similar to: Help needed in interpreting linear models

the study design of the data I have to analyse is simple. There is 1 control group (CTRL) and 2 different treatment groups (TREAT_1 and TREAT_2). The data also includes 2 covariates COV1 and COV2. I have been asked to check if there is a linear or quadratic treatment effect in the data. I created a dummy data set to explain my situation: df1 <- data.frame( Observation =

I have to answer the following question for a homework assignment. A researcher was interested in whether people taking part in sports at university made more money after graduating, taking into account the students' GPA. They sampled 200 alumni from a large university. The variables are: income (income 10 years after graduating), sports (1 if they did sports, 0 if they did not), and GPA (the

Dear R-users, I am a dentist (so forgive me if my question looks stupid) and came across a problem when I did simulations to compare a few single level and two level regressions. The simulations were interrupted and an error message came out like 'Error in MEestimate(lmeSt, grps) : Singularity in backsolve at level 0, block 1'. My collegue suggested that this might be due to my codes

Hola Manuel lo he tratado de hacer pero me sale Error: unexpected string constante in: "anova(a,as,test=Chisq") no tengo ni idea de por qué... Me resulta alucinante no poder contar ya con pvals.fnc. ¿Será imposible hacerse con ello? Saludos, Miguel -------------------------------------------- El vie, 13/6/14, Manuel Azcárate <mazcarategarcia en gmail.com> escribió:

Hello, I've written a simple (although probably overly roundabout) function to test whether two regression slope coefficients from two linear models on independent data sets are significantly different. I'm a bit concerned, because when I test it on simulated data with different sample sizes and variances, the function seems to be extremely sensitive both of these. I am wondering if

Existe discusión sobre el uso de los p-valores en modelos mixtos. Como se ha dicho antes, para mi lo más adecuado es comparar modelos mediante la función anova. Por Internet se puede encontrar un buen libro de Douglas Bates y en español, busca modelos mixtos con R de Luis Cayuela, enfocado hacia ecología, pero está muy bien El 13/06/2014 14:00, "Jorge I Velez"

Hi developers, After some investigation I have found there can be large discrepancies in the same object being saved as an external "xx.RData" file. The immediate repercussion of this is the possible increased size of your .RData workspace for no apparent reason. The function and its three scenarios below highlight these discrepancies. Note that the object being returned is exactly

This is a continuing issue with the one on the list a long time ago (I couldn't find a solution to it from the web): -------------------------------------------------------------------------- > Using a formula converted with as.formula with lme leads > to an error message. Same works ok with lm, and with > lme and a fixed formula. > > # demonstrates problems with lme and

Hi all, how can I remove objects from the workspace which starts with a certain pattern, e.g. lm (lm1, lm2, lm3 etc). Wildcards won?t work. Thanks, Sven -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the

> lm2 <- function(...) lm(...) > lm2(mpg ~ wt, data=mtcars) Call: lm(formula = ..1, data = ..2) Coefficients: (Intercept) wt 37.285 -5.344 > lm2(mpg ~ wt, weights=cyl, data=mtcars) Error in eval(expr, envir, enclos) : ..2 used in an incorrect context, no ... to look in Can anyone explain why this is happening? (Obviously this is a manufactured example, but it

Hi all! I'm fairly new to R and not too experienced with regression. Because of one or both of those traits, I'm not seeing why some terms are being dropped from my model when doing a regression using lm(). I am trying to do a regression on some experimental data d, which has two numeric predictors, p1 and p2, and one numeric response, r. The aim is to compare polynomial models in p1

This isn't a question about R, but I'm hoping someone will be willing to help. I've been looking at calibration plots in multiple regression (plotting observed response Y on the vertical axis versus predicted response [Y hat] on the horizontal axis). According to Frank Harrell's "Regression Modeling Strategies" book (pp. 61-63), when making such a plot on new data

Dear all, Please excuse my ignorance, but I am having difficulty with this, and am unable to find help on the website/Google. I have a series of explanatory variables that I am aiming to get parsimony out of. For example, if I have 10 variables, a-j, I am initially looking at the linear relationships amongst them: my.lm1 <- lm(a ~ b+c+d+e+f+g+h+i+j, data=my.data) summary(my.lm1) my.lm2

Using a formula converted with as.formula with lme leads to an error message. Same works ok with lm, and with lme and a fixed formula. # demonstrates problems with lme and as.formula demo<-data.frame(x=1:20,y=(1:20)+rnorm(20),subj=as.factor(rep(1:2,10))) demo.lm1<-lme(y~x,data=demo,random=~1|subj) print(summary(demo.lm1)) newframe<-data.frame(x=1:5,subj=rep(1,5))

Hola a todos, quería preguntaros un medio para obtener los valores p usando lmer. He tratado con pvals.fnc, que es lo que me habían recomendado, pero por algún motivo no está ya disponible etc. Ésta es la función que tengo, pero da las "t", sin los valores p. Aunque Baayen indica que valores por encima de 2 son significativos necesito saber las p. resultado = lmer(rt_ln ~ (fre_ln *

Although this should be trivial, I'm having a spot of trouble. I want to make a lattice plot of the format y1+y2 ~ x | grp but then fit a lm to each y variable and add an abline of those models in different colors. If the xyplot followed y~x|grp I would write a panel function as below, but I'm unsure of how to do that with y1 and y2 without reshaping the data before hand. Thoughts

Dear R-List users, Can anyone explain exactly the difference between Weights options in lm glm and gls? I try the following codes, but the results are different. > lm1 Call: lm(formula = y ~ x) Coefficients: (Intercept) x 0.1183 7.3075 > lm2 Call: lm(formula = y ~ x, weights = W) Coefficients: (Intercept) x 0.04193 7.30660 > lm3 Call:

R-help, I'm trying to test the significance of two regression lines , i.e. the significance of the slopes from two samples originated from the same population. Is it correct if I fit a liner model for each sample and then test the slope signicance with 'anova'. Something like this: lm1 <- lm(Y~ a1 + b1*X) # sample 1 lm2 <- lm(Y~ a2 + b2*X) # sample 2 anova(lm1, lm2)

Hi Folks, I'm analysing some data which, in its simplest aspect, has 3 factors A, B, C each at 2 levels. If I do lm1 <- lm(y ~ A*B) say, and then summary(lm1, corr=T) I get the correlation matrix of the estimated coeffcients with numerical values for the correlations (3 coeffs in this case). Likewise with 'glm' instead of 'lm'. However, if I do lm2 <- lm(y ~

Hello, I am using a simple linear model and I would like to get an AIC value. I came across both AIC() and extractAIC() and I am not sure which is best to use. I assumed that I should use AIC for a glm and extractAIC() for lm, but if I run my model in glm the AIC value is the same if I use AIC() on an lm object. What might be going on? Did I interpret these functions incorrectly? Thanks,