Question

1. The common cricket can be used as a crude thermometer. The colder the temperature, the slower the rate of

chirping. The table below shows the average chirp rate of a cricket at various temperatures.

A) Determine the equation of the regression line for the given data (You may use your calculator to find the line). Round values to the nearest hundredth (two decimals).

B) State and interpret the vertical intercept of the regression line in the context of this problem. Be sure to include units in your interpretation.

C) State and interpret the slope of the regression line in the context of this problem. Be sure to include units in your interpretation.

D) State and interpret the correlation coefficient, r, in context.

E) What does the least squares regression line predict that the temperature will be if a cricket is chirping at a rate of 2 chirps per second? Determine the residual when x = 2. Does the regression line over or under estimate the temperature when a cricket chirps at a rate of 2 chirps per second?

Chirp Rate, x (chirps per second)	2.4	3.8	3.7	2.0	3.4	3.9
Temperature, y (degrees fahrenheit)	61.9	83.6	71.8	47	65.1	75.8

Answer 1

X	Y	XY	X²	Y²
2.4	61.9	148.56	5.76	3831.61
3.8	83.6	317.68	14.44	6988.96
3.7	71.8	265.66	13.69	5155.24
2	47	94	4	2209
3.4	65.1	221.34	11.56	4238.01
3.9	75.8	295.62	15.21	5745.64
Ʃx =	Ʃy =	Ʃxy =	Ʃx² =	Ʃy² =
19.2	405.2	1342.86	64.66	28168.46

Sample size, n =	6
x̅ = Ʃx/n = 19.2/6 =	3.2
y̅ = Ʃy/n = 405.2/6 =	67.53333333
SSxx = Ʃx² - (Ʃx)²/n = 64.66 - (19.2)²/6 =	3.22
SSyy = Ʃy² - (Ʃy)²/n = 28168.46 - (405.2)²/6 =	803.9533333
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 1342.86 - (19.2)(405.2)/6 =	46.22

a)

Slope, b = SSxy/SSxx = 46.22/3.22 = 14.354037

y-intercept, a = y̅ -b* x̅ = 67.53333 - (14.35404)*3.2 = 21.600414

Regression equation :

ŷ = 21.60 + (14.35) x

b)

When chirps per second is zero, the line crosses the y axis.

c)

With a unit increase in chirp per second x the value of y changes by degree fahreheit of 14.35 on average.

d)

Correlation coefficient, r = SSxy/√(SSxx*SSyy) = 46.22/√(3.22*803.95333) = 0.9084

There is a positive and strong relationship between Chirp rate and Temperature.

e)

Predicted value of y at x = 2

ŷ = 21.6004 + (14.354) * 2 = 50.3085

Residual = y - ŷ = 47 - 50.3085 = -3.3085

The regression line over estimate the temperature when a cricket chirps at a rate of 2 chirps per second.