Confidence Intervals for One Variance

For a random sample $X_1, X_2, \ldots, X_n$ from a normal population with variance $\sigma^2,$ the standardized sample variance follows a chi-square distribution:

$$\frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{n-1}.$$

This is the pivot quantity we use to build a confidence interval for $\sigma^2.$

Let $\chi^2_{p,\,\nu}$ denote the upper-$p$ critical value of the chi-square distribution with $\nu$ degrees of freedom; that is, $P(\chi^2_\nu > \chi^2_{p,\,\nu}) = p.$

A $(1-\alpha)100\%$ confidence interval for $\sigma^2$ is

$$\left(\dfrac{(n-1)S^2}{\chi^2_{\alpha/2,\,n-1}},\;\dfrac{(n-1)S^2}{\chi^2_{1-\alpha/2,\,n-1}}\right).$$

Because the chi-square distribution is not symmetric, the two critical values differ in magnitude. The larger critical value sits in the denominator of the lower endpoint, and the smaller in the denominator of the upper endpoint.

For example, suppose a sample of size $n=8$ from a normal population gives $s^2 = 12.$ To find a 95% CI for $\sigma^2,$ we use $\alpha = 0.05$ and $\nu = n - 1 = 7{:}$

• $\chi^2_{0.025,\,7} = 16.013$ and $\chi^2_{0.975,\,7} = 1.690$
• $(n-1)s^2 = 7 \cdot 12 = 84$
• CI: $\left(\dfrac{84}{16.013},\;\dfrac{84}{1.690}\right) = (5.246,\;49.704)$

We are 95% confident that $\sigma^2$ lies between $5.246$ and $49.704.$