But that just gives me the prediction of Y for treatment A or B, not the ratio.
As I stated:
# I am interested in the relationship:
# Y(treatment =="B") / Y(treatment=="A") as a function of X, with a
confidence interval!
I can get the SE for either of them using predict.gam without a
problem, but I don't know how to get the CI for the ratio!
thanks,
remko
-------------------------------------------------
Remko Duursma
Research Lecturer
Hawkesbury Institute for the Environment
University of Western Sydney
Hawkesbury Campus, Richmond
Mobile: +61 (0)422 096908
www.remkoduursma.com
On Tue, Jun 28, 2011 at 2:09 PM, David Winsemius <dwinsem...@comcast.net> wrote:
>
> On Jun 27, 2011, at 10:45 PM, Remko Duursma wrote:
>
>> Dear R-helpers,
>>
>> I am trying to construct a confidence interval on a prediction of a
>> gam fit. I have the Wood (2006) book, and section 5.2.7 seems relevant
>> but I am not able to apply that to this, different, problem.
>>
>> Any help is appreciated!
>>
>> Basically I have a function Y = f(X) for two different treatments A
>> and B. I am interested in the treatment ratios : Y(treatment = B) /
>> Y(treatment = A) as a function of X, including a confidence interval
>> for this treatment ratio (because we are testing this ratio against
>> some value, across the range of X).
>>
>> The X values that Y is measured at differs between the treatments, but
>> the ranges are similar.
>>
>>
>> # Reproducible problem:
>> X1 <- runif(20, 0.5, 4)
>> X2 <- runif(20, 0.5, 4)
>>
>> Y1 <- 20*exp(-0.5*X1) + rnorm(20)
>> Y2 <- 30*exp(-0.5*X2) + rnorm(20)
>>
>> # Look at data:
>> plot(X1, Y1, pch=19, col="blue", ylim=c(0,max(Y1,Y2)), xlim=c(0,5))
>> points(X2, Y2, pch=19, col="red")
>>
>> # Full dataset
>> dfr <- data.frame(X=c(X1,X2), Y=c(Y1,Y2),
>> treatment=c(rep("A",20),rep("B",20)))
>>
>> # Fit gam
>> # I use a gamma family here although it is not necessary: in the real
>> problem it is, though.
>> gfit <- gam(Y ~ treatment + s(X), data=dfr, family=Gamma(link=log))
>>
>> # I am interested in the relationship:
>> # Y(treatment =="B") / Y(treatment=="A") as a function of X,
>
> Can't you use predict.gam?
>
> plot(predict(gfit, newdata=data.frame(X=rep(seq(0.4, 4, by=0.1), 2),
> treatment=c(rep("A",37),rep("B",37) ) ) )[1:37] )
> lines(predict(gfit3, newdata=data.frame(X=rep(seq(0.4, 4, by=0.1), 2),
> treatment=c(rep("A",37),rep("B",37) ) ) )[-(1:37)])
>
>> with a confidence interval!
>
> There is an se.fit argument to predict.gam().
>
>
> --
>
> David Winsemius, MD
> West Hartford, CT
>
>
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