similar to: t.test (PR#1086)

I am running a binomial glm with response variable the no of mites of two species y->cbind(mitea,miteb) against two continuous variables (temperature and predatory mites) - see below. My model shows overdispersion as the residual deviance is 48.81 on 5 degrees of freedom. If I use quasibinomial to account for overdispersion the dispersion parameter estimate is 2501139, which seems

I am running an lme model with the main effects of four fixed variables (3 continuous and one categorical – see below) and one random variable. The data describe the densities of a mite species – awsm – in relation to four variables: adh31 (temperature related), apsm (another plant feeding mite) awpm (a predatory mite), and orien (sampling location within plant – north or south). I have read

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

I am running a two-way anova with Type III sums of squares and would like to be able to understand what the different SS mean when I use different contrasts, e.g. treatment contrasts vs helmert contrasts. I have read John Fox's "An R and S-Plus Companion to Applied Regression" approach -p. 140- suggesting that treatment contrasts do not usually result in meaningful results with Type

I am analyzing a dataset on the effects of six pesticides on population growth rate of a predatory mite. The response variable is the population growth rate of the mite (ranges from negative to positive) and the exploratory variable is a categorical variable (treatment). The experiment was blocked in time (3 blocks / replicates per block) and it is unbalanced - at least 1 replicate per block. I am

I am analyzing a dataset on the effects of six pesticides on population growth rate of a predatory mite. The response variable is the population growth rate of the mite expressed as ln(Nfinal/Nstarting) of the mite, where N final the population of the mite at the end of the experiment and N starting the population of the mite at the beginning of the experiment. Each of the six treatments was ran

It's a bug in summary.aareg which no one found until now. What's wrong: If dfbeta=TRUE then there is a second estimate of variance calculated, labeled as test.var2. If maxtime is set, then both estimates of variance need to be recalculated by the summary routine. An incorrect if-then-else flow led it to look for test.var2 when it wasn't relevant. My test cases with maxtime also

Hi all, I've ventured into the world of nonparametric survival and I would like to use the "maxtime" option for printing and plotting my aareg fit. However, my fit does not have "test.var2" and this stops the print and plot when adding a maxtime. My code is as follows: Response<-Surv(Time,Event) Model<-aareg(Response~Factor1*Factor2)

Buenas, En R, como en la mayoría del software estadístico, no se utiliza ningún nivel de confianza sino que lo que se calcula es el p-valor asociado al contraste. De forma que cuanto más cerca de 0 esté el p-valor "menos credibilidad le damos a la hipótesis nula". Dicho mejor, debemos rechazar la hipótesis nula si el p-valor está por debajo de nuestro nivel de confianza. Por ejemplo,

I can think of many brute-force ways to do this outside of R, but was wondering if there was a simple/elegant solution within R instead. I have a table that looks something like the following: Factor1 Factor2 Value A 11/11/2009 5 A 11/12/2009 4 B 11/11/2009 7 B 11/13/2009 8 >From that I need to generate all permutations of Factor1 and Factor2 and force a 0 for any combination that doesn?t

Is names(model1$coef) what you're looking for? -----Original Message----- From: Kenneth Cabrera [mailto:krcabrer at epm.net.co] Sent: 29 November 2002 10:36 Cc: R-help at stat.math.ethz.ch Subject: [R] Obtaining the variable names of a glm object Hi, R users! Suppose I make a model like this:

Gracias Carlos. Yo también he visto el ejemplo que te pone chatGPT, pero la salida que te da no soy capaz de interpretarla. Os paso las ordenes y las respuestas de R de la propuesta de chatGPT Ejemplo.aov<- aov(P~TRAT+CORTE+REP) > summary (Ejemplo.aov) Df Sum Sq Mean Sq F value Pr(>F) TRAT 6 0.0028 0.00046 0.777 0.590 CORTE 2 0.5022 0.25110 424.542 <2e-16

Hello, as far as I see, R reports type I sums of squares. I'd like to get R to print out type III sums of squares. e.g. I have the following model: vardep~factor1*factor2 to get the type III sum of squares for factor1 I've tried anova(lm(vardep~factor2+factor1:factor2),lm(vardep~factor1*factor2)) but that didn't yield the desired result. Could anyone give me a hint how to proceed?

I'd like to get a long data set of minimum values from groups in another data set. The following almost does what I want. (Note, I'm using the word factor differently from it's meaning in R; bad choice of words) myframe = data.frame(factor1 = rep(1:2,each=8), factor2 = rep(c("a","b"),each=4, times=2), factor3 = rep(c("x","y"),each=2, times=4),

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

Dear R people, I have a couple of questions about post-doc analyses for 2 by 2 within subjects ANOVA. I conducted a psycholinguistic study that combined a 2 by 2 design and a latin square design. Specifically, I had 32 items each of which generated 4 conditions. Participants saw each of the 32 items only once: 8 in Condition A, 8 in B, 8 in C, and 8 in D. The table below serves as an example.

Hello R How do I refer to you in a publication? Many Thanks Ross Maller -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch

Hi All, I have a 4-dimensional data. I'm using barchart() function from lattice package. The R code and data are below - code includes one for stack=TRUE and other for stack=FALSE. I would like to present the data in another form which would be plotting Factor3 levels (P, Q, R, S) as two stacked bars (side by side). Like, for each level of Factor1 there should be two bars: first bar showing

I am attempting to replicate some of my experience from SAS in R and assume there are best methods for using a combination of summary(), subset, and which() to produce a subset of mean values by categorical or ordinal factors. within sas I would write proc means mean data=dataset; class factor1 factor2 var variable1 variable2; RUN; producing an output with means for each variable by factor

Full_Name: Mike Ulrich Version: 2.9 OS: Mac OSX Submission from: (NULL) (69.169.178.34) The help documentation for loadings() lists more then one parameter. The function call only expects one parameter. The digits, cutoff, and sort parameters are not used in the function. ## S3 method for class 'loadings': print(x, digits = 3, cutoff = 0.1, sort = FALSE, ...) ## S3 method for class