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 | 46 | 30 | 47 | 23 | 40 | 34 | 52 |
y | 30 | 20 | 25 | 13 | 29 | 17 | 21 | 14 |
Complete parts (a) through (e), given Σx = 299, Σy = 169, Σx^{2} = 11,943, Σy^{2} = 3861, Σxy = 5880, and
r ≈ −0.923.
(b) Verify the given sums Σx, Σy, Σx^{2}, Σy^{2}, Σxy, and the value of the sample correlation coefficient r. (Round your value for r to three decimal places.)
Σx = | |
Σy = | |
Σx^{2} = | |
Σy^{2} = | |
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
r^{2}. 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 r^{2}
to three decimal places. Round your answers for the percentages to
one decimal place.)
r^{2} = | |
explained | % |
unexplained | % |
(f) Suppose a car weighs x = 30 (hundred pounds). What
does the least-squares line forecast for y = miles per
gallon? (Round your answer to two decimal places.)
mpg
b)
ΣX = | 299.000 |
ΣY= | 169.000 |
ΣX^{2} = | 11943.000 |
ΣY^{2} = | 3861.000 |
ΣXY = | 5880.000 |
r = | -0.923 |
c)
X̅=ΣX/n = | 37.38 |
Y̅=ΣY/n = | 21.13 |
ŷ = | 42.365+(-0.568)x |
e)
coefficient of determination r^{2} = | 0.853 | |||
explained = | 85.3% | |||
unexplained= | 14.7% |
f)
predicted value =42.365+-0.568*30= | 25.33 |
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