similar to: unbalanced effects in aov

Hello, This one is very perplexing. I have teacher observation data, with factors teacher ID, observer ID, component, grade and subject. When I do this, aov(data=ratings.prin.22, rating ~ obsid.f + tid.f + subject.f + grade.f + comp.f) I get this: Terms: obsid.f tid.f grade.f comp.f Residuals Sum of Squares 306.23399 221.38173 1.70000 14.52831 279.05780 Deg. of

Full_Name: Tanya Logvinenko Version: 1.7.0 OS: Windows 2000 Submission from: (NULL) (132.183.156.125) For unbalanced design, I ran into problem with ANOVA (aov function). The sum of squares for only for the second factor and total are computed correctly, but sum of squares for the first factor is computed incorreclty. Changing order of factors in the formula changes the ANOVA table. For the

Greetings, The aov() function may mislabel the random effects as in the example below: Has anybody else noticed this? Cordially, Giles Crane, MPH, ASA, NJPHA gilescrane at verizon.net > m2 Call: aov(formula = y ~ ap + pe + Error(ju), data = d) Grand Mean: 77.50667 Stratum 1: ju Terms: ap Sum of Squares 4322.538 Deg. of Freedom 12 13 out of 25 effects not

Hi, I am trying to do a repeated measures ANOVA to determine if there is a significant difference between two sets of timecourse data. Each individual was given a single treatment and then measured for one variable for 10 days. Here is made-up example of what my data would look like:

Dear all, I'm analysing a split-plot experiment, where there are sometimes one or two values missing. I realized that if the data is slightly unbalanced, the effect of the subplot-treatment will also appear and be tested against the mainplot-error term. I replicated this with the Oats dataset from Yates (1935), contained in the nlme package, where Variety is on mainplot, and nitro on

Hi, This question is related to the bwplot issue I reported yesterday. I have a 3 factors (2x3x2) dataset that I collapsed into a 2 factors dataset (3x2 = sizexModality). For size==small, I have 2 observations per subject (Snr), for the other sizes only 1. I reckoned that aov (and underneath, lm) might handle this as it should, since the subjects are idendified, when I do > aov(

Hi all, I'm not sure that there is really a way to do this, but I thought I'd see if anyone knew. I have a file with 1 to n columns all named something like X1, X2, X3....Xn. I have another file that has in one column n number of rows. Each row has a number in it (not in order; the ordering of the numbers is important but it isn't in count order). Basically, I would like to order

Hi all, I'm new to R and have the following problem: I have a 2 factor design (a has 2 levels, b has 3 levels). I have an object kidney.aov which is an aov(y ~ a*b), and when I ask for model.tables(kidney.avo, se=T) I get the following message along with the table of effects: Design is unbalanced - use se.contrast() for se's but the design is NOT unbalanced... each fator level

Hi, I have an unbalanced dataset on which I would like to perform a one-way anova test using R (aov). According to Wannacott and Wannacott (1990) p. 333, one-way anova with unbalanced data is possible with a few modifications in the anova-calculations. The modified anova calculations should take into account different sample sizes and a modified definition of the average. I was wondering if the

Hello R-users, I am trying to obtain Type III SS for an ANOVA with subsampling. My design is slightly unbalanced with either 3 or 4 subsamples per replicate. The basic aov model would be: fit <- aov(y~x+Error(subsample)) But this gives Type I SS and not Type III. But, using the drop() option: drop1(fit, test="F") I get an error message: "Error in

Hi all I'm trying to complete a textbook example originally designed for SPSS in R, and I therefore need to find out how to compute an unbalanced ANOVA in R. I did a search on the mailinglist archives an found a post by Prof. Ripley saying one should use the lme function for (among other things) unbalanced ANOVAs, but I have not been able to use this object. My code gives me an error.. Why

R : Copyright 2001, The R Development Core Team Version 1.4.0 (2001-12-19) R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type `license()' or `licence()' for distribution details. R is a collaborative project with many contributors. Type `contributors()' for more information. Type `demo()' for some demos,

Hi all, Say I have the following data: a<-data.frame(col1=c(rep("a",5),rep("b",7)),col2=runif(12)) a_aov<-aov(a$col2~a$col1) summary(aov) Note that there are 5 observations for a and 7 for b, thus is unbalanced. What would be the correct way of doing anova for this set? Thanks, Sachin [[alternative HTML version deleted]]

I have a three-way unbalanced ANOVA that I need to calculate (fixed effects plus interactions, no random effects). But word has it that aov() is good only for balanced designs. I have seen a number of different recommendations for working with unbalanced designs, but they seem to differ widely (car, nlme, lme4, etc.). So I would like to know what is the best or most usual way to go about working

Dear all I was looking at the aov documentation page and came across the following which seems like a contradiction to me: " This provides a wrapper to |lm| for fitting linear models to balanced or unbalanced experimental designs." (I presume 'This' refers to aov) and "|aov| is designed for balanced designs, and the results can be hard to interpret without

I am using R 1.1 with Redhat 6.2 and RW 1.001 with Win98 (the upkey doesn't work on my IBM either as has been previously reported by others). The function aov doesn't return either the residuals or the residual degrees of freedom for split-plot designs. If you use the following code from Baron and Li's "Notes on the use of R for psycology experiments and questionnaires"

Dear R-list members, I've been struggling with the proper setup for analysing my data. I've performed a route choice experiment, in which participants had to make a choice at each junction for the next road. During the experiment they received traffic information, but also encountered two different accidents. They also made trips without accidents. What I'm interested in is to

Dear all, I need an help because I am really not able to find over internet a good example in R to analyze an unbalanced table with Anova with repeated measures. For unbalanced table I mean that the questions are not answered all by the same number of subjects. For a balanced case I would use the command aov1 = aov(response ~ stimulus*condition + Error(subject/(stimulus*condition)), data=scrd)

Dear all, I need to do a Manova but I have an unbalanced design. I have morphological measurements similar to the iris dataset, but I don't have the same number of measurements for all species. Does anyone know a procedure to do Manova with this kind of input in R? Thank you very much, Naiara. -------------------------------------------- Naiara S. Pinto Ecology, Evolution and Behavior 1

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