To test the claim about the standard deviation of weights of women basketball players, we will use the Chi-Squared test for variance. The formula for the test statistic is given by:
\[ \chi^2 = \frac{(n - 1) s^2}{\sigma_0^2} \]
Where:
- \( n \) is the sample size (8 in this case)
- \( s \) is the sample standard deviation (7.5 lb)
- \( \sigma_0 \) is the hypothesized population standard deviation (9 lb)
-
First, we need to calculate \( s^2 \) (the sample variance) and \( \sigma_0^2 \) (the hypothesized population variance):
\[ s^2 = (7.5)^2 = 56.25 \]
\[ \sigma_0^2 = (9)^2 = 81 \]
-
Substitute the values into the Chi-Squared formula:
\[ \chi^2 = \frac{(8 - 1) \cdot 56.25}{81} = \frac{7 \cdot 56.25}{81} \]
\[ \chi^2 = \frac{393.75}{81} \approx 4.861 \]
Now that we have calculated the test statistic, we can compare it to the critical value from the Chi-Squared distribution to determine if we can reject the null hypothesis. However, we are interested only in the calculated test statistic.
So the test statistic is:
\[ \chi^2 \approx 4.861 \]
Thus, the correct answer is:
D) 4.861