similar to: Chi-Square Test Disagreement

Hello Everyone, I am trying to excess the inbuit .Fortran and .C codes of R. Can any one help me in that. For example in kmeans clustering the algorithms are written in .Fortran I want to access them and see the .Fortran syntax of the codes. Can any one help me how can I do that? Thanx, Nitin Kumar On Thu, Nov 27, 2008 at 12:00 PM, <r-help-request@r-project.org> wrote: > Send R-help

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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,

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

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

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

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

Hello, I found this in the logs of a Dovecot v1.0-test29 test server (OpenBSD 2.9, as mentioned in a previous email): Jul 24 05:37:47 myhostname dovecot: IMAP(username): mbox sync: UID inserted in the middle of mailbox /home/username/.mailbox (235 > 234) Jul 24 05:37:52 myhostname dovecot: IMAP(username): file mbox-sync-rewrite.c: line 357 (mbox_sync_read_and_move): assertion failed:

FreeBSD 4.10-RELEASE, dovecot-1.0-test29 crashes on sort and thread commands: Jul 25 15:12:08 owl dovecot: imap-login: Login: dima [81.19.64.101] Jul 25 15:12:26 owl dovecot: IMAP(dima): file index-mail-headers.c: line 408 (index_mail_get_header): assertion failed: (ret != 0) Jul 25 15:12:26 owl dovecot: child 20384 (imap) killed with signal 6 (gdb) bt #0 0x281e0fc4 in kill () from

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 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 am getting "Chi-squared approximation may be incorrect in: chisq.test(x)" with the data bleow. Frequency distribution of number of male offspring in families of size 5. Number of Male Offspring N 0 518 1 2245 2 4621 3 4753 4 2476 5

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,

Hello everybody, I'm seeing some strange behavior on R 1.8.1 on Intel/Linux compiled with gcc 3.2.2. The p-value calculated from the chisq.test function is incorrect for some input values: > chisq.test(matrix(c(0, 1, 1, 12555), 2, 2), simulate.p.value=TRUE) Pearson's Chi-squared test with simulated p-value (based on 2000 replicates) data: matrix(c(0, 1, 1,

Hi, I want to do a chi square test and I have two tab delimited text files with Expected and Observed values to compare. Each file contains only the values and are 48 rows by 116 columns. I have managed to do something with them, but I don't think it is right as I got a p value of 1. In this case I used the read.table() function to read the values from the files. But I don't know if

Hi, I want to determine the confidence interval on the sum of two sigma's. Is there an easy way to do this in R? I guess I have to use some sort of chisquare convolution algorithm??? Thanx, Roy -- The information contained in this communication and any atta...{{dropped}}

How i can perform a Pearson's Chi-squared Test in this data set: | Outcome -----------------+-----------+----------------------------------+ Treatment | Sex | None |Some | Marked | Total -----------------+------------+--------+--------+-------------+ Active | Female | 6 | 5 | 16 | 27

Dear all, I have some trouble understanding the chisq.test function. Take the following example: set.seed(1) A <- cut(runif(100),c(0.0, 0.35, 0.50, 0.65, 1.00), labels=FALSE) B <- cut(runif(100),c(0.0, 0.25, 0.40, 0.75, 1.00), labels=FALSE) C <- cut(runif(100),c(0.0, 0.25, 0.50, 0.80, 1.00), labels=FALSE) x <- table(A,B) y <- table(A,C) When I calculate the test statistic by hand

Hi R-users, I would like to find the goodness of fit using Chi-suare test for my data below: xobs=observed data, xtwe=predicted data using tweedie, xgam=predicted data using gamma > xobs <- c(223,46,12,5,7,17) > xtwe <- c(217.33,39,14,18.33,6.67,14.67) > xgam <- c(224.67,37.33,12.33,15.33,5.33,15) > chisq.test(xobs, xtwe = xtwe, rescale.p = TRUE) Error in chisq.test(xobs,

a warning message appears when i use the chisq.test ,but it doesnt appear everytime, why? "Warning message: Chi-squared approximation may be incorrect in: chisq.test(matrix(c(20.1, 18.5, 2.6, 32.9, 23.5, 5.4), nc = 2)) " why does the warning message appear, please? Thank you very much here is the data which I have tried appear and not appear warning message have warning message