Simply, Fisher's exact test doesn't use a chi-square value or degrees of freedom. And it doesn't use asymptotic values for the p-values. So these cells are left blank in the output.
I had requested Fisher’s Exact Test. This was the 11th test I had run and the only one without the Fisher’s output. I am trying to work out if this is an SPSS issue or if I’m doing something wrong Sal Mangiafico
Rachel Casper , can you send SPSS output that isn't blank for the chi-sq for Fisher exact? As Sal Mangiafico says, this would be the odd thing.
Daniel Wright This is the test I ran before the one above, I haven't changed anything on my input except the variables, though have noticed that Fisher's test is termed slightly different? Thank you!
The test has changed and it is no longer for a 2x2 contingency table. That is not chi-sq value. According to SPSS: "The result labeled as Fisher's Exact test in the output is in fact the Fisher-Freeman-Halton Test. Note that a chi-square value is not calculated or printed for this test, although it was requested by the chi-square check box or the CHISQ keyword. An exact 2-sided significance is printed." https://www.ibm.com/support/pages/fisher-exact-test-rxc-table-fisher-freeman-halton-test
Daniel Wright thank you. I don’t want chi-square just Fisher’s as even in the 2x2 my cell count is
As said above, Fisher's is a probability, it doesn't produce a statistic like the chi-square (though see the ps). You use it when the margins are fixed, so that you know in advance what the row and column totals are. ps. I don't use SPSS, but if you post the contingency table it would be interesting to see if this is what R calls the hybrid Fisher procedure. For example:
Code:
fisher.test(matrix(c(2,4,8,2,12,8),nrow=3),hybrid=TRUE)
Output:
Fisher's Exact Test for Count Data hybrid using asym.chisq. iff (exp=5, perc=80, Emin=1)
data: matrix(c(2, 4, 8, 2, 12, 8), nrow = 3)
p-value = 0.3853
alternative hypothesis: two.sided
With hybrid=FALSE (the default) it just runs the probability test. So that value but with a non-central parameter not equal to zero. I am curious now so I hope you post the data!
Rachel Casper , the output is showing exactly* what it should: the p-value for the Fisher's exact test.
The chi-square value is still valid (if that's what you want for some reason), even with small cell counts, but the p-value for the chi-square may not be.
____________
* No pun intended ?
The values are being generated yes. Value (27.591) and df (1) are the same as Pearson Chi-square, but Fisher's test adjusts the p-value above, that it is
Victor Guimarães , maybe you can help me understand your answer, as it contradicts most of the other answers here. Since you answered after these (so as a responsible human being would have read them) it would have been useful for you to say why you disagree with them (if you do) and are putting forward a very different perspective (if you are). Are you claiming that the chi-sq value and the df for the Fisher "are the same as [for the] Pearson Chi-sq" test? And then the Fisher procedure just "adjusts the p-value"? This is a different reading of, for example, his tea tasting description of this. Are you referring to his disagreement with Pearson about the dfs?
Daniel Wright Sal Mangiafico thank you both for your help! Think I was too immersed in what I was doing that I hadn’t considered I would only be reporting the p-values for Fishers! The previous outputs with values on the Fishers row had thrown me and my common sense!! Really appreciate your help
Daniel Wright
I meant that I report the Pearson values (value and df), but I specify that the p-value is Fisher's. This I define in the caption of the table and methodology. As mentioned earlier, this model is for 2x2 tables. I do not disagree with the comments of previous colleagues, on the contrary, I agree.
Victor Guimarães , since the chi-sq value and df are for one statistical test, and the p-value is for a different one, you should not pretend they are for the same thing.
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