Here are summary statistics for randomly selected weights of newborn girls: $n=205, \bar{x}=32.8 \mathrm{hg}, s=6.7 \mathrm{hg}$. Construct a confidence interval estimate of the mean. Use a $99 \%$ confidence level. Are these results very different from the confidence interval $31.3 \mathrm{hg}< \mu< 35.3 \mathrm{hg}$ wth only 14 sample values, $\bar{x}=33.3 \mathrm{hg}$, and s $=25 \mathrm{hg}$ ?

What is the confidence interval for the population mean $\mu$ ?
$31.6 \mathrm{hg}< \mu< 34.0 \mathrm{hg}$ (Round to one decimal place as needed)
Are the results between the two confidence intervals very different?
A. Yes, because one confidence interval does not contain the mean of the other confidence interval.
B. No, because the confidence interval limits are similar,
C. Yes, because the confidence interval limits are not similar.
D. No, because each confidence interval contains the mean of the other confidence interval.

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: \(\boxed{\text{D. No, because each confidence interval contains the mean of the other confidence interval.}}\)