#### Answer

Reject the null hypothesis.

#### Work Step by Step

$H_{0}:\sigma_1=sigma_2,$ $H_{a}:\sigma_1\ $ is more than $ \sigma_2$. :$F=\frac{s_1^2}{s_2^2}=\frac{10.6383^2}{5.2129^2}=4.1467.$ The critical values by the table with df=min(11,11)=11: $f_{0.05}$ is between 2.7876 and 2.8536. If the value of the test statistic is in the rejection region, reject the null hypothesis. 2.8536 is less than 4.1467, hence we reject the null hypothesis.