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 confidence level \(c = 0.90\), the significance level \(\alpha=1 - c=1 - 0.90 = 0.10\).
The critical values are \(\chi_{1-\frac{\alpha}{2}}^{2}\) and \(\chi_{\frac{\alpha}{2}}^{2}\).
\(\frac{\alpha}{2}=0.05\) and \(1-\frac{\alpha}{2}=0.95\).
Using a \(\chi^{2}\) - distribution table or technology, \(\chi_{0.95,16}^{2}=7.962\) and \(\chi_{0.05,16}^{2}=26.296\)

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

The formula for the confidence interval for the population variance is \(\frac{(n - 1)s^{2}}{\chi_{\frac{\alpha}{2}}^{2}}\leq\sigma^{2}\leq\frac{(n - 1)s^{2}}{\chi_{1-\frac{\alpha}{2}}^{2}}\)
We know that \(n = 17\), \(s = 37\), so \(s^{2}=37^{2}=1369\)
Substitute the values into the formula:
\(\frac{(17 - 1)\times1369}{26.296}\leq\sigma^{2}\leq\frac{(17 - 1)\times1369}{7.962}\)
First, calculate \(\frac{16\times1369}{26.296}=\frac{21904}{26.296}\approx833.05\)
Second, calculate \(\frac{16\times1369}{7.962}=\frac{21904}{7.962}\approx2751.07\)

Answer:

The confidence interval for the population variance is \((833.05,2751.07)\)