similar to: multiple comparisons in nested design

Dispongo de un experimento en el que cinco tratamientos ha sido aplicados a cinco grupos de voluntarios. En cada grupo había tres personas y a cada persona se le tomaron 3 medidas, En total dispongo de 45 medidas, pero evidentemente no son independientes entre sí. Si no tomo en cuenta que las medidas de la misma persona son más parecidas entre sí (bloques anidados) estaría incurriendo en

I'm new using R im trying to do a tukey test, but when i see the results the p value results in NA im guessing its because i have missing values but im not sure how to fix it AnovaModel.2 <- aov(area ~ trat, data=apilados) > summary(AnovaModel.2) Df Sum Sq Mean Sq F value Pr(>F) trat 4 11847 2961.76 9.9905 1.500e-06 *** Residuals 76 22531 296.46

Hi, I run a maanova analysis and found this message error: Error in ma.svd(X, 0, 0) : 0 extent dimensions I did a google search and found this: \item ma.svd: function to compute the sigular-value decomposition of a rectangular matrix by using LAPACK routines DEGSVD AND ZGESVD. \item fdr: function to calculate the adjusted P values for FDR control. I did a search for LAPACK and

Hi, I'm building Tukey's table using qtukey function. It happens that I can't get the values of Tukey's one degree of freedom and also wanted to eliminate the first column. The program is: Trat <- c(1:30) # number of treatments gl <- c(1:30, 40, 60, 120) # degree freedom tukval <- matrix(0, nr=length(gl), nc=length(Trat)) for(i in

Hi all, I'm working with a dataset from 10 treatments, each treatment with 30 subjects, each subject measured 5 times. The plot of the dataset suggests that a 3-parameter logistic could be a reasonable function to describe the data. When I try to fit the model using gnls I got the message 'Step halving factor reduced below minimum in NLS step'. I´m using as the initial values of the

Hi, I used the commands below to make Hartley's table, but some values are NA. require(SuppDists) trat = seq(2, 15, 1) gl = seq(2, 40, 1) har = matrix(0, nr=length(gl), nc=length(trat)) for(i in 1:length(gl)) for(j in 1:length(trat)) har[i,j] <- qmaxFratio(.95, df=gl[i], k=trat[j]) rownames(har) <- gl colnames(har) <- trat head(har) The output (head): 2

Hi, I try to plot bars with different colours in a barchart graphic. My idea is make that all X-Levels from trat var with different colour (grey scale). I search for a solution but dont find any. Any help? Thanks dados <- structure(list(Medias = c(0.994169096209855, 0.99416342412449, 0.974683544303797, 0.954430379746835, 0.669047619047619, 0.999999998475569, 0.994163424124514,

I want to specify two factors and their interaction as random effects using the function lme(). This works okay when I specify these terms using the function Error() within the function aov(), but I can't get the same model fitted using lme(). The code below illustrates the problem. a <- factor(rep(c(1:3), each = 27)) b <- factor(rep(rep(c(1:3), each = 9), times = 3)) c <-

Dear all How do I access individual elements of a "summary.aov" object? > data(iris) > AnovaModel.1 <- aov(Sepal.Length ~ Species, data=iris) > tmp <- summary(AnovaModel.1) > tmp Df Sum Sq Mean Sq F value Pr(>F) Species 2 63.2 31.6 119 <2e-16 *** Residuals 147 39.0 0.3 --- Signif. codes: 0 '***' 0.001 '**' 0.01

Hi, I did the analysis of variance of a split-plot and the effect of treatment was significant. I would like compare treatment means using Tukey. I can't extract the mean square to apply HSD.test to use in agricolae package. anava = aov(ganhos ~ Blocos + Trat*Supl + Error(Blocos/Trat)) names(anava) summary(anava) require(agricolae) HSD.test(ganhos, Trat, df, MSerror, alpha = 0.05)

Dear all I have found that the two "equivalent" commands do not produce the same results. 1. (I wrote this command by hand, this is what I would do usually) >summary(aov(eduyrs ~ cntry * edf, data=ESS1)) Df Sum Sq Mean Sq F value Pr(>F) cntry 1 257 256.65 21.2251 4.243e-06 *** edf 4 11010 2752.42 227.6296 <

Hi, How I could improve this graphic? http://www.divshare.com/download/10754700-f81 I would like to write groups labels in each panel and override the labels from object. I am try this code: xyplot(percentagem.mortos~tempo|trat, data=bio.ens, type="a", ? ? ? auto.key=list(points=FALSE, lines=TRUE, columns=3), ? ? ? ylim=c(0,100),scales = list(x = list(at = c(48, 72, 96), labels ? ?

Hi R-users I'm trying to do multivariate analysis of variance of a experiment with 3 treatments, 2 variables and 5 replicates. The procedure adopted in SAS is as follow, but I'm having difficulty in to implement the contrasts for comparison of all treatments in R. I have already read manuals and other materials about manova in R, but nothing about specific contrasts were found in them,

Hi: I'm using function lmer from package lme4, and I get this message: " There were 12 warnings (use warnings() to see them)" So I checked them: Warnings 1 to 11 said: 1: optim returned message ERROR: ABNORMAL_TERMINATION_IN_LNSRCH in: "LMEoptimize<-"(`*tmp*`, value = structure(list(maxIter = 50, ... and Warning 12 said: 12: IRLS iterations for glmm did

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,

Hi all, I have a little problem. I make an weibull survival analysis using the survival package. It,s OK, them I have the functions. I plot this funcions with curve(). I want to make a plot with the real survival points (proportion of alive x time) and them add the curves to points. I have the time to dead, the censor data and my trataments. To analysis the model is: model1 <-

Hi, I make a upgrade to R 2.1.0 and in some analysis I give an error: anova(model,test="F") Analysis of Deviance Table Model: binomial, link: logit Response: landing/total Terms added sequentially (first to last) Df Deviance Resid. Df Resid. Dev F Pr(>F) NULL 16 105.079 trat 1 93.149 15 11.930 93.15 Warning message:

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

Hola Compañeros, Estoy tratando de "contar" utilizando la función tapply y length y obtengo un error que no entiendo. Agradecería su ayuda. Las líneas son las siguientes: UNO <- subset (datos, CRIH2008 == "VERDADERO") dim (UNO) Abund2008 <- tapply (UNO$Cons, list(UNO$Site, UNO$Trat), length, na.rm=T) Y obtengo este error: Error en FUN(X[[1L]], ...) : 2 arguments

Hi, I am trying to run an analysis with the package maanova and I am not getting success. I suppose that I am wrong on set up the formula, so the issue may not be related to R, properly. I have two varieties of plants (V1 and V2). A group of each ones were treated and another was not treated. After treatment, in three different time RNA was collected from treated and from not treated plants for