similar to: multidimensional Fisher or Chi square test

Hello, I originally posted this on the stats stack exchange site, but given its focus on R software, it was removed -- so I figured I'd post here. I'm having trouble interpreting a change in effect direction and significance when I add an interaction term to my glmer() model. *Part 1* I ran an experiment in which participants made categorical decisions (out of two categories) in one of

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 list, I had some confusion regarding what function too use in order too relate results from spec.pgram() too a chi-square distribution. The documentation indicates that the PSD estimate can be approximated by a chi-square distribution with 2 degrees of freedom, but I am having trouble figuring out how to do it in R, and figuring out what specifically that statement in the documentation

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

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

Hello, I've tried to analyze some data with a CMH test. My 3 dimensional contingency tables are 2x2xN where N is usually between 10 and 100. The problem is that there may be 2 strata with opposite counts (the 2x2 contigency table for these are reversed), producing opposite odds ratios that cancle out in the overall statistics. These opposite counts are very important for my analysis, since

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}}

Dear R experts, I was wondering if there are any R functions that give the tail area of a sum of chisquare distributions of the type: a_1 X_1 + a_2 X_2 where a_1 and a_2 are constants and X_1 and X_2 are independent chi-square variables with different degrees of freedom. Thanks, Klaus -- "Feel free" - 5 GB Mailbox, 50 FreeSMS/Monat ...

Sir, I have a problem here while applying chisquare test to the following Data ( below the subject of this mail) ...when I wanted to test the significance using three different free statistical packages, here R, EpiInfo and OpenEpi. *Only OpenEpi accepts the test based on Cochran's Recommendations. * R says " chi squared approximation may be incorrect." Does it mean the same as

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 R users I have serveral binary variables (e.g., X1, X2, X3, X4, X5, X,6, and X7) and one continuous variable (e.g., Y1). I combined these variables using data.frame() mydata <- data.frame(X1,X2,X3,X4,X5,X6,X7,Y1) after that, I sorted this data.frame rank.by.Y1<-order(mydata[,8]) sorted.mydata<-mydata[rank.by.Y1,] after that, I replaced Y1's values with values ranging from 1

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

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

Full_Name: Chien-yu Peng Version: 2.0.1 OS: Windows XP Professional Submission from: (NULL) (140.109.72.181) Dear all: Although I don't know you, I am thankful for your help. When I use the function mantelhaen.test for R x C x K (R, C > 2) table, the output is not the same as SAS's. I don't know that the result consist with one of SAS's. But it works correctly for 2

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

Hello, I'd like to use DBI to store lm objects in a database. I've to analyze many of linear models and I cannot store them in a single R-session (not enough memory). Also it'd be nice to have them persistent. Maybe it's possible to create a compact binary representation of the object (the kind of format created created by "save"), so that one doesn't need to write

Dear listers, I am trying to compute the exact conditional test given strata margins of a 2 by 2 by K array using the mantelhaen.test function to get a common odds ratio estimate. The estimate for the test on the following data is 0, which in my opinion dosen't make any sense. x <- array(c(53, 6098, 1006, 4521, 63, 8070, 1163, 6137), dim=c(2,2,2)) x , , 1 [,1] [,2] [1,] 53 1006

Hi The short version of my questions is this: How can I run a chi-square test over a matrix (table) to get the distanaces between rows and then run a SingleLinkage (or other fusion algorithm over the resulting table? ------------ The long-version of my question: My data consists of different data of different countries so I have stuff like how many people can read, write in X,Y,Z countries

Dear Developers: I have a question about memory management in R 2.2.1 and am wondering if you would be kind enough to help me understand what is going on. (It has been a few years since I have done software development on Windows, so I apologize in advance if these are easy questions.) ------------- MY SYSTEM ------------- I am currently using R (version 2.2.1) on a PC running Windows 2000

Hi, I am a newbie with R, so I hope my question isn't too stupid. I am trying to generate dose-response curves using the "drfit" package. I have formatted my CSV files to the correct format, and have no trouble running drfit to get a summary of my data. The problem is that when I try to use "drplot" to graph my data I get an error. The message is: Error in