similar to: Doubts about mixed effect models

I have some data I would like to model which involves choice of food by dung beetles. There are a number of experiments where in each case, there are five choices. Overall there are more than 5 different foods being compared (including a placebo) and different experiments use different comparisons. The problem is a generalization of Bradley-Terry but it differs from some generalizations in

Una distribución binomial negativa podría ser adecuada. De hecho, una extensión de la binomial negativa, como la distribución de Waring generalizada, termina convergiendo a una distribución binomial negativa. Puedes ver que la mejora en la bondad del ajuste de la binomial negativa con respecto a la Poisson es sustancial (te muestro también las salidas.). datos<-c(1280, 1262, 1290, 1321,

Hello I?m need fitt growth curve with data length-age. I want to evaluate which is the function that best predicts my data, to do so I compare the Akaikes of different models. I'm now need to evaluate if changing the initial values changes the parameters and which do not allow to estimate the model. To do this I use the function nls(); and I randomize the initial values (real positive number).

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)

Hi R masters! I need a Help with plot confidence intervals for one equation. I use library gplots and plotCI command in this script: require(gplots) ano <-1980:2002 rf<-exp(91.37162-0.04720281*ano) ciw.f<-sqrt(1.766073e-08) plotCI(ano,rf,uiw=ciw.f) But in the graph not shown the errors bar and I have this error msg zero-length arrow is of indeterminate angle and so skipped

Hi, How to order the levels os factor not by alphabetic order but by mean of Y. Somethink like this: I have this alphabetic order: > levels(pH) [1] "alto" "baixo" "medio" the order by mean os yvar is: > sort(tapply(Riqueza,pH,mean)) baixo medio alto 11.56667 20.00000 26.80000 How to make the levels of pH ordered by this mean to the result to

I'm trying to make a glmm to identify the relationship between insect species richness with fragment size, isolation and time (different years). I already tried to analyse it using poisson distribution error, but I always face with the following warning: *glm.fit: fitted probabilities numerically 0 or 1 occurred * This is probably hapenning because my dataset has a lot of zeros. So, what

Request For Translators<g> So what I've gathered from Wikipedia is that the Greek language does not use semicolons as a separator for enumerations: http://en.wikipedia.org/wiki/Semicolon#Greek_and_Church_Slavonic | In Greek and Church Slavonic, a semicolon indicates a question, | similar to a Latin question mark.[2] To indicate a long pause or | separate sections, each with commas

Hi I'm doing a glm analysis and I have two doubts (at least :) 1) When I run the function it gives a lot of warnings (see below) what they mean ? (may be I'm ignorant about this analysis ...) glm.poisson<-glm(log(Jkij+1)~fac.ano+fac.tri+fac.icesr+fac.mat+fac.ano:fac.icesr+fac.ano:fac.tri,family=poisson()) warnings() 40: non-integer x = 1.252763 41: non-integer x = 1.864785 42:

dear list! i have run several glm analysises to estimate a mean rate of dung decay for independent trials. i would like to compare these results statistically but can't find any solution. the glm calls are: dung.glm1<-glm(STATE~DAYS, data=o_cov, family="binomial(link="logit")) dung.glm2<-glm(STATE~DAYS, data=o_cov_T12, family="binomial(link="logit")) as

Estimados Les consulto por lo siguiente, en R y ubuntu, ambos en actualizados, tengo el siguiente mensaje (utilizando rstudio), no hay problemas porque realizando click se soluciona, pero antes no me hacía esa pregunta. ¿Es posible saltar esa pantalla? En la linea 13 de R no hay nada, ese fragmento de código es: # generate the tex files, one for each hospital in df library(knitr)

Hola, Vaya desastre al mandar la cabecera de la base de datos. Al enviarlo, se veía bien, pero está visto que por el camino la cosa se torció. Siguiendo el consejo de Javier, he subido un pequeño fragmento al DropBox. He quitado algunas especies para no hacerlo muy largo. Espero que ésta vez no haya problemas. https://www.dropbox.com/s/q7zla50oq7owg8k/CPUE.csv?dl=0 Un saludo y gracias Juan

Hi, I'm trying to make a model in order to know wich factors got?s influence in the intensity of a infection, but just in the individuals who's got this infection. In my data I've got a variable called "prevalence" with 2 levels: 1.- Infected individual 0.- Non infected So what i'm trying to do is a subset in a model like this,

Hi all, I have a problem with an object inside a formula in a function that I make. I create an object named qvartemp. This object is OK, look the levels: > print(levels(qvartemp)) [1] "baixomedio" "alto" Now I create a new object with a new formula: > new.form <- as.formula(gsub(qvar,"qvartemp",as.expression(mma$formula))) Look this new object: >

I currently have a program that automates 2-way ANOVA on a series of endpoints, but before the ANOVA is carried out I want the code to test the assumptions of normality and equal variance and report along with each anova result in the output file.  How can I do this? I have pasted below the code that I currently use.   library(car) numFiles = x #

Hola a todos, Os pongo aquí debajo un pequeño fragmento de código. Resulta que dispongo de una CADENA que tiene dos columnas. Una con palabras y la otra que dice si es bueno o malo. En otra variable tengo una LISTA. Quiero cruzar todas las palabras de la lista con todas las de la cadena y cuando alguna coincida, que me indique en qué fila está y también si la palabra es "buena" o

Hi, I have this data set: obitoss = c( 5.8,17.4,5.9,17.6,5.8,17.5,4.7,15.8, 3.8,13.4,3.8,13.5,3.7,13.4,3.4,13.6, 4.4,17.3,4.3,17.4,4.2,17.5,4.3,17.0, 4.4,13.6,5.1,14.6,5.7,13.5,3.6,13.3, 6.5,19.6,6.4,19.4,6.3,19.5,6.0,19.7) (dados = data.frame( regiao = factor(rep(c('Norte', 'Nordeste', 'Sudeste', 'Sul', 'Centro-Oeste'), each=8)), ano =

Dear R-Experts. I collected different datas about Nitrogen content (mg/ml) in Dung. The dung was eighter fresh (day=0) or had different ages (15,29,47) to observe nutrient changes over time. Now I like to draw a boxplot. boxplot(nmgml~day) abline((nmgml~day) The Problem is, that the boxplot considers the day values as groups and not as time series (neighter when the days are numeric or

Dear R-Experts, Dear friends of dung. I have a statistical Problem, to which nobody I asked could give me an answer. Maybe you can. I was in the African-Savanna and made a Dung-Monitoring. This means I walked randomly over the field and for every Dung-Event I found I noted following parameters: Species (Waterbuck, Giraffe, Reedbuck); Age-Category (less than a week, more than a week & less

Hi Alexander Wuerstlein Thank for the information. Now I agree that it's better to save the socket in /tmp/ I checked the source code and found that it is hard-coded. /* Allocate a buffer for the socket name, and format the name. */ auth_sock_dir = xstrdup("/tmp/ssh-XXXXXXXXXX"); It would be nice if openssh provides an option to overwrite this default. Regards Tran