I was recently asked this question about Chi-square tests. This question comes up a lot, so I thought I’d share my answer.
I have to compare two sets of categorical data in a 2×4 table. I cannot run the chi-square test because most of the cells contain values less than five and a couple of them contain values of 0. Is there any other test that I could use that overcomes the limitations of chi-square?
And here is my answer:
1. The assumption of the Chi-square test is not that the observed value in each cell is greater than 5. It is that the expected value in each cell is greater than 5. (The expected value for each cell is row total*column total/overall total). Often when the observed values are low, the totals are too, so they overlap a lot, but not always.
2. If you still find that your expected values are too low, use a Fisher’s Exact test. Not all standard statistical software will do one for a 2×4 table, but it can be done. StatXAct will do it. Great software if you can get access to it.
(If you’re wondering, SPSS will only do a Fisher’s Exact for a 2×2 table).
Hi. I am using spss. What can I do if I have one cell with an expected count less than 5? Which test should I perform, since spss wont allow me to do it with a 4×2 table? Thank you a lot of your response in advance…
I have a question regarding the validity of using chi-square test. Say if I want to look at the writing of two groups of people (native speaker vs. non-native speaker) and see if there’s association between the two groups in terms of a set of words (11 words). I intend to use normalized frequencies (normal frequency of each word per thousand words) because the usage of a certain word is sensitive to text length. And some of the normal frequencies are very small (<0.2) regardless that the raw frequencies are not (actually pretty large, such ), can I still use chi-square test of independence? Thanks!
there is one 2*2 table and in which one cell contain less than 5 count i.e 1. can be possible for Chi-square.
seasons greetings!
all my observations are less than 5 so i used fisher’s exact test but it showing that in place of ‘f’ value i.e ” Cannot be computed because there is insufficient memory”. can you please suggest any other procedure to check chi square test. my thesis topic is smart phone effect on psychological aspects related to social science. Thanks
Can I use Chi square when the expected count is less than 5 but the observed ones are higher than 5?
Hi Angel,
No, but you can when it’s the other way around.
I have a crosstab 2×2 with a cell having 0 value. So should I use what kinds of test for this crosstab.
Positive Negative
positive negative
male 38 6
female 30 0
hi all,
below is my data, I want generate chisq p-value for 5 tests together. I have data for 3 tests (1,2,3) but don’t have counts for test 4 and 5. these should include in the model to get pvalue. any one could suggest what I have to do?
data test;
test=1; trt=1; count=6;output;
test=1; trt=2; count=6;output;
test=2; trt=1; count=10;output;
test=2; trt=2; count=6;output;
test=3; trt=1; count=8;output;
test=3; trt=2; count=6;output;
test=4; trt=1; count=0;output;
test=4; trt=2; count=0;output;
test=5; trt=1; count=0;output;
test=5; trt=2; count=0;output;
run;
proc sort data=test;by test trt;run;
proc freq data=test order=data;
table test*trt/chisq;
weight count/zeros;
run;
*when i run this code i am getting below warning not getting chisqure pvalue;
Statistics for Table of test by trt
Row or column sum zero. No statistics computed for this table.
Sample Size = 42
Iam using graph pad prism and for 2×2 contingency table iam using fisher exact test because i have total sample size lees than 100 but at some times it gives very odd results that are not acceptable like
26 4
83 18
it gives significant p value for these 4 samples out of total sample size of one city
I have a crosstab 3×2 with a cell having 0 value. So should I use what kinds of test for this crosstab.
Positive Negative
4 17
6 20
0 13
Hello Karen. Thank you for your usefull input. I have a crosstab of 4×2 with more than 20% of expected frequencies lower than 5 and I am currently using SPSS. I found on the internet that we can use Fisher’s exact test on crosstabs bigger than 2×2 on SPSS, as a Fisher-Freeman-Halton Exact Test. Is it the same as running the first test? Once again, thanks for your help.
Thanks, Karen! Great explanation. This helps a lot.
I use STATA, which does to chi-squared for 2×2 tables, and also will provide expected cell frequencies so you can check your numbers.
tab var1 var2, expected chi
tab var1 var2, expected exact
In many cases, Fisher’s exact test can be too conservative. The mid-p quasi-exact test or N-1 chi-square may be good alternatives.
nice one.. this cleared my mind: observed versus expected..
i experienced having 0 values as well but my expected is of course more than 5..
Excellent. 🙂
thanks dear writer of the main text for introducing StatXAct and also Andreas for introducing R. Would somebody kindly tell me how can i access these softwares? by the way i have a 3×9 table, with two main variables and one variable as COUNT, i have to run Chi-square but several cells would contain zeros and i get a warning from SPSS and i’m afraid my results would not be true, what can i do if i dont get access to those mentioned softwares u introduced? i wonder of a great helpful hero would leave me a message on maryam_na_li@yahoo.com as within few days i’ve to submit my thesis and i’m still stuck with this data!!
Karen – Yes I believe so. Any 2-dimensional matrix will do – regardless of size.
Try:
v1 <- c(2,4,1,5,2,1,7,8,2)
v2 <- c(2,3,6,1,2,5,1,7,4)
v <- cbind(v1,v2)
fisher.test(v)
Sincerely
Thanks. I haven’t used R yet, but should probably start. I used to use SPlus, which I understand is the same or very similar, so I’m sure it wouldn’t be much of a stretch. But I find that most of my clients use either SPSS or SAS. I suspect R is going to infiltrate the market more and more, though.
Fisher exact was the reason I started learning R. R is free and the comand is simply fisher.test(name of contingency table)
Andreas–will R do a Fisher’s on any size table (within reason)?
Karen