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 29 47 23 40 34 52 y 31 18 27 13 29 17 21 14 Complete parts (a) through (e), given Σx = 295, Σy = 170, Σx2 = 11,617, Σy2 = 3950, Σxy = 5794, and r ≈ −0.951. (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 = 38 (hundred pounds). What does the least-squares line forecast for y = miles per gallon? (Round your answer to two decimal places.) mpg

Answer #1

a.

b.

*X Values*

∑ = 295

Mean = 36.875

∑(X - M_{x})^{2} = SS_{x} = 738.875

*Y Values*

∑ = 170

Mean = 21.25

∑(Y - M_{y})^{2} = SS_{y} = 337.5

*X and Y Combined*

*N* = 8

∑(X - M_{x})(Y - M_{y}) = -474.75

*R Calculation*

r = ∑((X - M_{y})(Y - M_{x})) /
√((SS_{x})(SS_{y}))

r = -474.75 / √((738.875)(337.5)) = -0.951

c.

Sum of *X* = 295

Sum of *Y* = 170

Mean *X* = 36.875

Mean *Y* = 21.25

Sum of squares (*SS _{X}*) = 738.875

Sum of products (

Regression Equation = ŷ =

ŷ = -0.643

d.

e. r= -0.951

r^2=0.904

Explained variation is 90.4%

Unexplained variation is 9.6%

f. For x=38,

ŷ = (-0.643**38)* + 44.943=20.51

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