A simple random sample from a population with a normal distribution of 98 body temperatures has $\bar{x}=98.90^{\circ} \mathrm{F}$ and $\mathrm{s}=0.63^{\circ} \mathrm{F}$. Construct a $99 \%$ confidence interval estimate of the standard deviation of body temperature of all healthy humans.
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${ }^{\circ} \mathrm{F}< \sigma< \square^{\circ} \mathrm{F}$
(Round to two decimal places as needed.)

Step 1 ：We are given a simple random sample from a population with a normal distribution of 98 body temperatures. The sample mean is \(\bar{x}=98.90^\circ F\) and the sample standard deviation is \(s=0.63^\circ F\). We are asked to construct a 99% confidence interval estimate of the standard deviation of body temperature of all healthy humans.

Step 2 ：To construct a confidence interval for the standard deviation, we can use the chi-square distribution. The formula for the confidence interval is: \[\sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}}}<\sigma<\sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}}}\] where: n is the sample size, s is the sample standard deviation, and \(\chi^2_{\alpha/2, n-1}\) and \(\chi^2_{1-\alpha/2, n-1}\) are the chi-square critical values for degrees of freedom n-1 and significance level \(\alpha\).

Step 3 ：In this case, n=98, s=0.63, and \(\alpha\)=0.01 (because the confidence level is 99%).

Step 4 ：We need to find the chi-square critical values. These can be found in a chi-square table or calculated using a statistical software or programming language. For this problem, the chi-square critical values are \(\chi^2_{\alpha/2, n-1} = 64.88\) and \(\chi^2_{1-\alpha/2, n-1} = 136.62\).

Step 5 ：Substituting these values into the formula, we get the lower and upper bounds of the confidence interval as 0.53 and 0.77 respectively.

Step 6 ：Thus, the 99% confidence interval estimate of the standard deviation of body temperature of all healthy humans is \(\boxed{0.53^\circ F<\sigma<0.77^\circ F}\).