I think there is an error in this website which calculates a confidence interval for Variance Please consider the following website:
 http://www.kean.edu/~fosborne/bstat/06evar.html

In this website, they calculate s to be about $0.391868$ but in the calculations
they use $0.391868$ for the value of $s^2$. Therefore, I feel they are wrong but I am not sure. Therefore, I am hoping somebody could either confirm that I am wrong or tell me where I went wrong.
Thanks
Bob
 A: Comment:  In R statistical software the computations are as follows.
 x = c(6.0, 6.4, 7.0, 5.8, 6.0, 5.8, 5.9, 6.7, 6.1, 6.5, 6.3, 5.8)
 n = length(x);  v = var(x);  s = sd(x);  n;  v;  s
 ## 12              # sample size
 ## 0.1535606       # sample variance
 ## 0.3918681       # sample standard deviation

 ci.var = (n-1)*v/qchisq(c(.975,.025), n-1);  ci.var
 ## 0.07706035 0.44268294       # 95% CI for pop variance
 ci.sd = sqrt(ci.var);  ci.sd
 ## 0.2775975 0.6653442         # 95% CI for pop SD

The computation is based on the fact that $(n-1)S^2/\sigma^2 \sim Chisq(n - 1).$ First, we find a CI for $\sigma^2$.
Then if a CI for $\sigma$ is required, take square roots of
the endpoints of the CI for $\sigma^2,$ as in the fine Answer by @Math1000. 
Unfortunately, it seems to be a frequent mistake in
textbooks and software manuals to confuse the sample
variance $S^2$ and the sample standard deviation $S,$
and it seems you have found an example. (There is no excuse for such mistakes, but I am sympathetic. I just
repaired this kind of error in one of my other posts on this site an
hour ago.)
