similar to: Create 2x2 table from summary data and run chi square test.

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

Can anyone tell me how to calculate a mid-p value for a chi-squared test in R? Many thanks, Andrew Wilson

Dear UseRs, I'm running a chi-squared test where the expected matrix is the same as the observed, after rounding. R reports a X-squared of zero with a p value of one. I can justify this because any other result will deviate at least as much from the expected because what we observe is the expected, after rounding. But the formula for X-squared, sum (O-E)^2/E gives a positive value. What

I don't quite understand the difference between the two methods for performing a chi-squared test on contingency tables: summary(table()) and chisq.test() They may different results. E.g.: aa <- gl(2, 10) bb <- as.factor(c(1,2,2,2,1,2,1,2,2,2,1,2,2,2,1,1,1,2,1,1)) aa <- c(aa, aa) bb <- c(bb, bb) table(aa, bb) summary(table(aa, bb)) chisq.test(aa, bb) Could somebody give me

Full_Name: nobody Version: 2.2.0 OS: any Submission from: (NULL) (219.66.34.183) 2 x 2 table, such as > x [,1] [,2] [1,] 10 12 [2,] 11 13 > chisq.test(x) Pearson's Chi-squared test with Yates' continuity correction data: x X-squared = 0.0732, df = 1, p-value = 0.7868 but, X-squared = 0.0732 is over corrected. when abs(a*d-b*c) <= sum(a,b,c,d), chisq.value

Hello I have the matrix a<-matrix(c(2,1,0,1,2,2,1,5,7,2,5,12),nrow=6) a [,1] [,2] [1,] 2 1 [2,] 1 5 [3,] 0 7 [4,] 1 2 [5,] 2 5 [6,] 2 12 Suppose that in the first row we have 3 men of England, 2 with hair, and 1 no In the second we have 6 italian men, 1 with hair and 5 no ... I want to find if there is a dependence between men withouth hair and

I was asked by my boss to do an analysis on a large data set, and I am trying to convince him to let me use R rather than SPSS. I think Sweave could make my life much much easier. To get me a little closer to this goal, I ran my analysis through R and SPSS and compared the resulting values. In all but one case, they were the same. Given the matrix [,1] [,2] [1,] 110 358 [2,] 71 312 [3,]

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

Dear R help I have been trying to conduct a chi square test on two groups of variables to test whether there is any relationship between the two sets of variables chisq.test(oxygen, train) Pearson's Chi-squared test data: oxygen X-squared = 26.6576, df = 128, p-value = 1 > chisq.test(oxygen) Pearson's Chi-squared test data: oxygen X-squared = 26.6576, df = 128,

Hi All, I woul like to ask you a couple of questions on chisq.test. First, I have 40 flies, 14 males and 26 females and I want to test for an a priori hypothesis that the sex ratio is 1:1 sex<-c(14,26) pr<-c(1,1)/2 chisq.test(se, p=pr, correct=TRUE) Chi-squared test for given probabilities data: sex X-squared = 3.6, df = 1, p-value = 0.05778 If my calculations are correct, this is

Is Pearson's Chi-Square test for contingency tables asymptotically unbiased for large tables (large degrees of freedom) regardless of the expected values in each cell? The rule of thumb is that Pearson's Chi-square should not be used when large numbers of cells have expected values < 5. However, I compared the results on 4x4 contingency tables for R's chisq.test using chi-square

An organization has asked me to comment on the validity of their recent all-employee survey. Survey responses, by geographic region, compared with the total number of employees in each region, were as follows: > ByRegion All.Employees Survey.Respondents Region_1 735 142 Region_2 500 83 Region_3 897 78

Hello all, Let say I have 2-way contingency table: Tab <- matrix(c(8, 10, 12, 6), nr = 2) and the Chi-squared test could not reject the independence: > chisq.test(Tab) Pearson's Chi-squared test with Yates' continuity correction data: Tab X-squared = 1.0125, df = 1, p-value = 0.3143 However I want to get all possible contingency tables under this independence

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

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, I was using the mantelhaen.test function (2x2 J tables for conditional independence testing). I noticed that the option for the continuity correction (correction=T or correction=F) sometimes made a difference while sometimes it did give the same results regardless of correction=T or correction=F. Does the continuity correction apply only in certain cases or in certain strata with some

Hi, I am trying to test for pairwise associations between genotypes ( Rows=individuals, Columns =genes, data are up to 4 genotypes per gene, some with 2,3 or 4) where each chisquare comparison is different depending on the genes tested. The test is the observed multilocus (across columns for each individual) genotypes vs the expectation, which is the product of the individual frequency for each

Hello! Very sorry for a probably very simple question - I looked but did not find an answer in the archives. I have a table "counts" (below) that shows counts by Option within each of my 2 groups. However, my groups have different sizes (N1=255 and N2=68). Table "prop" shows the resulting proportions within each group. I would like to compare the proportions in 2 groups using

Hello, I'm trying to calculate a chi-squared test to see if my data are different from the theoretical distribution or not: chisq.test(rbind(c(79, 52, 69, 71, 82, 87, 95, 74, 55, 78, 49, 60),c(80,80,80, 80, 80, 80, 80, 80, 80, 80, 80, 80))) Pearson's Chi-squared test data: rbind(c(79, 52, 69, 71, 82, 87, 95, 74, 55, 78, 49, 60), c(80, 80, 80, 80, 80, 80, 80, 80, 80, 80, 80,

Full_Name: foo ba baz Version: R2.2.0 OS: Mac OS X (10.4) Submission from: (NULL) (219.66.32.183) chisq.test(matrix(c(9,10,9,11),2,2)) Chi-square value must be 0, and, P value must be 0 R does over correction when | a d - b c | < n / 2 &#65292;chi-sq must be 0