similar to: Split Plot and Tukey

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, 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

Dear professors and collegues I need to perform a analysis of dates from a nested experimental design. From "Bioestatical Analysis" of Zar "Bimetry of Sokal" & Rohlf "Design and Analysis of Experiments" of Montgomery I have: Sum (mean(x)_i - mean(x)_T)2 / (a-1) -> var(epsilon) + n sigma2_B + n b (sum alfa_i)2 / (a-1) Sum (mean(x)_ij - mean(x)_i)2 /

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 build a graph taking into account the times: 1, 2,4,6,8,10,12,15,18,21,24,28,32 and 48. Be that the scale of the X axis does not look right. It seems equidistant. (graph attached) What changes have I to do in the following commands so that the scale be correct? interaction.plot(Tempo, Trat, Valor, ylim=c(0, 2), las=1, lty=c(1,2,3,4), lwd=3, bty='l',

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

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 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,

?Hi: I always use subset the same way but now is returning 0 rows. What's wrong with the way I am subsetting? library(ggplot2) structure(list(first = c(38.2086, 43.1768, 43.146, 41.8044, 42.4232, 46.3646, 38.0813, 40.0745, 40.4889, 38.6246, 40.2826, 41.6056, 34.5353, 40.0768), second = c(43.3295, 42.4326, 38.8994, 37.0894, 42.3218, 46.1726, 39.1206, 41.2072, 42.4874, 40.2657, 38.7766,

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 there, I am trying to run LSD.test(model) I used the following commands: attach(model) m1<- glm(ttl.m ~ site+year, family=quasipoisson, data= model) df<-df.residual(m1) MSerror<-deviance(m1)/df The following command did not work: comparison<- LSD.test (ttl.m, site+year, df, MSerror, alpha = 0.05, group=FALSE) I get an error message: Error: could not find function

I'm trying to create a graph where different cells of a grid (a shapefile) will be painted with a color share scale, where the most easy way is to use gray(). Can I somehow get a vector (gradient) of colors, a vector of colors with other methods but gray()? I'm doing this until now quad_N_sp <-

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