similar to: Unexpected behaviour when testing for independence with multiple factors

>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

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)

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 =

Warranty on Accuracy, Precision, Legality, ... of R in Research (These questions may well have been raised.) What is the implied warranty of using R for research & publications, consulting, etc.? Alternately, how does one obtain such a warranty? Your answers will be much appreciated. Perhaps you can point me to some websites which discussed this subject in the past. Thanks & regards

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

Can anybody tell me if there is an R implementation of the Fligner-Policello robust rank test? Thanks, Elisabetta

> 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

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

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.

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:

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 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

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

Subject: Eliminate repeated components from a vector X-Mailer: VM 7.00 under 21.4 (patch 6) "Common Lisp" XEmacs Lucid Reply-To: fjmolina at lbl.gov FCC: /home/f/.xemacs/mail/sent Does anyone know how I can eliminate repeated elements from a vector?

Hello, Does anybody know if the outscale option of randomForest yields the standarized version of the outlier measure for each case? or the results are only the raw values. Also I have notice that this measure presents very high variability. I mean if I repeat the experiment I am getting very different values for this measure and it is hard to flag the outliers. This does not happen with two other

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 am trying to get this code to work, but as I get to the predict command, it displays an error due to the length of the data sets from the removal of the NA's. Here is the data, and the code that I am using so far, if you run it, you'll see the error pop up.... please help me to get around this problem. Thanks in advance. Location Dist. Size low1 .5 10.5 low2 .5 23 low3 .5 NA