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. White Blood Counts of Women $99 \%$ confidence; $n=146, s=1.96(1000$ cells $/ \mu \mathrm{L})$. $\mathrm{df}=\square$ (Type a whole number.) $\chi_{\mathrm{L}}^{2}=$ (Round to two decimal places as needed.) $\chi_{\mathrm{R}}^{2}=$ (Round to two decimal places as needed.) The confidence interval estimate of $\sigma$ is $\square(1000$ cells $/ \mu \mathrm{L})<\sigma<\square(1000$ cells/ $\mu \mathrm{L})$. (Round to two decimal places as needed.)

Step 1 ：Given that the sample size, n = 146 and the sample standard deviation, s = 1.96, we can calculate the degrees of freedom as df = n - 1 = 146 - 1 = 145.

Step 2 ：Using a chi-square distribution table or a statistical calculator, with the degrees of freedom as 145 and the confidence level as 99%, we find the critical values as \(\chi_{L}^{2} = 104.89\) and \(\chi_{R}^{2} = 192.61\).

Step 3 ：We can then calculate the confidence interval estimate of \(\sigma\) using the formula \(\sqrt{\frac{(n-1)s^2}{\chi_{R}^{2}}} < \sigma < \sqrt{\frac{(n-1)s^2}{\chi_{L}^{2}}}\). Substituting the given values, we get \(\sqrt{\frac{(145)(1.96)^2}{192.61}} < \sigma < \sqrt{\frac{(145)(1.96)^2}{104.89}}\), which simplifies to \(1.70 < \sigma < 2.30\).

Step 4 ：So, the degrees of freedom is \(\boxed{145}\), the critical value \(\chi_{L}^{2}\) is \(\boxed{104.89}\), the critical value \(\chi_{R}^{2}\) is \(\boxed{192.61}\), and the confidence interval estimate of \(\sigma\) is \(\boxed{1.70}(1000 \text{ cells } / \mu \mathrm{L}) < \sigma < \boxed{2.30}(1000 \text{ cells } / \mu \mathrm{L})\).