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: Determine the degrees of freedom

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

Step2: Find the critical values

For a \(99\%\) confidence interval (\(c = 0.99\)), the significance level \(\alpha=1 - c=1 - 0.99 = 0.01\). The critical values are \(\chi_{1-\frac{\alpha}{2},df}^2\) and \(\chi_{\frac{\alpha}{2},df}^2\).
\(\frac{\alpha}{2}=0.005\) and \(1-\frac{\alpha}{2}=0.995\).
Using a chi - square distribution table or technology, \(\chi_{0.995,28}^2 = 11.524\) and \(\chi_{0.005,28}^2=55.676\).

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

The formula for the confidence interval for the population variance is \(\frac{(n - 1)s^{2}}{\chi_{\frac{\alpha}{2},df}^2}<\sigma^{2}<\frac{(n - 1)s^{2}}{\chi_{1-\frac{\alpha}{2},df}^2}\).
Substitute \(n = 29\), \(s^{2}=7.29\), \(\chi_{0.005,28}^2 = 55.676\), and \(\chi_{0.995,28}^2=11.524\) into the formula.
\(\frac{(29 - 1)\times7.29}{55.676}<\sigma^{2}<\frac{(29 - 1)\times7.29}{11.524}\)
First, calculate \((n - 1)s^{2}=28\times7.29 = 204.12\)
Then, \(\frac{204.12}{55.676}\approx3.67\) and \(\frac{204.12}{11.524}\approx17.71\)

Answer:

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