Anyone who has been outdoors on a summer evening has probably heard crickets. Did you know that it is possible to use the cricket as a thermometer? Crickets tend to chirp more frequently as temperatures increase. This phenomenon was studied in detail by George W. Pierce, a physics professor at Harvard. In the following data, x is a random variable representing chirps per second and y is a random variable representing temperature (°F).
x | 20.2 | 15.0 | 20.7 | 18.3 | 17.3 | 15.5 | 14.7 | 17.1 |
y | 86.8 | 71.2 | 94.3 | 82.3 | 79.6 | 75.2 | 69.7 | 82.0 |
x | 15.4 | 16.2 | 15.0 | 17.2 | 16.0 | 17.0 | 14.4 |
y | 69.4 | 83.3 | 79.6 | 82.6 | 80.6 | 83.5 | 76.3 |
Complete parts (a) through (e), given Σx = 250, Σy = 1196.4, Σx^{2} = 4217.26, Σy^{2} = 96,047.22, Σxy = 20,089.69, and r ≈ 0.844.
(a) Draw a scatter diagram displaying the data.
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(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) What is the predicted temperature when x = 17.8 chirps
per second? (Round your answer to two decimal places.)
°F
Part a)
Part b)
ΣX = 250
ΣY = 1196.4
ΣX * Y = 20089.69
ΣX2 = 4217.26
ΣY2 = 96047.22
r = 0.844
Part c)
X̅ = Σ( Xi / n ) = 250/15 = 16.67
Y̅ = Σ( Yi / n ) = 1196.4/15 = 79.76
Equation of regression line is Ŷ = a + bX
b = 2.959
a =( Σ Y - ( b * Σ X) ) / n
a =( 1196.4 - ( 2.9587 * 250 ) ) / 15
a = 30.448
Equation of regression line becomes Ŷ = 30.448 + 2.959 X
Part e)
Coefficient of Determination
R^2 = r^2 = 0.712
Explained variation = 0.712* 100 = 71.2%
Unexplained variation = 1 - 0.712* 100 = 28.8%
Part f)
When X = 17.8
Ŷ = 30.448 + 2.959 X
Ŷ = 30.448 + ( 2.959 * 17.8 )
Ŷ = 83.12
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