To construct a 95% confidence interval for the population standard deviation $\sigma$, we use the chi-square distribution since the sample standard deviation $s$ is given. The formula for the confidence interval is:

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

Step-by-step Solution:

1. Given Data:

   • Sample size: $n = 18$
   • Sample standard deviation: $s = 20$
   • Degrees of freedom: $df = n - 1 = 18 - 1 = 17$
   • Confidence level: 95% → $\alpha = 0.05$ → $\alpha/2 = 0.025$

2. Find critical chi-square values:

   • From chi-square tables or software:
     • $\chi^2_{\alpha/2} = \chi^2_{0.025, 17} = 30.191$
     • $\chi^2_{1-\alpha/2} = \chi^2_{0.975, 17} = 8.672$

3. Compute the confidence interval:

   • The formula for the variance $\sigma^2$ is: $\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$
   • Plugging in values for the variance: $\frac{(17)(20^2)}{30.191} = \frac{6800}{30.191} \approx 225.17$ $\frac{(17)(20^2)}{8.672} = \frac{6800}{8.672} \approx 783.87$
   • Taking square roots to get $\sigma$: $\sqrt{225.17} \approx 15.01, \quad \sqrt{783.87} \approx 29.98$

4. Confidence Interval for $\sigma$: $(15.01, 29.98)$

Final Answer:

D. (15.01, 29.98)

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Would you like more details on confidence intervals or other related topics?

Expanding Questions:

1. What are the uses of confidence intervals for population parameters in real-world applications?
2. How is the chi-square distribution different from other probability distributions?
3. Can confidence intervals be used for other statistics (e.g., mean or proportion)?
4. What happens to the confidence interval if the sample size increases?
5. Why do we use degrees of freedom ($n-1$) instead of just $n$ in the chi-square method?

Tip:

For small sample sizes, always double-check the critical values from the chi-square table to avoid errors.