Construct the confidence interval for the population variance for the given values. Round your answers to one decimal place.

n=11
, s2=48.7
, and c=0.95

To construct the confidence interval for the population variance, we use the chi-square distribution. The formula for the confidence interval for population variance \( \sigma^2 \) is:

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

Where:
- \( n \) is the sample size
- \( s^2 \) is the sample variance
- \( \alpha = 1 - c \) is the significance level
- \( \chi^2_{\frac{\alpha}{2}} \) and \( \chi^2_{1-\frac{\alpha}{2}} \) are the critical values from the chi-square distribution

### Step-by-Step Solution:

1. Given values:
   - \( n = 11 \)
   - \( s^2 = 48.7 \)
   - Confidence level \( c = 0.95 \), so \( \alpha = 1 - 0.95 = 0.05 \)

2. Find the degrees of freedom:
   \[
   df = n - 1 = 11 - 1 = 10
   \]

3. Find the critical values for the chi-square distribution at \( \alpha/2 \) and \( 1-\alpha/2 \). For a 95% confidence interval and 10 degrees of freedom:
   - \( \chi^2_{\frac{0.05}{2}} = \chi^2_{0.025} = 20.483 \) (right-tail critical value)
   - \( \chi^2_{1-\frac{0.05}{2}} = \chi^2_{0.975} = 3.247 \) (left-tail critical value)

4. Calculate the confidence interval for the population variance:
   - Lower bound:
     \[
     \frac{(n-1) \cdot s^2}{\chi^2_{0.025}} = \frac{(11-1) \cdot 48.7}{20.483} = \frac{10 \cdot 48.7}{20.483} = 23.8
     \]
   - Upper bound:
     \[
     \frac{(n-1) \cdot s^2}{\chi^2_{0.975}} = \frac{(11-1) \cdot 48.7}{3.247} = \frac{10 \cdot 48.7}{3.247} = 150.0
     \]

### Conclusion:

The 95% confidence interval for the population variance is:

\[
(23.8, 150.0)
\]

Let me know if you'd like further details or have any questions!

### Related Questions:
1. How do we find the critical values for different confidence levels in chi-square distribution?
2. How is the formula for confidence interval of population standard deviation derived?
3. Can we calculate the confidence interval for population variance with unequal sample sizes?
4. What would the interval look like for a 99% confidence level?
5. How does increasing the sample size affect the confidence interval?

**Tip:** The width of a confidence interval decreases as the sample size increases. This makes the estimate of population variance more precise.

Construct the confidence interval for the population variance for the given values. Round your answers to one decimal place. n=11, s^2=48.7, and c=0.95.

Calculate the 95% confidence interval for the population variance with sample size n=11, sample variance s^2=48.7, using the chi-square distribution. This step-by-step guide provides critical values and the final interval, (23.8, 150.0).

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