Question

Do heavier cars really use more gasoline? Suppose a car is chosen at random. Let x be the weight of the car (in hundreds of pounds), and let y be the miles per gallon (mpg).

x	27	43	31	47	23	40	34	52
y	30	20	25	13	29	17	21	14

Complete parts (a) through (d), given Σx = 297, Σy = 169, Σx2 = 11,737, Σy2 = 3861, Σxy = 5845, and r ≈ −0.944.

(a) Verify the given sums Σx, Σy, Σx2, Σy2, Σxy, and the value of the sample correlation coefficient r. (Round your value for r to three decimal places.)

Σx =

Σy =

Σx2 =

Σy2 =

Σxy =

r =

(b) Find x, and y. Then find the equation of the least-squares line ŷ = a + bx. (Round your answers for x and y to two decimal places. Round your answers for a and b to three decimal places.)

x=

y=

ŷ = + x

(c) Find the value of the coefficient of determination r2. What percentage of the variation in y can be explained by the corresponding variation in x and the least-squares line? What percentage is unexplained? (Round your answer for r2 to three decimal places. Round your answers for the percentages to one decimal place.)

r2 =

explained = %

unexplained = %

(d) Suppose a car weighs x = 40 (hundred pounds). What does the least-squares line forecast for y = miles per gallon? (Round your answer to two decimal places.)

= mpg

Answer 1

ΣX = 297
ΣY = 169
ΣX * Y = 5845
ΣX2 = 11737
ΣY2 = 3861

Part a)

r = -0.944

part b)

X̅ = Σ( Xi / n ) = 297/8 = 37.12
Y̅ = Σ( Yi / n ) = 169/8 = 21.12

Equation of regression line is Ŷ = a + bX


b = -0.604
a =( Σ Y - ( b * Σ X) ) / n
a =( 169 - ( -0.6037 * 297 ) ) / 8
a = 43.536
Equation of regression line becomes Ŷ = 43.5358 - 0.6037 X.

part c)

Coefficient of Determination
= 0.891
Explained variation = 0.891* 100 = 89.1%
Unexplained variation = 1 - 0.891* 100 = 10.9%

part d)

When X = 40
Ŷ = 43.536 + -0.604 X
Ŷ = 43.536 + ( -0.604 * 40 )
Ŷ = 19.38