Questions: Determine the critical values for the confidence interval for the population standard deviation from the given values. Round your answers to three decimal places. n=18 and α=0.01

Solution Steps

To find the confidence interval for the variance of a population with an unknown mean, we use the formula:

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

Given:

• Sample size \( n = 18 \)
• Sample variance \( s^2 = 10.0 \)
• Significance level \( \alpha = 0.01 \)

The degrees of freedom \( df = n - 1 = 17 \).

The confidence interval for the variance is calculated as:

\[ \left(\frac{17 \times 10.0}{\chi^2_{0.005}}, \frac{17 \times 10.0}{\chi^2_{0.995}}\right) \]

Using the chi-square distribution table, we find the critical values:

• \( \chi^2_{0.005} \approx 34.805 \)
• \( \chi^2_{0.995} \approx 6.908 \)

Thus, the confidence interval for the variance is:

\[ \left(\frac{170}{34.805}, \frac{170}{6.908}\right) = (4.759, 29.839) \]

To convert the variance confidence interval to a standard deviation confidence interval, we take the square root of each bound:

\[ \left(\sqrt{4.759}, \sqrt{29.839}\right) \]

Calculating these values gives:

\[ (2.1815, 5.4625) \]

Final Answer