similar to: Find if there is independence

hi, is there a way of calculating of measuring dependence between two categorical variables. i tried using the chi square test to test for independence but i got error saying that the lengths of the two vectors don't match. Suppose X and Y are two factors. X has 5 levels and Y has 7 levels. This is what i tried doing >temp<-chisq.test(x,y) but got error "the lengths of the two

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

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(        

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

Hi, Can someone set me straight as to how to write formulas in R to indicate: complete independence [A][B][C] joint independence [AB][C] conditional independence [AC][BC] nway interaction [AB][AC][BC] ? For example, if I have 4 factors: hair colour, eye colour, age, sex does > mosaicplot( frequency ~ hair + eye + age + sex) mean that the model fitted is of complete independence of

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

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

R 2.12 windows 7 I am summary data that I would like to make into a 2x2 table representing counts positive vs. negative counts: 28/289 20/276 My table should look something like the following: group1 group2 Positive 28 20 Negative 289 276 How can a (1) create the 2x2 table (2) run a chi square test on the table? I have tried the following code, but I

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

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

Hello there, I have two groups (men and women) and I know per group how many of them smoke or don't smoke (women 40 of 200; men 100 of 300). I would like to know how I can compare in R if men and women differ significantly in their smoking. However, because there are more men in the sample than women I cannot just compare the number of smokers and non-smokers in both groups, right?! (I would

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

I have an adjacency matrix and I want to obtain a matrix of the minimum paths between the nodes. Thank you Alessandro Ambrosini -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !)

I'm trying to launch a visual basic game,i have all the neccesery libraries installed. On launch i get this Code: err:ntdll:RtlpWaitForCriticalSection section 0x7bcbf8e0 "../../../wine/dlls/ntdll/loader.c: loader_section" wait timed out in thread 003c, blocked by 0034, retrying (60 sec) err:ntdll:RtlpWaitForCriticalSection section 0x78037130 "?" wait timed out in thread

Dear List, If any of observed and/or expected data has less than 5 frequencies, then chisq.test (Pearson's Chi-squared Test for Count Data from package:stats) gives warning messages. For example, x<-c(10, 14, 10, 11, 11, 7, 8, 4, 1, 4, 4, 2, 1, 1, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1) y<-c(9.13112391745095, 13.1626482033341, 12.6623267638188, 11.0130706413029, 9.16415925139016,

The first: if I have a vector as (1,1,3,2,1,1), which is the command that gives all the positions of the min value? From the vector of the example a would like to obtain a new vector as (1,2,5,6) that give me all the positions of the minimum value 1. The second: if I have a matrix "A" and I want to obtain a new matrix deleting a column or a row of A, what have I to do? Thank you.

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

The cells are interpreted as counts, so by scaling you're analyzing a different experiment (one with fewer observations). So the chi-squared value will change (the terms (O-E)^2/E in the statistic scale linearly ignoring rounding and "Yates' continuity correction"). The chisq.test on the original data is a test of association. Conventionally you decide ahead of time on a

Hello! Two questions: 1: I have to import a matrix of adjacency from a file of a software that is not R (for example "bloc notes" of Windows). The problem is that the matrix is not in the explicit form as 0 1 1 1 0 0 1 0 0 but it is a scattered matrix where in each row there are two nodes that have a direct path. The matrix is a b a c b a c a For example, the first row

Hello. Given a vector 1 3 4 2 8 9 5 I want to obtain a vector with all 0 except in the second position and in the fifth, where the numbers are the same of the first vector. The new vector must be 0 3 0 0 8 0 0 Thank you very much. Excuseme but my mind is out of order. Alessandro -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read