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 30 44 33 47 23 40 34 52 y 33 21 22 13 29 17 21 14 Complete parts (a) through (e), given Σx = 303, Σy = 170, Σx2 = 12,123, Σy2 = 3950, Σxy = 6040, and r ≈ −0.853.

(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 = 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 = 32 (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 = 303,

Σy = 170,

Σx2 = 12,123,

Σy2 = 3950,

Σxy = 6040, and

r ≈ −0.853

c)

xbar=37.88

ybar=21.25

yhat=44.597+(-0.616)*x

e)

r2 = 0.728

explained=72.8%

unexplained=27.2%

f)

forecast for y =44.597+(-0.616)*32=24.89