A data set includes 103 body temperatures of healthy adult humans having a mean of $98.1^{\circ} \mathrm{F}$ and a standard deviation of $0.56^{\circ} \mathrm{F}$. 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?
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What is the confidence interval estimate of the population mean $\mu$ ?
(Round to three decimal places as needed.)

Step 1 ：We are given the sample mean \(\bar{x} = 98.1\), standard deviation \(s = 0.56\), and sample size \(n = 103\). We are asked to construct a 99% confidence interval for the population mean.

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 our confidence level and degrees of freedom (n-1), \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step 3 ：As the sample size is large (n > 30), the t-distribution approximates the standard normal distribution, and we can use the z-score for a 99% confidence interval, which is 2.576.

Step 4 ：We calculate the standard error as \(\frac{s}{\sqrt{n}} = \frac{0.56}{\sqrt{103}} = 0.055\).

Step 5 ：We then calculate the lower and upper bounds of the confidence interval as \(\bar{x} - z \times \text{standard error} = 98.1 - 2.576 \times 0.055 = 97.958\) and \(\bar{x} + z \times \text{standard error} = 98.1 + 2.576 \times 0.055 = 98.242\) respectively.

Step 6 ：\(\boxed{\text{The 99% confidence interval estimate of the population mean is (97.958, 98.242).}}\)

Step 7 ：Given that 98.6 does not fall within this interval, the sample suggests that using 98.6 as the mean body temperature may not be accurate.

Step 8 ：\(\boxed{\text{The sample suggests that using 98.6 as the mean body temperature may not be accurate.}}\)