Question

complete the following activities with these cricket chirp data.

Temperature (F degree) 69.7, 93.3, 84.3, 76.3, 88.6, 82.6, 71.6, 79.6

Chirp in 1 minute 882, 1188, 1104, 864, 1200, 1032, 960, 900

1. Find the regression models (linear and quadratic) for the above data

a. What is the equation for the line of best fit in y=mx+b form?

b. What is the equation for the best fitting quadratic model?

2. Use your models (not the actual data) and calculators to predict the number of chirps at each of the three temperatures.

	Linear	Quadratic
69.7
76.3
88.6

Answer 1

1)

(a) Sum of x= 646
Sum of y = 8130
Mean x = 80.75
Mean y = 1016.25
Sum of squares (SS_X) = 462.1
Sum of products (SP) = 6747.9

x	y	x-MX = x-80.75	y-My=y-1016.25	(x-Mx)2	(x-Mx)(y-My)
69.7	882	-11.05	-134.25	122.1025	1483.4625
93.3	1188	12.55	171.75	157.5025	2155.4625
84.3	1104	3.55	87.75	12.6025	311.5125
76.3	864	-4.45	-152.25	19.8025	677.5125
88.6	1200	7.85	183.75	61.6225	1442.4375
82.6	1032	1.85	15.75	3.4225	29.1375
71.6	960	-9.15	-56.25	83.7225	514.6875
79.6	900	-1.15	-116.25	1.3225	133.6875
Sum of X = 646	Sum of Y = 8130			Sum = 462.1	Sum = 6747.9

y = mx + b

Here m = SP/SS_X = 6747.9/462.1 = 14.60268

b = (Mean of y) - m(Mean of x) = 1016.25-14.60268*80.75 = -162.91668

Hence required linear equation is :

y = 14.60268 x - 162.91668

(b) y = A+Bx+Cx²

Here C =

B =

A = (Mean of y)-B(Mean of x) -C(Mean of x²)

We get y = 0.446x² -57.708x +2742.823

2) Linear :y = 14.60268 x - 162.91668

For 69.7, y = 14.60268(69.7)-162.91668 = 855 chirps

For 76.3, y = 14.60268(76.3)-162.91668 = 951 chirps

For 88.6, y = 14.60268(88.6)-162.91668 = 1131 chirps

Quadratic : y = 0.446x² -57.708x +2742.823

For 69.7, y = 0.446(69.7)²-57.708(69.7)+2742.823 = 887 chirps

For 76.3, y = 0.446(76.3)²-57.708(76.3)+2742.823 = 936 chirps

For 88.6, y = 0.446(88.6)²-57.708(88.6)+2742.823 = 1131 chirps