similar to: fisher.test - can I use non-integer expected values?

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 have a text mining project and currently I am working on feature generation/selection part. My plan is selecting a set of words or word combinations which have better discriminant capability than other words in telling the group id's (2 classes in this case) for a dataset which has 2,000,000 documents. One approach is using "contrast-set association rule mining" while the

Hi, I want to analyse a contigency table (3 x 12) with a fisher.test beacause there are cells that are less than 5. ?mmen Anken Baf Belchen H?chi Hof Porti R?m Schmutz Sch?n Sissa Tann class14 7 26 150 2 46 68 126 66 3 31 7 61 class24 7 6 55 5 49 71 93 90 1 18 16 79 class34 1 1 4 3 19 8 29 61

Dear list! I would like to calculate "chisq.test" on simple data set with 70 observations, but the output is ''Warning message:'' Warning message: In chisq.test(tabele) : Chi-squared approximation may be incorrect Here is an example:         tabele <- matrix(c(11, 3, 3, 18, 3, 6, 5, 21), ncol = 4, byrow = TRUE)         dimnames(tabela) <- list(        

i have a data frame(dat) which has many variables.and i use the following script to get the crosstable. >danx2<-c("x1.1","x1.2","x1.3","x1.4","x1.5","x2","x4","x5","x6","x7","x8.1","x8.2","x8.3","x8.4","x11",

TOPIC My question regards the philosophy behind how R implements corrections to chi-square statistical tests. At least in recent versions (I'm using 2.13.1 (2011-07-08) on OSX 10.6.8.), the chisq.test function applies the Yates continuity correction for 2 by 2 contingency tables. But when used as a goodness of fit test (GoF, aka likelihood ratio test), chisq.test does not appear to implement

Hello all, I would like to know what the difference is between chisq.test and fisher.test when using the Monte Carlo method with simulate.p.value=TRUE? Thank you -- View this message in context: http://r.789695.n4.nabble.com/Difference-in-Monte-Carlo-calculation-between-chisq-test-and-fisher-test-tp2322494p2322494.html Sent from the R help mailing list archive at Nabble.com.

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),

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

Hi, In the chisq.test(), if the expected frequency for some categories is <5, there will be a warning message which says Warning message: Chi-squared approximation may be incorrect in: chisq.test(x, p = probs) I am wondering whether there are some methods to get rid of this mistake... Seems the ?chisq.test() doesn''t provide more options to solve this problem. Or, the only choice is

Recently, I noted a post and replies on R-Help from Professor Marc Feldesman regarding a cross tabulation function that generates row and column proportions, marginal values, expected cell values and tests for independence presumably similar in a fashion to the output of the S-Plus crosstabs() function or the SAS Proc Freq. Martin Maechler had posted some code in reply for folks to update and

Dear all, I’m having an awkward problem in R. When I write the command Fisher.test(school.data,workspace=2e+07), where school.data is the matrix corresponding to the data set, I get the error message: FEXACT error 7. LDSTP is too small for this problem. Try increasing the size of the workspace. Increasing the workspace: Fisher.test(school.data,workspace=1e+10), I

R Help on 'chisq.test' states that "if 'simulate.p.value' is 'TRUE', the p-value is computed by Monte Carlo simulation with 'B' replicates. In the contingency table case this is done by random sampling from the set of all contingency tables with given marginals, and works only if the marginals are positive... In the

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 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

Hi There, There are 300 boy students and 100 girl students in a class. One interesting question is whether boy is smarter than girl or not. first given the exam with a difficulty level 1, the number of the student who got A is below 31 for boy, 10 for girl. Then we increase the difficulty level of the exam to level 2, the number of the student who got A is below 32 for boy, 10 for girl. We

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

Dear all, I have a dataset with qualitative variables (factors) and I want to test the null hypothesis of independance between two variables for each pair by using appropriate tests on contingency tables. I first applied chisq.test and obtained dependance in almost all cases with extremely small p-values and warning messages. > chisq.test(table(data$ins.f, data$ins.st))$p.val [1]

Hello, I am trying to see whether there has been a significant difference in whether people experienced damages from wildlife in two different years. I therefore have two columns: year 1: yes no no no yes yes no year 2: no yes no yes I wanted to do a chisq.test, but if I enter it this way: chisq.test(year1, year2) I get the error saying the columns are two different lengths. So then I tried

I want to calculate chi square test of goodness of fit to test, Sample coming from Poisson distribution. please copy this script in R & run the script The R script is as follows ########################## start ######################################### No_of_Frouds<- c(4,1,6,9,9,10,2,4,8,2,3,0,1,2,3,1,3,4,5,4,4,4,9,5,4,3,11,8,12,3,10,0,7) N <- length(No_of_Frouds) # Estimation of