From: Christos Hatzis <christos_at_silicoinsights.com>

Date: Wed 03 May 2006 - 05:24:47 EST

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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Wed May 03 06:38:47 2006

Date: Wed 03 May 2006 - 05:24:47 EST

I think you got it right.

The mean of the (weighted) sum of a set of random variables is the (weighted) sum of the means and its variance is the (weighted) sum of the individual variances (using squared weights). Here you don't have to worry about weights.

So what you proposed does exactly this.

-Christos

-----Original Message-----

From: r-help-bounces@stat.math.ethz.ch

[mailto:r-help-bounces@stat.math.ethz.ch] On Behalf Of Bill Szkotnicki
Sent: Tuesday, May 02, 2006 2:59 PM

To: 'R-Help help'

Subject: [R] predict.lm

I have a model with a few correlated explanatory variables.
i.e.

> m1=lm(y~x1+x2+x3+x4,protdata)

and I have used predict as follows:

> x=data.frame(x=1:36)

*> yp=predict(m1,x,se.fit=T)
**> tprot=sum(yp$fit) # add up the predictions tprot
*

tprot is the sum of the 36 predicted values and I would like the se of that
prediction.

I think

> sqrt(sum(yp$se.fit^2))

is not correct.

Would anyone know the correct approach?

i.e. How to get the se of a function of predicted values (in this case sum)

Thanks, Bill

R-help@stat.math.ethz.ch mailing list

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R-help@stat.math.ethz.ch mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Wed May 03 06:38:47 2006

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