On 10-09-29 10:04 AM, Sam wrote: > Hi Ben and list, > > Sorry to be a pain! I have followed your code, and modified it - >
**You should not use type="response" here.** The point is that the (symmetric) confidence intervals are computed on the link/linear predictor scale, and then inverse-link-transformed (in this case, along with the fitted values) -- they then become asymmetric. >> pp <- predict(model1,se.fit=TRUE, type = "response") >>> etaframe <- >> + with(pp,cbind(fit,lower=fit-1.96*se.fit,upper=fit+1.96*se.fit)) >>> pframe <- plogis(etaframe) >>> pframe > > My response variable is 0 or 1, the probabilities are all high above my cut-off point of 0.445, i think this may have something to do with > >> you may need to multiply by the binomial N as >> appropriate. > > However how do i multiply if my response is 0 or 1? > if 'size=1' (i.e. Bernoulli trials) then you would be multiplying by 1, so don't bother. > > On 29 Sep 2010, at 13:44, Ben Bolker wrote: > > > ## from ?glm > counts <- c(18,17,15,20,10,20,25,13,12) > outcome <- gl(3,1,9) > treatment <- gl(3,3) > d.AD <- data.frame(treatment, outcome, counts) > glm.D93 <- glm(counts ~ outcome + treatment, family=poisson, > data=d.AD) > > ## predict on 'link' or 'linear predictor' scale, with SEs > pp <- predict(glm.D93,se.fit=TRUE) > etaframe <- > with(pp,cbind(fit,lower=fit-1.96*se.fit,upper=fit+1.96*se.fit)) > pframe <- exp(etaframe) ## inverse-link > ## picture > pframe <- as.data.frame(cbind(obs=d.AD$counts,pframe)) > library(plotrix) > plot(pframe$obs,ylim=c(5,30)) > with(pframe,plotCI(1:9,fit,li=lower,ui=upper,col=2,add=TRUE)) > > If you're using a binomial model you need 'plogis' rather than 'exp' > as your > inverse link, and you may need to multiply by the binomial N as > appropriate. > > On 10-09-29 06:07 AM, Sam wrote: > > I am looking to do the same but am a bit confused > > >> and apply the inverse link function for your model. > > > i understand up to this point and i understand what this means, > > however i am unsure why it needs to be done and how you do it - i.e > > i use family="binomial" is this wrong if i use this method to > > calculate 95% CI? > > > Thanks > > > Sam On 28 Sep 2010, at 14:50, Ben Bolker wrote: > > > zozio32 <remy.pascal <at> gmail.com> writes: > > >> > >> > >> Hi > >> > >> I had to use a glm instead of my basic lm on some data due to > >> unconstant variance. > >> > >> now, when I plot the model over the data, how can I easily get > >> the 95% confidence interval that sormally coming from: > >> > >>> yv <- predict(modelVar,list(aveLength=xv),int="c") > >>> matlines(xv,yv,lty=c(1,2,2)) > >> > >> There is no "interval" argument to pass to the predict function > >> when using a glm, so I was wondering if I had to use an other > >> function > >> > > > You need to use predict with se=TRUE; construct the confidence > > intervals by computing predicted values +- 1.96 times the standard > > error returned; and apply the inverse link function for your > > model. > > > If heteroscedasticity is your main problem, and not a specific > > (known) non-normal distribution, you might consider using the gls > > function from the nlme package with an appropriate 'weights' > > argument. > > > ______________________________________________ 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. > > ______________________________________________ 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. ______________________________________________ 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.
