In Professor Friedman’s econom­ ics course, the correlation between the students’ total scores prior to the final examination and their final- examination scores is r = 0.6. The pre-exam totals for all students in the course have mean 280 and stan­ dard deviation 30. The final-exam scores have mean 75 and standard deviation 8. Professor Friedman has lost Julie’s final exam but knows that her total before the exam was 300. He decides to predict her final- exam score from her pre-exam total.

This problem revolves around predicting a value using the correlation coefficient, which suggests that we can use a **linear regression model**. Here, we are tasked with predicting Julie’s final exam score based on her pre-exam score, using the information given about the relationship between the two.

The general form of a **linear regression equation** for prediction is:

\[
\hat{y} = \bar{y} + r \cdot \frac{s_y}{s_x} \cdot (x - \bar{x})
\]

Where:
- \( \hat{y} \) is the predicted value (Julie's final exam score).
- \( \bar{y} \) is the mean of the final exam scores.
- \( r \) is the correlation coefficient.
- \( s_y \) is the standard deviation of the final exam scores.
- \( s_x \) is the standard deviation of the pre-exam scores.
- \( x \) is the observed value of the pre-exam score (Julie's pre-exam score).
- \( \bar{x} \) is the mean of the pre-exam scores.

### Given Data:
- \( r = 0.6 \) (correlation between pre-exam and final-exam scores)
- \( \bar{x} = 280 \) (mean of pre-exam scores)
- \( s_x = 30 \) (standard deviation of pre-exam scores)
- \( \bar{y} = 75 \) (mean of final exam scores)
- \( s_y = 8 \) (standard deviation of final exam scores)
- Julie’s pre-exam score \( x = 300 \)

### Step-by-Step Solution:
1. **Calculate the difference between Julie's pre-exam score and the mean pre-exam score:**
   \[
   x - \bar{x} = 300 - 280 = 20
   \]

2. **Calculate the scaling factor using the correlation and the standard deviations:**
   \[
   r \cdot \frac{s_y}{s_x} = 0.6 \cdot \frac{8}{30} = 0.6 \cdot 0.2667 = 0.16
   \]

3. **Multiply the difference by the scaling factor:**
   \[
   0.16 \cdot 20 = 3.2
   \]

4. **Add this result to the mean of the final exam scores to get Julie’s predicted final exam score:**
   \[
   \hat{y} = 75 + 3.2 = 78.2
   \]

Thus, **Julie’s predicted final exam score is 78.2**.

### Would you like any further details or have any questions on the process?

---

Here are 5 related questions to expand your understanding:
1. How would Julie’s predicted score change if her pre-exam total had been 310 instead of 300?
2. What would the predicted score be if the correlation \( r \) were higher or lower?
3. Can you explain the significance of the correlation coefficient in this context?
4. What is the role of standard deviations in the regression equation?
5. How accurate is this prediction, and what factors could affect its precision?

**Tip:** When using linear regression, the closer the correlation is to 1 or -1, the stronger the linear relationship and the more accurate the predictions will be.

In Professor Friedman’s econom­ics course, the correlation between the students’ total scores prior to the final examination and their final- examination scores is r = 0.6. The pre-exam totals for all students in the course have mean 280 and stan­dard deviation 30. The final-exam scores have mean 75 and standard deviation 8. Professor Friedman has lost Julie’s final exam but knows that her total before the exam was 300. He decides to predict her final- exam score from her pre-exam total.

Learn how to predict Julie’s final exam score in an economics course using linear regression. The problem involves a correlation coefficient of 0.6 between pre-exam and final exam scores, with given means and standard deviations. Step-by-step solution included. Ideal for college-level statistics or economics students.

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