similar to: How to use Subpopulation data?

R helpers   Please help me combine the simulated data to a form of table where: Hypermarket have 10 rows, supermarket have 15 rows,......., spazashops with 35 rows.   Hypermarket <- rnorm(10, mean=20000, sd=7000) Supermarket <- rnorm(15, mean=12000, sd=4000) Minimarket   <- rnorm(20, mean=10000, sd=4000) Cornershop  <- rnorm(20, mean= 8000, sd=3000) Spazashop   <- rnorm(35, mean=

Hi   Out of curiosity, how can i determine the minimum required sample size of this data in a combined form say 100 entries which r functions can i use?   R helpers   Please help me combine the simulated data to a form of table where: Hypermarket have 10 rows, supermarket have 15 rows,......., spazashops with 35 rows.   Hypermarket <- rnorm(10, mean=20000, sd=7000) Supermarket <-

Sorry about the subject --- On Thu, 24/9/09, KABELI MEFANE <kabelimefane@yahoo.co.uk> wrote: From: KABELI MEFANE <kabelimefane@yahoo.co.uk> Subject: Re: [R] Multiply Normal Curves To: R-help@r-project.org Date: Thursday, 24 September, 2009, 11:48 AM R -helpers   i have been trying to do this problem without must success,i managed to do a graph for x, but it is not what i want to

R -helpers   i have been trying to do this problem without must success,i managed to do a graph for x, but it is not what i want to define. I have also been able to do simple rendom sample. If possible can someone help here is the code for the graph pleae help   data.frame(ID=c(1,2,3),mu=c(10000,34000,50000),sigma=c(2000,3000,5000)) curve(dnorm(x,mean=parms$mu[1],sd=parms$sigma[1]),from=2000,

R-helpers   A curious question: Can you make suggestions  as to  what to use in R for the data from a sample of the following:   Hypermarket <- matrix(rnorm(100, mean=50000, sd=5000)) Supermarket <- matrix(rnorm(400, mean=34000, sd=3000)) Minimarket  <- matrix(rnorm(1000, mean=10000,sd=2000)) Cornershop  <- matrix(rnorm(1500, mean=2500, sd=500)) Spazashop   <- matrix(rnorm(2000,

R -helpers   i have been trying to do this problem without must success,i managed to do a graph for x, but it is not what i want to define,(i want to specify number of observations as well). I have also been able to do simple rendom sample.   data.frame(ID=c(1,2,3),mu=c(10000,34000,50000),sigma=c(2000,3000,5000)) curve(dnorm(x,mean=parms$mu[1],sd=parms$sigma[1]),from=2000, to=80000,

Hello R Experts, I kindly request your assistance on figuring out how to get a stratified random sampling proportional to 100. Below is my r code showing what I did and the error I'm getting with sampling::strata # FIRST I summarized count of records by the two variables I want to use as strata Library(RODBC) library(sqldf) library(sampling) #After establishing connection I query the data

R-list   I am sorry for asking this stupid question, but i have been running in circles. I want to randomly generate a scaling point of between 1 and 10, for say hundred entries, where the first 10% percent is has rates between 2 and 7, the next 15% 3 and 7, 20% between 3 and 9, 20% between 3 and  10, 35% between 5 and 10. The problem is that i can only generate the usual 100 using runif function

Hi, i read something about MECOSA3 (G.Arminger) which is a software to : "Analysis of finite mixtures of conditional Lisrel models" I'm very interested to identify heterogeneous subpopulations and ask me are similar functions doing something in R, too ??? I know SEM and perhaps with some additions/modifications it is possible doing this general mean- and covariance structures with

Dear all, I have to select 122 stratified random samples from a population of >3900 cells. I have 41 strata and I have to draw a different number of samples from them(between 2 and 8). I have tried to apply the funtion strata following the instruction in the manual: strata(dataframe, stratanames=NULL, size, method=c("srswor"), pik,description=TRUE) but I get the error

Dear All   I hope you can help me with this small problem. I want to draw a normal distribution line to this data: p<-rnorm(100, mean=50000, sd=3000) hist(p)   Kabeli [[alternative HTML version deleted]]

Hello, I'm running R 2.10.1 on Windows Vista. I'm selecting a random sample of several hundred items out of a larger population of several thousand. I realize there is srswor() in package sampling for exactly this purpose, but as far as I can tell it uses the native PRNG which may or may not be random enough. Instead I used the random package which pulls random numbers from random.org,

Dear All ? I know that you do not have to help me but please do, i am new to R as a CPI compiler, i just need to do a sample to see which sampling method best works in different situations, therefore since this is for practice purposes nobody will finance a real project thats why i need you to help me direct me as to how simulate data (just direct me,not 100% help). See my attachment for problem

Hi, is there a way to analyze subpopulations (e.g. women over 50, those who answered "yes" to a particular question) in a survey using Survey package? Other packages (e.g. Stata, SUDAAN) do this with a subpopulation option to identify the subpopulation for which the analysis shoud be done. I did not see this option in the Survey package. Is there another way to do this?

Hallo, I'm trying to sample a matrix with simple random sampling without replacement but seem to have a problem with the matrix length. Both sample() and srswor() use length() which returns the number of columns in matrix. This means that to function it seems that the sample size exceeds the matrix length. I need to sample the whole matrix for there are auxiliary variables I need for further

Hello, We have written a program (below) to model the effect of a covariate on observed values of a response variable (using only 80% of the rows in our dataframe) and then use that model to calculate predicted values for the remaining 20% of the rows. Then, we compare the observed vs. predicted values using a linear model and inspect that model's coefficients and its R2 value. We wish

 Dear All   let me go one step further by asking you if you could help me show that the distribution of this data in normal.  have a little idea (by trial and error) but i seem to not fully understand how its done.   H<-rnorm(100, mean=50000, sd=3000) par(las=1) hist(H, breaks=seq(40000, 60000, 1000), freq=F) f<- function(x) exp(-(x-5000)^2/18000000)/sqrt(18000000*pi)  x<- seq(40000,

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 =

OS X R 4.3.3 Colleagues I have created objects using the Surv function in the survival package: > FIT.1 Call: survfit(formula = FORMULA1) n events median 0.95LCL 0.95UCL SUBDATA$ARM=1, SUBDATA[, EXP.STRAT]=0 18 13 345 156 NA SUBDATA$ARM=2, SUBDATA[, EXP.STRAT]=1 13 5 NA 186 NA SUBDATA$ARM=2, SUBDATA[, EXP.STRAT]=2 5

Thanks a lot, so simple so efficient! I will study more the grep command I did not know. Thanks! Francesca Pancotto > Il giorno 19 ott 2017, alle ore 12:12, Enrico Schumann <es at enricoschumann.net> ha scritto: > > df[grep("strat", row.names(df)), ] [[alternative HTML version deleted]]