Question

a data set includes 106 body temperatures of healthy adult humans having a mean of 98.9°f and a standard deviation of 0.62°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 t 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 μ?
$\square$°f < μ < $\square$°f
(round to three decimal places as needed )

Explanation:

Step1: Identify the formula for confidence interval

For a confidence interval for the population mean when the population standard deviation is unknown (we use sample standard deviation \( s \)), the formula is \( \bar{x} \pm t_{\alpha/2, n - 1} \times \frac{s}{\sqrt{n}} \), where \( \bar{x} \) is the sample mean, \( t_{\alpha/2, n - 1} \) is the t - critical value, \( s \) is the sample standard deviation, and \( n \) is the sample size.

Given: \( n = 106 \), \( \bar{x}=98.9^{\circ}\text{F} \), \( s = 0.62^{\circ}\text{F} \), confidence level \( = 99\% \), so \( \alpha=1 - 0.99 = 0.01 \), \( \alpha/2=0.005 \), and degrees of freedom \( df=n - 1=106 - 1 = 105 \).

Step2: Find the t - critical value

Using a t - distribution table or calculator, for \( df = 105 \) and \( \alpha/2=0.005 \), \( t_{0.005,105}\approx2.623 \) (we can use the t - distribution table approximation or a calculator. For large degrees of freedom, the t - value is close to the z - value for the same significance level, but since \( n = 106 \), we use t - distribution).

Step3: Calculate the margin of error (E)

The margin of error \( E=t_{\alpha/2, n - 1}\times\frac{s}{\sqrt{n}} \)

Substitute the values: \( E = 2.623\times\frac{0.62}{\sqrt{106}} \)

First, calculate \( \sqrt{106}\approx10.2956 \)

Then, \( \frac{0.62}{10.2956}\approx0.0602 \)

Then, \( E=2.623\times0.0602\approx0.1579 \)

Step4: Calculate the confidence interval

Lower limit: \( \bar{x}-E=98.9 - 0.1579 = 98.7421\approx98.742 \)

Upper limit: \( \bar{x}+E=98.9+0.1579 = 99.0579\approx99.058 \)

Answer:

\( 98.742^{\circ}\text{F}<\mu<99.058^{\circ}\text{F} \)