Elementary Statistics (12th Edition)

$H_{0}:\sigma=0.15$. $H_{a}:\sigma < 0.15.$ Hence the value of the test statistic: $X^2=\frac{(n-1)s^2}{\sigma^2}=\frac{(36-1)^20.09^2}{0.15^2}=12.6.$ 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:12.6<22.501, hence we reject the null hypothesis. Hence we can say that there is sufficient evidence to support that the volumes have a standard deviation less than 0.15.