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Unformatted text preview: . i.e.,
∑(Y – Yc)2 < ∑ (Y – Ai)2
Where Ai = corresponding values of any other straight line.
(iii) The lines of regression (best fit) intersect at the mean
values of the variables X and Y, i.e., intersecting point is
x, y .
9.4 Methods of Regression Analysis:
The various methods can be represented in the form of chart
given below:
Regression methods Graphic
(through regression lines) Algebraic
(through regression equations) Scatter Diagram Regression Equations
(through normal equations) Regression Equations
(through regression coefficient)
221 9.4.1 Graphic Method:
Scatter Diagram:
Under this method the points are plotted on a graph paper
representing various parts of values of the concerned variables.
These points give a picture of a scatter diagram with several points
spread over. A regression line may be drawn in between these
points either by free hand or by a scale rule in such a way that the
squares of the vertical or the horizontal distances (as the case may
be) between the points and the line of regression so drawn is the
least. In other words, it should be drawn faithfully as the line of
best fit leaving equal number of points on both sides in such a
manner that the sum of the squares of the distances is the best.
9.4.2 Algebraic Methods:
(i)
Regression Equation.
The two regression equations
for X on Y; X = a + bY
And for Y on X; Y = a + bX
Where X, Y are variables, and a,b are constants whose
values are to be determined
For the equation, X = a + bY
The normal equations are
∑X = na + b ∑Y and
∑XY = a∑Y + b∑Y2
For the equation, Y= a + bX, the normal equations are
∑Y = na + b∑ X and
∑XY = a∑X + b∑X2
From these normal equations the values of a and b can be
determined.
Example 1:
Find the two regression equations from the following data:
X:
Y: 6
9 2
11 10
5
222 4
8 8
7 Solution:
Y
X2
Y2
XY
9
36
81
54
11
4
121
22
5
100
25
50
8
16
64
32
7
64
49
56
40
220
340
214
Regression equation of Y on X is Y = a + bX and the
normal equations are
X
6
2
10
4
8
30 ∑Y = na + b∑X
∑XY = a∑X + b∑X2
Substituting the values, we get
40 = 5a + 30b … (1)
…
214 = 30a + 220b …. (2)
…
Multiplying (1) by 6
240 = 30a + 180b…. (3)
…
(2) – (3)
 26 = 40b
26
=  0.65
40
Now, substituting the value of ‘ b’ in equation (1)
40 = 5a –...
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 Winter '08
 Moshiri
 Business

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