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Yates correction

4 messages · Jomy Jose, David L Carlson, Marc Schwartz +1 more

Jomy Jose
# 
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 ?

Jose
David L Carlson
# 
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.
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 
  1.641969

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

Jose


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Marc Schwartz
# 
Hi,

In general, statistical questions that are more conceptual in nature, which is the case here, are generally frowned upon on this list, since they are considered off-topic here.

That being said, increasingly, both the Fisher Exact test and the use of the Yates correction to the Chi-Square test are being challenged as being overly conservative in small sample situations. The same goes for the common recommendation of switching to the Fisher Exact Test when there are any expected cell values of <5, which is what is generating the "Chi-squared approximation may be incorrect" in the examples below, since one of your expected cell values is 4.94.

There was a paper by Campbell back in 2007 that discussed this:
  
Chi-squared and Fisher?Irwin tests of two-by-two tables with small sample recommendations
http://onlinelibrary.wiley.com/doi/10.1002/sim.2832/abstract

and he has a web site here with additional resources:

http://www.iancampbell.co.uk/twobytwo/twobytwo.htm

Even using his 'n-1' variant of the test with his online calculator on the above web site, you end up with a p value of 0.5, which is close to the p value for the uncorrected chi-square (chisq.test() with correct = FALSE) of 0.4925. Thus, none of these cases results in a "statistically significant" p value of <=0.05. Not that you should be p value hunting anyway here. 

The whole p value discussion is further from being on topic here, but under none of these hypothesis tests would you reject the null.

Further consultation with a local statistical expert would seem prudent.

Regards,

Marc Schwartz
Rolf Turner
#
On 21/02/17 23:47, Jomy Jose wrote:
(a) With a p-value of 0.7477 there is no evidence against the null 
hypothesis no matter how you slice it.

(b) To assuage your trepidations, use "simulate.p.value=TRUE".

E.g.

    chisq.test(M,simulate.p.value=TRUE,B=9999)

Note that the value of X-squared that is returned is "of course" the 
same as what you'd get by setting correct=FALSE. I got a p-value of 
0.7178; you will get something slightly different, since a simulated 
p-value is random, but it will be about 0.71 or 0.72.

Bottom line:  Don't reject H_0!!!

cheers,

Rolf Turner