Listed in the accompanying data table are student evaluation ratings of courses and professors, where a rating of 5 is for "excellent." Assume that each sample is a simple random sample obtained from a population with a normal distribution.
a. Use the 93 course evaluations to construct a $98 \%$ confidence interval estimate of the standard deviation of the population from which the sample was obtained.
b. Repeat part (a) using the 93 professor evaluations.
c. Compare the results from part (a) and part (b).
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a. $< \sigma< $
(Round to two decimal places as needed.)

Step 1 ：Assume that each sample is a simple random sample obtained from a population with a normal distribution.

Step 2 ：Use the 93 course evaluations to construct a 98% confidence interval estimate of the standard deviation of the population from which the sample was obtained.

Step 3 ：Repeat the previous step using the 93 professor evaluations.

Step 4 ：Compare the results from the two previous steps.

Step 5 ：To construct a confidence interval for the standard deviation, we can use the chi-square distribution. The formula for the confidence interval is given by: \[\sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}}}\] where: n is the sample size, s is the sample standard deviation, and \(\chi^2_{\alpha/2, n-1}\) and \(\chi^2_{1-\alpha/2, n-1}\) are the chi-square values for the degrees of freedom (n-1) and the significance level \(\alpha\).

Step 6 ：However, the data is not provided in the question. Therefore, we cannot generate the code to solve this problem.

Step 7 ：If the data was provided, we would first calculate the sample standard deviation, then use the chi-square distribution to calculate the confidence interval for the standard deviation.

Step 8 ：Final Answer: Without the data, we cannot calculate the confidence interval for the standard deviation. \(\boxed{\text{Data not provided}}\)