Questions: The values listed below are waiting times (in minutes) of customers at the Jefferson Valley Bank, where customers enter a single waiting line that feeds three teller windows. Construct a 95% confidence interval for the population standard deviation of the waiting times. 7.77 .7 6.56 .66 .76 .87 .17 .37 .47 .7

Solution Steps

The mean \( \mu \) of the waiting times is calculated as follows:

\[ \mu = \frac{\sum x_i}{n} = \frac{19.03}{10} = 1.9 \]

The sample variance \( s^2 \) is computed using the formula:

\[ s^2 = \frac{\sum (x_i - \mu)^2}{n-1} = 7.81 \]

The sample standard deviation \( s \) is the square root of the variance:

\[ s = \sqrt{7.81} = 2.8 \]

The confidence interval for the variance of a single population with unknown population mean is given by:

\[ \left(\frac{(n - 1)s^2}{\chi^2_{\alpha/2}}, \frac{(n - 1)s^2}{\chi^2_{1 - \alpha/2}}\right) \]

Substituting the values:

\[ CI = \left(\frac{(10 - 1) \times 7.81}{\chi^2_{\alpha/2}}, \frac{(10 - 1) \times 7.81}{\chi^2_{1 - \alpha/2}}\right) \]

This results in:

\[ CI = (3.7, 26.03) \]

Final Answer