# A situation where to consider row totals in a $2\times 2$ contingency table fixed or not?

In a prospective study, we draw a sample of size 100 where two sttributes A and B were present.

Our goal is to assess whether there is any association between these two attributes. The data looks like following:

Table 1.
A
Present    Absent|
Present      x1         x2 |
B                            |
Absent      x3         x4 |
----------------------------|---
100


Attribute A can be divided into two parts, pathogenic A and non-pathogenic A. Now, with the same sample of 100, we test whether there is any association between attribute B and pathogenic A.

Table 2.
PATHOGENIC A
Present    Absent|
Present      y1         y2 |
B                            |
Absent      y3         y4 |
----------------------------|---
100


Since in both tables, we have used the same sample of size 100 and the attribute B is common, hence

$$(x1+x2)=(y1+y2), \quad\text{and}\quad (x3+x4)=(y3+y4),$$ that is, the corresponding row totals are same.

In the first table, the number of sample where B is present or absent was not known, that is, the row totals were not fixed. A data analyst didn't saw table 2, before conducting the test of table 1. But after conducting the test of table 1, if the analyst comes to table 2, will he consider the row totals fixed?

But if he would see table 2 before table 1, then he would consider the row totals of table 2 not fixed.

Fixing the marginal totals... means that you conducted an experiment in which they where explicitly fixed.

This does not seam to be your case (at least not according to the explanation).

It is about the data gathering process. Just redoing some other analysis on the same data does not change the process by which it was gathered and it does not fix the marginal totals (although you might want to look at multiple comparisons ).

Note the difference between the two following:

• You select a fixed total number (total n=100) and then you started to measure among those groups the presence of B and A

• You select a fixed number $$x$$ of B present and a fixed number $$y$$ of B absent (with $$x+y=100$$) and then you started to collect among those groups the presence of A

By explicitly expecting (using) a certain marginal total for B you have been changing the statistical model.

• In the model with the fixed marginals the parameters for the distribution (which is hyper-geometric if the null hypothesis, independent columns and rows) are known.

• In the model with the not-fixed marginals the parameters for the distribution of the observations in the cells are not known. These parameters are estimated by using the marginal totals, but they are themselves variables.

A test like the Fisher exact test, does not consider these as variables and only calculates the distribution as if one had thes marginals fixed.

The ratios that you observe will be different.

When you did not fix the marginals then you can use the Bernard's test

• Very clear explanation. Thank you. Could you please tell me how much the result can be distorted if someone conducts a "Boschloo's exact test with binomial distribution", when he is actually supposed to perform "Boschloo's exact test with MULTINOMIAL distribution"? Jul 13, 2018 at 7:45