It is not clear to me whether we can consider the Fisher's exact test as a "parametric" or "non-parametric" one. My gut feeling is that it should be defined as "parametric" as it involves a fully specified distribution (the hypergeometric). If so, however, I would not able to find an example of a non-parametric test for 2×2 contingency tables, which makes me wonder whether the distinction can be useful at all in this case.
Solved – Is the Fisher’s exact test “parametric” or “non-parametric”
contingency tablesfishers-exact-testnonparametric
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The only problem with applying Fisher's exact test to tables larger than 2x2 is that the calculations become much more difficult to do. The 2x2 version is the only one which is even feasible by hand, and so I doubt that Fisher ever imagined the test in larger tables because the computations would have been beyond anything he would have envisaged.
Nevertheless, the test can be applied to any mxn table and some software including Stata and SPSS provide the facility. Even so, the calculation is often approximated using a Monte Carlo approach.
Yes, if the expected cell counts are small, it is better to use an exact test as the chi-squared test is no longer a good approximation in such cases.
You can turn the question around. Since the ordinary Pearson $\chi^2$ test is almost always more accurate than Fisher's exact test and is much quicker to compute, why does anyone use Fisher's test?
Note that it is a fallacy that the expected cell frequencies have to exceed 5 for Pearson's $\chi^2$ to yield accurate $P$-values. The test is accurate as long as expected cell frequencies exceed 1.0 if a very simple $\frac{N-1}{N}$ correction is applied to the test statistic.
Campbell, I. Chi-squared and Fisher-Irwin tests of two-by-two tables with small sample recommendations. Statistics in Medicine 2007; 26:3661-3675. (abstract)
...latest edition of Armitage's book recommends that continuity adjustments never be used for contingency table chi-square tests;
E. Pearson modification of Pearson chi-square test, differing from the original by a factor of (N-1)/N;
Cochran noted that the number 5 in "expected frequency less than 5" was arbitrary;
findings of published studies may be summarized as follows, for comparative trials:
Yates' chi-squared test has type I error rates less than the nominal, often less than half the nominal;
The Fisher-Irwin test has type I error rates less than the nominal;
K Pearson's version of the chi-squared test has type I error rates closer to the nominal than Yate's chi-squared test and the Fisher-Irwin test, but in some situations gives type I errors appreciably larger than the nominal value;
The 'N-1' chi-squared test, behaves like K. Pearson's 'N' version, but the tendency for higher than nominal values is reduced;
The two-sided Fisher-Irwin test using Irwin's rule is less conservative than the method doubling the one-sided probability;
The mid-P Fisher-Irwin test by doubling the one-sided probability performs better than standard versions of the Fisher-Irwin test, and the mid-P method by Irwin's rule performs better still in having actual type I errors closer to nominal levels.";
strong support for the 'N-1' test provided expected frequencies exceed 1;
flaw in Fisher test which was based on Fisher's premise that marginal totals carry no useful information;
demonstration of their useful information in very small sample sizes;
Yates' continuity adjustment of N/2 is a large over-correction and is inappropriate;
counter arguments exist to the use of randomization tests in randomized trials;
calculations of worst cases;
overall recommendation: use the 'N-1' chi-square test when all expected frequencies are at least 1; otherwise use the Fisher-Irwin test using Irwin's rule for two-sided tests, taking tables from either tail as likely, or less, as that observed; see letter to the editor by Antonio Andres and author's reply in 27:1791-1796; 2008.
Crans GG, Shuster JJ. How conservative is Fisher's exact test? A quantitative evaluation of the two-sample comparative binomial trial. Statistics in Medicine 2008; 27:3598-3611. (abstract)
...first paper to truly quantify the conservativeness of Fisher's test;
"the test size of FET was less than 0.035 for nearly all sample sizes before 50 and did not approach 0.05 even for sample sizes over 100.";
conservativeness of "exact" methods;
see Stat in Med 28:173-179, 2009 for a criticism which was unanswered
Lydersen S, Fagerland MW, Laake P. Recommended tests for association in $2\times 2$ tables. Statistics in Medicine 2009; 28:1159-1175. (abstract)
...Fisher's exact test should never be used unless the mid-$P$ correction is applied;
value of unconditional tests;
see letter to the editor 30:890-891;2011
Best Answer
tl;dr: Fisher's Exact Test is nonparametric in the sense that it does not assume that the population is based on theoretical probability distributions (normal/geometric/exponential etc.), but that the data itself reflects the parameters, which is why it proceeds with the assumption that the row/col totals are fixed.
Fisher's exact test, as its name suggests, gives the exact p-value rather than an estimation based on a particular sampling distribution thought to be aligning with the variable(s).
If you have two or more variables, all categorical/nominal, and your data consists of independent observations, then you can already intuitively create a cross-tabulation to assess conditional frequencies (akin to how you would want to see overlaps in a Venn diagram). For instance, say your independent variable is gender (M/F/O) and the dependent variable is party allegiance (D/R/I).
Now let's say we do not know the probability distribution of either variable, which means that we can't just plug the data into any parametric test. (In the classical FET where it's only a 2x2 (two dichotomous variables) which you know are binomially distributed, you could proceed using the hypergeometric distribution to estimate the p-value.)
Fisher's exact test directly gives us the probability of finding a result as extreme as the one we have. In other words, it reflects how far our observed frequencies are from the expected frequencies. If gender is truly independent of party membership, then there ought to be roughly uniform distribution. (Aside: you can use the 1 sample K-S test here to test for uniform distribution.)
But Fisher's test takes all the discrete values <= the observed ones, calculates their probabilities, and adds them up to give you the p-value, which you then compare to your alpha (probability of a Type I error, i.e. mistakenly rejecting the null hypothesis of there being no association between gender and party membership).
NB that although the FET is used as a recourse to the cross-tabbed chi square test when the sample size is low, the FET has its own assumptions -- I'd use it only for MECEly organised data such that the variables are 'really' nominal in a fundamental sense and not contrived for simplicity's sake (e.g. biological sex is 'truly' nominal if we use the usual definitions, whereas 'treatment status' must never be taken to be a true nominal variable) and where the individual instances are independently recorded.
For an actual rigorous idea of what the FET entails mathematically, take a look at Weisstein's neat definition -- http://mathworld.wolfram.com/FishersExactTest.html.