Listed below are student evaluation ratings of courses, where a rating of 5 is for "excellent." The ratings were obtained at one university in a state. Construct a confidence interval using a 99% confidence level. What does the confidence interval tell about the population of all college students in the state? 3.9, 2.9, 4.1, 4.6, 3.2, 4.2, 3.7, 4.6, 4.5, 4.1, 4.0, 3.6, 3.5, 4.1, 3.7 What is the confidence interval for the population mean μ? <<(Round to two decimal places as needed.) What does the confidence interval tell about the population of all college students in the state? Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. OA. We are confident that 99% of all students gave evaluation ratings between and (Round to one decimal place as needed.) OB. The results tell nothing about the population of all college students in the state, since the sample is from only one university. OC. We are 99% confident that the interval from to actually contains the true mean evaluation rating. (Round to one decimal place as needed.)

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### Constructing a 99% Confidence Interval for Student Evaluation Ratings

**Scenario:**
Listed below are student evaluation ratings of courses, where a rating of 5 is for "excellent." These ratings were obtained at one university in a state. Our objective is to construct a confidence interval using a 99% confidence level to understand what this interval tells us about the population of all college students in the state.

**Given Data:**
The student evaluation ratings are as follows:
3.9, 2.9, 4.1, 4.6, 3.2, 4.2, 3.7, 4.6, 4.5, 4.1, 4.0, 3.6, 3.5, 4.1, 3.7

**Task:**
1. **Determine the confidence interval for the population mean (μ).**

2. **What does the confidence interval tell about the population of all college students in the state?**
   - Select the correct choice below and fill in the answer box (if necessary) to complete your choice.

**Options:**
A. We are confident that 99% of all students gave evaluation ratings between ____ and ____.
   - *Note: Round to one decimal place as necessary.*

B. The results tell nothing about the population of all college students in the state, since the sample is from only one university.

C. We are 99% confident that the interval from ____ to ____ actually contains the true mean evaluation rating.
   - *Note: Round to one decimal place as necessary.*

**Steps to Solve:**

1. **Calculate the sample mean (\(\bar{x}\))** and **sample standard deviation (s)**:
   - \(\bar{x} = \frac{\sum{x_i}}{n}\)
   - \(s = \sqrt{\frac{\sum{(x_i - \bar{x})^2}}{n-1}}\)

2. **Determine the critical value (z\(_{\alpha/2}\)) for a 99% confidence interval**.

3. **Calculate the margin of error (E)**:
   - \(E = z\(_{\alpha/2}\) \cdot \frac{s}{\sqrt{n}}\)

4. **Determine the confidence interval (CI)**:
   - \(CI = \bar{x} \pm E\)

**Interpre
Transcribed Image Text:### Constructing a 99% Confidence Interval for Student Evaluation Ratings **Scenario:** Listed below are student evaluation ratings of courses, where a rating of 5 is for "excellent." These ratings were obtained at one university in a state. Our objective is to construct a confidence interval using a 99% confidence level to understand what this interval tells us about the population of all college students in the state. **Given Data:** The student evaluation ratings are as follows: 3.9, 2.9, 4.1, 4.6, 3.2, 4.2, 3.7, 4.6, 4.5, 4.1, 4.0, 3.6, 3.5, 4.1, 3.7 **Task:** 1. **Determine the confidence interval for the population mean (μ).** 2. **What does the confidence interval tell about the population of all college students in the state?** - Select the correct choice below and fill in the answer box (if necessary) to complete your choice. **Options:** A. We are confident that 99% of all students gave evaluation ratings between ____ and ____. - *Note: Round to one decimal place as necessary.* B. The results tell nothing about the population of all college students in the state, since the sample is from only one university. C. We are 99% confident that the interval from ____ to ____ actually contains the true mean evaluation rating. - *Note: Round to one decimal place as necessary.* **Steps to Solve:** 1. **Calculate the sample mean (\(\bar{x}\))** and **sample standard deviation (s)**: - \(\bar{x} = \frac{\sum{x_i}}{n}\) - \(s = \sqrt{\frac{\sum{(x_i - \bar{x})^2}}{n-1}}\) 2. **Determine the critical value (z\(_{\alpha/2}\)) for a 99% confidence interval**. 3. **Calculate the margin of error (E)**: - \(E = z\(_{\alpha/2}\) \cdot \frac{s}{\sqrt{n}}\) 4. **Determine the confidence interval (CI)**: - \(CI = \bar{x} \pm E\) **Interpre
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