similar to: Is it safe? Cochran etc

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

Hola Manuel lo he tratado de hacer pero me sale Error: unexpected string constante in: "anova(a,as,test=Chisq") no tengo ni idea de por qué... Me resulta alucinante no poder contar ya con pvals.fnc. ¿Será imposible hacerse con ello? Saludos, Miguel -------------------------------------------- El vie, 13/6/14, Manuel Azcárate <mazcarategarcia en gmail.com> escribió:

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

Existe discusión sobre el uso de los p-valores en modelos mixtos. Como se ha dicho antes, para mi lo más adecuado es comparar modelos mediante la función anova. Por Internet se puede encontrar un buen libro de Douglas Bates y en español, busca modelos mixtos con R de Luis Cayuela, enfocado hacia ecología, pero está muy bien El 13/06/2014 14:00, "Jorge I Velez"

Hi, This report deals with p-values coming from chisq.test using the simulate.p=TRUE option. The issue is numerical accuracy and was brought up in previous bug reports 3486 and 3896. The bug was considered fixed but apparently was only mostly fixed. Just the typical problem of two values that are mathematically equal not ending up numerically equivalent. Consider this series of three 2x2

Hola a todos, quería preguntaros un medio para obtener los valores p usando lmer. He tratado con pvals.fnc, que es lo que me habían recomendado, pero por algún motivo no está ya disponible etc. Ésta es la función que tengo, pero da las "t", sin los valores p. Aunque Baayen indica que valores por encima de 2 son significativos necesito saber las p. resultado = lmer(rt_ln ~ (fre_ln *

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

I have a print method for a set of statistical tests, vcdExtra::CMHtest, for which I'd like to have more sensible printing of pvalues, as in print.anova(). [Testing this requires the latest version of vcdExtra, from R-Forge **|install.packages("vcdExtra", repos="http://R-Forge.R-project.org")|** ] With my current print method, I get results like this, but all Prob values

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 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,

Full_Name: Reginaldo Constantino Version: 2.8.0 OS: Ubuntu Hardy (32 bit, kernel 2.6.24) Submission from: (NULL) (189.61.88.2) For many tables, chisq.test with simulate.p.value=TRUE gives a p value that is obviously incorrect and inversely proportional to the number of replicates: > data(HairEyeColor) > x <- margin.table(HairEyeColor, c(1, 2)) >

Dear all, I am trying to validate a model by comparing simulated output values against observed values. I have produced a simple X-y scatter plot with a 1:1 line, so that the closer the points fall to this line, the better the 'fit' between the modelled data and the observation data. I am now attempting to quantify the strength of this fit by using a statistical test in R. I am no

I know this is something simple that I cannot do because I do not yet "think" in R. I have a data frame has a variable participation (a factor), and several other factors. I want a chisq test (no contingency tables) for participation vs all of the other factors. In SPSS I would do: CROSSTABS /TABLES= (my other factors) BY participation /FORMAT=NOTABLES /STATISTICS=CHISQ

I'm using R1.7.0 runing with Win XP. Thanks, ...Tao ???????????????????????????????????????????????????????? >x [,1] [,2] [1,] 149 151 [2,] 1 8 >t(x) [,1] [,2] [1,] 149 1 [2,] 151 8 >chisq.test(x, simulate.p.value=T, B=100000) Pearson's Chi-squared test with simulated p-value (based on 1e+05 replicates) data: x X-squared = 5.2001, df =

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,

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

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

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

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,

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