On 01/05/2013, at 06:09 AM, Antonio Silva wrote:
Thanks Zoltan R R-sig list members
I still with some doubts.
Is there a way to calculate in R the results for the equation
X2= n*[|ad-bc|-(p/2)]^2 / [(a+b)*(c+d)*(a+c)*(b+d)]
(see Legendre & Legendre pg 295 eq. 7.6)
This should be doable in vegan::designdist() function. Set 'abcd = TRUE' and you can directly use terms like you defined them above. I won't give the equation here as I don't know what is the equation you used: Your previous message had a different equation than this one here with some mix up with 'n' and 'p'. The designdist() function calculates (dis)similarities between rows. You must transpose (t()) your data if you want to have dissimilarities between columns. The code is pure R so that you can see how to do these calculations yourself.
Cheers, Jari Oksanen
for the following data:
ST01 ST02 ST03 ST04 ST05 ST06 ST07 ST08 ST09 SP1 1 1 1 1 1 0 0 1 1 SP2 1 1
0 1 0 1 0 1 0
We have a=4, b=3, c=1, d=1, N=9 and the 2x2 table for this example is
SP2
Presence Absent
SP1 Presence 4 3
Absent 1 1
Following the Chi square formula I got 110.25 / 280 = 0.3937
I have tried many things, without success.
Thanks in advance for any suggestion.
Antonio Olinto
2013/4/25 Zoltan Botta-Dukat <botta-dukat.zoltan at okologia.mta.hu>
Dear Antonio,
Try this:
chisq.test(table(sp1,sp2))
Best wishes
Zoltan
On Wed, Apr 24, 2013 at 6:02 PM, Antonio Silva <aolinto.lst at gmail.com
wrote:
Hi,
I'm trying to use Pearson's Chi-squared to verify the dependence among
species distribuion.
I have a dataframe with the presence/absence data of two species in a
number of sample units
The equation I'm using is:
X2= p*(|ad-bc|-p/2)^2 / ((a+b)*(c+d)*(a+c)*(b+d))
where a is the number of double presence (1-1), b is the number of 1-0,
is the number of 0-1 and d is the number of 0,0 (double absence)
p is a+b+c+d
Is there a function to caltulate it using R? I could not understand how
use chisq.test function for this.
Thanks in advance.
Antonio Olinto
--
Zolt?n BOTTA-Duk?t
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