To test the claim that the standard deviation of all customer waiting times is greater than 3.5 minutes, you can use the chi-square test for standard deviation. The test statistic is calculated with the following formula:
\[ \chi^2 = \frac{(n - 1)s^2}{\sigma_0^2} \]
Where:
- \( n \) = sample size
- \( s \) = sample standard deviation
- \( \sigma_0 \) = claimed population standard deviation
Given:
- \( n = 15 \)
- \( s = 4.8 \) minutes
- \( \sigma_0 = 3.5 \) minutes
First, calculate \( s^2 \) and \( \sigma_0^2 \):
\[ s^2 = (4.8)^2 = 23.04 \] \[ \sigma_0^2 = (3.5)^2 = 12.25 \]
Now, substitute these values into the chi-square formula:
\[ \chi^2 = \frac{(15 - 1) \times 23.04}{12.25} = \frac{14 \times 23.04}{12.25} \]
Calculating \( \chi^2 \):
\[ \chi^2 = \frac{322.56}{12.25} \approx 26.3336 \]
Rounding this to three decimal places:
\[ \chi^2 \approx 26.334 \]
Thus, the test statistic is approximately \( 26.331 \).
The answer is:
C) 26.331