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 42 31 47 23 40 34 52

y 33 21 24 13 29 17 21 14

Complete parts (a) through (e), given Σx = 296, Σy = 172, Σx2 = 11,652, Σy2 = 4042, Σxy = 5917, and r ≈ −0.911. (a) Draw a scatter diagram displaying the data.

(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 =2 Σy =3 Σx2 =4 Σy2 =5 Σxy =6 r =7

(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 = ___ = 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 = 39 (hundred pounds). What does the least-squares line forecast for y = miles per gallon? (Round your answer to two decimal places.) ________mpg

Answer 1

	X	Y	X * Y	X2	Y2
	27	33	891	729	1089
	42	21	882	1764	441
	31	24	744	961	576
	47	13	611	2209	169
	23	29	667	529	841
	40	17	680	1600	289
	34	21	714	1156	441
	52	14	728	2704	196
Total	296	172	5917	11652	4042

r = ( ( N * \Sigma XY) - \Sigma X \Sigma Y) / (\sqrt{ (N \Sigma (X^{2}) - (\Sigma X)^{2} )(N \Sigma (Y^{2}) - (\Sigma Y)^{2} })
r = ( ( 8 * 5917 ) - 296 * 172 ) / ( \sqrt{ ( 8 * 11652 ) - ( 296 )^{2} )( 8 * 4042 ) - ( 172 )^{2} })
r = -0.911

X̅ = Σ( Xi / n ) = 296/8 = 37
Y̅ = Σ( Yi / n ) = 172/8 = 21.5

Equation of regression line is Ŷ = a + bX


b = -0.639
a =( Σ Y - ( b * Σ X) ) / n
a =( 172 - ( -0.6386 * 296 ) ) / 8
a = 45.127
Equation of regression line becomes Ŷ = 45.1271 - 0.6386 X

Coefficient of Determination
R² = r² = 0.830

Explained variation = 0.83* 100 = 83%
Unexplained variation = 1 - 0.83* 100 = 17%

When X = 39
Ŷ = 45.127 + -0.639 X
Ŷ = 45.127 + ( -0.639 * 39 )
Ŷ = 20.21