similar to: GLM questions

Dear all, I'm using the latest version of the package lme4 and R version 2.7.2 (2008-08-25). After I run the model, I get the results of the model (cf. below). Then, I run an ANOVA using the "anova" function and I get the following message "Error in anova(lmer1) : Calculated PWRSS for a LMM is negative". I went trough the R-mailing list and a similar error message was

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).

Hi everyone! I have some doubts about mixed effect models and I hope someone could help me. I´m trying to analyze a dataset coming from samples of dung beetles in the same forest fragments along 3 consecutive years (1994, 1995 and 1996) and 14 years after (2010). I sampled dung beetles in 18 different fragments with different sizes and different degrees of isolation. My aim is to determine

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

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)

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

?tryCatch -- Sent from my phone. Please excuse my brevity. On October 20, 2017 7:37:12 AM PDT, Evangelina Viotto <evangelinaviotto at gmail.com> wrote: >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

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, When I run the Aggregate function in repeat loop, I got the following error: Error in sort.list(unique.default(x), na.last = TRUE) : 'x' must be atomic for 'sort.list' Have you called 'sort' on a list? How to solve this error. Anyone can help me. Urgent. Thanks. Meenakshi. -- View this message in context:

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 =

Hi, I am trying to convert a colleague from using SPSS to R, but am having trouble generating a result that is similar enough to a bootstrapped cox regression analysis that was run in SPSS. I tried unsuccessfully with bootcens, but have had some success with the bootcov function in the rms package, which at least generates confidence intervals similar to what is observed in SPSS. However, the

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 #

Hello everybody, we´re using Red Hat Kickstarts for some systems. On every new kickstart we´ve to delete the client certificate first on the master. Ist there a best practise to renew the certificate or delete it remotely on the master? kind regards, Ano -- You received this message because you are subscribed to the Google Groups "Puppet Users" group. To view this discussion on

Dear all here is my data called "rglp" structure(list(vzorek = structure(1:17, .Label = c("179/1/1", "179/2/1", "180/1", "181/1", "182/1", "183/1", "184/1", "185/1", "186/1", "187/1", "188/1", "189/1", "190/1", "191/1", "192/1",

Hi If I run puppetd on a client with -l console I see things like: ... warning: //Node[default]/packages/ notice: ... err: ... ... Here puppet makes a distinction between errors, notice and warning. But it I run puppet without the -l option, I see the entries in /var/ log/messages. But there puppet makes no distiction between errors, warnings etc....: Nov 19 10:26:12 puppetd[28676]: Starting

Hello, How I use the function as.svrepdesign without memory.size problems? desenho_npc_JK <- as.svrepdesign(desenho_npc,type="JKn") Error: cannot allocate vector of size 161.3 Mb In addition: Warning messages: 1: Reached total allocation of 1022Mb: see help(memory.size) 2: Reached total allocation of 1022Mb: see help(memory.size) 3: Reached total allocation of 1022Mb:

Hola, Tengo un data set con información de transferencias monetarias por paciente provincia y modalidad de atención. Necesito hacer en una misma operación una suma de dinero por provincia y modalidad y a su vez una cuenta de pacientes por provincia y modalidad. Esto funciona bien: transferencias<-becas_modalidades %>% group_by(PROVINCIA,Modalidad) %>% summarize(dinero =

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,

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

Hi, I apologize for a portuguese message at this group, but I am trying to reach brazilians who read this great group. It''s about the release of the first Ruby on Rails book for the brazilian audience. Here it goes: Galera, finalmente terminei de escrever o livro e entreguei o material para a editora esta semana, agora resta esperar at? que eles nos digam quando ser? a data do