Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation $\sigma$. Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation. To find the standard deviation of the diameter of wooden dowels, the manufacturer measures 19 randomly selected dowels and finds the standard deviation of the sample to be $s=0.16$. Find the $95 \%$ confidence interval for the population standard deviation $\sigma$. $0.13<\sigma<0.22$ $0.15<\sigma<0.21$ $0.12<\sigma<0.24$ $0.11<\sigma<0.25$ - Previous Next II app.honorlock.com is sharing your screen. Stop sharing Webcam

Step 1 ：Given values are: sample size \(n = 19\), sample standard deviation \(s = 0.16\), and confidence level \(0.95\).

Step 2 ：Calculate alpha, which is \(1 - \text{confidence level} = 1 - 0.95 = 0.05\).

Step 3 ：Calculate degrees of freedom, which is \(n - 1 = 19 - 1 = 18\).

Step 4 ：Calculate chi-square values for the lower and upper limits of the confidence interval. The lower limit chi-square value is \(8.23\) and the upper limit chi-square value is \(31.53\).

Step 5 ：Calculate the confidence interval for the population standard deviation \(\sigma\). The lower limit is \(\sqrt{(df \times s^2) / \text{chi2 upper}} = \sqrt{(18 \times 0.16^2) / 31.53} = 0.12\) and the upper limit is \(\sqrt{(df \times s^2) / \text{chi2 lower}} = \sqrt{(18 \times 0.16^2) / 8.23} = 0.24\).

Step 6 ：Final Answer: The \(95 \%\) confidence interval for the population standard deviation \(\sigma\) is \(\boxed{0.12<\sigma<0.24}\).