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?

---

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!