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