Bonferroni, HSD, or Holm don't address the strength of association.

Are you asking about a post-hoc test ? Or a strength of association --- effect size statistic ?

I wonder if Ashok Ayer is asking for advice about how to decompose an overall Chi2 with df=p into p smaller tables, each with df=1? Here is an article that (apparently) explains how to do that:

Article Decomposing Chi-Square: A Forgotten Technique

I don't know if that article mentions it, but note that the orthogonal components add up to the overall Chi2 exactly (as they should) when you use the likelihood ratio Chi2, but not when you use Pearson's Chi2.

EDIT. Here is another resource on decomposition of the overall Chi2, if that is indeed the interest.

• https://rpubs.com/wgalvord/147800

As Salvatore was suggesting, you may want to clarify your question. Are you wondering what to do after a chi-square test proves significant for a table larger than 2x2?

Kind regards

I don't know the exact answer. But if a table is larger than 2X2 and significant association is seen, we may not clearly know among multiple parameters which is associated and which is not. My guess is here, the variables are qualitative so it may be clumsy to test the effects of two among many on some results, thus post-hoc is not necessary.

If we opt to do, however, my idea is take two variables at a time and repeat the test. Which means convert into 2X2 table and find result.

Once again, I am adding an answer just for discussion and I don't really know the statistical complexity. Thank you

Dear Gianmarco Alberti that is precisely the issue for which I seek a solution. Dear Sal Mangiafico I am looking for a more in-depth analysis of the significant outcomes obtained from the Chi-square test across multiple groups. Regards

Ashok Ayer Typically, one can consider the adjusted standardised residuals. I am not referring to the standarised residuals, but on the adjusted version which are "truly" standardised. Essentially, they help you pinpoint where (i.e., in which cell) a significant departure from independence lies. Typically, you should look up for adjusted stnd residuals whose absolute value is larger than 1.96 (corresponding to an alpha value of at least 0.05). Some authors suggest applying some correction for multiple comparisons, among which Bonferroni, Sidak, or other methods.

I have put together an R package ('chisquare') which does all that for you (and more). You can find info about my package here on RG.

Hope this helps

I agree with the comments by Gianmarco Alberti , and that chisquare package looks very helpful.

If it is the case where you want to break down a larger table "groupwise" or "pairwise", it's possible to make all possible 2 x N tables, and perform a chi-square analysis on each pair. In some cases, this kind of analysis makes sense.

The following example is R. It uses a dataset from the chisquare package, moment corrected standardized residuals from the chisquare package, and pairwise chi-square tests from the rcompanion package.

socialclass = read.table(header=TRUE, text="

Neurotic Depressed PersonalityDisorder Schizophrenic

SocClass1 45 25 21 18

SocClass2 10 45 24 22

SocClass3 17 21 18 18

")

library(chisquare)

chisquare(socialclass)$mom.corr.stand.resid

### Neurotic Depressed PersonalityDisorder Schizophrenic

### SocClass1 4.673 -2.376 -0.915 -1.277

### SocClass2 -4.361 3.141 0.477 0.427

### SocClass3 -0.576 -0.788 0.553 1.051

library(data.table)

SocialClass = as.matrix(socialclass)

library(rcompanion)

CT = pairwiseNominalIndependence(SocialClass, fisher=FALSE, gtest=FALSE)

CT

### Comparison p.Chisq p.adj.Chisq

### 1 SocClass1 : SocClass2 3.11e-06 9.33e-06

### 2 SocClass1 : SocClass3 7.93e-02 7.93e-02

### 3 SocClass2 : SocClass3 4.99e-02 7.48e-02

cldList(p.adj.Chisq ~ Comparison, data=CT)

### Group Letter MonoLetter

### 1 SocClass1 a a

### 2 SocClass2 b b

### 3 SocClass3 ab ab

Thank you Sal Mangiafico for the follow up, and for pointing out the pairwise comparisons of 2xN tables. I have always read about that, but never had the time (and/or will) to dig deeper into it.

I will give your function a try!

Cari saluti

Here is another article that followers of this thread might find helpful.

Sharpe, Donald (2015). Your Chi-Square Test is Statistically Significant: Now What? Practical Assessment, Research & Evaluation, 20(8).

Link: https://scholarworks.umass.edu/cgi/viewcontent.cgi?article=1269&context=pare

Just because the results are so difficult to difficult to read on this forum, I attached an image of the results from the analysis I presented above.

I think in a case like this, examining both the residuals and the pairwise tests is very helpful in understanding the results.

Jumping down to the compact letter display, it appears that the distribution of the diagnoses is different between SocClass1 and SocClass2. But that SocClass3 is not different from either of the others.

Looking at the residuals, it's clear that it's the differences in counts in Neurotic and Depressed is driving the difference between SocClass1 and SocClass2.

I think you could say that SocClass1-Neurotic and SocClass2-Depressed were higher than "expected". And SocClass1-Depressed and SocClass2-Neurotic were lower than "expected".

I assume you are trying to decompose the overall association (the X^2).

You haven't told us if it is a square table with comparable categories or not. Different analyses are often used for these. I will assume not.

1. Look at the standardized residuals. You can plot these using correspondence analysis on these residuals, and this is part of most big stats programs.

2. Association models. The RC1 model, where a score is estimated for each of the row and column category, often can account for much of the deviation. See books on log-linear modeling.

3. I wouldn't recommend testing pairs of categories as either the numbers will increase rapidly and/or you are basing these on the data that you observed.

Mantel hanzel chi-square by splitting the table