4 votes

Fisher's exact test when a category has a 0

My answer to your initial question is that you shouldn't worry about the message, because (1) it is telling you that the least-populated expected count is greater than 5 (the table in your question ...
Ben Bolker's user avatar
2 votes
Accepted

Modification of Pearson's Chi-square test

As I know the Pearson test uses: $$X_i = \frac{n_i - np_i}{\sqrt{np_i}}$$ and not the $X_i$ that you mentioned. To prove the result of the Pearson test, let $\widehat{\boldsymbol{\theta}}_n = \frac1n \...
Kroki's user avatar
  • 270
2 votes

Chi square test of independence with year as a categorical variable?

If you run logistic with glm, you get: ...
BenP's user avatar
  • 587
1 vote

Pearson chi squared test vs deviance test in GLM

Generally, the G-test is useful in case of categorical data, but when we see the concept of "Goodness of fit", the more powerful test is chi-square test as compared to the LRT. Also I would ...
Awais Munir's user avatar

Only top scored, non community-wiki answers of a minimum length are eligible