To test the hypothesis that the waiting times with a single line vary less than the waiting times with separate lines, we can use the chi-square test for variance.
Step 1: Define the Hypotheses
- Null Hypothesis (\(H_0\)): The variance of waiting times with a single line is greater than or equal to the variance of waiting times with separate lines.
- Alternative Hypothesis (\(H_a\)): The variance of waiting times with a single line is less than the variance of waiting times with separate lines.
Step 2: Identify the Parameters
- Population variance with separate lines (\(\sigma_0^2\)) = (5.7)^2 = 32.49 minutes²
- Sample standard deviation with a single line (\(s^2\)) = (4.9)^2 = 24.01 minutes²
- Sample size (n) = 29
Step 3: Calculate the Test Statistic
The test statistic for the chi-square test for variance is calculated using the formula:
\[ \chi^2 = \frac{(n-1) \cdot s^2}{\sigma_0^2} \]
Plugging in the values:
\[ \chi^2 = \frac{(29-1) \cdot 24.01}{32.49} \] \[ \chi^2 = \frac{28 \cdot 24.01}{32.49} \approx \frac{672.28}{32.49} \approx 20.692 \]
Step 4: Determine the Critical Value
Since this is a one-tailed test (we are testing if the variance is less), we will use the \(\chi^2\) distribution with \(n-1 = 28\) degrees of freedom.
Using a chi-square distribution table or calculator, we find the critical value for \(\alpha = 0.025\) and 28 degrees of freedom:
- Critical value (\( \chi^2_{\text{critical}} \)) is approximately 15.308.
Step 5: Conclusion
Now we can compare the test statistic with the critical value:
- Test statistic: \(\chi^2 = 20.692\)
- Critical value: \(\chi^2 = 15.308\)
Since \(20.692 > 15.308\), we fail to reject the null hypothesis.
Final Answers:
Test statistic: \(\chi^2 = 20.692\). Critical value: \(\chi^2 = 15.308\).
Therefore, the correct answer is:
B) Test statistic: χ² = 20.692. Critical value: χ² = 15.308.