Questions: 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 a. Use the 93 course evaluations to construct a 99% confidence interval estimate of the standard deviation of the population from which the sample was b. Repeat part (a) using the 93 professor evaluations c. Compare the results from part (a) and part (b).

Solution Steps

For the course evaluations, we have:

• Mean \( \mu = \frac{\sum x_i}{n} = \frac{386.0}{89} \approx 4.34 \)
• Variance \( \sigma^2 = \frac{\sum (x_i - \mu)^2}{n-1} = 0.11 \)
• Standard Deviation \( \sigma = \sqrt{0.11} \approx 0.33 \)

The confidence interval for the variance of a single population with unknown population mean is given by:

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

Substituting the values:

\[ CI = \left(\frac{(89 - 1) \times 0.11}{\chi^2_{\alpha/2}}, \frac{(89 - 1) \times 0.11}{\chi^2_{1 - \alpha/2}}\right) \approx (0.08, 0.17) \]

For the professor evaluations, we have:

• Mean \( \mu = \frac{\sum x_i}{n} = \frac{394.7}{89} \approx 4.43 \)
• Variance \( \sigma^2 = \frac{\sum (x_i - \mu)^2}{n-1} = 0.1 \)
• Standard Deviation \( \sigma = \sqrt{0.1} \approx 0.32 \)

The confidence interval for the variance of a single population with unknown population mean is given by:

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

Substituting the values:

\[ CI = \left(\frac{(89 - 1) \times 0.1}{\chi^2_{\alpha/2}}, \frac{(89 - 1) \times 0.1}{\chi^2_{1 - \alpha/2}}\right) \approx (0.07, 0.15) \]

The confidence intervals for the variances are:

• Course Evaluations CI: \( (0.08, 0.17) \)
• Professor Evaluations CI: \( (0.07, 0.15) \)

Final Answer