similar to: ANOVA and Levene's test in nested model

Dear All, I am willing to test homogenity of variances using Levene's test. I know how to do it with simple model, but I am lost when dealing with a little bit more complex model. I have response variable (y) and four factors (a, b, c, d). One of these four factors (d) is nested within another factor (c). In addition, I would like to take into account only 2nd degree interactions in my

Hi, All! To calculate Levene's Test for Homogeneity of Variance I use R Commander, and this is the output: > levene.test(Dataset$age, Dataset$sex) Levene's Test for Homogeneity of Variance Df F value Pr(>F) group 1 0.8739 0.3567 33 I am not sure what means "Pr(>F)"? Can anyone explain/translate this? Regards, Iurie Malai Department of Psychology and

> 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

Hello all, I am low down on the learning curve of R but so far I have had little trouble using most of the packages. However, recently I have run into a wall when it comes to bootstrapping a Levene's test (from the car package) and thought you might be able to help. I have not been able to find R examples for the "boot" package where the test statistic specifically uses a

Dear All I am trying to use Levene's test (of package car), but I do not understand quite well how to use it. '?levene.test' does not unfortunately provide any example. My data are in a data frame and correspond to 4 factors plus response. Could someone please give me an example about how to use the command levene.test(y, group) ? Thanks in advance, Paul

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, I try to run simulation of levene's test to find the p-value but the error of replacement has length zero occur, could anyone help me to fix this problem? asim <- 1000 pv<-rep(NA,asim) for(i in 1:asim) {print(i) set.seed(i) g1 <- rnorm(20,0,2) g2 <- rnorm(20,0,2) g3 <- rnorm(20,0,2) x <- c(g1,g2,g3) group<-as.factor(c(rep(1,20),rep(2,20),rep(3,20))) library(Rcmdr)

Dear Contributors, I am asking help on the way how to solve a problem related to loops for that I always get confused with. I would like to perform the following procedure in a compact way. Consider that p is a matrix composed of 100 rows and three columns. I need to calculate the sum over some rows of each column separately, as follows: fa1<-(colSums(p[1:25,])) fa2<-(colSums(p[26:50,]))

Dear contributors I have tried this experiment: x<-c() for (i in 1:12){ x[i]<-list(cbind(x1[i],x2[i])) #this is a list of 12 couples of time series I am using to perform a test } # that compares them 2 by 2 # ################# #trace statistic test<-data.frame() cval<-array( , dim=c(2,3,12)) for (i in 2:12){ for (k in 1:2){ for (j in 1:3){ result[k,j,i]<-

Dear Contributors, I am back asking for help concerning the same type of dataset I was asking before, in a previous help request. I needed to sum data over subsample of three time series each of them made of 100 observations. The solution proposed were various, among which: db<-p dim( db ) <- c(25,4,3) db2 <- apply(db, c(2,3), sum) db3 <- t(apply(db2, 1, function(poff)

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Please can you help me I have the following problem: I have selected an lm model through the step procedure which visualize for each step the AIC value; then I have calculated for the initial model and the selected one the AIC using the funnction AIC. The results are different.What's happened? Emilia Rocco Dipartimento di Statistica "G. Parenti" Università di Firenze e-mail:

Works perfectly. Thank you. -Hugh Rand -----Original Message----- From: Spencer Graves [mailto:spencer.graves at pdf.com] Sent: Sunday, January 15, 2006 6:41 PM To: Rand, Hugh Cc: 'r-help at lists.R-project.org' Subject: Re: [R] trouble with extraction/interpretation of variance structure para meters from a model built using gnls and varConstPower How about this: >

Dear all, I need to compute tensor product of B-spline defined over equi-spaced break-points. I wrote my own program (it works in a 2-dimensional setting) library(splines) # set the break-points Knots = seq(-1,1,length=10) # number of splines M = (length(Knots)-4)^2 # short cut to splineDesign function bspline = function(x) splineDesign(Knots,x,outer.ok = T) # bivariate tensor product of

Hi, I'm trying to compare models, one of which has all parameters fixed using offsets. The log-likelihoods seem reasonble in all cases except the model in which there are no free parameters (model3 in the toy example below). Any help would be appreciated. Cheers, Jarrod x<-rnorm(100) y<-rnorm(100, 1+x) model1<-lm(y~x) logLik(model1) sum(dnorm(y, predict(model1),

I have been using gnls with the weights argument (and varConstPower) to specify a variance structure for curve fits. In attempting to extract the parameters for the variance model I am seeing results I don't understand. When I simply display the model (or use "summary" on the model), I get what seem like reasonable values for both "power" and "const". When I

Dear R users, i'm using a custom function to fit ancova models to a dataset. The data are divided into 12 groups, with one dependent variable and one covariate. When plotting the data, i'd like to add separate regression lines for each group (so, 12 lines, each with their respective individual slopes). My 'model1' uses the group*covariate interaction term, and so the coefficients

Dear R-users, I learned today that there exists an interesting topic in numerical analysis names "best polynomial approximation" (BSA). Given a function f the BSA of degree k, say pk, is the polynomial such that pk=arginf sup(|f-pk|) Although given some regularity condition of f, pk is unique, pk IS NOT calculated with least square. A quick google tour show a rich field of research

Hello, I am trying to plot a few correlograms on the same figure, with the function corrgram() from the package corrgram. However, the function does not seem to use the base graphic system, as setting out the multiple figure layout with, e.g., par(mfrow=c(2, 2,)) does not work. Does anybody know a workaround for this? Many thanks in advance for any advice best giuseppe -- Giuseppe Pagnoni,

Dear R users, Thanks in advance. I am Deb, Statistician at NSW Department of Commerce, Sydney. I am using R 2.9.1 on Windows XP. This has reference to the package “MCMCpack”. My objective is to select a better model using various alternatives. I have provided here an example code from MCMCpack.pdf. The matrix of Bayes Factors is: model1 model2 model3 model1 1.000 14.08