similar to: Fisher's test

Is anyone aware of a fast way of doing fisher's exact test for a series of 2 x 2 tables in R? The fisher.test is really slow if n1=1000 and n2 = 1000. -- Thanks, Jim. [[alternative HTML version deleted]]

Hi folks, Forgive me if this question is a trivial issue. I was doing a series of Fishers' exact test using the fisher.test function in stats package. Since the counts I have were quite large (c(64, 3070, 2868, 4961135)), R suggested me to use *other algorithms* for the test which can be specified through the 'control' argument of the fisher.test function as I understood. But where

A simple question - using the following fishers test it appears that the P value is significant, but the CI includes 1. Is this result correct? > data.50p10min <- matrix(c(16,15, 8, 24),nrow=2) > fisher.test(data.50p10min) Fisher's Exact Test for Count Data data: data.50p10min p-value = 0.03941 alternative hypothesis: true odds ratio is not equal to 1 95

I have a matrix say, 1 4 23 30 and I want to find the previously attainable fisher's exact test p-value. Is there a way to do this in R? -- Thanks, Jim. [[alternative HTML version deleted]]

Hello R help members. I came across today with an article on Barnard's exact test (http://www.cytel.com/Papers/twobinomials.pdf), that is supposed to give a more powerful fisher.test - Because it doesn't assume that we know the row and column totals are in advance. Any pointers to such a function ? Thanks, Tal -- ---------------------------------------------- My contact information:

In an effort to select the most appropriate number of clusters in a mixture analysis I am comparing the expected and actual membership of individuals in various clusters using the Fisher?s exact test. I aim for the model with the lowest possible p-value, but I frequently get p-values below 2.2e-16 and therefore does not get exact p-values with standard Fisher?s exact tests in R. Does anybody know

I want to ascertain the basis of the table ranking, i.e. the meaning of "extreme", in Fisher's Exact Test as implemented in 'fisher.test', when applied to RxC tables which are larger than 2x2. One can summarise a strategy for the test as 1) For each table compatible with the margins of the observed table, compute the probability of this table conditional on the

Full_Name: Martha Nason Version: 2.0.1 OS: Windows XP Submission from: (NULL) (137.187.154.154) I am running simulations using fisher's test on 2 x c tables and a very small p.value from fisher's test (<2.2e-16) is returned as a negative number. Code follows. > set.seed(0) > nreps.outer <-7 > pvalue.fisher <- rep(NA,nreps.outer) > > population1 <- c(

Hi, I'm trying to run Fisher's Exact test on the data below, but I'm not sure how interpret the data shown. Can someone tell me what this is saying? Looking at the numbers it should be that there's no significant difference between the HDL and LDL, but a p-value of 1 seems high. Is the low value in the LDL unbound making the test unstable and should I be using an alternative? Best

Hello everyone, Can someone tell me exactly how to generate data from a null distribution for the fisher exact test? I know I have to use the hypergrometric but exactly what commands do I use? Jim [[alternative HTML version deleted]]

Hi all, I?m fitting GLM?s and I can?t interprete the coefficients when I run a model with interaction terms. When I run the simpliest model there is no problem: Model1<-glm (Fishes ~ Year + I(Year^2) + Kind.Geographic + Kind.Fishers + Zone.2 + Hours + Fishers + Month, family = poisson(log)) # Fishes, Year, Hours, and Fishers are numeric, Kind.Geographic, Kind.Fishers, Zone.2 and

Appology in advance if this has been asked and answered. I am getting different answers using Fisher's exact test in the package ctest. For example: x_cbind(c(8, 22), c(31, 29)) fisher.test(x) Fisher's Exact Test for Count Data data: x p-value = 0.04024 alternative hypothesis: two.sided However, in SAS or other packages on the net the answer I get is p-value= 0.02664. Is this

Hello, I'm using R to apply Fishers exact test to a whole pile of contingency tables, and I've run into the bug shown below. regards, Francis -- > dat1 = matrix(c(0,1,0,1,0,0,1,0,0,0,1,0,1,0,0,1,0,0,0,1,0,1,0,0,1,0,0,0,0, 1,1,0,0,1,0,0,1,0,0,1,0,0,3,0,0,2,0,0,1,0,0,0,1,0,0,2,0,0, 2,0,0,1,0,1,0,0,3,0,0,2,0,0,0,1,0,5,0,0,0,1,0,1,0,0), nrow=3) >

Dear all, how can I "reshape"/"cast" the following matrix 00;01;10;11 John.Mike;123;313;12;31 John.Jim;54;57;39;36 John.Steve;135;47;47;74 Mike.Jim;63;37;27;16 Mike.Steve;15;15;5;61 Jim.Steve;6;10;34;35 into a set of stacked 2x2 contingency tables 0;1 John;123;12 Mike;313;31 John;54;39 Jim;57;36 John;135;47 Steve;47;16 ... so that the "fisher.test" and

Hi, R-help, I have a group of data from RNA-seq want to be analyzed by Fisher's exact test in R. I want to compare the significant difference of about 30,0000 individuals in two different samples, and I have no idea how to use R, so could you please give me some suggestions or the scripts for Fisher's exact test? Thank you very much. Best, Guanfeng Wang [[alternative HTML version

I have two vectors (a and b) with counts of animals and wanted to calculate fisher's alpha: library(vegan) a <- c(2043, 1258, 52, 1867, 107, 1624, 2, 157, 210, 402, 5, 107, 267, 2, 13683) b <- c(2043, 1258, 52, 1867, 107, 1624, 2, 157, 210, 402, 5, 107, 267, 2, 3000) fisher.alpha(a) fisher.alpha(b) fisher.alpha(a) gave the following warnings: > fisher.alpha(a) [1] 1.572964

Dear friends, My dataset have many zeros, so i must use fisher exact test . Unfortunately, the fisher.test(stats) function fail to do it. Anybody knows how to do the fisher exact test with many zeros in the dataset? My dataset is: a<-matrix(c(0,1,0,0,0,0,1,0,1,0,0,0,0,1,0,1,1,0,2,1,5,1,1,6,4,4,1,17,2,8,5,7,1,1,24,3,6,1,1,3,2,16,7,4,0,2,4,0,17,0,1,0,0,0,1,2),nrow=8,byrow=TRUE) data.frame(a)

# is there a bug in the calculation of the odds ratio in fisher.test? # Nicholas Horton, nhorton at smith.edu Fri Jan 22 08:29:07 EST 2010 x1 = c(rep(0, 244), rep(1, 209)) x2 = c(rep(0, 177), rep(1, 67), rep(0, 169), rep(1, 40)) or1 = sum(x1==1&x2==1)*sum(x1==0&x2==0)/ (sum(x1==1&x2==0)*sum(x1==0&x2==1)) library(epitools) or2 = oddsratio.wald(x1, x2)$measure[2,1] or3 =

Timothy, I believe you are mistaken. Fisher's exact test give the correct answer even in the face of small expected values for the cell counts. Pearson's Chi-square approximates Fisher's exact test and can give the wrong answer when expected cell counts are low. Chi-square was developed because it is computationally "simple". Fisher's exact test, particularly with tables

Hi, I have a table of observed counts for various genetic markers. Instead of doing Fisher's exact test for each marker one at a time and recording the P value manually, is there a script to go through the whole list and generate the P value column automatically? An example of my data: Counts_CHB and Counts_AA are the observed counts for one allele. 2N_CHB and 2N_AA are the total number