Answer
There is sufficient evidence to support that the weights have a standard deviation less than 0.023.
Work Step by Step
$H_{0}:\sigma=0.023$. $H_{a}:\sigma < 0.023.$ Hence the value of the test statistic: $X^2=\frac{(n-1)s^2}{\sigma^2}=\frac{(37-1)^20.01648^2}{0.023^2}=18.483.$ 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:18.483<22.501, hence we reject the null hypothesis. Hence we can say that there is sufficient evidence to support that the weights have a standard deviation less than 0.023.
