Question 5 0.1 pts 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. A sociologist develops a test to measure attitudes about public transportation, and 27 randomly selected subjects are given the test. Their mean score is 76.2 and their standard deviation is 21.4 . Construct the $95 \%$ confidence interval for the standard deviation, $\sigma$, of the scores of all subjects. $17.5<\sigma<27.8$ $16.9<\sigma<29.3$ $17.2<\sigma<27.2$ $16.6<\sigma<28.6$ - Previous Next II app.honorlock.com is sharing your screen. Stop sharing Hide Webcam

Step 1 ：First, we need to calculate the chi-square values for the given degree of confidence and degrees of freedom. The degree of confidence is 95%, so \( \alpha = 0.05 \). The degrees of freedom are \( n-1 = 27-1 = 26 \).

Step 2 ：We can use the chi-square distribution to calculate the chi-square values. The lower limit of the confidence interval is given by \( \sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}}} \) and the upper limit is given by \( \sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}}} \).

Step 3 ：Substituting the given values, we get \( \chi^2_{lower} = 41.92317009635392 \) and \( \chi^2_{upper} = 13.843904982007603 \).

Step 4 ：Next, we calculate the lower and upper limits of the confidence interval. The lower limit is \( \sqrt{\frac{(n-1)s^2}{\chi^2_{lower}}} = 16.852851335356355 \) and the upper limit is \( \sqrt{\frac{(n-1)s^2}{\chi^2_{upper}}} = 29.32723656706524 \).

Step 5 ：The lower limit of the confidence interval is approximately 16.85 and the upper limit is approximately 29.33. These values are rounded to the same number of decimal places as the sample standard deviation, which is 1 decimal place.

Step 6 ：Therefore, the confidence interval for the population standard deviation is \( \boxed{16.9<\sigma<29.3} \).