The standard deviation of a simple random sample of 15 customer waiting Times is found to be 4.8 minutes. Find the test statistic to test the claim that the standard deviation of all customer waiting Times is greater than 2.5 minutes. Use a 0.01 significance level
To test the claim that the standard deviation of all customer waiting times is greater than 2.5 minutes, we can use the Chi-Square test statistic.
The test statistic for testing the claim that the population standard deviation is greater than a specified value is calculated as:
\[ \chi^2 = \frac{(n-1) s^2}{\sigma_0^2} \]
Where:
- \( n \) is the sample size (15 in this case)
- \( s \) is the sample standard deviation (4.8 minutes)
- \( \sigma_0 \) is the hypothesized population standard deviation (2.5 minutes)
Plugging in the values, we get:
\[ \chi^2 = \frac{(15-1) \times 4.8^2}{2.5^2} \]
\[ \chi^2 = \frac{14 \times 23.04}{6.25} \]
\[ \chi^2 = \frac{322.56}{6.25} \]
\[ \chi^2 \approx 51.61 \]
The critical value for a Chi-Square test with 14 degrees of freedom and a significance level of 0.01 is approximately 28.71.
Since our calculated test statistic (51.61) is greater than the critical value (28.71), we reject the null hypothesis that the population standard deviation is less than or equal to 2.5 minutes at a 0.01 significance level. This means we have enough evidence to suggest that the population standard deviation is greater than 2.5 minutes.