1
$\begingroup$

I don't have the minimum required data of 5 in each cell for a chi-squared test, but my contingency table is greater than 2x2, so I can't use a Fisher's exact. Would it help to use a chi-squared test set to either exact or Monte Carlo to prevent it from making the assumption of normal distribution?

I'm looking at trauma patterns created by five different tools; six categories of trauma have been defined. I would like to see whether the patterns of trauma (by occurrence of each type) differ significantly from one another. N=67, but not all tools have all types of trauma.

Improve this question
$\endgroup$
10
  • 2
    $\begingroup$ See Given the power of computers these days, is there ever a reason to do a Chi-squared test rather than Fisher's exact test? $\endgroup$
    – Andy W
    Commented Aug 20, 2012 at 14:56
  • 1
    $\begingroup$ One thing to note is that the 'requirement' is expected frequencies <5, not raw counts. Also, what software are you using? $\endgroup$
    – gung - Reinstate Monica
    Commented Aug 20, 2012 at 15:12
  • 1
    $\begingroup$ if you have less than 5 expected frecuencies, you can try exact tests, an excellent software (of course r will do it too) is StatXact $\endgroup$
    – AnastD
    Commented Aug 20, 2012 at 15:31
  • 2
    $\begingroup$ @gung Although the rule of thumb deals with expected frequencies, the problem of sparse cells is also identified by the actual sparseness of the cells and so having raw numbers below 5 is a concern. $\endgroup$
    – Michael R. Chernick
    Commented Aug 20, 2012 at 15:32
  • 1
    $\begingroup$ @MichaelChernick, I was simply referring to the discussion on the linked question. I was not myself asking that question. It is certainly related to the OP's original question, which is only a 2X2 contingency table (which would seem to make the comment about computation time a bit moot, at least in this circumstance). $\endgroup$
    – Andy W
    Commented Aug 20, 2012 at 15:48

1 Answer 1

Reset to default
3
$\begingroup$

There is a form of Fisher's test that is applicable to general rxc contingency tables. Use it or other alternatives to chi square in situations like yours where the chi square approximation is likely to be poor.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks for the responses! Most of my expected counts are under five, so I've used FET as suggested, using SPSS. $\endgroup$
    – Kendra
    Commented Aug 21, 2012 at 19:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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