Explore BrainMass
Share

# Relationship between three Contingency Tables

Two contingency tables are related by a common attribute C. Use any relationships between the cell entries, such as, derived probabilities to form a third contingency table directly relating attributes A and B. You may solve for the third table either as a frequency table with cell entries in counts [0..N], or a relative probability table with cell entries [0.0..1.0].

Assume the total number of observations represented in all three tables is the same. It may not be necessary to show the solution for all the cell entries of the new third contingency table, if it is clear that the remaining cells could be derived from, for example, the marginals or some other means as long as the method is described. The solution is intended to be generalizable and calls for the symbolic solution of at least one cell in the third contingency table.

#### Solution Preview

Attached is one way of solving this problem with the given conditions

Two 2x2 contingency table come from the same sample:
Given a first 2x2 contingency table with attributes A and C,
Attribute A
A not A (Marginal
totals)
Attribute C C cell
'a1' cell
'b1' Number of
C's
not C cell
'c1' cell
'd1' Number of
not C's
(Marginal
totals) Number of
A's Number of
not A's Total N
and a second 2x2 contingency table with attributes B and C:
Attribute B
B not B (Marginal
totals)
Attribute C C cell
'a2' cell
'b2' Number of
C's
not C cell
'c2' cell
'd2' Number of
not C's
(Marginal
totals) Number of
B's Number of
not B's Total N
derive the cell values for a third 2x2 contingency table with attributes A and B:
Attribute A
A not A (Marginal
totals)
Attribute B B cell
'a3' cell
'b3' Number of
B's
not B cell
'c3' cell
'd3' Number of
not B's
(Marginal
totals) Number of
A's Number of
not A's Total N

METHOD 1:

I will make some assumptions and continue:
First, we calculate the marginal on both Tables 1 and 2 as shown below.

A not A (Marginal
totals)
Attribute C C a1 b1 a1+b1
not C
c1
d1 c1+d1
(Marginal
totals) a1+c1 b1+d1 N1

Attribute B
B not B (Marginal
totals)
Attribute C C a2 b2 a2+b2
not C
c2
d2 c2+d2
(Marginal
totals) a2+c2 b2+d2 N2
To derive the cell values for a third 2x2 contingency table with attributes A and B, we have:
We begin ...

#### Solution Summary

Two contingency tables are related by a common attribute C. Use any relationships between the cell entries, such as, derived probabilities to form a third contingency table directly relating attributes A and B. You may solve for the third table either as a frequency table with cell entries in counts [0..N], or a relative probability table with cell entries [0.0..1.0].

Assume the total number of observations represented in all three tables is the same. It may not be necessary to show the solution for all the cell entries of the new third contingency table, iff it is clear that the remaining cells could be derived from, for example, the marginals or some other means as long as the method is described. The solution is intended to be generalizable and calls for the symbolic solution of at least one cell in the third contingency table.

\$2.19