similar to: Extract p-value from linear model

Hello R community, I have a question regarding correlation and regression analysis. I have two variables, x and y. Both have a standard deviation of 1; thus, correlation and slope from the linear regression (which also must have an intercept of zero) are equal. I want to probe two particular questions: 1) Is the slope significantly different from zero? This should be easy with the lm

Dear all, I would like to ask how to extract the p-value for the whole model from summary(lm). This didn't help a lot summary.lm summary(lm(speed~dist, cars)) Thanks a lot! [[alternative HTML version deleted]]

Dear all, Does anybody know the probable reason why <= gives false when it should give true? These two variables are of the same type, and everything works in the cycle but then it stops when they are equal. this is the output result > Rk[47] <= RB[21] [1] FALSE > Rk[47] [1] 0.002842007 > RB[21] [1] 0.002842007 Thanks a lot. [[alternative HTML version deleted]]

Dear all, I cannot find to explicitly get the R-squared or adjusted R-squared from summary(lm()) Thanks a lot! [[alternative HTML version deleted]]

Dear all, does anybody ever encountered the problem with diagnostic plots? x<-rnorm(100) y<-rnorm(100) plot(lm(x~y)) It gives the following message "Waiting to confirm page change..." and nothing happens. Thanks a lot! [[alternative HTML version deleted]]

Dear all, Are there any functions in R to reverse the order of the string. smth like reverse("abc") to get "cba"? Thanks a lot. [[alternative HTML version deleted]]

Dear all, might seem and easy question but I cannot figure it out. floor(100*(.58)) [1] 57 where is the trick here? And how can I end up with the right answer? Thanks a lot everybody for your help. Trafim [[alternative HTML version deleted]]

Dear all, My question how is it possible to define a matrix A with 10 rows 1 column, so that its elements are vectors of undefined length. I need to have a possibility later to add elements like A[1,1] <- c(A[1,1],3,4,5) Thanks a lot for the help! [[alternative HTML version deleted]]

Dear all, I have a question is there a possibility to plot points with different symbols like stars, crosses? I looked at different types for plot command and didn't find anything like that. Thanks a lot. [[alternative HTML version deleted]]

Dear all, I would like to ask you if there is a possibility in R to give the names to graphs which are not const. For example, How to name each plot, or to add notes like a=x[i], b=y[i] in this cycle x<-c(1,2,3,4,5,6) y<-c(7,8,9,10,11,12) par(mfrow=c(2,3)) for (i in 1:6){ plot(x,y) } Thank you for your time and help! [[alternative HTML version deleted]]

Hello list, I have a linear regression ctl <- c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14) trt <- c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69) group <- gl(2,10,20, labels=c("Ctl","Trt")) weight <- c(ctl, trt) reg <- lm(weight ~ group) sreg <- summary(reg) and I would like to dump exactly what I see on the console with

Hi, I am trying to extract the coefficients matrix from a gls summary. Contrary to the lm function, the command fit$coefficients returns only the estimates of the model, not the whole matrix including the std errors, the t and the p values. example: ctl <- c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14) trt <- c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69) group <-

Hi, perhaps this question was answered previously however I could not find them. My problem is how how to extract a particular statistic from the result given by lm(). For e.g. ctl <- c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14) trt <- c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69) group <- gl(2,10,20, labels=c("Ctl","Trt")) weight <- c(ctl, trt) >

Dear all, does anyone know how to define the structure of the required samples using function createDataPartition, meaning proportions of different types of variable in the partition? Smth like this for iris data: createDataPartition(y = c(setosa = .5, virginica = .3, versicolor = .2), times = 10, p = .7, list = FALSE) Thanks a lot for your help. Regards, Trafim [[alternative HTML version

Hi, I'm newbie with R and don't really know how to add a median line to each of the groups that is not all the plot long. Here is a small working code that i have adapted for my purpose. If somebody could tell me how to draw median lines on each group and not all plot long. ctl <- c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14) trt <-

Hello list, I am used to give a lot of attention to the standardized regression coefficients, which in SPSS are listed automatically. Is there alternative to running the last two lines in the following example to get all the information? ctl <- c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14) trt <- c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69) summary( lm(ctl ~ trt) )

Hi everybody, Does anyone know what problem may be with this test. I am applying 5 different normality tests and use p-values for them, but for some reason S-W gives me NA, while sample size is 100. Any ideas? Thanks a lot! [[alternative HTML version deleted]]

Dear all, I have an error in the simple prediction function for lm(). Maybe someone experienced the same? xma <- matrix(data = 0, nrow = 100, ncol = 2) xma[, 1] <- rnorm(100) xma[, 2] <- rchisq(100, df = 3) m1 <- lm(xma[, 1] ~ xma[, 2]) predict(m1, as.data.frame(seq(-13, 13, 0.5))) Thanks a lot, Trafim [[alternative HTML version deleted]]

Dear Everyone in list: I am writing some codes to automate the process of fitting linear models where the names of variables of models are produced and stored in character vectors. But I have problems to pass the vectors to the lm( ) because I don't know how to strip the quotation marks automatically. Here are the codes of the example of lm( ): ## Annette Dobson (1990) "An Introduction

<r-help@r-project.org> Hi, R users, How can I export an R object as a .txt file? As an example I have the result from a regression and I need to save this object in a .txt file ctl <- c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14) trt <- c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69) group <- gl(2,10,20, labels=c("Ctl","Trt")) weight <- c(ctl,