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


> On Feb 21, 2017, at 7:40 AM, David L Carlson <dcarl...@tamu.edu> wrote:
> 
> 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 
>  1.641969
> 
> 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-boun...@r-project.org] On Behalf Of Jomy Jose
> Sent: Tuesday, February 21, 2017 4:48 AM
> To: r-help@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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