R help - Oct 2005 - predictive interval in nlme

Suppose I have the following data:

y x id
44 0 104
48 58 104
48 55 204
47 105 204
41 275 206
18 67 209
.......

I fit the model
>fit=lme(y~x+I(x^2),random=~1|id)
Now I want to make a prediction plot:
>time=seq(0,300,len=100)
>plot(predict(fit,data.frame(x=time),level=0))
Very fine. It gives me the prediction curve based on
the model. My further request is to make a confidence
bands around the curve. I guess I can derive its
mathematical form analytically and implement it
myself. But I just hope some experts can point out one
simple way in R to avoid my redundant work.

Thanks!

This is a difficult problem.  After several false starts with 
RSiteSearch, I finally got 15 hits from 'RSiteSearch("predict lme with 
confidence")', including the following 3 which looked to me like they 
might help with your question:

http://finzi.psych.upenn.edu/R/Rhelp02a/archive/27860.html
This outlines the mathematics required to do this conditional on the 
variance components estimates, and includes a comment from Doug Bates 
saying, "The new version of the lme4 package has a method for the
"vcov"
generic. This method returns" the "Sig" matrix required to do
this
computation.

http://finzi.psych.upenn.edu/R/Rhelp02a/archive/42666.html
Suggests using estimable() from the gmodels part of the gregmisc package.

http://finzi.psych.upenn.edu/R/Rhelp02a/archive/44675.html
This includes a comment by Doug Bates saying, 'That's not currently 
implemented in lme. It's on the "To Do" list but it is not very
close to
the top.'

	  It also might be interesting and useful to check the accuracy of the 
conditioning using Monte Carlo.

	  Spencer Graves

Cunningham Kerry wrote:
> Suppose I have the following data:
> 
> y x id
> 44 0 104
> 48 58 104
> 48 55 204
> 47 105 204
> 41 275 206
> 18 67 209
> .......
> 
> I fit the model
> 
> 
>>fit=lme(y~x+I(x^2),random=~1|id)
> 
> 
> Now I want to make a prediction plot:
> 
> 
>>time=seq(0,300,len=100)
>>plot(predict(fit,data.frame(x=time),level=0))
> 
> 
> Very fine. It gives me the prediction curve based on
> the model. My further request is to make a confidence
> bands around the curve. I guess I can derive its
> mathematical form analytically and implement it
> myself. But I just hope some experts can point out one
> simple way in R to avoid my redundant work.
> 
> Thanks!
> 
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