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To understand the correlation of bivariate frequency one has to divide their attention toward marginal and joint distribution of variables. Let us consider this following table with two variables X and Y:

                  X

Y

                            Classes
                        Midpoints

X1         X2…………………………… Xn

Classes

Mid Points

Y1

Y2

:

:

:

Ym

 

 

 

 

 

 

 

 

 

f (x, y)

g (y1)=∑ₓ f(x, y1)

g (y2)=∑ₓ f(x, y2)

 

 

 

 

 

g (ym)=∑ₓ f(x, ym)

 

 

 

 

 h(X1)………………………… h(Xn)

= ∑y f (x1, y)       = ∑y f (xn, y)

N = ∑ₓ∑y f (x, y)

 

x̅ = 1/N ∑xi h (xi),  y̅ = 1/N ∑yj g (yj)

Or

= 1/N. ∑xh (x)

= 1/N ∑y g(y)

σ2 = 1/N ∑x2 h(x) – (x̅)2, σ2y = 1/N ∑y2 g (y) – (y̅)2

Cov (x,y) = 1/N∑x y xy f (x,y) – x̅.y̅

So we get,

rxy = Cov. (x,y)

σx σy

Here a special note should be made on large data where this two way frequency table is very advantageous.

Calculate the correlation of coefficient of this table:

                                  X                                Y                  0 – 8                 8 – 16             16 – 24
1 -5

5 – 9

9 – 13

2

3

2

0

2

5

4

2

1

Solution:

y                                                  x        Mid                       Values

4                     12                   20

                          g(y)
Mid                     3

Values                7

11

2                     0                      4

3                     2                      2

2                     5                      1

                             6

7

8

                         h(x) 7                     7                      7                            21

 

x̅ = 1/21 [(4) (7) + (12) (7) + (20) (7)] = 12

y̅ = 1/21 [(3) (6) + (7) (7) + (11) (8) ] = 7.38

σx2  = 1/21 [(16) (7) + (144) (7) + (400) (7)] –(12)2 = 42.67

σy2  = 1/21 [(9) (6) + (49) (7) + (121) (8)] –(7.38)2 = 10.54

Cov (x,y) = 1/21 ∑x y xy f (x,y) – x̅.y̅

= 1764/21 – (12) (7.38) = -4.56

The correlation of coefficient is:

rxy  =  -4.56/√42.67. √10.54 = -0.22.

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