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 $80 \%$ confidence; $n=147, s=1.98(1000$ cells $/ \mu \mathrm{L})$.

Step 1 ：Given that the sample size, \(n = 147\), and the sample standard deviation, \(s = 1.98(1000)\) cells/\(\mu L\), we can calculate the degrees of freedom using the formula \(df = n - 1\). This gives us \(df = 147 - 1 = 146\).

Step 2 ：The significance level, \(\alpha\), is calculated as \(\alpha = 1 - 0.80 = 0.20\).

Step 3 ：The critical values \(\chi_{L}^{2}\) and \(\chi_{R}^{2}\) can be found using a chi-square distribution table. These values correspond to the values of the chi-square distribution that cut off the upper and lower \(\alpha/2\) tails of the distribution. For this problem, \(\chi_{R}^{2} = 124.57\) and \(\chi_{L}^{2} = 168.28\).

Step 4 ：The confidence interval estimate of \(\sigma\) (the population standard deviation) is given by the formula: \(\sqrt{\frac{(n-1)s^{2}}{\chi_{R}^{2}}} \leq \sigma \leq \sqrt{\frac{(n-1)s^{2}}{\chi_{L}^{2}}}\). Substituting the given values, we find that the confidence interval estimate of \(\sigma\) is \([1844.26, 2143.53]\) (in units of cells/\(\mu L\)).

Step 5 ：\(\boxed{\text{Final Answer: The number of degrees of freedom is } 146. \text{ The critical values are } \chi_{L}^{2} = 168.28 \text{ and } \chi_{R}^{2} = 124.57. \text{ The confidence interval estimate of } \sigma \text{ is } [1844.26, 2143.53] \text{ (in units of cells/} \mu L\text{).}}\)