Chapter 3: Q R3.5. (page 214)

What’s my grade? In Professor Friedman’s economics course, the correlation
between the students’ total scores prior to the final examination and their final exam scores is r = 0.6. The pre-exam totals for all students in the course have a mean of 280 and a standard deviation of 30. The final exam scores have a mean of 75 and a standard deviation of 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.
a. Find the equation for the least-squares regression line Professor Friedman should use to make this prediction.
b. Use the least-squares regression line to predict Julie’s final exam score.
c. Explain the meaning of the phrase “least squares” in the context of this question.
d. Julie doesn’t think this method accurately predicts how well she did on the final exam. Determine r2. Use this result to argue that her actual score could have been much higher (or much lower) than the predicted value.

Short Answer

Expert verified

Part (a) $\overset{⏜}{y} = 30.2 + 0.16 x$

Part (b) $78.2$ is Julie’s exam score.

Part (c) The squared disparities between the actual final exam score and the anticipated final exam score are minimized using the least-squares regression line.

Part (d) $r^{2} = 0.36 = 36 %$

Step by step solution

Part (a) Step 1: Given information

$r = 0.6 x = 280 y = 75 s_{x} = 30 s_{y} = 8$

Part (a) Step 2: Explanation

Thus the slope and the constant will be calculated as:

$b = r \frac{s_{y}}{s_{x}}$

$= 0.6 \times \frac{8}{30} = 0.16$

$a = y - b x = 75 - 0.16 \times 280 = 30.2$

Thus the regression line will be as:

$y = a + b x \Rightarrow \overset{⏜}{y} = 30.2 + 0.16 x$

Part (b) Step 1: Calculation

The regression line is:

$\overset{⏜}{y} = 30.2 + 0.16 x$

Thus, the Julie’s final exam score be as:

$\overset{⏜}{y} = 30.2 + 0.16 x = 30.2 + 0.16 (300) = 78.2$

Part (c) Step 1: Calculation

The least-squares regression line is the one that reduces the squared residuals to the smallest possible value. The residuals are the disparities between the observed a the projected values. As a result, the least-squares regression line minimizes the squared differences between the actual final exam score and the predicted final exam score.

Part (d) Step 1: Explanation

The value $r^{2}$ is calculated as:

$r^{2} = {0.6}^{2} = 0.36 = 36 %$

This suggests that the regression line can account for $36 %$ of the variation between the variables. Because the regression line only explains $36 %$ of the variation in the final exam score, the final result could diverge significantly from the regression line's anticipated values.