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 95% 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 \(df=n - 1\), where \(n = 20\), so \(df=20 - 1=19\).
For a \(95\%\) confidence level, \(\alpha=1 - 0.95 = 0.05\), and \(\frac{\alpha}{2}=0.025\), \(1-\frac{\alpha}{2}=0.975\).
From the \(\chi^{2}\) - distribution table, \(\chi_{L}^{2}=\chi_{0.025,19}^{2}=8.907\), \(\chi_{R}^{2}=\chi_{0.975,19}^{2}=32.852\).
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: \(\frac{(20 - 1)\times10.24}{32.852}\leq\sigma^{2}\leq\frac{(20 - 1)\times10.24}{8.907}\).
First, \(\frac{19\times10.24}{32.852}=\frac{194.56}{32.852}\approx5.9\).
Second, \(\frac{19\times10.24}{8.907}=\frac{194.56}{8.907}\approx21.8\).

Answer:

\((5.9,21.8)\)