search for: means

I am trying to write a function that will solve a simple system of nonlinear equations for the parameters that describe the beta distribution (a,b) given the mean and variance. mean = a/(a+b) variance = (a*b)/(((a+b)^2) * (a+b+1)) Any help as to where to start would be welcome. -- Scott Story Graduate Student MSU Ecology Department 319 Lewis Hall Bozeman, Mt 59717 406.994.2670 sstory at

Dear R-help listers, I want to add 95% confidential interval as vertical lines to x axe in density plot. I have found the library(hdrcde) can do this work, but I do not know how to handle functions of this library when I used ggplot2 to draw the graph. Thank you in advance. The data and codes followed: # dummy data factor<-rep(c("Alice","Jone","Mike"),each=100)

Dear R-help listers, Under the helps from the ggplot2 list, I have set up method of drawing a graph with multiple density plots arranged one by one in one page. Now, I want to add 95% confidential interval as vertical lines to x axe in density plot. I have found the library(hdrcde) can do this work, but I do not know how to handle functions of this library when I used ggplot2 to draw the graph.

Hi there: I have a question about generating mean value of a data.frame. Take iris data for example, if I have a data.frame looking like the following: --------------------- Sepal.Length Sepal.Width Petal.Length Petal.Width Species 1 5.1 3.5 1.4 0.2 setosa 2 4.9 3.0 1.4 0.2

Dear all, I know there have been various questions posted over the years about loops but I'm afraid that I'm still stuck. I am using Windows XP and R 2.9.2. I am generating some data using the multivariate normal distribution (within the 'mnormt' package). [The numerical values of sanad and covmat are not important.] > datamat <-

Dear All: I run a logistic regression (using lrm in the Design package), and after that, I use the command "summary" to get the marginal effects of each variable. But one strange thing happens on my binary dependent variable: The marginal effect of it jumping from 0 to 1 is 1.77. I believe the marginal effect of binary variable x1 has interpretation should be P(Y=1|x1=1,

I am new to R and have two scripts written slightly different but should to relatively the same thing but my lack of experience with the program I can not figure out the what I need to do to correct it. The first script gives me a consistent mean fold change values with every run but can generate negative p values for some. For the second version of the script, the fold changes seem to be very

Dear Sirs, I am Professor at Indira Gandhi Krishi Vishwavidyalaya, Raipur, Chhattisgarh, India. While taking classes, I found the *by() *function producing following error when I use FUN=mean or median and some other functions, however, FUN=summary works. Given below is the output of the example I used on a built-in dataset "mtcars", along with error message reproduced herewith: >

I have 3 columns of numbers, and i want to find the mean of each separately, and then the mean of all of the means. ive tried: av3daysB<- mean(mean(avr1), mean(avr2), mean(avr3)) av3daysB but it only gives me the mean of the first column (avr1) Any suggestions? Thanks, Emilija

How can I have both printout and saved results at the same time. The subroutine first return "out" and the printout gets printed, but not saved. I then run the "invisible" line. Results got saved and accessible but no printout. How can I have both printout and also have the results saved? Thank you! > dstat4 <- function(data,digits=3){ +?? Mean??? <-

Dear All, I am trying to generate some Pareto random variates using the inverse method. This is really straightfoward and my R function looks as below : pareto <- function(c, a, cnt=1000) { u <- runif(cnt) x <- (c / ((u ^ (1 / a)))) mean.theo <- ((c * a) / (a - 1)) mean.gen <- mean(x) cat('Pareto mean : theoritical', mean.theo, 'generated', mean.gen,

Hello everyone, I am using *mlogit* to analyse my choice experiment data. I have *3 alternatives* for each individual and for each individual I have *9 questions*. I have a response from *516 individuals*. So it is a panel of 9*516 observations. I have arranged the data in long format (it contains 100 columns indicating different variables and identifiers). In mlogit I tried the following

...9752620740 Email: akhileshsingh.igkv at gmail.com On Fri, Apr 15, 2016 at 3:06 AM, Adrian Du?a <dusa.adrian at unibuc.ro> wrote: > I think you are not using the best function for what your intentions are. > Try: > > > by(data=mtcars, INDICES=list(as.factor(mtcars$am)), FUN=colMeans) > : 0 > mpg cyl disp hp drat wt > qsec vs > 17.1473684 6.9473684 290.3789474 160.2631579 3.2863158 3.7688947 > 18.1831579 0.3684211 > am gear carb > 0.0000000 3.2105263 2.73684...

...all groups # sb2 = Between group variance # sw2 = Within group variance # r = Corelation coefficient # z = standard normal variate cutoff value for binomial rv # Simulate the random effects A = rnorm(n=k, mean=0, sd=sqrt(sb2)) # Work out the mean of each group of data means = matrix(mu, nrow=k,ncol=1) for(row in 1:k){ means[row] = means[row,1] + A[row]} # Initializing the data vectors g=c=c1=mu1=numeric(k*m) # Now generate the data one group at a time. ind=0 for(u in 1:k){ for(replicate in 1:m){ ind = ind + 1 x=r...

Hi Dear all, I have some gene expression data samples from different tissue types ----------------------------------------------- - 120 samples from blood (B) - 20 samples from Liver (L) - 15 samples from Kidney (K) - 6 samples from heart (H) ----------------------------------------------- All the samples are from different individuals, so there are in total 161 individuals from which the DNA was

...ly appreciated. I'm working now on a similar problem, but instead of a balanced ANOVA, I have an unbalanced one. The first part of the question was: You assume that the within-population standard deviations all equal 9. You set the Type 1 error rate at á = .05. You presume that the population means will have the following values: uA = 17.5, uB = 19, uC = 25, and uD = 20.5. You intend to run 80 subjects in all, with equal n's across all 4 groups. You plan on conducting a one-way ANOVA. Compute your power to reject the null hypothesis under these conditions. I did: within.var = 9 ^ 2 mean...

...4594 3890 4096 4019 4204 M447T90 26553 27647 26889 26751 26929 Now I'm trying to transform each value column-wise to make columns to all have the same mean with: df * mean(mean(df)) / mean(df). I just can't get my head around this: mean(df) gives me the correct column means vector, and mean(mean(df)) gives me the correct total mean. The above operation works correctly for individual rows, i.e. if I do df[1,]*mean(mean(df))/mean(df) Can someone tell me what I am doing wrong?? Thanks!

...mean.bins <- function(data) { # For the vincentizing procedures in vincentize() and binify(), # it made sense to check the data array/vector/matrix. Here, # we now just need to check that data is a matrix. if ( !is.matrix(data) ) { stop("The data is not in matrix form."); } means <- c(); bins <- dim(data)[1]; for (i in 1:bins) { means <- c(means, mean(data[i,])); } # return a vector of means. means; } bins.factor <- function(data, bins) { if ( !is.data.frame(data) ) { stop("data is not a data frame."); } source('Ratcliff.r', lo...

I would like to generate pseudo-random numbers between two numbers using R, up to a given distribution, for instance, rnorm. That is something like rnorm(HowMany,Min,Max,mean,sd) over rnorm(HowMany,mean,sd). I am wondering if dnorm(runif(HowMany, Min, Max), mean, sd) is good. Any idea? Thanks. -james

Dear group, Is there a way I could avoid the sort of duplication illustrated below? i.e., I have the same dplyr::summarise function on different group_by arguments. So I'd like to create a single summarise function that could be applied to both. My attempt below fails. df <- data.frame(matrix(rnorm(40), 10, 4), f1 = gl(3, 10, labels = letters[1:3]), f2 =