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.