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Does the chi squared test work if the variables have different levels? Here's an example

x <- factor(sample(c('a', 'b', 'c'), 100, replace = T))
y <- factor(ifelse(x == 'a', 'd', 'e'))

The contingency table includes zeros:

table(x, y)

   y
x    d  e
  a 29  0
  b  0 39
  c  0 32

Running chisq.test(x, y) gives a very small p-value, so x and y are not independent. This is the correct result, but I thought one of the assumptions behind chisq.test is that there needs to be no zeros in the contingency table (point 6 in this paper).

What's going on here? To me, it looks like chisq.test shouldn't work, since not all of the assumptions are met, but the result from the test is correct?

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    $\begingroup$ The paper doesn't say what you say that it says... $\endgroup$
    – Glen_b
    Mar 2, 2020 at 2:28

1 Answer 1

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First, the assumption is about the cells of the expected table. None of the cells in your expected table would be 0.

Second, "not working" doesn't mean it won't give a correct answer. When assumptions fail then you lack the full guarantee of the things that have been proven to be true. But you could still get a factually accurate result if you use the method while flouting the assumptions.

That said, the assumption of no zeroes in the expected counts is necessary because of how the chi-square statistic is calculated. If you had actually failed this assumption then running the test would have thrown an error.

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