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.
$c = 0.90$, $s = 37$, $n = 17$
the confidence interval for the population variance is ().
(round to two decimal places as needed.)

Explanation:

Step1: Determine the degrees of freedom and critical values

The degrees of freedom \(df=n - 1=17-1 = 16\).
For a \(90\%\) confidence interval (\(c = 0.90\)), the significance level \(\alpha=1 - c=1 - 0.90 = 0.10\).
\(\alpha/2=0.05\) and \(1-\alpha/2 = 0.95\).
Using a \(\chi^{2}\) - distribution table or technology, \(\chi_{L}^{2}=\chi_{1-\alpha/2,df}^{2}=\chi_{0.95,16}^{2}=7.962\) and \(\chi_{R}^{2}=\chi_{\alpha/2,df}^{2}=\chi_{0.05,16}^{2}=26.296\)

Step2: Calculate the confidence interval for the population variance \(\sigma^{2}\)

The formula for the confidence interval for the population variance is \(\frac{(n - 1)s^{2}}{\chi_{R}^{2}}\leq\sigma^{2}\leq\frac{(n - 1)s^{2}}{\chi_{L}^{2}}\)
We know that \(n = 17\), \(s = 37\), so \(s^{2}=37^{2}=1369\) and \(n - 1=16\)
Substitute the values:
\(\frac{16\times1369}{26.296}\leq\sigma^{2}\leq\frac{16\times1369}{7.962}\)
\(\frac{21904}{26.296}\leq\sigma^{2}\leq\frac{21904}{7.962}\)
\(833.05\leq\sigma^{2}\leq2751.07\)

Answer:

\((833.05,2751.07)\)