Step 1 :Given that the sample size \(n=205\), the sample mean \(\overline{x}=32.8\) hg, and the sample standard deviation \(s=6.7\) hg, and we want a \(99\%\) confidence level. The z-score for a \(99\%\) confidence level is approximately \(2.576\).
Step 2 :We use the formula for a confidence interval for a population mean, which is given by: \(\overline{x} \pm z \frac{s}{\sqrt{n}}\)
Step 3 :Substituting these values into the formula, we get: \(32.8 \pm 2.576 \frac{6.7}{\sqrt{205}}\)
Step 4 :Calculating the margin of error gives: \(32.8 \pm 1.2\)
Step 5 :So, the \(99\%\) confidence interval for the population mean is \(31.6\) hg \(< \mu <\) \(34.0\) hg.
Step 6 :Comparing this to the given confidence interval of \(31.3\) hg \(< \mu <\) \(35.3\) hg, we can see that the two intervals are not very different. Both intervals contain the mean of the other interval, and the limits are similar.
Step 7 :Therefore, the answer is: \(\boxed{\text{D. No, because each confidence interval contains the mean of the other confidence interval.}}\)