Fisher exact test is said to be used with a total sample (n) < 1000, whereas chi-squared test should be used when each category (/cell in a contingency table) >=5. What if you have an mxn contingency table where the total sample size > 1000 but some of the cells have 0 or 1 sample size?
Solved – Fisher and chi-squared assumptions/limitations not met
assumptionschi-squared-testcontingency tablesfishers-exact-test
Related Solutions
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
One solution would be to use a bootstrap test as an approximation to a permutation test. Permutation tests are exact and most powerful; in this case there are too many permutations to calculate every one of them, so you'd approximate the test with the bootstrap.
Basically, you:
1) Calculate your test statistic, label it $T_0$, on the actual data, say for illustrative purposes the same chi-square statistic you've already calculated,
2) Construct 1,000 or 10,000 or so ("many") random contingency tables under the assumption the null hypothesis is true, and for each one calculate the chi-square statistic, label them $T_1 \dots T_B$.
3) Compare your test statistic's value $T_0$ with the the test statistic values $T_1 \dots T_B$ from the randomly-generated contingency tables, and see what fraction are more extreme than $T_0$; this gives you a bootstrap p-value.
We are approximating the distribution of the test statistic under the null hypothesis by randomly generating a lot of values for the test statistic under the null hypothesis; this lets us estimate the p-value associated with the value of the statistic we actually observed.
I can't help you with the SPSS part of this, unfortunately.
Here's a reference which I've found helpful in the past: Permutation, Parametric, and Bootstrap Tests of Hypotheses (Good).
Best Answer
How far above 1,000 is your sample size? If it's not far above 1,000, you can use Fisher's exact test - it's simply recommended that you don't because of computational limitations.
If Fisher's exact test is too computationally intensive and you need the chi-square test, would try to "bin" the variables differently. That is, collapse categories until you have at least 5 in each cell. You could, alternatively, use Yates' correction to account for the undercounts in certain cells.