I have two between-subject factors, each with two levels (so 4 conditions). Furthermore, I have one dependent variable (qualitative), also consisting of two levels.
Now I want to make pairwise comparisons (so I have 6 chi-squared tests in total). Is there any way I can control type-1-errors? In the literature I saw they often calculated interaction with a chi-squared test. Is this the way to do it, and if so, how do I do it?
I can work with both SPSS and MATLAB.
An example of what I want to accomplish:
Independent chi-squared tests analyzing the interaction between the experimental conditions and the participant’s compliance with the request were performed. With the participants who first accepted to participate in the study, significant interaction was found [x2(1, N = 199) = 21.06, p<0.001, r = 0.30]. Pairwise comparisons revealed that the control condition was significantly different from the FITD condition [40.0% vs. 60.0%, x2 (1, N = 100) = 4.00, p<0.05, ϕ = 0.20] and the FITD-‘‘but you are free . . .’’ condition [40.0% vs. 56.0%, x2 (1, N = 100) = 14.92, p<.001, ϕ = 0.36] but not with the ‘‘but you are free . . .’’ condition [40.0% vs. 60.0, x2 (1, N = 100) = 2.56, nonsignificant (ns), ϕ = 0.16]. No statistical differences were found between the FITD condition and the ‘‘but you are free . . .’’ condition [60.0% vs. 56.0, x2 (1, N = 100) = 0.16, ns, ϕ = 0.04] and between the FITD condition and the FITD-‘‘but you are free . . .’’ condition [60.0% vs. 78.0%, x2 (1, N = 100) = 3.78, p = .06, ϕ = 0.19]. However, a significant difference was found between the ‘‘but you are free . . .’’ condition and the FITD -‘but you are free . . .’’ condition [56.0% vs. 78.0%, x2 (1, N = 100) = 5.47, p<0.02, ϕ = 0.23].
Reference:
Guéguen, N. Meineri, S. Martin, A. & Grandjean, I. (2010). The combined effect of the foot in-the-door technique and the “but you are free” technique: an evaluation on the selective sorting of household wastes. Ecopsychology. 2(4), 231 – 237. doi: 10.1089/eco.2009.0051
Best Answer
Typical Type I error inflation corrections are done on the cutoff p-value used to reject the null and independent of the kind of test. Therefore, you don't need special ones for chi-square, ANOVA, t, etc.
But in your case they appear unnecessary. The published paper is spelunking into the results using faulty logic (see Gelman and Stern 2006). With 2x2 tests all of those subsequent additional examinations are not only unnecessary but don't assess the issue at hand. An interaction in a 2x2 is the direct test that the effect of A in B1 is different from the effect of A in B2 (or vice versa). Additional tests of simple effects don't assess that and therefore do nothing to prove the point that they're attempting to make.