similar to: t-test behavior given that the null hypothesis is true

Hi Ted, yes this was the problem. Thank you very much. best idaios On Wed, Jan 9, 2013 at 4:51 PM, Ted Harding <Ted.Harding@wlandres.net>wrote: > Ah! You have aqssigned a parameter "equal.var=TRUE", and "equal.var" > is not a listed paramater for t.test() -- see ?t.test : > > t.test(x, y = NULL, > alternative = c("two.sided",

Hey I'm not really sure what I should put on here, but I am having trouble with my R code. I am trying to get the p-values, R^2s etc for a number of different groups of variables that are all in one dataset. This is the code: #Stand counter st<-1 #Collections stands<-numeric(67) slopes<-numeric(67) intercepts<-numeric(67) mses<-numeric(67) rsquares<-numeric(67)

Hello, I have some discrete pvalues and I would like to sort them. Then add random noise so that they are ordered the same way as the original pvalues. Off course I don't want any pvalues less than 0 or greater than 1. Any ideas on how to do this in R? -- Thanks, Jim. [[alternative HTML version deleted]]

Full_Name: Jason Polak Version: R version 2.5.0 (2007-04-23) OS: Xubuntu 7.04 Submission from: (NULL) (137.122.144.35) Dear R group, I have noticed a strange anomaly with the shapiro.test() function. Unfortunately I do not know how to calculate the shapiro test P values manually so I don't know if this is an actual bug. So, to produce the results, run the following code: pvalues = 0; for

Hi all, I have some microarray data on 40 samples that fall into two groups. I have a value for 480k probes for each of those samples. I performed a t test (rowttests) on each row(giving the indices of the columns for each group) then used p.adjust() to adjust the pvalues for the number of tests performed. I then selected only the probes with adj-p.value<=0.05. I end up with roughly 2000

How do we get 2-tailed p-values for the rlm summary? I'm using the following: > fit <- rlm(oatRT ~ oatoacData$erp, psi=psi.bisquare, maxit=100, na.action='na.omit') > fitsum <- summary(fit, cor=F) > print(fitsum) Call: rlm(formula = oatRT ~ oatoacData$erp, psi = psi.bisquare, maxit = 100, na.action = "na.omit") Residuals: Min 1Q Median

It might help if you provided the code you used. It's possible that you didn't use direction="backward" in stepAIC(). Or if you did, it was still running, so whatever else you try will still be slow. The statement "R provides only the pvalues for each level" is wrong: look at the anova() function. Bob On 29 June 2017 at 11:13, Beno?t PELE <benoit.pele at

Hello, i am a newby on R and i am trying to make a backward selection on a binomial-logit glm on a large dataset (69000 lines for 145 predictors). After 3 days working, the stepAIC function did not terminate. I do not know if that is normal but i would like to try computing a "homemade" backward with a repeated glm ; at each step, the predictor with the max pvalue would be

Hola, Bueno, puedes hacer el cálculo de una forma mucho más compacta y rápida. Esta forma es especialmente recomendable cuando tienes muchas columnas y muchas filas. > library(data.table) > myDT <- as.data.table(mtcars) > myDTlong <- melt(myDT, measure.vars=1:ncol(myDT)) > myDTlong[ , list(p_value = shapiro.test(value)$p.value, v_stat = shapiro.test(value)$statistic) , by

ok I created a matrix C with A B C A1 B1 C1 .................................. ................................. the columns contain the gene expression values.. I ran the following t.test: apply(C, 1, function(x) t.test( x[1:3], x[4,6] )$p.value ) which outputs out 16063 pvalues (the number of rows) I just want to output 3 pvalues showing if A's column is different from A1

I'm wondering if R has functions to return pvalues with given x-squares and df. I googled for it but couldn't seem to find one. Appreciate any helps here. An example: df=4, x<- c(33.69, 32.96, 30.90) which are the statistic for chi-square, I'd like to get the corresponding pvalues for each values in x. Thanks. -Karen [[alternative HTML version deleted]]

Hi folks, I'm doing ANOVA with multiple comparisons. I've used both the TukeyHSD function and the multcomp procedure. In both cases, I get some tantalizing results such as this... e = aov(lm(d.all[,(n+4)] ~ d.all[,4]) TukeyHSD(e) Tukey multiple comparisons of means 95% family-wise confidence level Fit: aov(formula = lm(d.all[, (n + 4)] ~ d.all[, 4], contrast = C)) $`d.all[,

Hi All, I have run many regression analyses (14000 +) and want to collect the coefficients and pvalues into an excel file. I can get the statements below to work up to step 4. I can printout the regressionresults (sample output below). So my hope is to run something like step 5 and 6 and put the pvalues (and then coefficients) into an excel file. Can anyone suggest what I am doing wrong or a

Hello, I am trying to do an ANOVA on a microarray data set consisting of 22690 elements. The ANOVA is fine, but when I try to put the data in a frame in order to exporting it, I get a stack overflow. I have found documentation on dynamic memory in R, but not on how to increase the stack size. The code I'm using is below. If anyone has any suggestions for a workaround here, I'd

Hello everyone, I am trying to estimate the parameter b. I have Y and X1 which I know and they are both random. However, I also have X2 which I don't know and is also random. I want to estimat b from the model: Y = b*X1 + ( 1 - b ) * X2 Can anyone offer some suggestions. The values of Y and X1 are both pvalues so they are constrained in (0,1). -- Thanks, Jim. [[alternative HTML version

I used the multcomp package sometime back for doing multiple comparisons. I see that it has been updated and the methods like simint are no longer supported. When I run the program it prompts to me to use glht. How do I get the lower and upper conf int and the pValues using glht? Does anyone have an example? Thanks ../Murli [[alternative HTML version deleted]]

Hi, I would like to derive p-values for pair-wise comparison (Tukey's) of effects when the response is a count. I am trying a test case where y ~ Po( lambda(x) ). x has three levels : A, B and C with lambda(x) = 10, 20 and 20 respectively. Hence, p-values for the contrast C - B should distribute uniformally. I have implemented this test case as below but do not get uniform

My two matrices are roughly the sizes of m1 and m2. I tried using two apply and cor.test to compute the correlation p.values. More than an hour, and the codes are still running. Please help to make it more efficient. m1 <- matrix(rnorm(100000), ncol=100) m2 <- matrix(rnorm(10000000), ncol=100) cor.pvalues <- apply(m1, 1, function(x) { apply(m2, 1, function(y) { cor.test(x,y)$p.value

Hello list, I have a similar issue as this post http://tolstoy.newcastle.edu.au/R/e6/help/09/04/11438.html#options2 and I used the suggestion provided by Jorge with modifications to my data do.call(c,lapply(your_list_with_the_t_tests,function(x) x$p.value)) but I am getting the following error after excuting the code B<-by(eo,eo$PlateID, function(.sub) t.test(mcp1~Self_T1D,data=.sub,

I made an ANOVA with repeated mesures (aov(Mesure~Distance*Genre*Correct+Error(Sujet/(Distance*Genre*Correct)), data)) and I would like to extract the pvalues. The output is: ----------------------------------------------------------- Error: Sujet Df Sum Sq Mean Sq F value Pr(>F) Residuals 21 97.082 4.623 Error: Sujet:Distance Df Sum Sq Mean Sq F value Pr(>F) Distance