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 bodytemperatures for the entire population.

Short answer

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

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 interval estimate of  σ

The degrees of freedom is computed as,

$$\begin{gathered} df=n-1 \\ =106-1 \\ =105 \end{gathered}$$

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 localid="1648108100252" ${\chi }_L^2=78.5364$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,

localid="1648108270737" $$\begin{gathered} \sqrt{\frac{\left(n-1\right)s^2}{{\chi }_R^2}}<\sigma <\sqrt{\frac{\left(n-1\right)s^2}{{\chi }_L^2}} \\ \sqrt{\frac{\left(106-1\right){0.62}^2}{135.247}}<\sigma <\sqrt{\frac{\left(106-1\right){0.62}^2}{78.5364}} \\ 0.55<\sigma <0.72 \end{gathered}$$

Therefore, the 95% confidence interval estimate of the standard deviation of the body temperatures for the entire population is $0.55{}^{\text{o}}\text{F}<\sigma <0.72{}^{\text{o}}\text{F}$.