# Confidence Interval Estimate of Variance

The variance of a population can be estimated using the chi-square variate, as explained in previous tutorials. Unlike the $t$ and $Z$ distributions, the value of the chi-square variate is defined only for positive values. At a level of significance $\alpha$ the ${t_{\alpha /2}}$ or ${Z_{\alpha /2}}$ are those values of the variate which give an area $\alpha /2$ in the right tail. Also $– {t_{\alpha /2}}$ or $– {Z_{\alpha /2}}$ are the values of the variate which give an area $\alpha /2$ in the left tail. Similarly $\chi _{\alpha /2}^2$ is the value of the variate which gives an area $\alpha /2$ in the right tail of the chi-square distribution (curve). However, the value of chi-square which gives an area $\alpha /2$ in the left tail is denoted by $\chi _{1 – \frac{\alpha }{2}}^2$, because chi-square cannot be negative.

As an example, at a level of significance of $\alpha = 0.1$ the $\chi _{0.95}^2$ and $\chi _{0.05}^2$ will contain 90 percent of the area. Similarly, $\chi _{0.975}^2$ and $\chi _{0.025}^2$ will contain 95 percent of the area. We can, therefore state as follows:
$P\left[ {\chi _{1 – \frac{\alpha }{2}}^2 < {\chi ^2} < \chi _{\frac{\alpha }{2}}^2} \right] = 1 – \alpha$

Replacing ${\chi ^2} = \frac{{\left( {n – 1} \right){S^2}}}{{{\sigma ^2}}}$ in the middle of the above statement, we get
$P\left[ {\chi _{1 – \frac{\alpha }{2}}^2 < \frac{{\left( {n – 1} \right){S^2}}}{{{\sigma ^2}}} < \chi _{\frac{\alpha }{2}}^2} \right] = 1 – \alpha$

Dividing both sides of the inequality by $\left( {n – 1} \right){S^2}$, we have
$P\left[ {\frac{{\chi _{1 – \frac{\alpha }{2}}^2}}{{\left( {n – 1} \right){S^2}}} < \frac{1}{{{\sigma ^2}}} < \frac{{\chi _{\frac{\alpha }{2}}^2}}{{\left( {n – 1} \right){S^2}}}} \right] = 1 – \alpha$ Now, invert the whole inequality, the inequality signs would be reversed, i.e., $P\left[ {\frac{{\left( {n – 1} \right){S^2}}}{{\chi _{1 – \frac{\alpha }{2}}^2}} > {\sigma ^2} > \frac{{\left( {n – 1} \right){S^2}}}{{\chi _{\frac{\alpha }{2}}^2}}} \right] = 1 – \alpha$

Alternatively, we can also write
$P\left[ {\frac{{\left( {n – 1} \right){S^2}}}{{\chi _{\frac{\alpha }{2}}^2}} < {\sigma ^2} < \frac{{\left( {n – 1} \right){S^2}}}{{\chi _{1 – \frac{\alpha }{2}}^2}}} \right] = 1 – \alpha$

Thus, the $100\left( {1 – \alpha } \right)$ percent lower and upper confidence limits of the population variances are
$\frac{{\left( {n – 1} \right){S^2}}}{{\chi _{\frac{\alpha }{2}}^2}}$ and $\frac{{\left( {n – 1} \right){S^2}}}{{\chi _{1 – \frac{\alpha }{2}}^2}}$

Here $\left( {n – 1} \right)$ are the degrees of freedom and the values of $\chi _{\frac{\alpha }{2}}^2$ and $\chi _{1 – \frac{\alpha }{2}}^2$ are obtainable from the chi-square table against $\left( {n – 1} \right)$ degrees of freedom and the appropriate level of significance. Also, ${S^2}$ is the sample variance given by the formula i.e., ${S^2} = \frac{{\sum {{\left( {{X_i} – \overline X } \right)}^2}}}{{n – 1}}$.

A short cut form of the same formula may be stated as ${S^2} = \frac{{\sum {X^2} – n{{\overline X }^2}}}{{n – 1}}$

Example: A random sample of 9 individuals measured 62, 63, 65, 61, 65, 64, 66, 67 and 63 inches in height. Construct a 95 percent confidence interval estimate for the population variance.

Solution: Given that
$\begin{gathered} n = 9,\,\,\,\,\,\,\,\sum X = 576,\,\,\,\,\,\,\,\overline X = 64 \\ \sum {X^2} = 36894,\,\,\,\,\,\,\,\alpha = 0.05 \\ \end{gathered}$

Using the short cut formula for ${S^2}$, we have
${S^2} = \frac{{\sum {X^2} – n{{\overline X }^2}}}{{n – 1}} = 3.75$

Also, consulting the chi-square table against 8 degrees of freedom,
$\begin{gathered} \chi _{0.025}^2 = 17.535 \\ \chi _{0.975}^2 = 2.180\\ \end{gathered}$

The lower limit of the interval $= \frac{{\left( {n – 1} \right){S^2}}}{{\chi _{0.025}^2}} = \frac{{8\left( {3.75} \right)}}{{17.535}} = 1.7108$

The upper limit of the interval $= \frac{{\left( {n – 1} \right){S^2}}}{{\chi _{0.975}^2}} = \frac{{8\left( {3.75} \right)}}{{2.180}} = 13.7615$

Thus, the required interval estimate is
$P\left( {1.71 < {\sigma ^2} < 13.76} \right) = 0.95$