I have 8 IV and 1 DV, they are all dichotomous and categorical data, so I am using Cross-tabulation Chi-square test. So 8 tests conducted. All test have at least one expected count less than 5. So I am interpreting fisher's exact test and not Chi-square. 3 out of 8 has fisher's exact test p=1. The groups are not identical. And the p value of chi-square of those 3 test are different from each other (although they have the same p value for fisher's exact test which is 1). I think there is something weird going on. Why 3 tests have p=1? What could be the problem?
If you report the observed frequencies for the 2x2 tables, others will be able to see if they can duplicate your results.
Also, if all expected counts are 1 or greater, consider using the N-1 Chi-square rather than Fisher's exact test (aka., the Fisher-Irwin test). See Ian Campbell's website for more info:
- http://www.iancampbell.co.uk/twobytwo/twobytwo.htm
HTH.
The tests are about the hypothesis that the odds ratio is 1. The observed odds ratio may differ more or less from 1. The p-value is the probability of "at least as extreme" deviations from 1 than what is observed.
If the observed odds ratio is 1 (what can happen, and presumably happend in your cases), the observed deviation is zero. So the probability is 1 (=it is sure) that any other hypothetical sample will give a deviation that is at least as large as the one observed.
You can get p=1 with any test. It happens when the observed test statistic is just identical to the test statistic under the null hypothesis. For instance, if you want to test the mean difference between two samples x = (1, 5, 6) and y = (3, 5, 4), the t-test will give p=1, because the observed mean difference is zero, and this is the null hypothesis of the test.
Jochen Wilhelm , Please can you further explain these 2 sentences because the odds ratio are 1 like you said. " If the observed odds ratio is 1 (what can happen, and presumably happend in your cases), the observed deviation is zero. So the probability is 1 (=it is sure) that any other hypothetical sample will give a deviation that is at least as large as the one observed."?
Also, I did not quite understand how it is possible that 3 independent tests have the exact p value. The 3 IV data-set are not identical. Do you think there is something wrong in my data?
Jochen Wilhelm I have attached the corresponding output of the 3 tests.
Ok, the OR is not exactly 1, but there is no other 2x2 table with the given row sums and column sums that would be closer to 1 than the one observed. This explains p = 1.
Take the first example. The row sums are 11 and 16, the col sums are 19 and 8. Fishers exact test is based on hoding these sums fixed. Your observed table is one out of 9 possible 2x2 tables with these marginal sums. One, for instance, is (3,8),(16,0). Another is (4,7),(15,1). If you calculate the ORs of all these 9 possible tables, you get:
0.000, 0.038, 0.119, 0.277, 0.583, 1.212, 2.700, 7.778, Infinity
Note that the 6th value is the one observed in your sample. This is only 0.212 off from H0 (from the value 1.0). All others are even further off. So your observed value is the "least extreme" possible. The probability of a deviation of at least 0.212 (or more) is sure (this is true for all 9 possible tables). This is why p = 1.
Ahmad Daboussi, given that all expected counts are > 1 and that the marginal totals do not appear to be fixed in advance (by design), I will again recommend that you use the N-1 Chi-square rather than the Fisher-Irwin test. See Ian Campbell's website and simulation study in Statistics in Medicine for details.
- http://www.iancampbell.co.uk/twobytwo/twobytwo.htm
And note that when you have a 2x2 table, the test of linear-by-linear association in the SPSS output is equivalent to the N-1 Chi-square.
- https://pubmed.ncbi.nlm.nih.gov/26576745/
HTH.
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