similar to: Probabilities greather than 1 in HIST

Hi, in order to save space for a publication, it would be nice to have a combined scatter and density plot similar to what is shows on http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=78 I wonder if anybody perhaps has already developed code for this and is willing to share. This is the reproducible code for the histogram version obtained from the site: def.par <-

Dear R people: I would like to superimpose a normal curve on a histogram. I've seen this example in a book, somewhere. I know that you draw the hist, get the mean and sd of the data set, but then I'm stuck. Could you help, please? Thanks! Erin hodgess at uhddx01.dt.uh.edu -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

Dear all, For two sets of random variables, say, x <- rnorm(1000, 10, 10) and y <- rnorm(1000. 3, 20). Is there any way to overlay the histograms (and density curves) of x and y on the plot of y vs. x? The histogram of x is on the x axis and that of y is on the y axis. The density curve here is to approximate the shape of the distribution and does not have to have area 1. Thank you

. I am looking for a histogram or box plot with the adding normal distribution curve I think that must be possible, but I am not able to find out how to do. Regards Knut

Hi, there I'm trying to plot what is returned from a call to tapply, and can't figure out how to do it. My guess is that it has something to do with the inclusion of row names when you ask for the values you're interested in, but if anyone has any ideas on how to get it to work, that would be stellar. Here's some example code: y1<-rnorm(40, 2) x1<-rep(1:2, each=20)

Hello, I have written a function that demonstrates the CLT by generating samples following the exponential distribution, calculating the means, plotting the histogram, and drawing the limiting normal curve as an overlay. I have the title of each histogram state the sample size and rate (1/theta) for the exponential (the output is actually 4 histograms), but I can't get the greek letter theta

I am having issues with the following: (muhat = 1/n^2(sum of all the xi's) ) essentially if xbar = the sample mean, muhat = sample mean but square the n. Question: Use R to run a Monte Carlo simulation which compares the finite-sample performance of xbar and muhat. Specifically generate 1000 samples n=30 from a standard normal distribution. For each sample calculate xbar and muhat. I have

I am new to R. I want to graph group data using a "Traditional Bar Chart with Standard Error Bar", like the kind shown here: http://samiam.colorado.edu/~mcclella/ftep/twoGroups/twoGroupGraphs.html Is there a simple way to do this? So far, I have only figured out how to plot the bars using barplot. testdata <- scan(, list(group=0,xbar=0,se=0)) 400 0.36038 0.02154 200 0.35927

Greets, My data looks like: > p3.18 s xbar subgroup 1 0.84 12.2 1 2 1.64 11.2 2 3 2.07 10.6 3 4 2.49 12.2 4 5 0.84 11.2 5 ... Using the command > qcc(p3.18$xbar,type="xbar",sizes=5,center=mean(p3.18$xbar),std.dev=mean(p3.18$s)/0.94,title="X-bar Chart for Paper Sheet Length Data") I get the x-bar chart I expect. However,

Dear Rusers, I would like to color different sections of a histogram different colors. I have an example that was done by "brute force" given below. Has anyone implemented something like this in general? If not, any suggestions/pointers on how to write a general function to do so would be most appreciated. Alan-

Hi Gents, I need to get the control limits from qcc function. As follows: qcc(MDI, type = "xbar.one") Call: qcc(data = MDI, type = "xbar.one") xbar.one chart for MDI Summary of group statistics: Min. 1st Qu. Median Mean 3rd Qu. Max. 0.3266 0.4249 0.4371 0.4333 0.4451 0.4858 Group sample size: 1 Number of groups: 383 Center

Hi Folks, I have data that looks like this: freq gender xBar 1000 m 2.32 1000 f 3.22 2000 m 4.32 2000 f 4.53 3000 m 3.21 3000 f 3.44 4000 m 4.11 4000 f 3.99 I want to plot two lines (with symbols) for the two groups "m" and "f". I have tried the following: plot(xBar[gender=="m"]~freq[gender=="f"]) followed by

Dear R users, I'm trying to extract from a given matrix (GROUP) the coordinates of the vertices of the different groups (i.e. 3, 7, 1 . . .) to plot the polygons to delineate the area in which each group "wins" and colour it diferentially. I can make a simple point plot, but I would like to add polygons with full colored area. The example is with a 5x5 matrix, but I'm working

Hi, I've had a quick look through the package list, and unless I've missed something, I can't seem to find anything that will do I-MR / Xbar-R / Xbar-S control charts ? Assuming there is something out there, can anyone point me in the right direction ? Thanks ! TIm

Hello there, I'm still struggling with the *apply commands. I have 5 people with id's from 10 to 14. I have varying amounts (nrep) of repeated outcome (value) measured on them. nrep <- 1:5 id <- rep(c("p1", "p2", "p3", "p4", "p5"), nrep) value <- rnorm(length(id)) I want to create a new vector that contains the sum of the

Hi, I wanted to understand exactly how acf and pacf works, so I tried to calculate ac and pac manually. For ac, I used the standard acf formula: acf(k) = sum(X(t)-Xbar)(X(t-k)-Xbar))/sum(X(t)-Xbar)^2. But for pac, I could not figure out how to calculate it by hand. I understand that in both R and EVIEWS, it is done using the Durbin-Levinson algorithm by the computer. However, I don't

Dear R users, I am stuck trying to figure out how to make a function return one value and print another.? Here is an example function: ################## eg <- function(x, digits=4) { xbar <- mean(x) sdx <- sd(x) value <- list(xbar, sdx) names(value) <- c("Mean of X", "SD of X") return(value)} ################## My current "solution" has been to

How do u calculated p values for a z test.. so far i ve done this A = read.table("cw3_data.txt") xbar = mean(A) s = 1 n = 20 mu = 0 z.test = (xbar-mu)/(s/sqrt(n)) p.value = pnorm(abs(z.test)) error = qnorm(0.99)*s/sqrt(n) left = xbar - error right = xbar + error and have got values off of it...but the values for p dont match up with other sites that i have used to check it

Dear Mac users, Hi, as you might have probably read the thread of "[R] Rsquared in summary(lm)" on May 10, it seems that Mac version of lm() seem to be working incorrectly. I enclose the script to produce the result both for lm() and manual calculation for a simple regression. Could you run the script and report with the version of R, so I don't have to go through every builds

Muchísimas gracias Carlos, de verdad que te agradezco la ayuda, pero no es lo que voy buscando. Quiero colocar en el eje de abscisas la secuencia temporal de los meses, es decir, agosto septiembre, octubre, etc? pero no las fechas de las toma de datos, sino que aparezca la marca de un mes, y la siguiente marca sea la del siguiente mes, etc?, y además que las muestras estén separadas de acuerdo con