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 :\(\boxed{\text{Final Answer: The confidence interval estimate of the population mean } \mu \text{ is } 98.044^\circ F < \mu < 98.356^\circ F. \text{ This suggests that the mean body temperature could very possibly be } 98.6^\circ F.}\)