[R] confidence interval of a average...

[Robert W. Baer, Ph.D.]

Sorry.  The last code line got destroyed by my emailer and should read:
> matlines(height,predict.lm(lm.vc,interval="p"),
+ lty=c(1,3,3),col=c('black','red','red'))


----- Original Message ----- 
From: "Robert W. Baer, Ph.D." <rbaer at atsu.edu>
To: <r-help at stat.math.ethz.ch>
Sent: Wednesday, November 24, 2004 4:56 PM
Subject: Re: [R] confidence interval of a average...


> It depends on whether you want to do 95% ocnfidence intervals on the
> predicition or the mean vital capacity.  Try the following and see if it
> gets you started:
> #Simulate data
> height=48:72
> vc=height*10+20*rnorm(72-48+1)
> # Do regression
> lm.vc=lm(vc~height)
>
> # Confidence interval on mean vc
> predict.lm(lm.vc,interval="confidence")
> #confidence interval on prediced vc
> predict.lm(lm.vc,interval="prediction")
>
> #plot everything
> plot(vc~height)
>
> matlines(height,predict.lm(lm.vc,interval="c"), lty=c(1,2,2),col='blue')
>
matlines(height,predict.lm(lm.vc,interval="p"),lty=c(1,3,3),col=c('black','r
> ed','red'))> Rob
> > ----------------------
> > Fom: "Duncan Harris" <dunc_harris at hotmail.com>
> > > I have a sample of lung capacities from a population measured against
> > > height.  I need to know the 95% CI of the lung capacity of a person of
> > > average height.
> > >
> > > I have fitted a regression line.
> > >
> > > How do I get a minimum and maximum values of the 95% CI?
> > >
> > > My thinking was that this has something to do with covariance, but
how?
> > >
> > > My other thinking was that I could derive the 0.975 (sqrt 0.95) CI for
> the
> > > height.  Then I could take the lower height 0.975 CI value and
calculate
> > > from that the lower 0.975 value from the lung capacity. And then do
the
> > same
> > > for the taller people.  That is bound to be wrong though.
> > >
> > > Dunc
> > >
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> >
>
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