search for: mean1

...1[1:uw]) uppw1<-mean(group1[(n1-uw+1):n1]) UWx<-uppw1-low1 vw<-floor(0.5*n1) low2<-mean(group1[1:vw]) uppw2<-mean(group1[(n1-vw+1):n1]) LWx<-uppw2-low2 gam.low<-g*(UWx/(UWx+LWx)) gam.upp<-g-gam.low gw1<-floor(n1*gam.low) gw2<-floor(n1*gam.upp) hw<-n1-gw1-gw2 trim.mean1<-1/hw*(sum(group1[(gw1+1):(n1-gw2)])) a<-((gw1+1)*((group1[gw1+1]-trim.mean1)^2))+sum(group1[(gw1+2):(n1-gw1-1)]-trim.mean1)^2+((gw1+1)*(group1[n1-gw1]-trim.mean1)^2) b<-(((gw1+1)*((group1[gw1+1]-trim.mean1)^2))+sum(group1[(gw1+2):(n1-gw2-1)]-trim.mean1)^2+((gw2+1)*(group1[n1-gw2]-trim.me...

...t I can not solve this. I think there has to be one simple solution to this, but I just can not find it. I have saved the data in excel (csv-format). Then I read the data in R e.g. >data <- read.csv2("example.csv", header=TRUE) I look the data and it looks ok. E.g >data Mean1 1 4.4332 2 8.5113 3 35.1624 4 9.1693 5 2.974 6 65.1578 7 43.2241 8 3.1278 9 5.3364 10 3.9767 However, this "Mean1" is categorical when it should be real numbers. > Mean1 [1] 4.4332 8.5113 35.1624 9.1693 2.974 65.1578 43.2241 3.1278 5.3364 Levels...

...and some of the data, is as follows (once the database has been imported from SQL, and outliers removed). Z1sub <- Z1[, c(2, 5, 7,11, 12, 13, 15, 16)] colnames(Z1sub) <- c("temp", "hum", "wind", "press", "rain", "s.rad", "mean1", "sd1" ) attach(Z1sub) names(Z1sub) Model1d <- lm(mean1 ~ hum*wind*rain + I(hum^2) + I(wind^2) + I(rain^2) ) summary(Model1d) Call: lm(formula = mean1 ~ hum * wind * rain + I(hum^2) + I(wind^2) + I(rain^2)) Residuals: Min 1Q Median 3Q Max -1230....

...air) were subjected to two treatments (treatment) and were measured for 12 successive days within each treatment (layingday). Overall 5*2*12=120 observations. I want to test the effect of treatment, time (layingday) and their interaction. I have done so through the ANOVA below: > bmc3<-aov(Mean1~treatment*layingday+Error(male.pair/treatment/layingday)) > summary(bmc3) Error: male.pair Df Sum Sq Mean Sq F value Pr(>F) Residuals 1 0.13850 0.13850 Error: male.pair:treatment Df Sum Sq Mean Sq treatment 1 0.60525 0.60525 Error: male.pair:treatment...

Hi all: I have a question about lm on t-test. data(sleep) I wanna perform t-test to test the difference between the 2 groups: I can use: t.test(extra~group) The t.test result shows that:t = -1.8608; mean1=0.75,mean2=2.33 But I still wanna use: summary(lm(extra~group)) Intercept=0.75,which is mean1,just the same as t.test. group2=1.58 means the difference of the 2 groups,so mean2=1.58+0.75=2.33,just the same as t.test. And some parameters of group2(t value,Pr) are the same as t.test,since group2 i...

Dear users, *******I have a function f to simulate data from a model (example below used only to show my problems) f<-function(n,mean1){ a<-matrix(rnorm(n, mean1 , sd = 1),ncol=5) b<-matrix(runif(n),ncol=5) data<-rbind(a,b) out<-data out} *********I want to simulate 1000 datasets (here only 5) so I use S<-list() for (i in 1:5){ S[[i]]<-f(n=10,mean1=0)} ******I have a very complicated function for estimation o...

Hello, users. Dear users, *******I have a function f to simulate data from a model (example below used only to show my problems) f<-function(n,mean1){ a<-matrix(rnorm(n, mean1 , sd = 1),ncol=5) b<-matrix(runif(n),ncol=5) data<-rbind(a,b) out<-data out} *********I want to simulate 1000 datasets (here only 5) so I use S<-list() for (i in 1:5){ S[[i]]<-f(n=10,mean1=0)} ******I have a very complicated function for estimation o...

...s with other scale on the left in the same plot. I need help to get the same x axis (centered) when overlay two plots (barplot2 and plotCI (or lineplot.CI) with errorbars), with diferent y axis. I use par (new=T) but the X axis names are not centered with the figures. An example of my data: mean1<-cbind(55.5,56.42,56.38,55.28) SD1<-cbind(10,9 ,7 ,3) x<-rbind(mean1,SD1) colnames(x)<-c("Autumn","Spring","Summer","Winter") rownames(x)<-c("MeanProts","SD") mean2<-cbind(10,15,9,8) SD2<-cbind(2,3 ,1 ,5) y<-...

I have a rather basic question. How can I get R to generate a ANOVA table and a f-value for a hypothesis test such as: Data: group1 values: 5.2 4.5 6.0 6.1 6.7 5.8 group2 values: 6.5 8.0 6.1 7.5 5.9 5.6 ... H0: mean1 = mean2 = mean3 = mean4 HA: at least two means different where I want to evaluate using a f test statistic? F = MSTr/MSE I'd like a table similar to one that Minitab would produce: Df Sum Sq Mean Sq F value mean1 mean2 mean3 Error Secondly, how can do a second evaluation using...

...s of trying around, I gave up: I have a 2-dimensional array, and I know how to split it into its rows and how to get the mean for every row using 'sapply'. But what I want is to calculate the mean over the first n rows, and then the second n rows, etc., so that I get a vector like: v == mean1(row 1:5), mean2(row6:10),... (trivial, you might say. I find it rather mind-boggling, though: I tried to get the mean from the array before splitting it, after splitting it, looping through it with for-loops...I feel like an idiot by now; looks like I missed a crucial point of how 'R' w...

...awn, some parts of them are, of course, visible even if they are drawn on a non visible "face". Any suggestion to avoid this problem ? Thank you Etienne Example : trans3d <- function(x,y,z, pmat) { tr <- cbind(x,y,z,1) %*% pmat list(x = tr[,1]/tr[,4], y= tr[,2]/tr[,4]) } mean1 = c(2,4) cov1 = matrix(c(0.7, 0.2,0.2,0.7), ncol=2) x = seq(0, 5, by = 0.1) y = seq(0,10, by= 0.1) z = matrix(nrow = length(x), ncol=length(y)) for(i in 1:length(x)) for(j in 1:length(y)) { z[i,j]=dmvnorm(c(x[i],y[j]),mean1,cov1) } pmat1 = persp(x, y, z, col= "red",shade = 0....

Dear users, *******I have a function f to simulate data from a model (example below used only to show my problem) f<-function(n,mean1){ a<-matrix(rnorm(n, mean1 , sd = 1),ncol=5) b<-matrix(runif(n),ncol=5) data<-rbind(a,b) out<-data out} *********I want to simulate 1000 datasets (here only 5) so I use S<-list() for (i in 1:5){ S[[i]]<-f(n=10,mean1=0)} ******I have a very complicated function for estimation o...

Dear All I want to generate variable with Bivariate Normal Distribution by use mean1 = a, variance1 = b, mean2 = c, variance2 = d, rho = e. How I can do this. Many Thanks. IRD [[alternative HTML version deleted]]

...index1 <- 1:5 > index2 <- 6:10 > d <- rowMeans(e[, index1]) - rowMeans(e[, index2]) > design <- model.matrix(~factor(eset$Key)) > fit <- lmFit(eset, design) > ebayes <- eBayes(fit) > sample <- row.names(ebayes) > Pvalue <- ebayes$p.value[,2] > Mean1 <- rowMeans(e[,index1]) > Mean2 <- rowMeans(e[,index2]) > sd1 <- apply(e[,index1], 1, "sd") > sd2 <- apply(e[,index2], 1, "sd") > csv <- read.csv(file="D:/Anne-Marie/Doctorat/puces ADN macrophages/puces rat/Annie Dube/Analyse/RaGene-1_0-st-v...

...en to whether my case is suitable for a wilcoxon test. After research about the appropriateness, I am pretty sure that a wilcoxon test is the right tool for my case. But how to compute the power of the test is still an unanswered question bothering me. The basic stats of my two paired samples are: mean1 = 0.0032, sd1 = 0.012 mean2 = 0.00042, sd2 = 0.0016 sample size = 366 Could anyone help? Thank you in advance. -- View this message in context: http://r.789695.n4.nabble.com/How-to-compute-the-power-of-a-wilcoxon-test-tp3815616p3815616.html Sent from the R help mailing list archive at Nabble.co...

..../R/numberof.r * ./R/poisgam.r * ./R/progress.meter.r * ./R/qnorm.appr.r * ./R/qres.r * ./R/rowSums.r * ./R/S.princomp.r * ./R/select.range.r * ./R/shapiro.wilk.test.r * ./R/smoothed.df.r * ./R/tri.r * ./R/tweedie.r * ./R/tweedief.r * ./R/weighted.mean1.r * ./R/zzz.R * ./R/weighted.median.r * ./R/plotCI.Rd * ./man/cbind.dimnames.Rd * ./man/chlib-internal.Rd * ./man/clean.na.Rd * ./man/eql.Rd * ./man/invgauss.Rd * ./man/my.table.Rd * ./man/negbingof.Rd * ./man/numberof.Rd * ./man/poisgam.Rd * ./ma...

Hi, I am trying to write a function; I want to subtract the mean of each class in level 2 from the mean of each class in level 1 and square the answer, eg..... level.1 level.2 observation 1 1 0.5 1 1 0.2 1 2 0.6 1 2 0.4 2 3

...0.5488882 sd = 0.2 sig.level = 0.01 power = 0.8 alternative = two.sided NOTE: n is number in *each* group S-PLUS 2000 Professional Release 2: > normal.sample.size(n1=5, n2=5, mean=0, sd1=0.2, sd2=0.2, power=0.8, alpha=0.01, alternative="two.sided") mean1 sd1 mean2 sd2 delta alpha power n1 n2 prop.n2 1 0 0.2 0.4322771 0.2 0.4322771 0.01 0.8 5 5 1 ------------------------------------------ Dr. J?rn Quedenau Coordinator Data Management Bioinformatics Metanomics GmbH & Co. KGaA Tegeler Weg 33, D-10589 Berlin, Germany Tel +49...

I have many files (0.4.dat, 0.5.dat, ...) of which I would like to calculate mean value and variance and save the output in a new file where each line shouldlook like: "0.4 mean(0.4.dat) var(0.4.dat)" and so on. Right now I got a a simple script that makes me unhappy: 1. I run it by "R --no-save < script.r > out.dat" unfortunately out.dat has all the original commands in

Hello, I am facing a problem with optimization in R from 2-3 weeks. I have some Gaussian mixtures parameters and I want to find the maximum in that *Parameters are in the form * mean1 mean2 mean3 sigma1 sigma2 sigma3 c1 c2 c3 506.8644 672.8448 829.902 61.02859 9.149168 74.84682 0.1241933 0.6329082 0.2428986 I have used optima and optimx to find the maxima, but it gives me value near by the highest mean as an output, for example 830 in the above...