Answer to Question #339413 in Statistics and Probability for small T

Question

A sample of 25 women had a variance in IQ scores of 62. A sample of 18 men had a variance of 72. Do the women have a smaller variance in IQ scores at the 0.01 level of significance? Assume both populations are normally distributed

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\sigma_1^2\ge\sigma_2^2$

$H_a:\sigma_1^2<\sigma_2^2$

This corresponds to a left-tailed test, for which a F-test for two population variances needs to be used.

Based on the information provided, the significance level is $\alpha = 0.01,$ $df_1=n_1-1=24$ degrees of freedom, $df_2=n_2-1=17$ degrees of freedom, and the rejection region for this left-tailed test test is $R = \{F: F < 0.3548\}.$

The F-statistic is computed as follows:

$$F=\dfrac{s_1^2}{s_2^2}=\dfrac{62}{72}=0.8611$$

Since from the sample information we get that $F = 0.8611 \ge0.3548= F_c,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population variance $\sigma_1^2$ is less than the population variance $\sigma_2^2,$ at the $\alpha = 0.01$ significance level.