In: Statistics and Probability
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 29 43 30 47 23 40 34 52 y 30 22 23 13 29 17 21 14
Complete parts (a) through (e), given Σx = 298, Σy = 169, Σx2 = 11,788, Σy2 = 3849, Σxy = 5906, and r ≈ −0.889.
(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 y hat = 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 hat = + x (
d) Graph the least-squares line. Be sure to plot the point (x, y) as a point on the line. WebAssign Plot WebAssign Plot WebAssign Plot WebAssign Plot
(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 = 45 (hundred pounds). What does the least-squares line forecast for y = miles per gallon? (Round your answer to two decimal places.) mpg
X | Y | X * Y | X2 | Y2 | |
29 | 30 | 870 | 841 | 900 | |
43 | 22 | 946 | 1849 | 484 | |
30 | 23 | 690 | 900 | 529 | |
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 | 298 | 169 | 5906 | 11788 | 3849 |
Part a)
Part b)
r = -0.889
Part c)
X̅ = Σ( Xi / n ) = 298/8 = 37.25
Y̅ = Σ( Yi / n ) = 169/8 = 21.12
Equation of regression line is Ŷ = a + bX
b = ( 8 * 5906 - 298 * 169 ) / ( 8 * 11788 - ( 298
)2)
b = -0.566
a =( Σ Y - ( b * Σ X) ) / n
a =( 169 - ( -0.5662 * 298 ) ) / 8
a = 42.215
Equation of regression line becomes Ŷ = 42.215 - 0.566 X
Part e)
Coefficient of Determination
R2 = r2 = 0.79
Explained variation = 0.79* 100 = 79%
Unexplained variation = 1 - 0.79* 100 = 21%
Part f)
When X = 45
Ŷ = 42.215 + -0.566 X
Ŷ = 42.215 + ( -0.566 * 45 )
Ŷ = 16.75