similar to: Ancova doesn't return test statistics

Dear members of the R help list, I want to do a hierarchical glm with binomial family but am unsure about how to write the syntax which involves nesting. I want to test whether the risk of being attacked by Herbivores for Meadowsweet plants is significantly dependent on the Distance to heterospecific source plants. Dependent variable = Herbivory (yes/no) Explanatory continuous variable =

When I perform a two-tailed sign test with the following simple syntax, binom.test(59,100) R returns a P-value (0.088) but nothing else. As I want the result for a one-tailed test I take P/2 = 0.044). However, the journal to which I've submitted my results requests the test statistics, not just the P-values. How can I make R return the test statistics? Best regards, Johan Stenberg, Umea

Hello, I am trying to create multiple matrices (to run a PVA) but can't import all of them from a .csv without the numbers treated as labels and not factors. I can enter the matrix slowly: Site05_96 <- matrix(c(0.07,0,0.03,0.00,NA,0.00, 0.09,0.166666667,0.31,0.42,NA,0.00, 0.00,0,0.00,0.00,NA,0.00, 0.00,0,0.00,0.00,NA,0.00,

Dear All, I am analysing a dataset on levels of herbivory in seedlings in an experimental setup in a rainforest. I have seven classes/categories of seedling damage/herbivory that I want to analyse, modelling each separately. There are twenty maternal trees, with eight groups of seedlings around each. Each tree has a TreeID, which I use as the random effect (blocking factor). There are two

Hi I compouted a multiple linear regression with repeated measures on one explanatory variable: BOLD peak (blood oxygenation) as dependent variable, and as independent variables I have: -age.group (binaray:young(0)/old(1)) -and task-difficulty measured by means of the reaction-time 'rt'. For 'rt' I have repeated measurements, since each subject did 12 different tasks. -> so

Hello R experts I have two linear regressions for sexes (Male, Female, Unknown). All have a good correlation between body length (response variable) and head length (explanatory variable). I know it is not recommended, but for a good practical reason (the purpose of study is to find a single conversion factor from head length to body length), the regressions need to go through the origin (0

Hi all, I am trying to set up a simple path analysis in the SEM package, but I am having some trouble. I keep getting the following error message or something similar with my model, and I'm not sure what I'm doing wrong: Error in solve.default(C) : system is computationally singular: reciprocal condition number = 2.2449e-20 In addition: Warning message: In sem.default(ram = ram, S = S,

R version 2.6.2 PowerBook G4 Hello R User, I try to perform an ANCOVA using the glm function. I have a dataset with continuous and categorical data (explanatory variables) and my response variable is also binary categorical. Fehler: NA/NaN/Inf in externem Funktionsaufruf (arg 4) Zus?tzlich: Warning messages: 1: In Ops.factor(y, mu) : - nicht sinnvoll f?r Faktoren (makes no sense for

Hi I am using R version 2.7.2. on a windows XP OS and have a question concerning an analysis of covariance with count data I am trying to do, I will give details of a scaled down version of the analysis (as I have more covariates and need to take account of over-dispersion etc etc) but as I am sure it is only a simple problem but I just can't see how to fix it. I have a data set with count

Hello all, I have two matrices, pop and pop2, each the same number of rows and columns that I want to compare for equality. I am concerned about efficiency in this operation. I've tried a few things without success so far. Doing something simple like: if (pop==pop2) { cat('equal') } else { cat('NOT equal') } results in the warning: 1: In if (pop == pop2) { : the

Hello all, I have the following function call to create a matrix of POP_SIZE rows and fill it with bit strings of size LEN: pop=create_pop_2(POP_SIZE, LEN) I have 3 questions: (1) If I did keep_pop[1:POP_SIZE] == pop[1:POP_SIZE] to keep a copy of the original data structure before manipulating 'pop' potentially, would this make a deep copy or just shallow? Ie

Hello. I've just started using R and am trying to figure out if the two codes snippets below have the same output gBest<-floor(runif(popsize,min=1,max=top)) velocity[i,j]<<-.4* velocity[i,j] + 1 * runif(1) * (pbestsVar[i,j] - popVar[i,j]) + 1 * runif(1) * (archiveVar[gBest,j] - popVar[i,j]) and for (i in 1:popsize) { for (j in 1:maxvar) { gBest<-sample(top,size=1)

Hi! as my subject says I am struggling with the different of a two-way ANOVA and a (two-way) ANCOVA. I found the following examples from this webpage: http://www.statmethods.net/stats/anova.html # One Way Anova (Completely Randomized Design) fit <- aov(y ~ A, data=mydataframe) # Randomized Block Design (B is the blocking factor) fit <- aov(y ~ A + B, data=mydataframe) # Two Way

Hi R-Users, I am working with sm.ancova (in the package sm) and I have two problems with the graph, which is automatically generated when sm.ancova() is run. 1-Besides of the fitted lines, the observed data appeared automatically in the graph. I prefer that only fitted lines appear. I check the sm.options, but I could not find the way that the observed data do not appear in the graph. 2-I

I have some data with two categorises plus/minus (p53) and a particular time (Time) and the outcome is a continuous vairable (Result). I set up a maximum model. ancova <- lm(Result~Time*p53) > summary(ancova) .. Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 0.05919 0.55646 0.106 0.916 Time -0.02134 0.01785 -1.195 0.241 p53plus

Hello, I am VERY new to R, just picking it up infact. I have got my head around the basics of ANOVA with post hoc tests but I am struggling with regression, especially with ANCOVAs. I have two sets of data, one of type A, one of type B. Both have been placed in a wind tunnel and sampled every week. The co variate is of course the days since the start. An example is day A B 0 10.0 10.0 7 9.0

I've run an ancova, edadysexo is a factor with 3 levels,and log(lcc) is the covariate (continous variable) I get this results > ancova<-aov(log(peso)~edadysexo*log(lcc)) > summary(ancova) Df Sum Sq Mean Sq F value Pr(>F) edadysexo 2 31.859 15.9294 803.9843 <2e-16 *** log(lcc) 1 11.389 11.3887 574.8081 <2e-16 ***

How to Ancova in R? I know this has got to be an FAQ, because I see it asked in the lists, but I haven't seen an answer to it. I see the R-sm has the ancova thing happening, but I kind of doubt that what I'm trying to do is "smoothing"... -- Pete Hurd phurd at uts.cc.utexas.edu http://www.zo.utexas.edu/research/phurd Section of Integrative Biology, University of Texas, Austin

Dear list members: I searched the R-help for packages to calculate power for an ANCOVA problem I have. I have found power.t.test, power.anova.test. But it seems that I can not found one with ANCOVA. I have two datasetsets with variables: univariate response(one data with continous response and one with 0 and 1), treatment(two levels), covariates(x1,x2,x3). I would appreciate your help. Tony

Hello,   Could you please help me in the following question: I have 16 persons 6 take 0.5 mg, 6 take 0.75 mg and 4 take placebo! Can I use the ANCOVA and t-test in this case? Is it possible in R?   Thank you in advance, Samuel [[alternative HTML version deleted]]