# Confidence of a variance estimate [duplicate]

Possible Duplicate:
Calculating required sample size, precision of variance estimate?

I would like to present some measure of how variable a particular phenomenon is. This phenomenon appears to be normally distributed, so a reasonable measure of variability might be variance. I've seen people report variances before, but I've always been bothered by how no margin of error is ever reported.

First half-question: Why do I never see people report margins of errors of variances? Or can you give me examples of where someone does? Or maybe there's another measure of variability that I should use?

Although I've never seen someone report a confidence measure of a variance, I feel like it would be incorrect not to report the standard error of the variability measure. So I considered how one might do that.

Cochran's theorem tells us that sample variance follows a $X^2$ distribution with $n-1$ degrees of freedom. Call the sample variance $s^2$.

$\frac{ns^2}{\sigma^2}\sim\chi^2_{n-1}$

Estimate the variance. Call that $\hat\sigma$.

$s^2\sim\frac{\hat\sigma^2}{n}\chi^2_{n-1}$

Second half-question: How do I get a confidence interval of a variance? Do I just compute the cdf of $\frac{\hat\sigma^2}{n}\chi^2_{n-1}$ within some bounds? And since $X^2$ is not symmetrical, is there a standard way of getting a two-sided confidence interval?

In case it's helpful, I post a formula for the standard error of a variance that I found in my code, although I don't think it's at all useful.

$Var(s^2)=\frac{2\sigma^4}{n-1}$

Approximate $\sigma$ as $\hat\sigma$, and we get

$\widehat{Var}(s^2)=\frac{2\hat\sigma^4}{n-1}$

$\widehat{SE}(s^2)=\sqrt{\frac{2\hat\sigma^4}{n-1}}$

• Sep 18, 2011 at 12:36
• I think my second half-question is the same as that one. Maybe someone could close this question as being a duplicate? Oct 10, 2011 at 16:50