similar to: Bartlett Test

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, 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 =

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)

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

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

Hi I needed some help with ANOVA I have a problem with My ANOVA analysis. I have a dataset with a known ANOVA p-value, however I can not seem to re-create it in R. I have created a list (zzzanova) which contains 1)Intensity Values 2)Group Number (6 Different Groups) 3)Sample Number (54 different samples) this is created by the script in Appendix 1 I then conduct ANOVA with the command >

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

Sorry i had a misprint in the appendix code in the last email Hi I needed some help with ANOVA I have a problem with My ANOVA analysis. I have a dataset with a known ANOVA p-value, however I can not seem to re-create it in R. I have created a list (zzzanova) which contains 1)Intensity Values 2)Group Number (6 Different Groups) 3)Sample Number (54 different samples) this is created by the

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:

Pat Burns makes a good point. -Peter -------- Original Message -------- Subject: Re: [R] Using seq_len() vs 1:n Date: Fri, 12 Feb 2010 09:01:20 +0000 From: Patrick Burns <pburns at pburns.seanet.com> To: Peter Ehlers <ehlers at ucalgary.ca> References: <4B746AEF.10900 at ucalgary.ca> If you want your code to be compatible with S+, then 'seq_len' isn't going to work.

Hi list, One of my co-workers found this problem with 'cor' in his code and I confirm it too (see below). He's using R 2.2.1 under Win 2K and I'm using R 2.3.0 under Win XP. =========================================== > R.Version() $platform [1] "i386-pc-mingw32" $arch [1] "i386" $os [1] "mingw32" $system [1] "i386, mingw32" $status

Hi I am very new to R and statistical programming in general. I am trying to reorder data from a .csv file. I have managed to import the data and create a zero matrix. I am now trying to fill the matrix. There seems to be some problem with this section of my code. I have highlighted the dodgy code in red. Please help if possible. ###################################################### ###########

I would like to use samp as a part of a filename that I can change. My source files are .csv files with date as the file name, and I would like to be able to type in the date (later perhaps automate this using list.files) and then read the csv and write the pdf automatically. I have tried different combinations with "" and () around samp, but I keep getting the error "object

Hello guys, I would like to ask for help to understand what is going on in "func2". My plan is to generalize "func1", so that are expected same results in "func2" as in "func1". Executing "func1" returns... 0.25 with absolute error < 8.4e-05 But for "func2" I get... Error in dpois(1, 0.1, 23.3065168689948, 0.000429064542600244,

Dear Rexperts: (R 2.0.1 on Windows XP Pro) Is the following problem unique to my setup? If it's a known problem, I didn't see it at http://bugs.r-project.org/ nor find discussion in the archives. plot(1:10) loc <- c(5, 6) text(loc, labels = "a") Produces expected results according to ?xy.coords. plot(1:10) loc <- list(x = 5, y = 6) text(loc, labels = "a") No

> 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

my reproducible example test<-structure(list(site = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L), .Label = c("A", "B", "C", "D", "E"), class = "factor"),

#I wish to find the "next non-NA" value within each row of a data-frame. #e.g. I have a data frame mydata. Rows 1, 2 & 3 have soem NA values. mydata <- data.frame(matrix(seq(20*6), 20, 6)) mydata[1,3:5] <-  NA mydata[2,2:3] <-  NA mydata[2,5] <-  NA mydata[3,6] <-  NA mydata[1:3,] #this loop accomplishes the task; I am tryign toi learn a "better" way for(i