[R] Yates correction

David L Carlson dcarlson at tamu.edu
Tue Feb 21 14:40:57 CET 2017

Use fisher.test(). Yates' correction compensates for a tendency for Chi-square to be overestimated in a 2x2 table, but Yates' can overcompensate, reducing Chi-square too much. It's main advantage was when computers were expensive and Fisher's Exact was hard to compute by hand.  You can see from the following that Fisher's Exact estimates the p-value as .717, a bit less than .7477.

> M <- matrix(c(8, 12, 4, 10), 2, 2)
> fisher.test(M)

        Fisher's Exact Test for Count Data

data:  M
p-value = 0.717
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
 0.3160571 9.7976232
sample estimates:
odds ratio 

An alternative would be to let chisq.test() use simulations to estimate the p-value:

> chisq.test(M, simulate.p.value=TRUE)

        Pearson's Chi-squared test with simulated p-value (based on 2000 replicates)

data:  M
X-squared = 0.471, df = NA, p-value = 0.7141

Which agrees pretty well with fisher.test(). The X-squared value of 0.471 is the uncorrected value so you can see that the Yates' correction reduced it substantially (to .1035).

David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-4352

-----Original Message-----
From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Jomy Jose
Sent: Tuesday, February 21, 2017 4:48 AM
To: r-help at r-project.org
Subject: [R] Yates correction

 I tried to do chi square test for the following observed frequencies
   A  B
A  8  4
B 12 10

R gave the following output:
        Pearson's Chi-squared test with Yates' continuity correction

data:  M
X-squared = 0.10349, df = 1, p-value = 0.7477

Warning message:
In chisq.test(M) : Chi-squared approximation may be incorrect

Whether this result can be relied or we have to use Fishers exact test ?


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