Question

A data set includes 106 body temperatures of healthy adult humans having a mean of 98.9 F and a standard deviation of 0.63'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? Click here to view a distribution table Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. What is the confidence interval estimate of the population mean? FF (Round to three decimal places as needed.) What does this suggest about the use of 98.6°F as the mean body temperature? Click to select your answer(s).

Answer 1

Solution :

The 99% confidence interval for population mean is given as follows :

$\large \left ( \bar{x}\pm t_{\left ( \frac{0.01}{2},n-1 \right )}\frac{s}{\sqrt{n}} \right )$

Where, x̄ is sample mean, s is sample standard deviation, n is sample size and t(0.01/2, n - 1) is critical t-value to construct 99% confidence interval.

We have,  x̄ = 98.9 °F, s = 0.63, n = 106

Using t-table we get, t(0.01/2, 106 - 1) = 2.6235

Hence, 99% confidence interval of the population mean is,

$\large \left ( 98.9\pm 2.6235\times\frac{0.63}{\sqrt{106}} \right )$

$\large =\left ( 98.9\pm 0.161 \right )$

$\large =\left ( 98.9 - 0.161, 98.9 + 0.161 \right )$

$\large =\left ( 98.739, 99.061 \right )$

The 95% confidence interval for population mean is,

98.739 °F < μ < 99.061 °F

Since, the lower limit of confidence interval is greater than 98.6 °F, therefore this suggest that the mean body temperature could be higher than 98.6 °F.