Solved – Post hoc chi-square test

chi-squared-testdyadic-datapost-hoc

I am looking at a $\chi^2$ (crosstabs) test looking at interactions between types of dyads in an animal group. My categories are Male-Male, Male-Female, and Female-Female dyads in the rows and "Were Seen to Interact" and "Were Never Seen to Interact".
The matrix looks something like:

21   7  
85   19  
77   1

There is a very significant p value. My question is, when running the post hoc tests, is it appropriate to compare each category to the sum of the other two (i.e. Male-Male, vs Male-Female+Female-Female) when looking for which category is driving the difference, or should I compare the categories to each other (Male-Male vs Male-Female, then Male-Male vs Female-Female)?

I'm looking for an answer to the question of whether one of the types of dyads are more likely to fail to interact. I think the statement "There are more Males-Female dyads that were never seen to interact than would be expected by chance" would be easier to comprehend than if I had to write out the comparisons of the three pairs. I just don't know if it's allowable.

Best Answer

Your chi-square test tells you that you have an interaction in your effects so the probability of subject interactions is dependent upon your dyad. Therefore, you're just checking to see which of those it's dependent upon and you can test. But you don't really need a test here... it's kinda obvious with the last row showing a 77:1 effect while the others are about or 3 or 4:1. If you want to do binomial test comparisons among the 3 perhaps just lower your alpha a little and don't try to conclude that the first two dyads are the same. You might also want to report a confidence interval for each individual dyad.

Another issue you have here is not one of post-hoc testing but one of the chi-square statistic being believed unreliable when the expected in a cell is less than 5 (for the male-male). However, this is probably not a problem.

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Solved – Transforming data for chi square — squaring negative value difference scores

You say you did paired t-tests on the original data, before dichotomizing it, and that males increased significantly from the old form to the new but the female change was not significant. Unfortunately, that can not be taken as showing that the male change was bigger than the female change. You need to do an independent-groups t-test on the two sets of change scores. (Better yet, you could replace all the t-tests by confidence intervals for the corresponding means and mean differences, which would give you more information.)

For the dichotomized data, the situation is similar.
You have two contingency table, one for males and one for females.

Males      
        Yes   No     Total
  Yes   Myy   Myn    My.
   No   Mny   Mnn    Mn.

Total   M.y   M.n    M.. = M = total number of Males

Females      
        Yes   No     Total
  Yes   Fyy   Fyn    Fy.
   No   Fny   Fnn    Fn.

Total   F.y   F.n    F.. = F = total number of Females

For each table, the analog of the paired t-test is the McNemar test,
http://en.wikipedia.org/wiki/McNemar%27s_test

I know of no simple standard test of the difference between the changes in endorsement rates, but if all of Myn, Mny, Myy+Mnn, Fyn, Fny, Fyy+Fnn are "large" then an asymptotic test might be justified.

Solved – Post-hoc test after a GLM with binary data

The problem you haveg is that are abusing (misusing) the formula notation. You never use a formula like this:

Periosna$Periostitis ~ Periosna$Cemetery

because, as you've found out it, it totally breaks the logic that you want to then use for subsequent operations.

You asked mcp() to set up Tukey contrasts for the covariate Cemetery, yet you fitted a model with covariate name Periosna$Cemetery, hence it quite rightly told you that a variable of the stated name was not involved in the model fit.

Instead, the call should have been

glm(Periostitis ~ Cemetery, data = Periosna, family = binomial)

noting that we use the variable names directly, as they exist in Periosna, and we tell R where to locate the variables in the formula via the data argument.

Now when you call glht() it should be able to find the Cemetery variable.

Best Answer

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Solved – Transforming data for chi square — squaring negative value difference scores

Solved – Post-hoc test after a GLM with binary data

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