FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 6.37Solution:
(1) ∑X = 350 N = 10
X X^35035
= N 10= =Y Y^31031
= N 10= =Therefore X - 35 and Y - 31 and deviations taken from actual mean
X - 35 = X
Y - 31= Y
Then ∑X^2 = 162, ∑Y^2 = 222
∑XY = 92
Two regression coefficients are
XY 2
b XY^92 0.41
= Y = 222 =YX 2
b XY^92 0.567 0.57
= X = 162 = =(2) r b xb= XY YX
= 0.41x 0.57
= 0.48
∴ Coefficient of correlation is 0.48
(3) Regression equation of X on Y
X X b (Y Y)− = XY −
X - 35 = 0.41 (Y-31)
X = 0.41Y -12.71+ 35
X = 0.41Y +22.29
Regression equation of Y on X.
Y Y b (X X)− = YX −
Y - 31 = 0.57 (X - 35)
Y = 0.57X - 19.95 + 31
Y = 0.57X + 11.05
Example 26 :
You are given the following information regarding a distribution—
N = 5, X = 10, Y = 20, ∑(X - 4)^2 = 100
∑(Y - 10)^2 = 160, ∑(X - 4) (Y - 10) = 80
Find two regression coefficients