Marginal independence has already been discussed before. Variables \(X\) and \(Y\) are marginally independent if:
\(\pi_{ij+}=\pi_{i++}\pi_{+j+}\)
They are marginally independent if they are independent within the marginal table. In other words, we consider the relationship between \(X\) and \(Y\) only and completely ignore \(Z\). Controlling for, or adjusting for different levels of \(Z\) would involve looking at the partial tables.
We saw this already with the Berkeley admission example. Recall from the marginal table between sex and admission status, where the estimated oddsratio was 1.84. We also had significant evidence that the corresponding oddsratio in the population was different from 1, which indicates a marginal relationship between sex and admission status.
boys.sas (boys.lst) or boys.R (boys.out) to answer this question. Hint: look for the chisquare statistic \(X^2=172.2\).
Question: How would you test the model of marginal independence between scouting and SES status in the boyscout example? See the filesRecall, that joint independence implies marginal, but not the other way around. For example, if the model (\(XY\), \(Z\)) holds, it will imply \(X\) independent of \(Z\), and \(Y\) independent of \(Z\). But if \(X\) is independent of \(Z\) , and \(Y\) is independent of \(Z\), this will NOT necessarily imply that \(X\) and \(Y\) are jointly independent of \(Z\).
If \(X\) is independent of \(Z\) and \(Y\) is independent of \(Z\) we can’t tell if there is an association between \(X\) and \(Y\) or not; it could go either way, that is in our graphical representation there could be a link between \(X\) and \(Y\) but there may not be one either.
For example, let’s consider a \(2\times2\times2\) table.
The following \(XZ\) table has \(X\) independent of \(Z\) because each cell count is equal to the product of the corresponding margins divided by the total (e.g., \(n_{11}=2=8(5/20)\), or equivalently, OR = 1)
C=1  C=2  Total  

A=1  2  6  8 
A=2  3  9  12 
Total  5  15  20 
The following \(YZ\) table has \(Y\) independent of \(Z\) because each cell count is equal to the product of the corresponding margins divided by the total (e.g., \(n_{11}=2=8(5/20)\)).
C=1 
C=2 
Total 


B=1 
2 
6 
8 
B=2 
3 
9 
12 
Total 
5 
15 
20 
The following \(XY\) table is consistent with the above two tables, but here \(X\) and \(Y\) are NOT independent because all cell counts are not equal to the product of the corresponding margins divided by the total (e.g., \(n_{11}=2\)), whereas \(8(8/20)=16/5\)). Or you can see this via the odds ratio, which is not equal to 1.
B=1  B=2  Total  

A=1  2  6  8 
A=2  6  6  12 
Total  8  12  20 
Furthermore, the following \(XY\times Z\) table is consistent with the above tables but here \(X\) and \(Y\) are NOT jointly independent of \(Z\) because all cell counts are not equal to the product of the corresponding margins divided by the total (e.g., \(n_{11}=1\), whereas \(2(5/20)=0.5\)).
C=1  C=2  Total  

A=1, B=1  1  1  2 
A=1, B=2  1  5  6 
A=2, B=1  1  5  6 
A=2, B=2  2  4  6 
Total  5  15  20 
Other marginal independence models in a threeway table would be, \((X,Y)\) while ignoring \(Z\), and \((Y,Z)\), ignoring any information on \(X\).