rispetto della normativa in materia di privacy ( DLGS n.196/'03)".
Stefano, I was aware of this option. I was assuming it was not ok to do fit +/- 1.96 se when you requested probabilities. If this is legitimate then all the better. Thanks! Troy On 6 August 2010 22:25, Guazzetti Stefano <Stefano.Guazzetti@ausl.re.it>wrote:> a closer look to the help on predict.glm will reveal that the function > accepts a 'type' argument. > In you case 'type = response' will give you the results in probabilities > (that it seems to be what you are looking for). > There also is an example on use of the 'type' argument at the end of the > page. > > Stefano > > -----Messaggio originale----- > Da: r-help-bounces@r-project.org > [mailto:r-help-bounces@r-project.org]Per conto di Troy S > Inviato: Friday, August 06, 2010 6:31 PM > A: Michael Bedward > Cc: r-help@r-project.org > Oggetto: Re: [R] Confidence Intervals for logistic regression > > > Michael, > > Thanks for the reply. I believe Aline was sgiving me CI's on coefficients > as well. > > So c(pred$fit + 1.96 * pred$se.fit, pred$fit - 1.96 * > pred$se.fit) gives me the CI on the logits if I understand correctly? > Maybe > the help on predict.glm can be updated. > > Thanks! > > On 6 August 2010 01:46, Michael Bedward <michael.bedward@gmail.com> wrote: > > > Sorry about earlier reply - didn't read your email properly (obviously :) > > > > You're suggestion was right, so as well as method for Aline below, > > another way of doing the same thing is: > > > > pred <- predict(y.glm, newdata= something, se.fit=TRUE) > > ci <- matrix( c(pred$fit + 1.96 * pred$se.fit, pred$fit - 1.96 * > > pred$se.fit), ncol=2 ) > > > > lines( something, plogis( ci[,1] ) ) > > lines( something, plogis( ci[,2] ) ) > > > > > > > > On 6 August 2010 18:39, aline uwimana <rwanuza@gmail.com> wrote: > > > Dear Troy, > > > use this commend, your will get IC95% and OR. > > > > > > logistic.model <- glm(formula =y~ x1+x2, family = binomial) > > > summary(logistic.model) > > > > > > sum.coef<-summary(logistic.model)$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-twigs@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. > > > > > > > [[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. > > Rispetta l'ambiente: Se non ti è necessario, non stampare questa mail. > > > "Le informazioni contenute nel presente messaggio di posta elettronica e in > ogni suo allegato sono da considerarsi riservate e il destinatario della > email è l'unico autorizzato > ad usarle, copiarle e, sotto la propria responsabilità, divulgarle. > Chiunque riceva questo messaggio per errore senza esserne il destinatario > deve immediatamente rinviarlo > al mittente cancellando l'originale. Eventuali dati personali e sensibili > contenuti nel presente messaggio e/o suoi allegati vanno trattati nel > rispetto della normativa > in materia di privacy ( DLGS n.196/'03)". > >[[alternative HTML version deleted]]
> I was aware of this option.? I was assuming it was not ok to do fit +/- 1.96 > se when you requested probabilities.? If this is legitimate then all the > better.I don't think it is. I understood that you should do the calculation in the scale of the linear predictor and then transform to probabilities. Happy to be corrected if that's wrong. Michael