Exercise 19 Page 197

a To perform a linear regression, we first have to enter the values into lists. Push

STAT

, choose Edit, and then enter the values in the first two columns.

To do a linear regression we push $STAT$ , scroll right to CALC, and then choose the fourth option in the list, LinReg.

We can see the equation for the line of best fit.

y = - 0.2 x + 20

b We can find the correlation coefficient

r

on the screen with linear regression results.

Therefore the correlation coefficient is approximately $r = - 0.968 .$ This tells us that correlation is both negative and very strong. We know that it is strong because it is extremely close to $- 1$ . A correlation of $- 1$ would be a direct correlation explained by a line that goes through all of the points.

c Since the data shows the price in thousands of dollars and mileage in thousands of miles, the slope of

- 0.2

means that for every

1000

miles, a car is decreasing in value by

$ 200 .

Meanwhile, the

y -

intercept has no interpretation because a used car cannot have the mileage equal to

0 .

d In order to estimate the mileage of a car that costs

$ 15500

we have to substitute

15.5

for

y

in the equation for the line of the best fit.

y = - 0.2 x + 20

Substitute

$y = 15.5$

15.5 = - 0.2 x + 20

SubEqn

$LHS - 20 = RHS - 20$

- 4.5 = - 0.2 x

MultEqn

$LHS \cdot (- 5) = RHS \cdot (- 5)$

22.5 = x

RearrangeEqn

Rearrange equation

x = 22.5

This means that the mileage of the car with a price of

$ 15500

is equal to

22500

mi.

e In order to estimate the price of a car with

6000

miles we have to substitute

6

for

x

in the equation for the line of the best fit.

y = - 0.2 x + 20

Substitute

$x = 6$

y = - 0.2 (6) + 20

MultNegPos

$(- a) b = - a b$

y = - 1.2 + 20

AddTerms

Add terms

y = 18.8

This means that the price of a car with

6000

miles is equal to

$ 18800 .

Subchapter links

Communicate Your Answer p.191

Monitoring Progress p.193-195

Exercises p.196-198