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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 \( (26.45, 64.40) \).

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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.

Use technology to construct the confidence intervals for the population variance σ^2 and the population standard deviation σ, assuming the sample is taken from a normally distributed population. Given: C = 0.99, s = 38, n = 19.

This problem involves constructing confidence intervals for the population variance (σ^2) and standard deviation (σ) with a sample standard deviation of 38, sample size of 19, and a confidence level of 0.99. Learn how to apply the chi-square distribution and calculate these intervals step-by-step.

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