Confidence Interval Estimate of Variance

The variance of a population can be estimated using the chi-square variate explained in pervious tutorials. Unlike the t and Z distributions the value of chi-square variate are 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 \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 would 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, 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}}

Where \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