The problem asks to construct the confidence intervals for the population variance $\sigma^2$ and the population standard deviation $\sigma$, given a confidence level $C = 0.99$, sample standard deviation $s = 38$, and sample size $n = 19$.

Step-by-Step Solution:

1. Confidence Interval for Population Variance $\sigma^2$: The confidence interval for the population variance is constructed using the chi-square distribution. The formula is:

   $\left( \frac{(n-1) s^2}{\chi^2_{\alpha/2}}, \frac{(n-1) s^2}{\chi^2_{1-\alpha/2}} \right)$

   Where:

   • $n = 19$ (sample size),
   • $s = 38$ (sample standard deviation),
   • $\alpha = 1 - C = 1 - 0.99 = 0.01$,
   • $\chi^2_{\alpha/2}$ and $\chi^2_{1-\alpha/2}$ are chi-square critical values for $\alpha/2$ and $1-\alpha/2$ with $n-1 = 18$ degrees of freedom.

2. Chi-Square Critical Values:

   • $\chi^2_{\alpha/2} = \chi^2_{0.005, 18} = 37.156$,
   • $\chi^2_{1-\alpha/2} = \chi^2_{0.995, 18} = 6.2658$.

3. Calculate the Confidence Interval for Variance: First, calculate $(n-1)s^2$:

   $(n-1)s^2 = 18 \times (38)^2 = 18 \times 1444 = 25992$

   Now, the confidence interval for variance $\sigma^2$ is:

   $\left( \frac{25992}{37.156}, \frac{25992}{6.2658} \right) = (699.53, 4148.91)$

4. Confidence Interval for Population Standard Deviation $\sigma$: To find the confidence interval for the population standard deviation $\sigma$, simply take the square root of the bounds for the variance:

   $\left( \sqrt{699.53}, \sqrt{4148.91} \right) = (26.45, 64.40)$

Final Answer:

• The confidence interval for the population variance $\sigma^2$ is approximately $(699.53, 4148.91)$.
• The confidence interval for the population standard deviation $\sigma$ is approximately $(26.45, 64.40)$.

Would you like further details or have any questions?

Here are five additional related questions to explore:

1. How does changing the confidence level affect the width of the confidence interval?
2. How are chi-square distributions used in hypothesis testing for variances?
3. What happens if the population is not normally distributed?
4. How would you calculate a confidence interval for the population mean instead of variance?
5. Why do we use the sample standard deviation instead of the population standard deviation?

Tip: The larger your sample size, the more precise your confidence intervals become.