A clinical trial was conducted to test the effectiveness of a drug used for treating insomnia in older subjects. After treatment with the drug, 20 subjects had a mean wake time of $96.8 \mathrm{~min}$ and a standard deviation of $44.9 \mathrm{~min}$. Assume that the 20 sample values appear to be from a normally distributed population and construct a $98 \%$ confidence interval estimate of the standard deviation of the wake times for a population with the drug treatments. Does the result indicate whether the treatment is effective?
Find the confidence interval estimate.
\[
\min < \sigma< \square \min
\]
(Round to two decimal places as needed.)

Step 1 ：Given that the sample size (n) is 20, the sample standard deviation (s) is 44.9 minutes, and the confidence level is 98%, which gives a significance level (alpha) of 0.02.

Step 2 ：We can use the chi-square distribution to find the confidence interval for the standard deviation. The formula for the confidence interval for a standard deviation is given by: \[\sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}}}<\sigma<\sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}}}\]

Step 3 ：Substituting the given values into the formula, we get the chi-square values for the lower and upper bounds as 7.632729647571471 and 36.19086912927004 respectively.

Step 4 ：Substituting these chi-square values into the formula, we get the lower and upper bounds of the confidence interval as 32.53296418203137 minutes and 70.84075832996265 minutes respectively.

Step 5 ：Thus, the 98% confidence interval estimate of the standard deviation of the wake times for a population with the drug treatments is \(\boxed{32.53 \min < \sigma < 70.84 \min}\).