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.


\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 \(\sigma\).