Question

A data set includes 108 body temperatures of healthy adult humans having a mean of 98.3degrees°F and a standard deviation of 0.69degrees°F. Construct a 99​% confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6°F as the mean body​ temperature? What is the confiedence interbal estimate of the population mean μ​?

Answer 1

Answer with explanation:

The confidence interval for population mean is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

Given : Sample size : n=106

Sample mean : $\overline{x}=98.3^{\circ}\ F$

Standard deviation : $\sigma= 0.69^{\circ}F$

Significance level : $\alpha: 1-0.99=0.01$

Critical value : $z_{\alpha/2}=2.576$

Then , 99​% confidence interval estimate of the mean body temperature of all healthy humans will be :-

$98.3\pm (2.576)\dfrac{0.69}{\sqrt{106}}\\\\\approx98.3\pm0.173=(98.3-0.173,\ 98.3+0.173)=(98.127,\ 98.473)$

Since 98.6 is higher than the upper limit of the confidence interval (98.473), then this suggest that the mean body temperature could be lower then 98.6 degrees.

The best estimate of population mean is always the sample mean. $\mu=\overline{x}=98.3^{\circ}\ F$