# Confidence Intervals For The Variance and Standard Deviation of a Normal Distribution

Posted by Beetle B. on Sun 16 July 2017

$$\newcommand{\Cov}{\mathrm{Cov}}$$ $$\newcommand{\Corr}{\mathrm{Corr}}$$ $$\newcommand{\Sample}{X_{1},\dots,X_{n}}$$

Let $$\Sample$$ be a random sample from a normal distribution. Then the r.v.

\begin{equation*} \frac{(n-1)s^{2}}{\sigma^{2}}=\frac{\sum\left(x_{i}-\bar{x}\right)^{2}}{\sigma^{2}} \end{equation*}

has a chi-squared distribution with $$n-1$$ degrees of freedom.

Denote $$\chi^{2}_{\alpha,\nu}$$ for the chi-squared critical value.

Do note that the chi-squared distribution is not symmetric. Also note that while the random variable depends on $$\sigma^{2}$$, the distribution does not.

So:

\begin{equation*} P\left(\chi^{2}_{1-\alpha/2,n-1}<\frac{(n-1)S^{2}}{\sigma^{2}}<\chi^{2}_{\alpha/2,n-1}\right)=1-\alpha \end{equation*}

Thus the $$100(1-\alpha)$$ CI for $$\sigma^{2}$$ is:

\begin{equation*} \frac{(n-1)S^{2}}{\chi^{2}_{\alpha/2,n-1}}<\sigma^{2}<\frac{(n-1)S^{2}}{\chi^{2}_{1-\alpha/2,n-1}} \end{equation*}

Take the square root to get the CI for $$\sigma$$.