#### Answer

There is sufficient evidence to support that the standard deviation of the population is less than 0.15.

#### Work Step by Step

$H_{0}:σ=0.15$. $H_{a}:σ <0.15.$ Hence the value of the test statistic: $X^2=\frac{(n−1)s^2}{σ^2}=\frac{(36-1)^2 0.0809^2}{0.15^2}=10.181.$ The critical value is the $X^2$ value corresponding to the found significance level, hence:$X_{1-0.05}^2=\frac{26.509+18.493}{2}=22.501$. If the value of the test statistic is in the rejection area, then this means the rejection of the null hypothesis. Hence:10.181<22.501, hence we reject the null hypothesis. Hence we can say that there is sufficient evidence to support that the standard deviation of the population is less than 0.15.