Questions: The values listed below are waiting times (in minutes) of customers at two different banks. At Bank A, customers enter a single waiting line that feeds three teller windows. At Bank B, customers may enter any one of three different lines that have formed at three teller windows. Answer the following questions, Bank A: 6.4, 6.6, 6.7, 6.8, 7.1, 72, 7.5, 79, 7.9, 7.9 Bank B: 4.3, 5.3, 5.8, 6.3, 6.6, 7.7, 7.7, 8.5, 9.4, 10.0 Construct a 99% confidence interval for the population standard deviation σ at Bank A. min < σBank A < max min (Round to two decimal places as needed)

Solution Steps

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

\[ \mu = \frac{\sum x_i}{n} = \frac{72.0000}{10} = 7.2 \]

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

\[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n-1} = 0.33 \]

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

\[ \sigma = \sqrt{0.33} \approx 0.58 \]

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:

\[ CI = \left(\frac{(10 - 1) \times 0.33}{\chi^2_{0.005}}, \frac{(10 - 1) \times 0.33}{\chi^2_{0.995}}\right) \]

This results in:

\[ CI = (0.13, 1.71) \]

Final Answer