similar to: What's the T-Value in fisher.test

Okay, so I'm fairly retarded, and asked a question about finding the T-Value in the Fisher Exact method. I suppose what I'm truly after can best be explained by the Biddle Consulting site that has a program setup to deal with this kind of thing. Unfortunately, it is not currently functioning, and has not been for some time. As a result, I'm trying to figure out how to do this on my

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

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)

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

The hybrid feature in fisher.test looks to me like an excellent way to analyze my two-way tables. The only problem is that it does not seem to be implemented. Am I right about this? An example is pasted below. I note that I get the warning message only when I shouldn't: for a 2x2 table hybrid seems to be ignored without warning. In no case does fisher.test seem to be checking Cochran

Hi, I noted a discrepancy between R and openepi when I ran a fisher test with the same matrix. In R: > a=matrix(c(1,2,6,17), nrow=2) > a [,1] [,2] [1,] 1 6 [2,] 2 17 > fisher.test(a, conf.int=T) Fisher's Exact Test for Count Data data: a p-value = 1 alternative hypothesis: true odds ratio is not equal to 1 95 percent confidence interval: 0.02061498

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

Dear All, I have a problem understanding the difference between the outcome of a fisher exact test and a chi-square test (with simulated p.value). For some sample data (see below), fisher reports p=.02337. The normal chi-square test complains about "approximation may be incorrect", because there is a column with cells with very small values. I therefore tried the chi-square with

Hi, I noted a discrepancy between R and openepi when I ran a fisher test with the same matrix. In R: > a=matrix(c(1,2,6,17), nrow=2) > a [,1] [,2] [1,] 1 6 [2,] 2 17 > fisher.test(a, conf.int=T) Fisher's Exact Test for Count Data data: a p-value = 1 alternative hypothesis: true odds ratio is not equal to 1 95 percent confidence interval: 0.02061498

Dear All, I am trying to find MLE by using "OPTIM" function. Difficult in differentiating some parameter in my objective function, I would like to use the returned hessian matrix to yield an estimate of Fisher's Information matrix. My question: Since the hessian is calculated by numerical differentiate, is it a reliable estimate? Otherwise I would have to do a lot of work to

I just found a bug in fisher.test(). This is rw1080, on windows XP. A p-value can certainly not be Inf, but: > religion Costumbres rel orig Religion Si Algunas veces Nunca católica 2121 4700 6234 prot/evan 100 216 2461 otra C 27 67 502 otra 0 0 14 > fisher.test(religion, workspace=2000000)

This mean First, I am no expert but I am analyzing some marketing data. I have information on two versions of the same site, and I have data on the number of times people filled out a form on each version of the site. Sample data: Site 1 Site 2 Filled out form 10 35 Did not fill out form 50

I'm wondering how odds ratio is computed. I thought that it is (n11/n12)/(n21/n22), but it is not what fisher.test() computes. Could somebody let me know? > n11=3 > n12=1 > n21=1 > n22=3 > > n1_=n11+n12 > n2_=n21+n22 > > n_1=n11+n21 > n_2=n12+n22 > > x=rbind(c(n11,n12),c(n21,n22)) > > threshold=dhyper(n11,n1_,n2_,n_1) >

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

Hello. I had a big collection of Web pages. Now I have this collection divided into clusters. Every page can be relevant or not. I made a table: relevant non relevant cluster1 1 20 cluster2 0 15 cluster3 3 35 . . . . . . . . . I cluster1 I have 21 Web pages, 1

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

fisher.test(matrix(c(1,20,246,6873),2),hybrid=F) Error in if (f(lower, ...) * f(upper, ...) >= 0) stop("f() values at end points not of opposite sign") : missing value where logical needed Thomas Lumley Assistant Professor, Biostatistics University of Washington, Seattle -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-devel mailing list --

i have 2 groups i want to compare: group A and group B each group contains let's say 20 patients i want to perform a Fisher Exact test on genotype distribution so, see if there is a sign diff in genotpe frequency/distribution (#AA, #AB, #BB) between group A and B not for 1, but for 1000 different genes my question: how should i build my table so i can do: test <-

This is a bad bug as reported by Robin Hankin, it is still in "R-patched" ... ##- From: Robin Hankin <r.hankin@auckland.ac.nz> ##- To: r-help@stat.math.ethz.ch ##- Subject: [R] possum sleeping: thanks and fisher.test() FEXACT error ##- Date: Thu, 13 Jun 2002 16:46:26 +1200 ## ..... ## Example slighlty modified (MM) d4 <- matrix(c(0, 0, 0, 0, 0, 0, 3, 0, 1, 0, 0, 0, 0, 0,

Hi to all is there a fisher's post hock test or fisher's combination test in R available? Maybe not -- its very easy to do it in excel ... Kind regards Knut