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
a.
b.
X Values
∑ = 295
Mean = 36.875
∑(X - Mx)2 = SSx = 738.875
Y Values
∑ = 170
Mean = 21.25
∑(Y - My)2 = SSy = 337.5
X and Y Combined
N = 8
∑(X - Mx)(Y - My) = -474.75
R Calculation
r = ∑((X - My)(Y - Mx)) /
√((SSx)(SSy))
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 (SSX) = 738.875
Sum of products (SP) = -474.75
Regression Equation = ŷ = bX + a
b = SP/SSX = -474.75/738.88 =
-0.643
a = MY - bMX = 21.25 -
(-0.64*36.88) = 44.943
ŷ = -0.643X + 44.943
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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