To find the test statistic for testing the claim that the standard deviation of all phone calls on a pay phone is less than 2.9 minutes, we will use the chi-squared test statistic for variance.
-
Set up the hypotheses:
- Null hypothesis (H0): \( \sigma \geq 2.9 \) (the standard deviation is 2.9 minutes or greater)
- Alternative hypothesis (H1): \( \sigma < 2.9 \) (the standard deviation is less than 2.9 minutes)
-
Calculate the test statistic: The test statistic for the chi-squared test for variance is given by: \[ \chi^2 = \frac{(n - 1) s^2}{\sigma_0^2} \] Where:
- \( n \) = sample size = 40
- \( s \) = sample standard deviation = 2.6
- \( \sigma_0 \) = hypothesized population standard deviation = 2.9
First, calculate \( s^2 \) (the sample variance) and \( \sigma_0^2 \) (the hypothesized variance): \[ s^2 = (2.6)^2 = 6.76 \] \[ \sigma_0^2 = (2.9)^2 = 8.41 \]
Now substitute these values into the test statistic formula: \[ \chi^2 = \frac{(40 - 1) \cdot 6.76}{8.41} \] \[ \chi^2 = \frac{39 \cdot 6.76}{8.41} \] \[ \chi^2 = \frac{263.64}{8.41} \] \[ \chi^2 \approx 31.35 \quad (rounding to two decimals) \]
-
Conclusion: The calculated chi-squared statistic is approximately 31.348.
Thus, the answer is:
B) 31.348