Answer:
The null and alternative hypotheses are:
[tex]H_{0}:\sigma_{men} = \sigma_{women}[/tex]
[tex]H_{a}:\sigma_{men} \neq \sigma_{women}[/tex]
Under the null hypothesis, the test statistic is:
[tex]F=\frac{\sigma^{2}_{men}}{\sigma^{2}_{women}}[/tex]
[tex]=\frac{19^{2}}{9^{2}}[/tex]
[tex]=\frac{361}{81}[/tex]
[tex]=4.46[/tex]
Therefore, the test-statistic is [tex]F=4.46[/tex]
Now the F critical value at 0.02 significance level for df1 = 10- 1 =9 and df2 = 16 - 1 =15 is:
[tex]F_{critical} = 3.303[/tex]
Since F statistic is greater than the F critical value, we therefore, reject the null hypothesis and conclude that there is sufficient evidence to support the claim that there is a difference in the variation in the listening times for men and women.
