Exercise 20 Page 207

Practice makes perfect

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 = 4.9 x - 37.7

b We can find the correlation coefficient

r

on the screen with linear regression results.

Therefore the correlation coefficient is approximately $r = 0.936 .$ This tells us that correlation is both positive 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 cost in thousands of dollars and the length in feet, the slope of

4.9

means that for every

1

foot, a sailboat increases in value by

$ 4900 .

Meanwhile, the

y -

intercept has no interpretation because a sailboat cannot have the length of

0

feet.

d In order to estimate the cost of a sailboat that is

20

feet long, we have to substitute

20

for

x

in the equation for the line of the best fit.

y = 4.9 x - 37.7

Substitute

$x = 20$

y = 4.9 (20) - 37.7

Multiply

y = 98 - 37.7

SubTerm

Subtract term

y = 60.3

This means that the cost of the sailboat with a length of

20

feet is equal to

$ 60300 .

e In order to estimate the length of a sailboat that costs

$ 147000,

we have to substitute

147

for

y

in the equation for the line of the best fit.

y = 4.9 x - 37.7

Substitute

$y = 147$

147 = 4.9 x - 37.7

AddEqn

$LHS + 37.7 = RHS + 37.7$

184.4 = 4.9 x

DivEqn

$LHS / 4.9 = RHS / 4.9$

x = 37.63265 \dots

RoundDec

Round to $1$ decimal place(s)

x \approx 37.6

This means that the length of a sailboat that costs

$ 147000

is approximately equal to

37.6

feet.

Subchapter links

Communicate Your Answer p.201

Monitoring Progress p.203-205

Exercise 20 Page 207

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