similar to: Testing year effects in lm()

Dear R-helpers, I have a linear model with a year effect (year is coded as a factor), i.e. the parameter estimates for each level of my year variable have significant P values (see some output below) and I am interested in testing: a) the overall effect of year; b) the significance of each year vis-a-vis every other year (the model output only tests each year against the baseline year). I'd

Hi, All. I have a logistic regression model that I have run. The question came up: which of these regressors is more important than another? (I'm using Design) Logistic Regression Model lrm(formula = iconicgesture ~ ST + SSP + magnitude + Condition + Expertise, data = d) Coef S.E. Wald Z P Intercept -3.2688 0.2854 -11.45 0.0000 ST 2.0871 0.2730 7.64

Hi all, I'd like to extract a value from an ANOVA table, but experience the following problem: ### This works: > s.pseudo <- summary(aov(m ~ block + mix*graz,data=split1)) > s.pseudo Df Sum Sq Mean Sq F value Pr(>F) block 2 1114.66 557.33 4.4296 0.04192 * mix 1 6.14 6.14 0.0488 0.82956 graz 2 1.45 0.72 0.0057 0.99427 mix:graz

Dear Rlist, Using an example in box.cox.powers() help, I have the following warning message. example: library(car) >attach(Prestige) > box.cox.powers(income) Box-Cox Transformation to Normality Est.Power Std.Err. Wald(Power=0) Wald(Power=1) 0.1793 0.1108 1.6179 -7.4062 L.R. test, power = 0: 2.7103 df = 1 p = 0.0997 L.R. test, power = 1: 47.261 df = 1 p = 0

Hi all, I have this figure: http://docs.google.com/Doc?id=df5zfsj4_103rjt2v4d5 created with the following steps: > x [1] 90.4 57.8 77.0 103.7 55.4 217.5 68.1 85.3 152.0 113.0 97.1 89.9 [13] 68.1 83.7 77.4 34.5 104.9 170.3 88.6 88.1 108.8 77.4 85.6 82.7 [25] 81.3 108.0 49.5 71.0 85.7 99.3 203.5 275.9 51.1 84.8 16.5 72.6 [37] 160.5 158.3 136.7 140.0 98.4 116.1

Hi all, I am trying to run a glm with mixed effects. My response variable is number of seedlings emerging; my fixed effects are the tree species and distance from the tree (in two classes - near and far).; my random effect is the individual tree itself (here called Plot). The command I've used is: mod <- glmer(number ~ Species + distance + offset(area) + (1|Plot), family = poisson)

Hi, I want to test for a contrast from a regression where I am regressing the columns of a matrix. In short, the following. X <- matrix(rnorm(50),10,5) Y <- matrix(rnorm(50),10,5) lm(Y~X) Call: lm(formula = Y ~ X) Coefficients: [,1] [,2] [,3] [,4] [,5] (Intercept) 0.3350 -0.1989 -0.1932 0.7528 0.0727 X1 0.2007 -0.8505 0.0520

Hi ! I've used the example given in the documentation for the prcomp function both in R and SPAD to compare the results obtained. Surprisingly, I do not obtain the same results for the coordinates of the principal composantes with these two softwares. using USArrests data I obtain with R : > summary(prcomp(USArrests)) Importance of components: PC1 PC2

Dear Users This is my dataset called mydata4. I want to sort the dataframe on the first column PxMid which is basically a column with dates. I've tried mydata4<-mydata4[order(mydata4$PxMid),] but it doesnt work. Could it be because these are dates? Please help I'm really stuck !! Thank you for your time. Regards Meenu PxMid EU0006MIndex.x DMSW1Curncy.x DMSW2Curncy.x DMSW3Curncy.x 1

---------------------------- Original Message ---------------------------- Subject: Different results from survreg with version 2.6.1 and 2.10.1 From: nathalcs at ulrik.uio.no Date: Fri, March 19, 2010 16:00 To: r-help at r-project.org -------------------------------------------------------------------------- Dear all I'm using survreg command in package survival.

Hi all, I have a dataframe of behavioral observations from 360 fish, each with 241 observation points(rows), which looks like this: > head(d) fish treatment tank trial video tid pid ang.chg abs.ac t len vel d2p x y 1 1 3 1 1 1 1 1 NA NA 0.0 0.000 NA NA 5.169 9.617 2

Hi R users, I need to create more than 20 figures (one for each group) in one page. I have a common legend for 20 figures using the facet_wrap. However the range of the values among the groups are very wide. For example one group has the value of 0 to 3, but the values of some of the groups has ranged from 0 to 20 so that when I used a single common legend for all 20 figures, I could not display

My setup: A SuperMicro 24-drive chassis with Intel dual-processor motherboard, three LSI SAS3081E controllers, and 24 SATA 2TB hard drives, divided into three pools with each pool a single eight-disk RAID-Z2. (Boot is an SSD connected to motherboard SATA.) This morning I got a cheerful email from my monitoring script: "Zchecker has discovered a problem on bigdawg." The full output is

Two time series questions: FITTING TRANSFER FUNCTIONS WITH LAGS: Consider the following toy example: > dates <- paste(11:21, "/01/2005", sep="") > Dates <- as.Date(dates, "%d/%m/%Y") > set.seed(1) > DF <- data.frame(date=Dates, y=rnorm(11), x=rnorm(11, 3)) > arima(DF$y, c(1,0,0), xreg=lag(DF$x, 1)) ar1 intercept lag(DF$x,

Dear all, I have a problem. I have m variables each of which has n observations. I want to calculate pairwise correlation among the m variables and store the values in a m x m matrix. It is extremely slow to use nested 'for' loops if m and n are large. Is there any efficient alternative to do this? Many thanks for your suggestions!! Bill

Hi all, I am trying to run Weibull PH model in R. Assume in the data set I have x1 a continuous variable and x2 a categorical variable with two classes (0= sick and 1= healthy). I fit the model in the following way. Test=survreg(Surv(time,cens)~ x1+x2,dist="weibull") My questions are 1. Is it Weibull PH model or Weibull AFT model? Call: survreg(formula = Surv(time, delta) ~ x1

Dears useRs, I have 2 factors, (for the sake of explanation - A and B), with 4 levels each. I've already fitted a negative binomial generalized linear model to my data, and now I need to split the factors in two distinct analysis of deviance table:  - A within B1, A within B2, A within B3 and A within B4  - B within A1, B within A2, B within A3 and B within A4 Here is a code that illustrates

Dear list, First my information: platform i386-pc-linux-gnu arch i386 os linux-gnu system i386, linux-gnu status major 2 minor 3.1 year 2006 month 06 day 01 svn rev 38247 language R version.string Version 2.3.1 (2006-06-01) Now my question: How is it possible that a command in an R script is not

Hello, I did a cross-validation using cvlm from DAAG package but wasn't sure how to assess the result. Does this result means my model is a good model? I understand that the overall ms is the mean of sum of squares. But is 0.0987 a good number? The response (i.e. gailRel5yr) has min,1st Quantile, median, mean and 3rd Quantile, and max as follows: (0.462, 0.628, 0.806, 0.896, 1.000, 2.400) ?

Hola, Manuel, No entiendo tu pregunta (la repito aqui para que sea mas explicito): hay alguna forma de comparar la probabilidad de supervivencias (en este caso anual) entre grupos sin utilizar un chi-cuadrado y un valor de P. Entiendo que lo que hace survdiff es comparar las curvas de supervivencia, pero yo quiero comparar la probabilidad de supervivencia entre grupos al final del estudio. Con