McNemar test for N*N contingency tables

  • Thread starter mnb96
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  • #1
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Main Question or Discussion Point

Hi,

I have a 4x4 contingency table that by visual inspection looks approximately symmetric, in the sense that for each entry aij we approximately have [itex]a_{ij} = a_{ji}[/itex] for [itex]i,j=1,2,3,4[/itex].
I would like to know how can I perform a statistical test for symmetry.

An attempt I made was to treat the whole contingency table as a 4x4 matrix [itex]O[/itex], and create its symmetrized version [itex]E=(O+O^T)/2[/itex]. At this point I could simply apply a Chi2 goodness-of-fit test (with 9 degrees of freedom) between the observed distribution [itex]O[/itex] and the expected one [itex]E[/itex].

What makes me suspicious is that I haven't found this kind of approach anywhere. Instead, only the simplest case of 2x2 tables is reported in the literature under the name of McNemar's test.

It's reasonable to think that if the generalization of McNemar's test was so trivial, then somebody else would have "invented" it. This makes me suspect that my approach is incorrect.
 

Answers and Replies

  • #2
713
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I might have found the answer to my question: Apparently the test I proposed already exists under the name of "Bowker test for symmetry" (see first formula on page 4).
 
  • #3
713
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...just the number of degrees of freedom in my original post is incorrect: it should be N(N-1)/2 = 6, i.e. the number of dof in a NxN symmetric matrix.
 

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