A data set includes 106 body temperatures of healthy adult humans having a mean of \( 98.9^{\circ} \mathrm{F} \) and a standard deviation of \( 0.63^{\circ} \mathrm{F} \). 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 higher than \( 98.6^{\circ} \mathrm{F} \). B. This suggests that the mean body temperature is lower than \( 98.6^{\circ} \mathrm{F} \). C. This suggests that the mean body temperature could very possibly be \( 98.6^{\circ} \mathrm{F} \).

To construct a \( 99\% \) confidence interval for the mean body temperature, we can use the formula for the confidence interval: \[ \bar{x} \pm z^* \left( \frac{s}{\sqrt{n}} \right) \] where: - \(\bar{x}\) is the sample mean, - \(z^*\) is the z-value corresponding to the desired confidence level, - \(s\) is the sample standard deviation, - \(n\) is the sample size. ### Step 1: Identify the values - Sample mean (\(\bar{x}\)) = \(98.9^{\circ} \mathrm{F}\) - Sample standard deviation (\(s\)) = \(0.63^{\circ} \mathrm{F}\) - Sample size (\(n\)) = \(106\) ### Step 2: Find the z-value for a \( 99\% \) confidence level For a \( 99\% \) confidence level, the z-value (from the standard normal distribution table) is approximately \( 2.576 \). ### Step 3: Calculate the standard error (SE) \[ SE = \frac{s}{\sqrt{n}} = \frac{0.63}{\sqrt{106}} \approx \frac{0.63}{10.2956} \approx 0.0612 \] ### Step 4: Calculate the margin of error (ME) \[ ME = z^* \cdot SE = 2.576 \cdot 0.0612 \approx 0.1575 \] ### Step 5: Construct the confidence interval \[ \text{Lower limit} = \bar{x} - ME = 98.9 - 0.1575 \approx 98.7425 \] \[ \text{Upper limit} = \bar{x} + ME = 98.9 + 0.1575 \approx 99.0575 \] ### Step 6: Round to three decimal places \[ 98.743^{\circ} \mathrm{F} < \mu < 99.058^{\circ} \mathrm{F} \] ### Conclusion The confidence interval estimate of the population mean \( \mu \) is: \[ 98.743^{\circ} \mathrm{F} < \mu < 99.058^{\circ} \mathrm{F} \] ### Interpretation of the results Now, regarding the use of \( 98.6^{\circ} \mathrm{F} \) as the mean body temperature: - The entire confidence interval \( (98.743, 99.058) \) is above \( 98.6^{\circ} \mathrm{F} \). - This suggests that the mean body temperature is likely higher than \( 98.6^{\circ} \mathrm{F} \). Thus, the correct answer is: **A. This suggests that the mean body temperature is higher than \( 98.6^{\circ} \mathrm{F} \).**