similar to: Breakdown of Treatment Sum of Squares

Dear R Core Team I've a proposal to improve drop1(). The function should change the contrast from the default ("treatment") to "sum". If you fit a model with an interaction (which ist not signifikant) and you display the main effect with drop1( , scope = .~., test = "F") If you remove the interaction, then everything's okay. There is no way to fit a

Hi, I am analyzing a repeated-measures Greco-Latin Square with the aov command. I am using aov to calculate the MSs and then picking by hand the appropriate neumerator and denominator terms for the F tests. The data are the following: responseFinger mapping.code Subject.n index middle ring little ---------------------------------------------------------------------------- 1 1

Hi everyone, Can someone explain me how to calculate SAS type III sums of squares in R? Not that I would like to use them, I know they are problematic. I would like to know how to calculate them in order to demonstrate that strange things happen when you use them (for a course for example). I know you can use drop1(lm(), test="F") but for an lm(y~A+B+A:B), type III SSQs are only

Please pardon an extremely naive question. I see related earlier posts, but no responses which answer my particular question. In general, I'm very confused about how to do variance decomposition with random and mixed effects. Pointers to good tutorials or texts would be greatly appreciated. To give a specific example, page 193 of V&R, 3d Edition, illustrates using raov assuming pure

Hello, I've got what I'd expect to be a pretty simple issue: I fit an aov object using multiple error strata, and would like some significance tests for the contrasts I specified. In this contrived example, I model some test score as the interaction of a subject's gender and two emotion variables (angry, happy, neutral), measured at entry to the experiment (entry) and later

useRs, A no doubt simple question, but I am baffled. Indeed, I think I once knew the answer, but can't recover it. The default contrasts for aov (and lm, and...) are contr.treatment and contr.poly for unordered and ordered factors, respectively. But, how does one invoke the latter? That is, in a data.frame, how does one indicate that a factor is an *ordered* factor such that

Hola compañeros de la lista, qué tal. Los molesto con la siguiente duda: Tengo un experimento con dos factores A y B, cada uno de los cuales tiene los siguientes niveles (que son concentraciones de dos hormonas vegetales aplicadas a plantas): niveles del factor A: 0, 0.2, 0.5, 1 niveles del factor B: 0, 0.1, 0.2, 0.5, 1 y mi variable de respuesta es continua, todo dentro del set de datos

Dear R-Community! The example "oats" in MASS (2nd edition, 10.3, p.309) is calculated for aov and lme without interaction term and the results are the same. But I have problems to reproduce the example aov with interaction in MASS (10.2, p.301) with lme. Here the script: library(MASS) library(nlme) options(contrasts = c("contr.treatment", "contr.poly")) # aov: Y ~

I think there are two bugs in aov() that shows up when the right hand side of `formula' contains both `-1' and an Error() term, e.g., aov(y ~ a + b - 1 + Error(c), ...). Without `-1' or `Error()' there is no problem. I've included and example, and the source of aov() with suggested fixes below. The first bug (labeled BUG 1 below) creates an extra, empty stratum inside

I believe you are right, but can you please explain why anyone would want to fit this model? It differs only in the coding from aov(y ~ a + b + Error(c), data=test.df) and merely lumps together the top two strata. There is a much simpler fix: in the line if(intercept) nmstrata <- c("(Intercept)", nmstrata) remove the condition (and drop the empty stratum later if you

R-help What are the strengths and weakness of 'aov' in 'car' package? My model looks something like this : library(car) aov(lm(fish.length~zone*area,data=my.data)) Thank you Luis Ridao Cruz Fiskiranns??knarstovan N??at??n 1 P.O. Box 3051 FR-110 T??rshavn Faroe Islands Phone: +298 353900 Phone(direct): +298 353912 Mobile: +298 580800 Fax:

Good morning, I used in R contr.sum for the contrast in a lme model: > options(contrasts=c("contr.sum","contr.poly")) > Septo5.lme<-lme(Septo~Variete+DateSemi,Data4.Iso,random=~1|LieuDit) > intervals(Septo5.lme)$fixed lower est. upper (Intercept) 17.0644033 23.106110 29.147816 Variete1 9.5819873 17.335324 25.088661 Variete2 -3.3794907 6.816101 17.011692 Variete3

Hi all, I have been trying to calculate Type III SS in R for an unbalanced two-way anova. However, the Type III SS are lower for the first factor compared to type I but higher for the second factor (see below). I have the impression that Type III are always lower than Type I - is that right? And a clarification about how to fit Type III SS. Fitting model<-aov(y~a*b) in the base package and

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

Hello R users, I am trying to run a three factor ANOVA on a data set with unequal sample sizes. I fit the data to a 'lm' object and used the Anova function from the 'car' package with the 'type=III' option to get type III sums of squares. I also set the contrast coding option to 'options(contrasts = c("contr.sum", "contr.poly"))' as

Hello, I believe the aov() function in R uses a "Type-I sum-of-squares" by default as against "Type-III". This is relevant for me because I am trying to understand ANOVA in R using my knowledge of ANOVA in SPSS. I can only reproduce the results of an ANOVA done using R through SPSS if I specify that SPSS uses a Type-I sum-of-squares. (And yes, I know that when the sample

Hi I have two questions regarding the meaning of intercept outputs of lm. Question 1: In data set 1 (a fully-balanced design), the line with (Intercept) contains the overall mean, and the estimates contain the differences from the overall mean (matching those from model.tables). But in data set 2, the line with the intercept does not correspond to the overall mean and the estimates don't

Hi everybody, I'm trying to construct contrasts for an ANOVA to determine if there is a significant trend in the means of my groups. In the following example, based on the type of 2x3 ANOVA I'm trying to perform, does the linear polynomial contrast generated by contr.poly allow me to test for a linear trend across groups? doi=data.frame( Group=c( rep(1, 5), rep(2, 5), rep(3, 5),

Hello R-help I don’t know how to interpret significance from the contr.poly() function . From the example below : how can I tell if data has a significant Linear/quadratic/cubic trend? > contr.poly(4, c(1,2,4,8))               .L         .Q          .C [1,] -0.51287764  0.5296271 -0.45436947 [2,] -0.32637668 -0.1059254  0.79514657 [3,]  0.04662524 -0.7679594 -0.39757328 [4,]  0.79262909 

I'm baffled. When I run make check after installing from source, I get a Error 2. From my understanding of how these things work, it would appear to be coming from this (as at the end of base-Ex.Rout.fail: > has.VR <- require(MASS, quietly = TRUE) Attaching package 'MASS': The following object(s) are masked from package:base : confint confint.lm nclass.FD nclass.scott