A lawn mower manufacturer is trying to determine the standard deviation of the life of one of its lawn mowers. To do this, it randomly selects 19 lawn mowers that were sold several years ago and finds that the sample standard deviation is 2.17 years. Construct a $99 \%$ confidence interval for the population standard deviation $\sigma$.

Step 1 ：We are given that the sample size is 19, the sample standard deviation is 2.17 years, and we are asked to construct a 99% confidence interval for the population standard deviation.

Step 2 ：We start by calculating the chi-square values for the lower and upper bounds of the confidence interval. The chi-square distribution is used because the standard deviation is a squared quantity and thus follows a chi-square distribution.

Step 3 ：The lower chi-square value is calculated as \(\chi^2_{\text{lower}} = \text{ppf}(\frac{\alpha}{2}, n - 1)\), where ppf is the percent point function (inverse of the cumulative distribution function) and \(\alpha\) is the significance level. Substituting the given values, we get \(\chi^2_{\text{lower}} = 6.26\).

Step 4 ：The upper chi-square value is calculated as \(\chi^2_{\text{upper}} = \text{ppf}(1 - \frac{\alpha}{2}, n - 1)\). Substituting the given values, we get \(\chi^2_{\text{upper}} = 37.16\).

Step 5 ：We then calculate the lower and upper bounds of the confidence interval for the population standard deviation using the formula \(\sqrt{\frac{(n - 1) \times s^2}{\chi^2}}\), where n is the sample size, s is the sample standard deviation, and \(\chi^2\) is the chi-square value.

Step 6 ：The lower bound is calculated as \(\sqrt{\frac{(19 - 1) \times 2.17^2}{37.16}} = 1.51\) years.

Step 7 ：The upper bound is calculated as \(\sqrt{\frac{(19 - 1) \times 2.17^2}{6.26}} = 3.68\) years.

Step 8 ：Thus, the 99% confidence interval for the population standard deviation is \(\boxed{[1.51, 3.68]}\) years.