I was reading this article (Kim and Cohen, 2010 [1]), and realized that the authors used chi-square test to test the significance of two-way ANOVA on page 541 (1st paragraph of "Results and Discussion" section), but used F-test for testing the significance of two-way ANOVA on page 543 (again, 1st paragraph of "Results and Discussion" section).

I understand that Chi-squared distribution and F distribution are closely related, but is it possible to use Chi-square test for testing whether "Between-variances" is significantly larger than "Within-variances"?

and if it is indeed possible to use Chi-square test for ANOVA, then what would be the rationale to choose the test instead of F-test?

[1]: Kim, Young-Hoon and Cohen, Dov (2010),
"Information, Perspective, and Judgments About the Self in Face and Dignity Cultures"
Pers Soc Psychol Bull; 36 (4); 537-550
DOI: 10.1177/0146167210362398


I believe your question relies on a mistaken premise.

On p543 they're doing ANOVA, just as they say (and as you suggest)

However, on p541 they're not doing ANOVA there (and they don't claim to be as far as I see); they're comparing proportions, for which some form of chi-squared test might well be suitable -- it's hard to say for sure exactly what they did from what's written there (I didn't read the whole paper but often it's hard to tell for sure even if you do), but I expect that it's a more-or-less uncontroversial use of a chi-squared test on a (possibly multidimensional) contingency table, though there would be some other possibilities involving proportions and chi-squared tests.

  • $\begingroup$ You're right. Upon second look I realized p541 Dependent Variable is a categorical variable, and thus the authors were conducting a chi-square test of independence. Thanks $\endgroup$ – Simonet Sep 15 '17 at 2:13
  • $\begingroup$ I agree that the dependent variable is categorical, though I don't necessarily conclude that it must be a chi-squared test of independence from that (while I think that's easily the most likely possibility, as I mentioned other possibilities exist) -- or do they explicitly state somewhere that it was a chi-squared test of independence that they performed? $\endgroup$ – Glen_b -Reinstate Monica Sep 15 '17 at 2:18
  • $\begingroup$ I see no explicit mention of 'chi-squared test of independence' in the paper. What are the other possibilities you presume as potential alternatives? $\endgroup$ – Simonet Sep 15 '17 at 2:24
  • $\begingroup$ A complete list of things they might have done might be a a bit long for a comment. One example would be a binomial GLM, which typically also results in a chi-squared (and if we're in a situation where you could use an ordinary Pearson test of independence, that overall test should correspond to a G-test of independence, I think) $\endgroup$ – Glen_b -Reinstate Monica Sep 15 '17 at 2:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.