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-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.

(a) Find the equation for the appropriate least-squares regression line for Professor Friedman’s prediction. Interpret the slope of this line in context.

(b) Use the regression line to predict Julie’s final exam score.

(c) Julie doesn’t think this method accurately predicts how well she did on the final exam. Determine $r^2$ Use this result to argue that her actual score could have been much higher (or much lower) than the predicted value.

$r=0.6$ ,

The Pre −exam:

mean $280$ and standard deviation $30$,

The final-exam:

mean $75$ and standard deviation $8$

A regression line is a straight line that shows how an explanatory variable $x$ affects a response variable $y$ You can use a regression line to forecast the value of $y$ for any value of $x$ by plugging this $x$ into the equation of the line.

The following is the formula for calculating the slope of the least-square regression line of Final test scores based on pre-exam scores: $$\begin{gathered} b=0.6\times \frac{8}{30} \\ =0.16 \end{gathered}$$

Also, the $Y$-intercept is calculated as follows:

$$\begin{gathered} a=\stackrel{⏜}{Y}-b\stackrel{⏜}{X} \\ =75-0.15x280 \\ =30.2 \end{gathered}$$

Hence, the equation of the least-squares line is: $Y=30.2+0.16X$

The rate at which the anticipated response $Y$ moves among the line as the explanatory variable $X$ changes is represented by the slope $b$ of the aforementioned regression line $(0.16)$ Specifically, when $X$ grows by one unit, $b$ is the projected change in $Y$

Here, $b$ is positive, indicating that as pre-exam totals rise by one mark, the final examination result rises by $0.16$ points. Therefore, the equation of the least-squares regression line is $Y=30.2+0.16X$

Julie's pre-exam total was $300$, so that's a given.

The following is how the estimated final examination score is calculated: $$\begin{gathered} \stackrel{⏜}{Y}=30.2+0.16X \\ =30.2+0.15\times 300 \\ =78.2 \end{gathered}$$

Julie’s predicted final examination score is $78.2$

The correlation between the totals from the pre-exam and the final examination result is 0.6.

That is, $r=0.6$

$$\begin{gathered} r^2={\left(0.6\right)}^2 \\ =0.36 \end{gathered}$$

That is, R-square=$36\%$

The least-squares regression line of the final examination score to the pre-exam total accounts for $36$ percent of the variation in the final examination score, according to the $R$-square value.

This means that the linear relationship for these data cannot account for $64\%$of the variation in final examination scores. Julie's final examination score may not be as accurate as Professor Friedman predicted. Hence, the value is R-square = $36\%$