91Ó°ÊÓ

Because the scatterplot's pattern is essentially linear and lacks substantial curvature. Furthermore, the points appear to vary little around the regression line, implying that the linear model is adequate. The residual pattern has no noticeable curvature and has nearly the same vertical spread throughout.

The correlation will be positive because the slope in the computer output is positive. As a result, it is stated in the computer output that,

$$\begin{gathered} r^2=83.7\% \\ =0.837 \end{gathered}$$

Thus the correlation will be as:

$$\begin{gathered} r=+\sqrt{r^2} \\ =+\sqrt{0.837} \\ =0.9149 \end{gathered}$$

There is a positive association. This indicates that age and mileage have a favorable association. And if it's near to one, it's powerful.

It is given in the computer output that:

$$\begin{gathered} a=3704 \\ b=12188 \end{gathered}$$

Thus the regression line will be as:

$$\begin{gathered} \stackrel{Áåœ}{y}=a+bx \\ ⇒\stackrel{Áåœ}{y}=3704+12188x \end{gathered}$$

Where $x$ is age and $y$ is mileage.

The regression line is:

$\stackrel{Áåœ}{y}=3704+12188x$

Thus the predicted value is :

$$\begin{gathered} \stackrel{Áåœ}{y}=3704+12188x \\ =3704+12188\left(6\right) \\ =76832 \end{gathered}$$

Thus the residual is as:

$$\begin{gathered} Residual=y−\stackrel{Áåœ}{y} \\ =65000−76832 \\ =−11863 \end{gathered}$$

This implies that the predicted mileage is $11863$above the actual mileage when the age is six years.

It is given in the computer output that:

The least-squares regression equation, which uses $x$ (vehicle age) to estimate $y$ (number of miles driven mileage), is normally wrong by $20870$ miles. The linear model connecting mileage to vehicle age accounts for $83.7$ percent of the variation in mileage.