91Ó°ÊÓ

$r=0.6$ ,

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: $$\begin{gathered} b=0.6×\frac{8}{30} \\ =0.16 \end{gathered}$$

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

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

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: $$\begin{gathered} \stackrel{Áåœ}{Y}=30.2+0.16X \\ =30.2+0.15×300 \\ =78.2 \end{gathered}$$

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 = $36\%$