Dear Alexandra,
Perhaps you can use a different approach but I think it works:
# Chi-suared CI
var.ci=function(x,alpha=0.05){
n=length(x)-sum(is.na(x))
s2=var(x,na.rm=T)
li=s2*(n-1)/qchisq(1-alpha/2,n-1)
ls=s2*(n-1)/qchisq(alpha/2,n-1)
c(li,ls)
}
# Example
set.seed(123)
y=rnorm(1000,25,2)
var.ci(y) # 3.610426 4.302965
# Bootstrap CI
var.bs=function(x,FUN=var,alpha=0.05,NSIM=1000){
temp=matrix(rep(x,NSIM),ncol=NSIM)
temp2=apply(temp,2,sample,replace=TRUE)
vars=apply(temp2,2,var)
quantile(vars,probs=c(alpha/2,1-alpha/2))
}
var.bs(y) # 3.618220 4.274773
I hope this helps,
--
JIV
On Mon, Apr 7, 2008 at 10:13 AM, Alexandra Ramos <aramos@fep.up.pt> wrote:
> Hi,
>
>
>
> Does anyone know an instruction to perform a test (or to construct a
> confidence interval) for the variance of a normal population
(distribution).
>
> The test statistic is (n-1)*S´^2/sigma^2 and has chi square (n-1)
> distribution.
>
>
>
> Thank you,
>
> Alexandra
>
>
>
>
>
>
> [[alternative HTML version deleted]]
>
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
[[alternative HTML version deleted]]