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	25	42	33	47	23	40	34	52
y	31	22	25	13	29	17	21	14

Complete parts (b) through (e), given Σx = 296, Σy = 172, Σx² = 11,676, Σy² = 4006, Σxy = 5924, and

r ≈ −0.932.

(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 = 42 (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 =	296.000
ΣY=	172.000
ΣX2 =	11676.000
ΣY2 =	4006.000
ΣXY =	5924.000
r =	-0.932

c)

X̅=ΣX/n =	37.00
Y̅=ΣY/n =	21.50
ŷ =	43.986+(-0.608)x

e)

coefficient of determination r2 =				0.868
explained =				86.8%
unexplained=				13.2%

f)

predicted value =43.986+-0.608*42=

18.45