similar to: Fligner-Policello robust rank test

Full_Name: Karel Zvara Version: 1.8.1 OS: MS Winows 2000 Submission from: (NULL) (195.113.30.163) The test statistics of the fligner.test (ctest package) depends on the order of cases: > fligner.test(count~spray,data=InsectSprays) Fligner-Killeen test for homogeneity of variances data: count by spray Fligner-Killeen:med chi-squared = 14.4828, df = 5, p-value = 0.01282 >

I've made fligner test with the same data, changing the orders of the variables, and this what i get > fligner.test(rojos~edadysexo*zona*ano*estacion) Fligner-Killeen test of homogeneity of variances data: rojos by edadysexo by zona by ano by estacion Fligner-Killeen:med chi-squared = 15.7651, df = 2, p-value = 0.0003773 > fligner.test(rojos~ano*edadysexo*zona*estacion)

Hello, I investigate survival until the following year (0,1) and I wish to test if the variance in survival for two or more groups are significantly different from each other. I read that the Fligner-Killeen test is a non-parametric test which is very robust against departures from normality but is it correct (valuable technique for publication) to use it on binary data? In other

That's not an R question but a stats question, but I wouldn't do it. For one thing: The variance of binary data is a function of the mean, so the research question is dubious in the first place. Secondly, the test is based on ranking and comparing absolute differences from the group median, which for binary data is generally 0 or 1, so all absolute differences will be 1.... Put

Hi, I'm a new user of R. My background is Electrical Engineering, so please bear with me if this is a silly question. I'm trying to assess whether the results of an experiment satisfy the hypothesis of homoscedasticity (my ultimate goal is to use ANOVA). The result of the experiment is mean delay (dT), which depends on three factors, topology, drift, and lambda. The first two factors are

I am testing the homogeneity of variances via bartlett.test and fligner.test. Using the following example, how should I interpret the p-value in order to accept or reject the null hypothesis ? set.seed(5) x <- rnorm(20) bartlett.test(x, rep(1:5, each=4)) Bartlett test of homogeneity of variances data: x and rep(1:5, each = 4) Bartlett's K-squared = 1.7709, df = 4, p-value =

> From: Peter Dalgaard BSA <p.dalgaard at biostat.ku.dk> > Date: 01 Sep 2000 09:54:59 +0200 > > Prof Brian D Ripley <ripley at stats.ox.ac.uk> writes: Important omission: specification from Murray Jorgensen The test that I was thinking of basically does an anova on a modified response variable that is the absolute value of the difference between an observation

In specific cases fligner.test() can produce a small p-value even when both groups have constant variance. Here is an illustration: fligner.test(c(1,1,2,2), c("a","a","b","b")) # p-value = NA But: fligner.test(c(1,1,1,2,2,2), c("a","a","a","b","b","b")) # p-value < 2.2e-16

Hello, I notice that when I do Levene's test to test equality of variances across levels of a factor, I get different answers in R and SPSS 16. e.g.: For the chickwts data, in R, levene.test(weight, feed) gives F=0.7493, p=0.5896. SPSS 16 gives F=0.987, p=0.432 Why this difference? Which one should I believe? (I would like to believe R :) Ravi -- View this message in context:

Hello, can someone please help me with coding a function for a bootstrap WMW test (package boot, R under Windows, version 1.6.2)?

Hello Using R 2.2.1 on a Windows machine. Has anyone programmed the Welch test for equality of variances? I tried RSiteSearch, but this gave references to t test and oneway.test, which are not quite what I need.....I need the Welch test itself, for use in a meta-analysis (to determine if variances are equal). TIA Peter Peter L. Flom, PhD Assistant Director, Statistics and Data Analysis

Hi All, is there Levene' test in R ? If not ,Could you give me some advice about Levene test with R? Thanks a lot! I am waiting for yours.

I have a bunch of benchmark measurements that look something like this: sample.1 0.0000066660 0.0000062500 0.0000058330 0.0000058330 0.0000058330 sample.2 0.0000058330 0.0000058330 0.0000058330 0.0000058330 0.0000058330 sample.3 0.0000062500 0.0000062500 0.0000070830 0.0000062500 0.0000066660 i.e each measurement take on one of a set of values. The set values isn't fixed, but

In specific cases fligner.test() can produce a small p-value even when both groups have constant variance. Here is an illustration: fligner.test(c(1,1,2,2), c("a","a","b","b")) # p-value = NA But: fligner.test(c(1,1,1,2,2,2), c("a","a","a","b","b","b")) # p-value < 2.2e-16

Hello, I'm quite new with R and so I would like to know if there is a command to calculate an integral. In particular I simulated a bivariate normal distribution using these simple lines: rbivnorm <- function(n, # sample size mux, # expected value of x muy, # expected value of Y sigmax, # standard deviation of

I'm looking for a port of Schafer's PAN module for multiple imputation of nested data. It is written in S-Plus, and I would like to use it in R. Any pointers most appreciated. Best wishes, Paul von Hippel

Hi R users! I have the following problem: how appropriate is my aov model under the violation of anova assumptions? Example: a<-c(1,1,1,1,1,1,1,1,1,1,2,2,2,3,3,3,3,3,3,3) b<-c(101,1010,200,300,400, 202, 121, 234, 55,555,66,76,88,34,239, 30, 40, 50,50,60) z<-data.frame(a, b) fligner.test(z$b, factor(z$a)) aov(z$b~factor(z$a))->ll TukeyHSD(ll) Now from the aov i found that my model

Hello, I?m starting with my PhD and I have to stop because i got a little knowledge in R and statistics. I?ve got a model of this kind: binary response variable: prevalence of infection (0/1) 3 categorical independent variables: sex, month and name of the area I was trying with a full model like this, before the simplification model<-aov(prevalencia~sex*month*area) but the Fligner test

>Ben Bolker <bolker <at> ufl.edu> writes: > >I would try > >fligner.test(dT ~ Topology:Drift:lambda) > >in response to: > >Javier Acuna <javier.acuna.o <at> gmail.com> writes: > > Hi, I'm a new user of R. My background is Electrical Engineering, so > please bear with me if this is a silly question. > > I'm trying to assess

how can I eliminate the influence of the festivities in a time series with daily data?I tried to remove them and replace their value with a value of interpolation using na.approx (). There is an alternative method? -- View this message in context: http://www.nabble.com/holidays-effect-tp21830785p21830785.html Sent from the R help mailing list archive at Nabble.com.