similar to: Statistical question: one-sample binomial test for clustered data

This is a follow-up to a query that was posted regarding some problems that emerge when running anova analyses for cox models, posted by Mathias Gondan: Matthias Gondan wrote: >* Dear List,*>**>* I have tried a stratified Cox Regression, it is working fine, except for*>* the "Anova"-Tests:*>**>* Here the commands (should work out of the box):*>**>*

Can anyone tell me how to calculate a mid-p value for a chi-squared test in R? Many thanks, Andrew Wilson

I was wondering, whether there is a way to have statistical significance test for cluster agreement. I know that I can use classAgreement() function to get Rand index, which will give me some indication whether the clusters agree or not, but it would be interesting to have a formal test. Thanks.

This may not actually be an R/Splus problem, but it started off that way ..... ===+===+===+===+===+===+===+===+===+===+===+===+===+===+===+===+===+===+=== Executive summary: ================== Simulations involving extreme value distributions seem to ``work'' when the underlying distribution is exponential(1) or exponential(2) == chi-squared_2, but NOT when the underlying distribution is

Dear List, I have tried a stratified Cox Regression, it is working fine, except for the "Anova"-Tests: Here the commands (should work out of the box): library(survival) d = colon[colon$etype==2, ] m = coxph(Surv(time, status) ~ strata(sex) + rx, data=d) summary(m) # Printout ok anova(m, test='Chisq') This is the output of the anova command: > Analysis of Deviance Table

I am appealing to the general collective wisdom of this list in respect of a statistics (rather than R) question. This question comes to me from a friend who is a veterinary oncologist. In a study that she is writing up there were 73 cats who were treated with a drug called piroxicam. None of the cats were observed to be subject to vomiting prior to treatment; 12 of the cats were subject to

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

Hi, Is anyone aware of a rank-based, non-paired test such as the Krustal-Wallis, that can accommodate weights? Alternatively, would it make sense to simulate a dataset by duplicating observations in proportion to their weight, and then using the Krustal-Wallis test? thanks! Dylan

Hi all, In my continuous transition of GLIM to R I try to make a glm with binomial errors. The data file have 3 vectors: h -> the factor that is ajusted (have 3 levels) d -> number of animais alive (the response) n -> total number of animals To test proportion of alive, make d/n. In GLIM: $yvar d$ $error binomial n$ $fit +h$ scale deviance = 25.730 (change = -9.138) at cycle 4

Dear R users, I noticed a problem in the anova command when applied on a single coxph object if there are missing observations in the data: This example code was run on R-2.6.1: > library(survival) > data(colon) > colondeath = colon[colon$etype==2, ] > m = coxph(Surv(time, status) ~ rx + sex + age + perfor, data=colondeath) > m Call: coxph(formula = Surv(time, status) ~ rx +

Hello, As I am quitte an ignorant user of R, excuse me for any wrongfull usage of all the terms. My question relates to the statistics behind the survdiff function in the package survival. My textbook knowledge of the logrank test tells me that if I want to compare two survival curves, I have to take the sum of the factors: (O-E)^2/E of both groups, which will give me the Chisq. If I calculate

Dear all Seems that puzzles always come in packs. I was asked to help with some statistics in blood analysis. (You can not refuse your wife's asks :-). She has contingency table for values IgVH mutation and ZAP expression. I can do chi-square test (in R) and get a results, and with some literature I can try explain them. However she found an article in which they use SPSS and use

Dear list, I have very little experience in dealing with proportions, i am sure this is a very simple question but i could find no suitable answer beyond doing a chi-sq test and then using the Marascuilo procedure as a post-hoc analysis. I am simply wanting to know if the proportions ( i.e the number of Yes / No) significantly differ between the cases and if so which cases are significantly

Dear R : Sorry for the off topic question, but does anyone know the reference for the -2 Ln Lambda following a Chi Square distribution, please? Possibly one of Bartlett's? Thanks in advance! Sincerely, Laura Holt mailto: lauraholt_983 at hotmail.com

Hello, I was trying to compare the performance of icc, gcc and llvm on the program almabench.c in Coyote Benchmark suite. Here is a line of code from the program. da = da + (ca[np][k] * cos(arga) + sa[np][k] * sin(arga)) * 0.0000001; gcc and icc are performing way better than llvm as they are using 'sincos' library function to compute the sin and the cos of the argument in a

Hello there, I have two data sets with 14 variables each, and wish to do the test for equality of their covariance matrices and mean vectors. Normally these tests should be done by chi square test (box provided) and Hotelling's T square test respectively. Which R functions could do this kind of test? I just find some functions could do for one dimension, but no for multidimension. Some one

Dear Tomas,Thank you.Regarding the "unnumbered" arguments, i.e. sprintf('%f %f', 1, 2, 3). This was the case I wanted to report, here a warning can be very useful.Regarding the "numbered" arguments, that is, sprintf('%$1f %$3f', 1, 2, 3). Here, omission of an argument might be intended, for example, in an application with support for multiple languages.

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'm trying to compute sample size requirements for a binomial exact test. we want to show that the proportion is at least 90% assuming that it is 95%, with 80% power so any asymptotic approximations are out of the questions. I was planning on using binom.test to perform the simple test against a prespecified value, but cannot find any functions for computing sample size. do any exist?

Dear list, Is anyone aware of a library for sample size calculation in R, similar to NQuery? I have to give a course in this area, and I would like to enable the students playing around with this. Best wishes, Matthias