Hi Aayush,
You are getting different results for fisher.test with
> mat2<-matrix(c(5,10,60,30),
nrow=2,dimnames=list(c("Dieting","Non-Dieting"),c("Men","Women")))
is because the first test used one-tailed (alternative="greater")
while the default without the alternative option is two-tailed.?? One-tailed has
more power, and should get a lower p-value if we select the correct option.??
For e.g. in the first option you used:
> mat <- matrix(c(10,5,30,60),
nrow=2,dimnames=list(c("Dieting","Non-Dieting"),c("Men","Women")))
> fisher.test(mat,alternative="greater")
??? Fisher's Exact Test for Count Data
data:? mat
p-value = 0.01588
alternative hypothesis: true odds ratio is greater than 1
95 percent confidence interval:
?1.319592????? Inf
sample estimates:
odds ratio
? 3.943534
> fisher.test(mat,alternative="two.sided")
??? Fisher's Exact Test for Count Data
data:? mat
p-value = 0.02063
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
? 1.10917 16.09195
sample estimates:
odds ratio
? 3.943534
Here, you selected the correct one-tailed, so the p-value got reduced compared
to two-tailed.
But, in the second case, the option is incorrect.? It shoud be
alternative="less" to get a pvalue of 0.01588.?
Null hypothesis: There is no association between gender and dietary habits.
Alternative hypothesis: There is an association between gender and dietary
habits (two-sided)
There is a positive association between gender and dietary habits (one-sided-
greater)
A.K.
----- Original Message -----
From: Aayush Raman <ayushraman at gmail.com>
To: r-help at r-project.org
Cc:
Sent: Friday, May 11, 2012 12:17 PM
Subject: [R] Fisher Test in R
Suppose we have the following data set:
? ? ? ? ? ? ? ? Men? ? Women
Dieting? ? ? ? 10? ? ? 30
Non-dieting? ? 5? ? ? 60
If I run the Fisher exact test in R then what does alternative = greater (or
less) imply? For example:
mat = matrix(c(10,5,30,60), 2,2)
fisher.test(mat,alternative ="greater")
I get the p-value = 0.01588 and odds ratio = 3.943534. Also, when I flip
the rows of the contingency table like this:
mat = matrix(c(5,10,60,30), 2,2)
fisher.test(mat,alternative ="greater")
then I get the p-value = 0.9967 and odds ratio = 0.2535796. But, when I run
the two contingency table without the alternative argument (i.e.,
fisher.test(mat)) then I get the p-value = 0.02063.
? 1. Could you please explain the reason to me?
? 2. Also, what is the null hypothesis and alternative hypothesis in the
? above cases?
? 3.
? Can I run the fisher test on a contingency table like this:
? mat = matrix(c(5000,10000,69999,39999), 2,2)
Thanks.
PS: I am not a statistician. I am trying to learn statistics so your help
(answers in simple English) would be highly appreciated.
??? [[alternative HTML version deleted]]
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