The following data represent the age (in weeks) at which babies first crawl based on a survey of 12 mothers. The data are normally distributed and $s=9.461$ weeks. Construct and interpret a $99 \%$ confidence interval for the population standard deviation of the age (in weeks) at which babies first crawl.
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52 & 31 & 44 & 35 & 38 & 27 \\
46 & 38 & 55 & 27 & 39 & 28
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Click the icon to view the table of critical values of the chi-square distribution.

Select the correct choice below and fill in the answer boxes to complete your choice.
(Use ascending order. Round to three decimal places as needed.)
A. There is $99 \%$ confidence that the population standard deviation is between $\square$ and $\square$.
B. If repeated samples are taken, $99 \%$ of them will have the sample standard deviation between $\square$ and $\square$.
C. There is a $99 \%$ probability that the true population standard deviation is between $\square$ and $\square$.

Step 1 ：The problem is asking for a confidence interval for the population standard deviation. To calculate this, we can use the chi-square distribution. The formula for the confidence interval is given by: \[\sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}}}\] where n is the sample size, s is the sample standard deviation, and \(\chi^2_{\alpha/2, n-1}\) and \(\chi^2_{1-\alpha/2, n-1}\) are the critical values of the chi-square distribution for degrees of freedom n-1 and significance level \(\alpha\).

Step 2 ：In this case, n=12, s=9.461, and \(\alpha=0.01\) (since we want a 99% confidence interval). We need to find the critical values of the chi-square distribution for 11 degrees of freedom and \(\alpha=0.01\).

Step 3 ：Using the chi-square distribution table or a statistical calculator, we find that the critical values are \(\chi^2_{\alpha/2, n-1} = 2.603221890515113\) and \(\chi^2_{1-\alpha/2, n-1} = 26.756848916469636\).

Step 4 ：Substituting these values into the formula, we find the confidence interval for the population standard deviation to be between \(\sqrt{\frac{(12-1)9.461^2}{2.603221890515113}} = 6.066188506544412\) and \(\sqrt{\frac{(12-1)9.461^2}{26.756848916469636}} = 19.448127420687378\).

Step 5 ：Rounding to three decimal places, we get the final answer: There is 99% confidence that the population standard deviation is between 6.066 and 19.448. So, the final answer is \(\boxed{6.066, 19.448}\).