To construct a confidence interval for the population variance $\sigma^2$ based on a sample, we can use the Chi-square distribution. The formula for the confidence interval for $\sigma^2$ is:

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

where:

• $n$ is the sample size,
• $s^2$ is the sample variance,
• $\chi^2_{\alpha/2}$ and $\chi^2_{1-\alpha/2}$ are the critical values from the Chi-square distribution for the given confidence level.

For a 90% confidence interval:

• $\alpha = 0.10$, so $\alpha/2 = 0.05$.
• We need to find the critical values $\chi^2_{0.05}$ and $\chi^2_{0.95}$ from the Chi-square distribution with $n-1$ degrees of freedom.

Part (b) - Confidence Interval for $\sigma^2$ with $n = 30$

Given:

• Sample variance $s^2 = 12.6$,
• Sample size $n = 30$,
• Confidence level is 90%.

Degrees of freedom $df = n - 1 = 30 - 1 = 29$.

From the Chi-square distribution table:

• $\chi^2_{0.05, 29} \approx 42.557$,
• $\chi^2_{0.95, 29} \approx 17.708$.

Using the formula for the confidence interval:

$\text{Lower bound} = \frac{(n-1) s^2}{\chi^2_{0.05}} = \frac{(30-1) \times 12.6}{42.557} \approx \frac{365.4}{42.557} \approx 8.59$

$\text{Upper bound} = \frac{(n-1) s^2}{\chi^2_{0.95}} = \frac{(30-1) \times 12.6}{17.708} \approx \frac{365.4}{17.708} \approx 20.63$

So, the 90% confidence interval for $\sigma^2$ when $n = 30$ is approximately:

• Lower bound: 8.59 (rounded to two decimal places),
• Upper bound: 20.63 (rounded to two decimal places).

Summary:

• (b) The lower bound is 8.59.
• The upper bound is 20.63.

Would you like further details on any of the steps or any additional explanations?

Here are 5 related questions to deepen your understanding:

1. How would the confidence interval change if the sample variance was larger?
2. How does the degrees of freedom affect the Chi-square distribution values?
3. Can you explain the concept of a confidence interval for the variance in simpler terms?
4. What is the relationship between the sample size and the width of the confidence interval?
5. How would you construct a 95% confidence interval for $\sigma^2$?

Tip: The Chi-square distribution is asymmetric, so the critical values for confidence intervals will not be equidistant from the center. Always use tables or software to find accurate values.