similar to: How to predict the mean and variance of the dependent variable after regression

Hi, folks linmod=y~x+z summary(linmod) The summary of linmod shows the standard error of the coefficients. How can we get the sd of y and the robust standard errors in R? Thanks! [[alternative HTML version deleted]]

Hi, folks, Here are the codes: ############## y=1:10 x=c(1:9,1) lin=lm(log(y)~x) ### log(y) is following Normal distribution x=5:14 prediction=predict(lin,newdata=x) ##prediction=predict(lin) ############### 1. The codes do not work, and give the error message: Error in eval(predvars, data, env) : numeric 'envir' arg not of length one. But if I use the code after the pound sign, it

I would like to prepare the data for barplot. But I only have the data frame now. x1=rnorm(10,mean=2) x2=rnorm(20,mean=-1) x3=rnorm(15,mean=3) data=data.frame(x1,x2,x3) If there a way to put data within a specific range? The expected result is as follows: range x1 x2 x3 -10-0 2 5 1 (# points in this

Hi, folks, As I understand, Least-squares Estimate (second-moment assumption) and the Method of Maximum Likelihood (full distribtuion assumption) are used for linear regression. I do >?lm, but the help file does not tell me the model employed in R. But in the book 'Introductory Statistics with R', it indicates R estimate the parameters using the method of Least-squares. However it

Dear all, Suppose my data frame is as follows: id price distance 1 2 4 1 3 5 ... 2 4 8 2 5 9 ... n 3 7 n 8 9 I would like to calculate the rank-order correlation between price and distance for each id. cor(price,distance,method = "spearman") calculate a correlation for all. Then I tried to use apply(data,list='id',cor(price , distance , method =

Hi, folks, runif (n,min,max) is the typical code for generate R.V from uniform dist. But what if we need to fix the mean as 20, and we want the values to be integers only? Thanks [[alternative HTML version deleted]]

Dear All, I am a beginner creating R packages. I followed the Leisch (2009) tutorial and the document ?Writing R Extensions? to write an example. I installed R 2.12.2 (I also tried R2.13.2), the last version of Rtools and the recommended packages in a PC with Windows 7 Home Premium. I can run R CMD INSTALL linmod in the command prompt and the R CMD check linmod. The following outputs are

Dear useRs, I have some trouble with the calculation of Cook's distance in R. The formula for Cook's distance can be found for example here: http://en.wikipedia.org/wiki/Cook%27s_distance I tried to apply it in R: > y <- (1:400)^2 > x <- 1:100 > lm(y~x) -> linmod # just for the sake of a simple example >

Full_Name: Anthony Gichangi Version: 1.90 OS: Windows XP Pro Submission from: (NULL) (130.225.131.206) The function rlnorm generates negative values for lognormal distribution. x- rlnorm(1000, meanlog = 0.6931472, sdlog = 1) Regards Anthony

I am reading Section 5 and 6 of http://cran.r-project.org/doc/contrib/Leisch-CreatingPackages.pdf It seems that I have to do the following two steps in order to make an R package. But when I am testing these package, these two steps will run many times, which may take a lot of time. So when I still develop the package, shall I always source('linmod.R') to test it. Once the code in

rlnorm takes two 'shaping' parameters: meanlog and sdlog. meanlog would appear from the documentation to be the log of the mean. eg if the desired mean is 1 then meanlog=0. So to generate random values that fit a lognormal distribution I would do this: rlnorm(N , meanlog = log(mean) , sdlog = log(sd)) But when I check the mean I don't get it when sdlog>0. Interestingly I

Apologize for not being clearer earlier. I would like to ask again. Thank Joris and Markleeds for response. Two examples: 1. Function 'var'. In R, it is the sum of square divided by (n-1) but not by n. (I know this in R class) 2. Function 'lm'. In R, it is the residual sum of square divied by (n-2) not by n, the same as in the least squares estimate. But the assumption following

Hi, folks, Please first look at the codes: plan_a=c('apple','orange','apple','apple','pear','bread') plan_b=c('bread','bread','orange','bread','bread','yogurt') value=1:6 data=data.frame(plan_a,plan_b,value) library(plyr) library(reshape) mm=melt(data, id=c('plan_a','plan_b'))

Hi there, I am stuck trying to solve what should be a fairly easy problem. I have a data frame that essentially consists of (ID, time as seqMonth, variable, value) and i want to find the regression coefficient of value vs time for each combination of ID and Variable. I have tried several approaches and none of them seems to work as i expected. For example, i have tried:

Dear R-Helpers Given a function like foo <- function(data,var1,var2,var3) { f <- formula(paste(var1,'~',paste(var2,var3,sep='+'),sep='')) linmod <- lm(f) return(linmod) } By typing foo(mydata,'a','b','c') I get the result of the linear model a~b+c. How can I rewrite the function so that the formula can be updated inside the function,

Hi, folks, I would like to copy the output of summary(lm) and anova (lm) in R to my word file. But the output will be a mess if I just copy after I call summary and anova. ##################### x=rnorm(10) y=rnorm(10,mean=3) lm=lm(y~x) summary(lm) Call: lm(formula = y ~ x) Residuals: Min 1Q Median 3Q Max -1.278567 -0.312017 0.001938 0.297578 1.310113

Hi, folks, Finally Friday~~ Here comes the question: x=c('germany','poor italy','usa','england','poor italy','japan') y=c('Spain','germany','usa','brazil','england','chile') s=1:6 z=3:8 test=data.frame(x,y,s,z) #Now I only concern the countries ('germany','england','brazil').

Hi! I 've a problem to have a lognorm distribution with mean=1 and var (or sigma)=1. rlnorm(1000,0,0) rlnorm(1000,1,1) rlnorm(1000,0,1) .... ? Can you help me?

Hi, Does anyone know what the mean value of a lognormal distribution in base-2 is? I am simulating stochastic population growth and if I were working in base-e, I would do:lambda <- 1.1 #multiplicative growth rates <- 0.6 #stochasticity (std. dev)lognormal <- rlnorm(100000, log(lambda) - (s^2)/2, s)## or lognormal <- exp( rnorm( 100000, log(lambda) - (s^2)/2,

Hello, I tried to fit a lognormal distribution by using optim. But sadly the output seems to be incorrect. Who can tell me where the "bug" is? test = rlnorm(100,5,3) logL = function(parm, x,...) -sum(log(dlnorm(x,parm,...))) start = list(meanlog=5, sdlog=3) optim(start,logL,x=test)$par Carsten. [[alternative HTML version deleted]]