[R] glm

Joris Meys jorismeys at gmail.com
Tue Jun 22 18:50:44 CEST 2010


On Tue, Jun 22, 2010 at 1:00 AM, Samuel Okoye <samuoko at yahoo.com> wrote:
> Hi,
>
> I have the following data
>
> data1 <- data.frame(count = c(0,1,1,2,4,5,13,16,14), weeks = 1:9,
>                     treat=c(rep("1mg",3),rep("5mg",3),rep("10mg",3)))
> and I am using
>
> library(splines)
>
> to fit
>
> glm.m <- glm(count~weeks)+as.factor(treat),family=poisson,data=data1)
>
> and I am interested in predicting the count variale for the weeks 10, 11 and
> 12 with treat 10mg and 15mg.

bad luck for you.

newdat <-data.frame(
    weeks=rep(10:12,each=2),
    treat=rep(c("5mg","10mg"),times=3)
    )

preds <- predict(glm.m,type="response",newdata=newdat,se.fit=T)
cbind(newdat,preds)

gives as expected :
Warning message:
In bs(weeks, degree = 3L, knots = numeric(0), Boundary.knots = c(1L,  :
  some 'x' values beyond boundary knots may cause ill-conditioned bases

  weeks treat       fit    se.fit residual.scale
1    10   5mg  5.934881  5.205426              1
2    10  10mg 12.041639  9.514347              1
3    11   5mg  4.345165  6.924663              1
4    11  10mg  8.816168 15.805171              1
5    12   5mg  2.781063  8.123436              1
6    12  10mg  5.642667 18.221007              1


Watch the standard errors on the predicted values. No, you shouldn't
predict outside your data space, especially when using splines. And
when interested in 15mg, well, you shouldn't put treatment as a factor
to start with.

Cheers
Joris

-- 
Joris Meys
Statistical consultant

Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control

tel : +32 9 264 59 87
Joris.Meys at Ugent.be
-------------------------------
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php



More information about the R-help mailing list