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Essentials of Mathematical Probability and Statistics

John Travis

Section 10.6 Interval Estimates - Confidence Interval for \(\sigma^2\)

Once again, you may need to approximate the population variance or standard deviation but only have the sample values available. One difference from the previous sections is that you are not dealing with an average of values (such as \(\overline{x}\) or \(\tilde{p}\)) but with the average of the squares of values. The Central Limit Theorem does not directly help you in this case but the following result (presented without proof) provides a solution.
To create a confidence interval for \(\sigma^2\) first consider an interval of the form
\begin{equation*} E_1 \lt \sigma^2 \lt E_2 \end{equation*}
and determine values for the boundaries so that the likelihood of this being true is high. For this case, since the chi-square distribution only has a positive domain and is not symmetrical, you will not expect to determine a symmetrical confidence interval. Therefore, consider
\begin{equation*} P (E_1 \lt \sigma^2 \lt E_2 ) = 1 - \alpha \end{equation*}
and by playing around with algebra you get
\begin{equation*} P \left ( \frac{E_1}{(n-1)S^2} \lt \frac{\sigma^2}{(n-1)S^2} \lt \frac{E_2}{(n-1)S^2} \right ) = 1 - \alpha \end{equation*}
or by inverting the inequality yields
\begin{equation*} P \left ( \frac{(n-1)S^2}{E_2} \lt \frac{(n-1)S^2}{\sigma^2} \lt \frac{(n-1)S^2}{E_1} \right ) = 1 - \alpha . \end{equation*}
Using the previous theorem, note that the inside variable can be replaced with a chi-square variable. If F is the distribution function for chi-square, then you get
\begin{equation*} F \left ( \frac{(n-1)S^2}{E_1} \right ) - F \left ( \frac{(n-1)S^2}{E_2} \right ) = 1 - \alpha . \end{equation*}
For a given value of \(\alpha\) there are many possible choices but often one often utilized is one in which
\begin{equation*} F(\chi^2_{1-\alpha/2} ) = F \left ( \frac{(n-1)S^2}{E_1} \right ) = 1 - \alpha / 2 \end{equation*}
and
\begin{equation*} F(\chi^2_{\alpha/2} ) = F \left ( \frac{(n-1)S^2}{E_2} \right ) = \alpha / 2. \end{equation*}
Using the inverse chi-square gives values for the expression on the inside and algebra can be used to solve for each of \(E_1, E_2\text{.}\) Indeed,
\begin{equation*} E_1 = \frac{(n-1)S^2}{\chi^2_{1-\alpha/2}} \end{equation*}
and
\begin{equation*} E_2 = \frac{(n-1)S^2}{\chi^2_{\alpha/2}} \end{equation*}
To determine appropriate values for \(\chi^2_{\alpha/2} \) and \(\chi^2_{1-\alpha/2} \) with equal probabilities in each tail, consider using the interactive cell below:
Given the data 570, 561, 546, 540, 609, 580, 550, 577, 585, determine a 95% confidence interval for \(\sigma^2\text{.}\)
Using the computational forumaula (or your calculator) gives \(s^2 \approx 479.5\text{.}\) Also, notice for n=9, the resulting interval will use a Chi-square variable with 8 degrees of freedom. Using the symmetric option, gives \(\chi_{0.025}^2 = 2.18\) and \(\chi_{0.975}^2 = 17.53\text{.}\) Therefore
\begin{equation*} E_1 = \frac{8 \cdot 479.5}{17.53} \approx 221.095 \end{equation*}
and
\begin{equation*} E_2 = \frac{8 \cdot 479.5}{2.18} \approx 1759.63. \end{equation*}
Hence, you are 95% certain that
\begin{equation*} 221.095 \lt \sigma^2 \lt 1759.63. \end{equation*}
By taking square roots you get
\begin{equation*} 14.87 \lt \sigma \lt 41.95. \end{equation*}
Notice, this interval is relatively wide which is a result both of the number of data values being relatively small (n=9) and the actual data values being relatively large and spread out.
Find the critical values \(\chi_L^2 = \chi_{1-\alpha/2}^2\) and \(\chi_R^2 = \chi_{\alpha/2}^2\) that correspond to \(99\)% degree of confidence and the sample size \(n = 22.\)
\(\chi_L^2 =\) \(\ \ \ \ \) \(\chi_R^2 =\)
Answer 1.
\(8.03365\)
Answer 2.
\(41.4011\)
Suppose that you have n=400 data values and suppose you have computed from those a sample variance of \(s^2 = 479.5\text{.}\) Then, the only change in the calculation is the two chi-square statistic values. For 95% but now with 399 degrees of freedom \(\chi_{0.025}^2 = 345.55\) and \(\chi_{0.975}^2 = 456.24\text{.}\)
Therefore
\begin{equation*} E_1 = \frac{8 \cdot 479.5}{456.24} \approx 419.3 \end{equation*}
and
\begin{equation*} E_2 = \frac{8 \cdot 479.5}{345.55} \approx 553.7. \end{equation*}
Hence, you are 95% certain that
\begin{equation*} 419.24 \lt \sigma^2 \lt 553.7. \end{equation*}
By taking square roots you get
\begin{equation*} 20.48 \lt \sigma \lt 23.53 \end{equation*}
which is a relatively tight confidence interval. Notice, these are also completely contained in the confidence intervals from the previous small n example.
Similar to above, another choice to estimate \(\sigma ^2\) is to use a one sided confidence interval. If you want to find one of these, continue as described above but just leave one endpoint off. Indeed,
\begin{equation*} \sigma^2 \lt E_2 \end{equation*}
can be determined using
\begin{equation*} F(\chi^2_{\alpha} ) = F \left ( \frac{(n-1)S^2}{E_2} \right ) = \alpha \end{equation*}
and
\begin{equation*} E_1 \lt \sigma^2 \end{equation*}
can be determined using
\begin{equation*} F(\chi^2_{1-\alpha} ) = F \left ( \frac{(n-1)S^2}{E_1} \right ) = 1 - \alpha. \end{equation*}
TBA
You can, of course, use the formula to work out the sample size needed in order to have a sufficiently narrow confidence interval
Find the minimum sample size needed to be \(99\)% confident that the sample variance is within \(10\)% of the population variance.
Answer.
\(1401\)
Finally, to determine a confidence interval for \(\sigma\text{,}\) proceed using the protocols described above and simply take the square root on the resulting interval.
TBA
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