Step 1 :Given the student evaluation ratings of courses at a university in a state, we are asked to construct a confidence interval using a 99% confidence level. The ratings are as follows: 3.9, 3.1, 3.8, 4.7, 2.8, 4.2, 3.5, 4.3, 4.3, 4.2, 4.1, 3.6, 3.3, 4.0, 3.6.
Step 2 :To find the confidence interval for the population mean, we first need to calculate the sample mean and the sample standard deviation. The formula for the confidence interval is given by: \(\bar{x} \pm Z \frac{s}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, Z is the Z-score which corresponds to the desired confidence level (for a 99% confidence level, the Z-score is approximately 2.576), s is the sample standard deviation, and n is the sample size.
Step 3 :Calculating the sample mean, we get \(\bar{x} = 3.83\).
Step 4 :Calculating the sample standard deviation, we get s = 0.51.
Step 5 :The sample size n is 15.
Step 6 :The Z-score for a 99% confidence level is approximately 2.576.
Step 7 :Substituting these values into the formula, we get the confidence interval as \((3.49, 4.17)\).
Step 8 :\(\boxed{\text{Final Answer: The 99% confidence interval for the population mean } \mu \text{ is approximately } (3.49, 4.17). \text{ This means that we are 99% confident that the true population mean of the student evaluation ratings falls within this interval.}}\)