Question

a data set includes 103 body temperatures of healthy adult humans having a mean of 98.0°f and a standard deviation of 0.65°f. construct a 99% confidence interval estimate of the mean body temperature of all healthy humans.
what is the confidence interval estimate of the population mean μ?
□°f<μ<□°f
(round to three decimal places as needed.)

Explanation:

Step1: Determine the critical value

For a 99% confidence interval, the significance level $\alpha = 1 - 0.99=0.01$. Then $\frac{\alpha}{2}=0.005$.
Since the sample size $n = 103$ (large - sample, $n>30$), we use the standard normal distribution $Z$.
Looking up in the standard normal table (or using a calculator), $z_{\frac{\alpha}{2}}=z_{0.005} = 2.576$.

Step2: Calculate the margin of error

The formula for the margin of error $E$ is $E = z_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}$.
Given $\bar{x}=98.0$, $s = 0.65$, $n = 103$.
Substitute the values: $E=2.576\times\frac{0.65}{\sqrt{103}}$.
First, calculate $\sqrt{103}\approx10.149$, then $\frac{0.65}{10.149}\approx0.064$.
$E = 2.576\times0.064\approx0.165$.

Step3: Construct the confidence interval

The confidence interval for the population mean $\mu$ is $\bar{x}-E<\mu<\bar{x} + E$.
Substitute $\bar{x}=98.0$ and $E = 0.165$.
We get $98.0-0.165<\mu<98.0 + 0.165$.

Answer:

$97.835^{\circ}F<\mu<98.165^{\circ}F$