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	46	30	47	23	40	34	52
y	30	20	25	13	29	17	21	14

Complete parts (a) through (e), given Σx = 299, Σy = 169, Σx² = 11,943, Σy² = 3861, Σxy = 5880, and

r ≈ −0.923.

(b) 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 =

(c) Find x, and y. Then find the equation of the least-squares line  = 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	=
	=	+  x

(d) Graph the least-squares line. Be sure to plot the point (x, y) as a point on the line.

(e) 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	%

(f) Suppose a car weighs x = 30 (hundred pounds). What does the least-squares line forecast for y = miles per gallon? (Round your answer to two decimal places.)
mpg

Answer 1

b)

ΣX =	299.000
ΣY=	169.000
ΣX2 =	11943.000
ΣY2 =	3861.000
ΣXY =	5880.000
r =	-0.923

c)

X̅=ΣX/n =	37.38
Y̅=ΣY/n =	21.13
ŷ =	42.365+(-0.568)x

e)

coefficient of determination r2 =				0.853
explained =				85.3%
unexplained=				14.7%

f)

predicted value =42.365+-0.568*30=

25.33