The variance of the mathematical scores of a sample of 21 high school girls was found to be 100 and that of a sample of 25 high school boys was 169.

Test the hypothesis that the scores of the high school boys are more varying that those of the high school girls at alpha = 0.05.

Thanks!

This appears to be a hypothesis test involving inferences concerning two variances.

Sample 1 (Girls): n = 21; variance = 100; df = n - 1 = 20
Sample 2 (Boys): n = 25; variance = 169; df = n - 1 = 24

Test statistic = sample 1 variance / sample 2 variance

You can use the F-distribution at .05 level using the above information for degrees of freedom. This will be your critical value to compare to the test statistic. If the test statistic exceeds the critical value from the table, the null will be rejected in favor of the alternate hypothesis and you can conclude a difference in variances. If the test statistic does not exceed the critical value from the table, then the null is not rejected and you cannot conclude a difference.

To test the hypothesis that the scores of the high school boys are more varying than those of the high school girls at alpha = 0.05, we can perform a hypothesis test by comparing the variances of the two populations.

The null hypothesis, denoted as H₀, assumes that the variances of the two populations are equal, while the alternative hypothesis, denoted as H₁, assumes that the variance of the high school boys' scores is greater than that of the high school girls' scores.

In this case, H₀: σ₁² = σ₂² (the variance of girls' scores is equal to the variance of boys' scores), and H₁: σ₁² < σ₂² (the variance of girls' scores is less than the variance of boys' scores).

To conduct the test, we can use the F-test, which compares the variances based on the F-statistic. The formula for the F-statistic is:
F = s₁² / s₂²

Where s₁² and s₂² are the sample variances for the girls and boys, respectively.

Let's calculate the F-statistic using the given information:

For girls:
Sample size (n₁) = 21
Sample variance (s₁²) = 100

For boys:
Sample size (n₂) = 25
Sample variance (s₂²) = 169

Plug these values into the F-statistic formula:
F = 100 / 169

To test the hypothesis, we need to compare the F-statistic with the critical value from the F-distribution table corresponding to our chosen significance level, alpha = 0.05.

The degrees of freedom for the numerator (girls' sample variance) are (n₁ - 1) = 21 - 1 = 20, and for the denominator (boys' sample variance) are (n₂ - 1) = 25 - 1 = 24.

Consulting the F-distribution table, the critical F-value at alpha = 0.05 with 20 degrees of freedom in the numerator and 24 degrees of freedom in the denominator is approximately 1.89.

If the calculated F-statistic is greater than the critical F-value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

Now, calculate the F-statistic and compare it to the critical value:

F = 100 / 169 ≈ 0.5911

Since the calculated F-statistic (0.5911) is less than the critical F-value (1.89), we fail to reject the null hypothesis.

Therefore, there is not enough evidence to conclude that the scores of the high school boys are more varying than those of the high school girls at alpha = 0.05.