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 26 47 30 47 23 40 34 52 y 30 20 23 13 29 17 21 14 Complete parts (a) through (e), given Σx = 299, Σy = 167, Σx2 = 11,983, Σy2 = 3765, Σxy = 5810, and r ≈ −0.909.
(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 = 40 (hundred pounds). What
does the least-squares line forecast for y = miles per
gallon? (Round your answer to two decimal places.)
mpg
b,c&d)
b&e)
explained = 82.69%
unexplained = 100-82.69 = 17.31 %
f)
when x = 40 , Y= 40.8434−0.5343(40) = 19.4714