Chapter 4.5, Problem 19E

a.

To determine

To write: An equation of line of best fit for given data and to further graph the points and plot the line.

Answer to Problem 19E

Equation of the of line of best fit is $y=-0.18x+19.7$ .

Explanation of Solution

Given Information: Data values of mileage, $x$ and price, $y$ is given.

Calculation:

Entering the data and using the linear regression feature in the graphing calculator,

  $\begin{array}{l}\text{LinReg}\\ \text{ y=ax+b}\\ \text{ a= -0}\text{.1836283185840708}\\ \text{ b= 19}\text{.716814159292035}\\ {\text{ r}}^2\text{= 0}\text{.9379169503063306}\\ \text{ r= -0}\text{.9684611248296602}\end{array}$

As $y=ax+b$ ,

Thus, equation of line of best fit is $y=-0.18x+19.7$ .

Graph:

  

b.

To determine

To identify and interpret: The correlation coefficient.

Answer to Problem 19E

  $r$ is $-0.96$ .

Data values have negative correlation.

Explanation of Solution

Given information: Graphing calculator display from the previous part.

Interpretation:

From the graphing calculator display in the previous part it can be inferred that correlation coefficient $r$ is $-0.96$ . Value of r is inclined towards $-1$ , this implies that data values have slight negative correlation and the points lie around a negative slope.

c.

To determine

To interpret: The slope and $y-\mathrm{int}ercept$ of the line of best fit.

Answer to Problem 19E

Slope is negative.

Line of best fit crosses the $y-axis$ at $19.7168$ , it represents the price of $\$19720$ for $0$ mileage.

Explanation of Solution

Given Information: Graphing calculator display from previous part.

Interpretation:

From the graphing calculator display shown in part $(a)$ , it can be inferred that slope $a$ is $-0.18$ , this implies that slope is negative. As the mileage increases, cost of car decreases. Therefore, car with more mileage can be bought cheaper than the one with less mileage.

  $y-\mathrm{int}ercept(b)$ is $19.71$ , this means that line of best fit crosses the $y-axis$ at $19.7168$ , so it represents the price of $\$19720$ for $0$ mileage

d.

To determine

To approximate: Mileage of automobile with cost $\$15500$ .

Answer to Problem 19E

The mileage would be $23.33$

  $miles$ .

Explanation of Solution

Given Information: Cost is $\frac{15500}{1000}=15.5$ .Equation from previous part.

Interpretation:

Substituting the value $y=15.5$ in equation derived in one of the previous parts,

  $\begin{array}{l}y=-0.18x+19.7\\ 15.5=-0.18x+19.7\\ -0.18x=-4.2\\ x=23.33\end{array}$

Thus, the mileage would be $23.33$

  $miles$ .

e.

To determine

To predict: Price of automobile with given mileage.

Answer to Problem 19E

Cost would be $\$18620.$

Explanation of Solution

Given Information: Mileage is $\frac{6000}{1000}=6$ . Equation from previous part.

Interpretation:

Substituting the value $x=6$ in equation derived in one of the previous parts,

  $\begin{array}{l}y=-0.18x+19.7\\ y=-0.18(6)+19.7\\ y=18.62\end{array}$

Thus, the cost would be $\$18620(=18.62\times 1000)$ .