# 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

$$P\left( {1.71 < {\sigma ^2} < 13.76} \right) = 0.95$$