Question

A simple random sample of 8 women basketball player’s weights has a standard deviation of 7.5 lb. Find the test statistic, using a 0.05 significance level, to test a claim that the standard deviation of all weights of women basketball players is less than 9 lb.

A) 5.833
B) 10.080
C) 5.556
D) 4.861

Answer 1

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)

1. First, we need to calculate \( s^2 \) (the sample variance) and \( \sigma_0^2 \) (the hypothesized population variance):

\[
s^2 = (7.5)^2 = 56.25
\]

\[
\sigma_0^2 = (9)^2 = 81
\]

2. 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**