A simple random sample from a population with a normal distribution of 106 body temperatures has $\bar{x}=98.80^{\circ} \mathrm{F}$ and $\mathrm{s}=0.67^{\circ} \mathrm{F}$. Construct an $80 \%$ confidence interval estimate of the standard deviation of body temperature of all healthy humans.
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(Round to two decimal places as needed.)

Step 1 ：We are given a simple random sample from a population with a normal distribution of 106 body temperatures. The sample mean is \(\bar{x}=98.80^\circ F\) and the sample standard deviation is \(s=0.67^\circ F\). We are asked to construct an 80% 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 given by: \[\sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}}}\] 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 \(\alpha/2\) and \(1-\alpha/2\) degrees of freedom respectively.

Step 3 ：In this case, we have n = 106, s = 0.67, and we want an 80% confidence interval, so \(\alpha = 1 - 0.80 = 0.20\). We need to find the chi-square critical values for \(\alpha/2 = 0.10\) and \(1-\alpha/2 = 0.90\) with n-1 = 105 degrees of freedom.

Step 4 ：Using the chi-square table, we find that the chi-square critical values are \(\chi^2_{0.10, 105} = 86.91\) and \(\chi^2_{0.90, 105} = 123.95\).

Step 5 ：Substituting these values into the formula, we find the lower and upper bounds of the confidence interval to be \(\sqrt{\frac{(106-1)(0.67)^2}{86.91}} = 0.62\) and \(\sqrt{\frac{(106-1)(0.67)^2}{123.95}} = 0.74\) respectively.

Step 6 ：Thus, the 80% confidence interval estimate of the standard deviation of body temperature of all healthy humans is \(\boxed{(0.62, 0.74)}\).