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I wish to simulate: Ho: Odds Ratio =1 H1 Odds Ratio <> 1 Can any of the R users show me how to generate this data for say 20% from Ho and 80% from H1? Thanks, Jim [[alternative HTML version deleted]]

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

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]]

The concrete problem is that I am refereeing a paper where a confidence interval is presented for the risk ratio and I do not find it credible. I show below my attempts to do this in R. The example is slightly changed from the authors'. I can obtain a confidence interval for the odds ratio from fisher.test of course === fisher.test example === > outcome <- matrix(c(500, 0, 500, 8),

Data from Fisher's paper: Confidence Limits for a Cross-Product Ratio. > y col1 col2 [1,] 10 3 [2,] 2 15 fisher.test(y) Fisher's Exact Test for Count Data data: y p-value = 0.0005367 alternative hypothesis: true odds ratio is not equal to 1 95 percent confidence interval: 2.753438 300.682787 sample estimates: odds ratio 21.30533 The crude odds

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) >

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

Hello: Please help me through my confusion. I am having trouble reconciling the difference between what I believe is the conventional definition of an odds ratio for a 2-by-2 table and the output produced by fisher.test() in R. Consider the following example: > Discrim <- matrix(c(1,10,24,17), + nr = 2, + dimnames = list(AGE = c('young', 'old'), +

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

# 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 =

I do not understand how to interpret this to find the T Value for the data. Is there a way to figure this out, or another function that will provide this for me using Fisher's Exact Test? The outcome of my data is listed below. data: DATA p-value = 0.1698 alternative hypothesis: true odds ratio is not equal to 1 95 percent confidence interval: 0.6026805 79.8309210 sample estimates:

Dear all, I want to simulate from the null distribution of the following 2 x 3 table, 2 5 10 4 8 5 I am using a chi-squared test. Anyone has any idea how to do this? -- Thanks, Jim. [[alternative HTML version deleted]]

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

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]]

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

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

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

Dear r-helpers, I would be very grateful if you could post the message below on the r-help discussion board. Thank you very much! Best Wishes, Pawel Hello R community, I am generating lots of results using the fisher.test function, testing many 2x2 tables of SNPs for association with a particular phenotype. A typical output of the fisher.test function would be (for example): data: data1

Dear all Seems that puzzles always come in packs. I was asked to help with some statistics in blood analysis. (You can not refuse your wife's asks :-). She has contingency table for values IgVH mutation and ZAP expression. I can do chi-square test (in R) and get a results, and with some literature I can try explain them. However she found an article in which they use SPSS and use

I have the following contingency table dat <- matrix(c(1,506,13714,878702),nr=2) And I want to test if their is an association between events A:{a,not(a)} and B:{b,not(b)} | b | not(b) | --------+-----+--------+ a | 1 | 13714 | --------+-----+--------+ not(a) | 506 | 878702 | --------+-----+--------+ I am worried that prop.test and chisq.test are not valid given the