[R] 95% confidence intercal with glm
Ben Bolker
bbolker at gmail.com
Wed Sep 29 16:27:40 CEST 2010
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 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.
>
>
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