Question

the waiting times (in minutes) of a random sample of 20 people at a bank have a sample standard deviation of 3.2 minutes. construct a confidence interval for the population variance $sigma^{2}$ and the population standard deviation $sigma$. use a 98% level of confidence. assume the sample is from a normally distributed population. what is the confidence interval for the population variance $sigma^{2}$? (round to one decimal place as needed.)

Explanation:

Step1: Determine the degrees of freedom and critical values

The degrees of freedom \(df=n - 1=20 - 1 = 19\).
For a \(98\%\) confidence level, \(\alpha=1 - 0.98 = 0.02\), and \(\frac{\alpha}{2}=0.01\).
From the \(\chi^{2}\) - distribution table, \(\chi_{L}^{2}=\chi_{1-\frac{\alpha}{2},df}^{2}=\chi_{0.99,19}^{2}=7.633\) and \(\chi_{R}^{2}=\chi_{\frac{\alpha}{2},df}^{2}=\chi_{0.01,19}^{2}=36.191\).
The sample variance \(s^{2}=(3.2)^{2}=10.24\).

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

The formula for the confidence interval of 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 the values: \(\frac{19\times10.24}{36.191}\leq\sigma^{2}\leq\frac{19\times10.24}{7.633}\).
First calculation: \(\frac{19\times10.24}{36.191}=\frac{194.56}{36.191}\approx5.4\).
Second calculation: \(\frac{19\times10.24}{7.633}=\frac{194.56}{7.633}\approx25.5\).

Answer:

\(5.4\leq\sigma^{2}\leq25.5\)