similar to: testing independence of categorical variables

using an example from r online help > state <- c("tas", "sa", "qld", "nsw", "nsw", "nt", "wa", "wa", "qld", "vic", "nsw", "vic", "qld", "qld", "sa", "tas", "sa", "nt", "wa", "vic",

Hi all, I am trying to connect Teradata with R using Windows. Due I need to install any specific package or what? I am bit clue-less. Can someone help. Regards, Saj _________________________________________________________________ [[alternative HTML version deleted]]

i want to make survival plots for a coxph object using survfit function. mod.phm is an object of coxph class which calculated results using columns X and Y from the DataFrame. Both X and Y are categorical. I want survival plots which shows a single line for each of the categories of X i.e. '4' and 'C'. I am getting the following error: > attach(DataFrame) >

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",

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

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

Before I reinvent the wheel, I have need for a relatively straightforward crosstabulation (2 x n) function. I know that R has table(), ftable(), xtabs(), and summary(xtabs()), but none of these produce a fully "tricked" out cross-tabulation with marginal totals, expected cell frequencies, and an array of statistics about the contingency table. Is there a more complete (something

hie all i am trying to carry out a categorical data analysis but my problem is that when in i use the chi squared test some of my expected values are less than 5. is there a test that can handle this situation. the data is not a 2*2 table. its more from the social sciences where you have from strongly agree to strongly disagree. i know i can collapse vthe tables but there is a loss of

I seem to be able to use expected values that are decimal (e.g., 1.33) when using chisq.test but not when using fisher.test. This happens when using an array/matrix as input. Fisher.test returns: Error in sprintf(gettext(fmt, domain = domain), ...) : invalid format '%d'; use format %s for character objects. Thus, it appears fisher.test is looking for integers only. I tried putting the

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, 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,

Assume that we collect below data : - subjects = 20 males + 20 females, every single individual is independence, and difference events = 1, 2, 3... n covariates = 4 blood types A, B, AB, O http://r.789695.n4.nabble.com/file/n4245397/CodeCogsEqn.jpeg ?m = hazards rates for male ?n = hazards rates for female Wm = Wn x ?, frailty for males, where ? is the edge ratio of male compare to female Wn =

https://bugs.freedesktop.org/show_bug.cgi?id=95188 Bug ID: 95188 Summary: GeForce 840M (GM108) troubles Product: xorg Version: git Hardware: Other OS: All Status: NEW Severity: normal Priority: medium Component: Driver/nouveau Assignee: nouveau at lists.freedesktop.org

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In SAS, for a two-way (or 3-way, stratified) table, the CMH option in SAS PROC FREQ gives 3 tests that take ordinality of the factors into account, for both variables, just the column variable or neither. Is there an equivalent in R? The mantelhaen.test in stats gives something quite different (a test of conditional independence for *nominal* factors in a 3-way table). e.g. I'd like to

. HI there, I'd like to show demonstrate how the chi-squared distribution works, so I've come up with a sample data frame of two categorical variables y<-data.frame(gender=sample(c('Male', 'Female'), size=100000, replace=TRUE, c(0.5, 0.5)), tea=sample(c('Yes', 'No'), size=100000, replace=TRUE, c(0.5, 0.5))) And I'd like to create a list of 100

Hello, I am working on my thesis and can't really figure out how to produce a reasonable graph from the output from my glm., I could just give the R-output in my results and then discuss them, but it would be more interesting if I could visualise what is going on. My research is how bees react to different fieldmargins, for this I have 4 different types of field margin (A,B,C & D) and

Hello there! I am still struggling with a binomial response over all categorical variables (some of them with 3 levels, most with 2 levels). After initial struggles with glm's (struggle coming from the data, not the actual analysis) I have decided to prefer contingency tables. I have my data such as: response:

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

In this years biostat teaching I will include Rcommander (it indeed simplifies syntax problems that makes students frequently miss the core statistical problems). But I could not find how to make a simple chisquare comparison between observed frequencies and expected frequencies (eg in genetics where you expect phenotypic frequencies corresponding to 3:1 in standard dominant/recessif