Question

In Exercises 9–16, assume that each sample is a simple

random sample obtained from a population with a normal distribution.

Body Temperature Data Set 3 “Body Temperatures” in Appendix B includes a sample of106 body temperatures having a mean of 98.20°F and a standard deviation of 0.62°F (for day 2at 12 AM). Construct a 95%confidence interval estimate of the standard deviation of the body

temperatures for the entire population.

Short answer

The 95% confidence interval estimate of the standard deviation of the body temperatures for the entire population is\(0.55{}^{\rm{o}}{\rm{F}} < \sigma < 0.72{}^{\rm{o}}{\rm{F}}\).

Step-by-step solution

Given information

The sample number of body temperatures is\(n = 106\).

The mean body temperature is 98.20.

The sample standard deviation is\(s = 0.62\).

The level of confidence is 95%.

Compute the confidence intervalestimate of \({\bf{\sigma }}\)

The degrees of freedom is computed as,

\(\begin{array}{c}df = n - 1\\ = 106 - 1\\ = 105\end{array}\)

The level of confidence is 95%, which implies that the level of significance is 0.05.

Using the Chi-square table, the critical values at 0.05 level of significance and at 105 degrees of freedom are\(\chi _L^2 = 78.5364\)and\(\chi _R^2 = 135.247\).

The 95% confidence interval estimate of the standard deviation of the body

temperatures for the entire population is computed as,

\(\begin{array}{c}\sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _R^2}}} < \sigma < \sqrt {\frac{{\left( {n - 1} \right){s^2}}}{{\chi _L^2}}} \\\sqrt {\frac{{\left( {106 - 1} \right){{0.62}^2}}}{{135.247}}} < \sigma < \sqrt {\frac{{\left( {106 - 1} \right){{0.62}^2}}}{{78.5364}}} \\0.55 < \sigma < 0.72\end{array}\)

Therefore, the 95% confidence interval estimate of the standard deviation of the bodytemperatures for the entire population is \(0.55{}^{\rm{o}}{\rm{F}} < \sigma < 0.72{}^{\rm{o}}{\rm{F}}\).