The p value by Fisher's exact test is usually more than the p calculated by Pearson Chi square; but in the SPSS output, sometimes, the p by Fisher is less than the p by Chi. Please explain.
Fisher's exact test (which was also developed independently by Irwin) assumes that both row and column totals are fixed (by design) in advance. When some of the marginal totals are not not fixed in advance (which is usually the case), the Fisher-Irwin test is conservative. That is just another way of saying that its p-value is systematically higher than it ought to be. For 2x2 tables where at least some of the marginal totals are free to vary, E.S. Pearson's N-1 Chi-square test is a better choice than the Fisher-Irwin test. For more info, see Ian Campbell's nice simulation study.
I see you included SPSS as a subject heading. In the Chi-square output from CROSSTABS in SPSS, the test of Linear-by-linear association for a 2x2 table is equivalent to the N-1 Chi-square.
HTH.
http://www.iancampbell.co.uk/twobytwo/twobytwo.htm
https://sites.google.com/a/lakeheadu.ca/bweaver/Home/statistics/notes/chisqr_assumptions
http://www.ncbi.nlm.nih.gov/pubmed/26576745
Please have a look at my recent paper on this issue. It provides a numerical answer to question posed here. Article :. The Tale of Cochran's Rule: My Contingency Table has so Many Expected Values Smaller than 5, What Am I to Do? https://amstat.tandfonline.com/doi/abs/10.1080/00031305.2017.1286260
Abstract: In an informal way, some dilemmas in connection with hypothesis testing in contingency tables are discussed. The body of the paper concerns the numerical evaluation of Cochran's Rule about the minimum expected value in r×c contingency tables with fixed margins when testing independence with Pearson's X2 statistic using the chi-squared distribution.
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