Question

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, Σx² = 4217.26, Σy² = 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, 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) What is the predicted temperature when x = 17.8 chirps per second? (Round your answer to two decimal places.)
°F

Answer 1

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