Use the given information to find the number of degrees of freedom, the critical values $\chi_{L}^{2}$ and $\chi_{R}^{2}$, and the confidence interval estimate of $\sigma$. It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution. Platelet Counts of Women $80 \%$ confidence; $n=25, s=65.1$. Click the icon to view the table of Chi-Square critical values. $d f=24$ (Type a whole number.) \[ \chi_{\mathrm{L}}^{2}= \] (Round to three decimal places as needed.)

Step 1 ：The degrees of freedom is calculated using the formula \(df = n - 1\), where \(n\) is the sample size. In this case, \(n = 25\), so \(df = 25 - 1 = 24\).

Step 2 ：The critical values \(\chi_{L}^{2}\) and \(\chi_{R}^{2}\) can be found using the Chi-Square distribution table. Since we are looking for an 80% confidence interval, we need to find the values that cut off the middle 80% of the distribution, leaving 10% in each tail. The critical values are \(\chi_{L}^{2} = 15.659\) and \(\chi_{R}^{2} = 33.196\).

Step 3 ：The confidence interval estimate of \(\sigma\) (the population standard deviation) can be calculated using the formula: \[\sqrt{\frac{(n-1)s^2}{\chi_{R}^{2}}} \leq \sigma \leq \sqrt{\frac{(n-1)s^2}{\chi_{L}^{2}}}\] where \(s\) is the sample standard deviation, \(n\) is the sample size, and \(\chi_{L}^{2}\) and \(\chi_{R}^{2}\) are the critical values from the Chi-Square distribution. The confidence interval estimate of \(\sigma\) is \((55.353, 80.595)\).

Step 4 ：Final Answer: The degrees of freedom is \(\boxed{24}\). The critical values are \(\chi_{L}^{2} = \boxed{15.659}\) and \(\chi_{R}^{2} = \boxed{33.196}\). The confidence interval estimate of \(\sigma\) is \(\boxed{(55.353, 80.595)}\).