similar to: re peated measures

Displaying 20 results from an estimated 90 matches similar to: "re peated measures"

2008 Apr 21
1
finding an unknown distribution
Hi, I need to analyze the influences of several factors on a variable that is a measure of fecundity, consisting of 73 observations ranging from 0 to 5. The variable is continuous and highly positive skewed, none of the typical transformations was able to normalize the data. Thus, I was thinking in analyzing these data using a generalized linear model where I can specify a distribution other than
2008 Mar 12
0
Re peated Measures (lme?)
Hello, I have a general data analysis question. I recently visited a lab where they are testing a new treatment and they had done the experiment several times on different dates. They repeated the experiment 3-5 times per day. And then for practical reasons they repeated the whole procedure for 5 days.(they wanted a large sample size but practically they couldn't handle more than 5-10
2009 Mar 06
2
Re peated ANOVA or nested ANOVA, or parallel one way ANOVA six times?
Hi, every body! I am a new comer for R, so my question would unavoidablely sounds stupid. Sorry! In my experiment, there are two type of soil ( soil F and soil D), each half of them were subjected to steam sterilize (result in FS and DS soil). A equal volume of soil from two of the four soil types (F, D, FS, DS) were mixed as follows: F+F, F+D, F+FS, F+DS, D+F, D+FS, D+DS, FS+DS (eight
2008 Aug 10
2
Differential Equations there use in R (population modeling)
e1 <- function(x,b,t){ d<-(x)*(b^t) plot(d) } e1(2, 2,seq(from=0, to=6, by=1)) Is there a way to do this with a change in time. I would like to use differential equations. I am trying to model a population with an initial value, fecundity per time step, and a death rate. The above simply shows an exponential growth rate. I would like to model species response like the famous
2013 Apr 06
1
Plotting a curve for a Holling Type III Functional Response
Hey, So I have a scatter plot and I am trying to plot a curve to fit the data based on a Holling Type III functional response. My function is this: nll2<-function(a,b) { conefun<-(a*DBH^2)/(b^2+DBH^2) nlls2<-dnbinom(x=cones ,size=DBH, mu=conefun,log=TRUE) -sum(nlls) } and my plot is this: plot (DBH,cones) DBH is on the x-axis and cones is on the y-axis. How do I get the curve
2011 Apr 04
1
Clarks 2Dt function in R
Dear Ben, you answerd to Nancy Shackelford about Clarks 2Dt function. Since the thread ended just after your reply, I would like to ask, if you have an idea how to use this function in R I defined it the following way: function(x , p, u) { (p/(pi*u))*(1+(x^2/u))^(p+1) } and would like to fit this one to my obeservational data (count) [,1] [,2] [1,] 15 12 [2,] 45 13 [3,]
2008 Mar 25
0
Mixed-effects models: question about the syntax to introduce interactions
hello everyone, I would like to as for advice for the use of ?lmer? (package ?lme4?) and writing the proper syntax to best describe my data using a mixed-effects model. I have just started to use these models, and although I have read some good examples (Extending the Linear Model with R, Faraway 2005; and the R book, Crawley 2007), I am still not sure of the syntax to test my hypothesis.
2013 Jun 11
1
calculo de poder estadistico en glm
Hola, Estoy tratando de calcular el poder estadistico de un GLM en R, programas estadisticos como SPSS lo puede calcular pero no eh encotrado nada sobre R. Alguien tiene alguna idea o sabe de literatura donde pueda encontrar ayuda? Luego tengo otra pregunta mas bien teorica sobre la funcion offset utilizada en negative poisson, por lo que eh entendido esta funcion posibilita trabajar con
2013 Jun 11
2
calculo de poder estadistico en glm
Estimado Matias, Reenvio a la lista. Es tarde aqui y hay que descansar. Quizas alguien en otra parte del mundo pueda darte una mano. En unas horas te envio mi opinion. Saludos, Jorge.- 2013/6/11 Matias Ledesma <> > Hola Jorge, > > Muchas gracias por las referencias, ya consegui el libro asi que me voy > aponer a leerlo aver si encuentro una solucion. > > La data
2013 Jun 12
3
calculo de poder estadistico en glm
Estimado Javier, Gracias por tus comentarios. Si, puede ser que me haya expresado en forma incorrecta. Mi idea fue tratar de ser lo mas conciso posible y limitarme al problema estadistico que tengo, por lo cual obvie mucha informacion. La data proviene de un monitoreo que se hace en el Mar Baltico anulamente hace mas de 20 años, debido a que la poblacion a decendido drasticamente en los
2013 Apr 12
2
model frame and formula mismatch in model.matrix()
Hello everyone, I am trying to fit the following model All X. variables are continuous, while the conditions are categoricals. model <- lm(X2
2006 Oct 25
1
Network card problem
Hi all, I''m trying to install Xen on a dell E520 computer. I''ve a network problem while rebooting on kernel vmlinuz-2.6.16.29-xen, all do work well, dom0 (ubuntu 6.06) start but I have no network at all. When booting on ubuntu, the NIC which is a Intel Gigabit Ethernet card drived by the e1000 modules does work correctly, but when the kernel xen is booted, trying to use the
2003 Aug 12
0
excellently
Wow. dada spam. Antranig Alfaro wrote: > humans bouts scrupulously taxied annette methacrylate exert crouching > exclude hubris meteorite aides acronym exogenous boater plead savoring > tapestries courtrooms taunter bodyweight poultry pokes brandishes > poems ama mightiness assam countless actinolite merges crave technic > accomplishers bombast creation bonaparte
2009 May 15
1
anova(cph(..) output
Hello, I am a beginner in R and statistics, so my question may be trivial. Sorry in advance. I performed a Cox proportion hazard regression with 2 categorical variables with cph{design}. Then an anova on the results. the output is > anova(cph(surv(survival, censor) ~ plant + leaf.age + plant*leaf.age, > Mpnymph) Wald Statistics Response: Surv(survival, censored)
2009 Sep 26
1
Multiple comparisons for coxph survival analysis model
Hello, all R-users! I am working on fitting a survival analysis model using the coxph function for Cox proportional hazards regression model. Data look like usual: ========================== group block death censor Group1 1 4 1 Group1 1 12 1 ... Group2 30 4 1 Group2 30 4 1 ... Group3 57 16
2008 Jul 02
0
question on dispersion parameter
Hi, I'm programming in R and below is a summary of a generalized linear model: ************************************************** *** Call: glm(formula = offspring ~ degdays, family = quasi(link = "log", variance = "mu"), data = fecundity) Deviance Residuals: Min 1Q Median 3Q Max -0.76674 -0.29117 -0.09664 0.15668 1.00800 Coefficients: Estimate Std. Error t value
2011 May 23
0
storing data from loops
Dear R-help list, I have a problem storing results from a bootstrap loop. What I doing is creating an empty matrix before the loop, then run the loop. Before the end of the loop I have a command line which defines which results inside the loop will have to go in the empty matrix. Here is the code I am using: eigen.values<-rep(NA,boots) #storage file for eigen values (1 for simulation)
2012 Jul 10
1
Revolutions blog: June Roundup
I write about R every weekday at the Revolutions blog: http://blog.revolutionanalytics.com and every month I post a summary of articles from the previous month of particular interest to readers of r-help. In case you missed them, here are some articles related to R from the month of June: The FDA goes on the record that it's OK to use R for drug trials: http://bit.ly/M0OoqA A review of
2001 Nov 05
1
Why doesn't outer work?
Hello I'm a population ecologist and use R for all my stats and modelling. Recently I have been using R to numerically solve integral projection models. This involves constructing several large matrices. The current code by Easterling (Size-specific sensitivity: Applying a new structured population model. Ecology, 2000, 81, 694-708) uses nested loops to construct the matrices. To speed up the
2013 Jun 11
0
calculo de poder estadistico en glm
Estimado Matías Ledesma Leí en la lista lo que usted desea realizar, en la misma expresa: La idea del estudio es mostrar un valor medio con un intervalo de confianza por estación de la frecuencia de malformaciones por individuo. Creo que se expreso en forma incorrecta, yo no se cuál es su estudio, yo no soy estadístico, soy veterinario, pero lo que usted expresa en su idea, aunque en