On Mar 19, 2012, at 22:32 , Dominic Comtois wrote:
> Thanks for your answer, much appreciated.
>
> This ain't trivial indeed. I worked my way through it, until I got a "non
> conformable arguments" error when trying to calculate the new standard error.
> Since I'm not following 100% what's happening, it's hard for me to figure out
> what I should do next.
>
> Here's an example with simulated data:
>
Had helped if you had at least got it working to the same level as you had from
the outset...
> x1 <- rbinom(100,1,.5)
> x2 <-
> factor(round(runif(100,1,5)),labels=c("cat1","cat2","cat3","cat4","cat5"))
> outcome <- rbinom(100,1,.2)
>
> model <- glm(outcome~x1+x2,family=binomial(logit))
>
Needs to be model1, I suppose
> newd <-
> data.frame(x1=factor(c(0,1)),x2=factor(c("cat1","cat2")),outcome=c(0,0))
Two problems here: x1 is not a factor in your model, and x2 has a different
level set
> M <- model.matrix(model1, data=newd)
There's a problem with my own suggestion: Apparently this construct drops
unused levels, even if the level set in newd is right. This seems to do the
trick for me:
M <- model.matrix(formula(model1), data=newd)
> V <- vcov(model1)
> contr <- c(-1,1) %*% M
> se <- contr %*% V %*% contr
Need t(contr) on the rightmost term
> OR.ci <- exp(pred2 - pred1 + qnorm(c(.025,.50,.975))*se)
This would be OK, had you actually computed pred1 and pred2. Might substitute
OR.ci <- exp(contr %*% coef(model1) + qnorm(c(.025,.50,.975))*se)
>
> Thanks for any additional hints,
>
> Dominic C.
>
>
> 2012/3/19 peter dalgaard <pda...@gmail.com>
>
> There's no trivial way since you need the covariance of pred2 and pred1 to
> calculate the variance of the difference.
>
> I think you can proceed somewhat like as follows (I can't be bothered to test
> it without a reproducible example to start from. You may need to throw in a
> few explicit t() and as.vector() here and there.)
>
> newd <- data.frame(age.cat=c(1,2),male=c(1,0),lowed=c(1,0))
> M <- model.matrix(model, data=newd)
> V <- vcov(model)
> contr <- c(-1,1) %*% M
> se <- contr %*% V %*% contr
>
> OR.ci <- exp(pred2 - pred1 + qnorm(c(.025,.50,.975))*se)
>
> (Sanity check: contr %*% coef(model) should be same as pred2 - pred1 )
>
> I'm not sure how general the model.matrix trick is. It works in cases like
>
> > mm <- glm(ff, data=trees)
> > model.matrix(mm, data=trees[1,])
> (Intercept) log(Height) log(Girth)
> 1 1 4.248495 2.116256
> attr(,"assign")
> [1] 0 1 2
>
> but I see that there are cases where a "data" argument may be ignored. If
> that is the case, then you may have to construct the "contr" vector by hand.
>
> --
> Peter Dalgaard, Professor
> Center for Statistics, Copenhagen Business School
> Solbjerg Plads 3, 2000 Frederiksberg, Denmark
> Phone: (+45)38153501
> Email: pd....@cbs.dk Priv: pda...@gmail.com
>
>
--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd....@cbs.dk Priv: pda...@gmail.com
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