91Ó°ÊÓ

Given In the question that We think that " $IQ$scores" will help explain "Grade point average". So $IQscore$is the explanatory variable and "Grade point average" is the response variable. Mean and standard deviation of these variables are given in the table.

localid="1649947222967" $$\begin{gathered} r=0.6337 \\ b=r\frac{Sy}{Sx} \\ a=\stackrel{¯}{Y}-b\stackrel{¯}{X} \end{gathered}$$

Using the least-squares line as a predictor of GPA from IQ, we must determine the equation.

According to a least-squares regression between IQ and GPA, its slope is:

$$\begin{gathered} b=0.06337×\frac{2.10}{13.17} \\ =0.1010 \end{gathered}$$

Since the least-squares line passes through$(\stackrel{¯}{X},\stackrel{¯}{Y)}$we use this information to find the intercept. :

localid="1649947236918" $$\begin{gathered} a=7.447-0.1010×108.9 \\ =-3.5519 \end{gathered}$$

Given in the question that GPA was records on $12$points scale withlocalid="1649437781844" $A+=12,A=11,A-=10,B+=9...D=1,F=0$. We have to find out that percent of the observed variation in these students’ GPAs can be explained by the linear relationship between GPA and IQ.

According to the least-squares regression line of $\mathrm{Y}$on $\mathrm{X}$, the coefficient of determination, $r^2$, represents the percentage of variation in $r$values.

Here,

$$\begin{gathered} r=0.6337 \\ {r}^2=0.4016 \end{gathered}$$

About $40.16$percent of the variation in $\mathrm{Y}$can be explained by the straight-line relationship between $y$and $X.$

$59.84$percent of the variance can't be explained by a linear relationship, which can be explained by individual variance.

Given in the question that One student has an IQ of $103$but a very low GPA of $0.53$we have to find and interpret the residual for this student

According to a least-squares regression between IQ and GPA, its slope is:

$\stackrel{Áåž}{Y}=-3.5519+0.1010X$

Calculating a student's GPA based on an IQ score$103$is

localid="1649947409386" $$\begin{gathered} \left(predicted\right)\stackrel{Áåž}{y}=-3.5519+0.1010\left(103\right) \\ =6.8511 \end{gathered}$$

However, for same student, observed GPA value is

localid="1649947416993" $$\begin{gathered} Y\left(observed\right)=0.53 \\ residual=observedY-predictedY \\ =Y-\hat{Y} \\ =0.53-6.8511 \\ =-6.3211 \end{gathered}$$

There is a negative residual for this data point, which shows below the line