Data Consistency
In order to find out whether the given data is consistent or not we have to apply a simple test. The test is to find out whether anyone or more of the ultimate class-frequencies is negative or not. If no one of the class- frequencies is negative then we can safely calculate that the given data is consistent (the frequencies do not conflict in any way with each other). On the other hand, if any of the ultimate class frequencies comes out to be negative the given data is inconsistent. Thus the most important and sufficient condition for the consistency of a set of independent class frequencies is that no ultimate class frequency is negative.
Illustration :
From the following two cases find out whether the data is consistent or not
Case I: (A) = 100 (B) = 150 (AB) = 60 N = 500
Case II: (A) = 100 (B) = 150 (AB) = 140 N = 500
Solution: case I
We are given (A) = 100 (B) =150 (AB) = 60 N = 500
Substituting the values in the square table
|
|
A
|
σ
|
total
|
|
B
|
(AB)60
|
(σB)90
|
(B)150
|
|
β
|
(Aβ)40
|
(??β)310
|
(β)350
|
|
Total
|
(A)100
|
(??) 400
|
N 500
|
From the table
(Aβ) = (A) - (AB) = 100 - 60 = 40
(αB) = (B) - (AB) = 150 - 60 = 90
(αβ) = (α) - (αB) = 40 - 90 = 310
since all the ultimate class frequencies are positive we conclude that the given data are consistent.
Case II: given values are (A) = 100 (B) =150 (AB) = 140 (N) = 500.
By putting these values in the nine-square table we can determine the missing value
|
|
A
|
σ
|
total
|
|
B
|
(AB)140
|
(σB)10
|
(B)150
|
|
β
|
(Aβ)-40
|
(??β)390
|
(β)350
|
|
Total
|
(A)100
|
(??) 400
|
N 500
|
From the table
(Aβ) = (A) - (AB) = 100 - 140 = - 40
(αB) = (B) - (AB) = 150 - 140 = 10
(αβ) = (α) - (αB) = 400 - 10 = 390
Since one of the ultimate class frequencies (Aβ) is negative and hence the given data are inconsistent.