91影视

It is given in the question the table in which the values are calculated. Thus, from table we have,

The value of $r^2$is given in the output as 鈥淩-Sq鈥�:

$r^2=77\%=0.77$

From this, we interpret that$77\%$ of the variation between the variables has been explained by the least squares regression line.

As we know that the least square regression line is as:

$\hat{y}=a+bx$

With the predicted mileage and the age.

The constant $a$is given in the row with 鈥淎ge鈥� and in the column with 鈥淐oef鈥�:

$a=-13832$

The slope $b$is given in the row with 鈥淎ge鈥� and in the column with 鈥淐oef鈥�:

$b=14954$

Then the least square equation then becomes:

$\hat{y}=-13832+14954x$

With $\hat{y}$the predicted mileage and $x$the age.

The point $(\stackrel{炉}{x},\stackrel{炉}{y})$lies on the least square regression line thus the mean mileage can be obtained by replacing $x$by $\stackrel{炉}{x}$in the least square equation and evaluating the expression:

$$\begin{gathered} \stackrel{炉}{y}=\hat{y} \\ =-13832+14954\stackrel{炉}{x} \\ =-13832+14954\left(8\right) \\ =105800 \end{gathered}$$

Thus, the mean mileage islocalid="1664182337396" $105800$.

In the question it is given the table of calculated values.

Thus, from that it is given the value of $s$, it is as:

$s=22723$

Thus, from this we interpret that the error made when predicting the mileage using the least square regression line is on average $22723$ miles.

It is not be reasonable to use the least square line to predict a car鈥檚 mileage from its age for a Council high school teacher because the least-squares line was determined using data of students. The data of students is not representative for the teachers and thus we cannot use the least-squares line to predict the mileage of cars of teachers.