Dear Gonçalo,
thanks for the additional information – I think I get now what you're trying to
do.
> On 27 Jun 2016, at 06:35, Gonçalo Ferraz wrote:
>
> probabilities in lpvec should be <=1, but it is not. The sum is something on
> the order of 1.48e-13.”
> It is absolutely right that lpvec do
Stefan,
I am sorry that I wasn’t more careful writing my question. I misrepresented my
problem when I wrote “the sum of the probabilities in lpvec should be <=1, but
it is not. The sum is something on the order of 1.48e-13.”
It is absolutely right that lpvec doesn’t contain probabilities, it con
Why do you want to do this? Why not simply use Fisher's exact test?
N <- 2178
N1 <- 165
N2 <- 331
J <- 97
ct <- rbind(c(J, N1-J), c(N2-J, N-N1-N2+J))
fisher.test(ct)
Background explanation:
- Your formula computes the log hypergeometric probability for a contingency
table as ct above, but with
> On 25 Jun 2016, at 16:13 , Gonçalo Ferraz wrote:
>
> PROBLEM: the sum of the probabilities in lpvec should be <=1, but it is not.
> The sum is something on the order of 1.48e-13.
Um, in which sense is 1.48e-13 not <=1 ???
--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Bus
Floating point numbers are rounded to 53 significant bits. When you
use logs of integers you are using
floating point numbers.
> .3 + .6 == .9
[1] FALSE
> (.3 + .6) - .9
[1] -1.110223e-16
See FAQ 7.31 for more discussion.
On Sat, Jun 25, 2016 at 10:13 AM, Gonçalo Ferraz wrote:
> Hi,
>
> I am w
Hi,
I am working on interactions between animals, studying whether animal 1 is
attracted to animal 2 (or vice-versa). I looked for the two animals in N=2178
sampling occasions, finding animal 1 a total of N1=165 times, and animal 2 a
total of N2=331 times. In J=97 occasions, I saw both animals
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