Question

use technology to construct the confidence intervals for the population variance \\( \sigma^{2} \\) and the population standard deviation \\( \sigma \\). assume the sample is taken from a normally distributed population.
\\( c = 0.99, s^{2} = 7.29, n = 29 \\)
the confidence interval for the population variance is ().
(round to two decimal places as needed.)

Explanation:

Step1: Find the degrees of freedom

The degrees of freedom \( df=n - 1\). Given \(n = 29\), so \(df=29-1 = 28\).

Step2: Find the critical values

For a confidence level \(c = 0.99\), the significance level \(\alpha=1 - c=1 - 0.99 = 0.01\). The critical values are \(\chi_{L}^{2}\) and \(\chi_{R}^{2}\). Using a chi - square distribution table or technology, \(\chi_{L}^{2}=\chi_{1-\frac{\alpha}{2},df}^{2}=\chi_{0.995,28}^{2}=11.524\) and \(\chi_{R}^{2}=\chi_{\frac{\alpha}{2},df}^{2}=\chi_{0.005,28}^{2}=55.676\).

Step3: Calculate the confidence interval for the population variance \(\sigma^{2}\)

The formula for the confidence interval for the population variance is \(\frac{(n - 1)s^{2}}{\chi_{R}^{2}}\leq\sigma^{2}\leq\frac{(n - 1)s^{2}}{\chi_{L}^{2}}\).
Substitute \(n = 29\), \(s^{2}=7.29\), \(\chi_{L}^{2}=11.524\), and \(\chi_{R}^{2}=55.676\) into the formula:
\(\frac{(29 - 1)\times7.29}{55.676}\leq\sigma^{2}\leq\frac{(29 - 1)\times7.29}{11.524}\)
First, calculate \(\frac{28\times7.29}{55.676}=\frac{204.12}{55.676}\approx3.67\)
Second, calculate \(\frac{28\times7.29}{11.524}=\frac{204.12}{11.524}\approx17.71\)

Answer:

The confidence interval for the population variance is \((3.67,17.71)\)