use technology to construct the confidence intervals for the population variance o^2 and the population standard deviation o.  Assume the sample is taken from a normally distributed population. c=0.95, s=33, n=15

To construct the confidence intervals for the population variance \( \sigma^2 \) and population standard deviation \( \sigma \), we need to use the chi-square distribution, given that the sample is taken from a normally distributed population.

### 1. **Confidence Interval for Population Variance (\( \sigma^2 \)):**

The confidence interval for the population variance \( \sigma^2 \) is calculated using the following formula:

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

Where:
- \( s^2 \) is the sample variance.
- \( n \) is the sample size.
- \( \alpha = 1 - c \) is the level of significance (for 95% confidence level, \( \alpha = 0.05 \)).
- \( \chi^2_{\alpha/2} \) and \( \chi^2_{1-\alpha/2} \) are the critical values from the chi-square distribution with \( n - 1 \) degrees of freedom.

In this case:
- \( s = 33 \), so \( s^2 = 33^2 = 1089 \).
- \( n = 15 \), so \( n - 1 = 14 \).
- \( \alpha = 1 - 0.95 = 0.05 \), so \( \alpha/2 = 0.025 \).

We need to find the critical values \( \chi^2_{0.025} \) and \( \chi^2_{0.975} \) for 14 degrees of freedom.

From the chi-square distribution table:
- \( \chi^2_{0.025, 14} = 26.119 \)
- \( \chi^2_{0.975, 14} = 5.629 \)

Now, substitute these values into the formula:

\[
\text{Lower bound} = \frac{(15 - 1) \times 1089}{26.119} = \frac{14 \times 1089}{26.119} \approx 583.84
\]

\[
\text{Upper bound} = \frac{(15 - 1) \times 1089}{5.629} = \frac{14 \times 1089}{5.629} \approx 2712.98
\]

Thus, the 95% confidence interval for the population variance \( \sigma^2 \) is:

\[
(583.84, 2712.98)
\]

### 2. **Confidence Interval for Population Standard Deviation (\( \sigma \)):**

The confidence interval for the population standard deviation \( \sigma \) is simply the square root of the confidence interval for the variance \( \sigma^2 \):

\[
\left( \sqrt{583.84}, \sqrt{2712.98} \right)
\]

\[
\text{Lower bound} = \sqrt{583.84} \approx 24.15
\]
\[
\text{Upper bound} = \sqrt{2712.98} \approx 52.09
\]

Thus, the 95% confidence interval for the population standard deviation \( \sigma \) is:

\[
(24.15, 52.09)
\]

### Summary:
- **Confidence Interval for \( \sigma^2 \) (Population Variance):** \( (583.84, 2712.98) \)
- **Confidence Interval for \( \sigma \) (Population Standard Deviation):** \( (24.15, 52.09) \)

Would you like further details on the steps or any specific clarifications?

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Here are 5 related questions you might consider:
1. How does changing the confidence level affect the confidence interval width?
2. What happens to the interval when the sample size increases?
3. How are the degrees of freedom in a chi-square distribution determined?
4. What is the relationship between the sample variance and the population variance?
5. How do confidence intervals for proportions differ from those for variances?

**Tip:** Always check that the population follows a normal distribution when using chi-square methods for variance!

use technology to construct the confidence intervals for the population variance o^2 and the population standard deviation o. Assume the sample is taken from a normally distributed population. c=0.95, s=33, n=15

Learn how to construct 95% confidence intervals for population variance and standard deviation using the chi-square distribution. With a sample size of 15 and a sample standard deviation of 33, this solution walks through each step to find the confidence intervals for both variance and standard deviation. Ideal for college-level statistics students.

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