A data set includes 109 body temperatures of healthy adult humans having a mean of $98.2^{\circ} \mathrm{F}$ and a standard deviation of $0.62^{\circ} \mathrm{F}$. Construct a $99 \%$ confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of $98.6^{\circ} \mathrm{F}$ 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 $\mu$ ?
$\square^{\circ} \mathrm{F}< \mu< \square^{\circ} \mathrm{F}$
(Round to three decimal places as needed.)
What does this suggest about the use of $98.6^{\circ} \mathrm{F}$ as the mean body temperature?
A. This suggests that the mean body temperature is lower than $98.6^{\circ} \mathrm{F}$.
B. This suggests that the mean body temperature could very possibly be $98.6^{\circ} \mathrm{F}$.
C. This suggests that the mean body temperature is higher than $98.6^{\circ} \mathrm{F}$.
Answer
\(\boxed{\text{Final Answer: The confidence interval estimate of the population mean } \mu \text{ is } 98.044^\circ F < \mu < 98.356^\circ F. \text{ This suggests that the mean body temperature could very possibly be } 98.6^\circ F.}\)
Step 1 ：Given that the sample size (n) is 109, the sample mean (\(\bar{x}\)) is 98.2, and the standard deviation (s) is 0.62. We are asked to construct a 99% confidence interval for the mean body temperature of all healthy humans.
Step 2 ：The formula for a confidence interval is \(\bar{x} \pm t \frac{s}{\sqrt{n}}\), where t is the t-score corresponding to the desired level of confidence.
Step 3 ：The degrees of freedom is calculated as n - 1, which is 108 in this case. The t-score for a 99% confidence interval with 108 degrees of freedom can be found in a t-distribution table, which is approximately 2.622.
Step 4 ：Substitute the given values into the formula, we get the margin of error as \(2.622 \times \frac{0.62}{\sqrt{109}}\), which is approximately 0.156.
Step 5 ：The confidence interval is then \(98.2 \pm 0.156\), which gives us the interval from 98.044 to 98.356.
Step 6 ：Comparing this interval to 98.6, we see that 98.6 is not within this interval.
Step 7 ：\(\boxed{\text{Final Answer: The confidence interval estimate of the population mean } \mu \text{ is } 98.044^\circ F < \mu < 98.356^\circ F. \text{ This suggests that the mean body temperature could very possibly be } 98.6^\circ F.}\)
