similar to: Y error bars, dates on the X-axis

Hello there I have a list, Y, and each component of that list is a real-valued function (that is, Y[[i]](u) returns a number). I was wishing to build the mean function and the first thing I thought of was Ybar<-function(u){ mean(Y[[1:n]](u)) } but obviously this doesn't work, since Y[[1:n]] is not allowed at all. Any elegant suggestions? I really would like to avoid something like

Please excuse me for asking such basic questions: Here is my code > y=c(1.21,0.51,0.14,1.62,-0.8,0.72,-1.71,0.84,0.02,-0.12) > ybar=mean(y) > ll=length(y); > ybarv=rep(ybar,ll) > x=1:ll > plot(x,ybarv,pch=1) > segments(x[1],ybar,x[ll],ybar) What I get is a collection of small circles, with a segment "on top" of the circles, which is almost what I want. But I

#See David Williams' book "Weighing the odds", p286 y <- c(1.21, 0.51, 0.14, 1.62, -0.8, 0.72, -1.71, 0.84, 0.02, -0.12) ybar <- mean(y) ylength <- length(y) ybarv <- rep(ybar, ylength) x <- 1:ylength plot(x,y,asp=1,xlab="position",ylab="ybar",type="n",ylim=c(-1,1)) segments(x[1], ybar, x[ylength], ybar) segments(x,ybarv,x,y)

Hi im trying to add error bars to my barplots, there very basic, i have a few grapghs where the y variable is different but on all the X variable is Age (Adult and Juvenile) however this is split into two levels so i have males and females, so my graph basically has four bars on it. I know how to add eror bars for instance when there is only one level eg lookng at the diffrence between male and

Hello: If anyone could guide me with this I would greatly appreciate it. Thanking you in advance for your assistance. Using a 3-level input factor alternative so that a function(below) can compute both a two-sided and one-sided p-values. Making the two-sided test the default. And produce output information about which alternative was tested. Where would I place the ifelse statement?

Dear friends. On p 95 in 3. ed. MASS a zero-truncated Poisson distribution is analyzed. I understand the probability distribution and expected mean. The Newton iteration scheme is Lam(m+1)=Lam(m)-[Lam(m)-Ybar(1-exp(-Lam(m))]/[1-Ybar*exp(-Lam(m)], and I suppose the latter part should be f(Lam(m))/f ' (Lam(m)) and f(Lam(m)) is Lam(m)/(1-exp(-Lam(m)), right ? But then f ' (Lam(m)) is

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

This was sent to me by someone on the R-list ( I don't know her ) but I don't have time to look at this right now so I told her I would send it to the R-list because she said it keeps getting bounced when she sends it. #======================================================================= ==================================================================== I am a bit confused with

Is this how a t hypothesis test is done when I don't have the actual data, but just the summarized statistics: > #Homework 9.2.6 [1] > n<-31 > xbar<-3.10 > s_x<-1.469 > m<-57 > ybar<-2.43 > s_y<-1.35 > s_pooled<- (((n-1)*s_x^2) + ((m-1)*s_y^2)) / (n + m - 2) > s_pooled [1] 1.939521 > t_obs <- (xbar - ybar) / (s_pooled * (sqrt(1/n + 1/m)))

All, I have just returned to R after a decade of absence, and it is good to see that R has become such a great success! I'm trying to bring Design of Experiments into some aspects of software performance evaluation, and to teach myself that, I picked up "Experiments: Planning, Analysis and Optimization" by Wu and Hamada. I try to reproduce an analysis in the book using lm, but

All, I've been using "sub" (subtitle) instead of "main" such that captions are below figures produced by xyplot. This works fine and captions are on a single line. However, when I try this for bar plots with error bars (altering the error.bars function form Crawley's The R Book, see below), the captions are split on more than 1 line. Is there a way to get the

Hello R-help, I am using R version 2.15.1. I upgraded from R version 2.13 a few months back. Previously, I was able to plot error bars on an xy scatter plot using the errbar function: errbar(RAEthylene$TIME,RAEthylene$AVE,RAEthylene$AVE+RAEthylene$STD,RAEthylene$AVE-RAEthylene$STD,add = TRUE,lty=2,pch=17); Today, I went to update my plot. However, in R version 2.15.1 I get error code

Hi, when doing as.double() on an object that is already a double, the object seems to be copied internally, doubling the memory requirement. See example below. Same for as.character() etc. Is this intended? Example: % R --vanilla > x <- double(1e7) > gc() used (Mb) gc trigger (Mb) max used (Mb) Ncells 234019 6.3 467875 12.5 350000 9.4 Vcells 10103774 77.1

hi all a technical question for those bright statisticians. my question involves ridge regression. definition: n=sample size of a data set X is the matrix of data with , say p variables Y is the y matrix i.e the response variable Z(i,j) = ( X(i,j)- xbar(j) / [ (n-1)^0.5* std(x(j))] Y_new(i)=( Y(i)- ybar(j) ) / [ (n-1)^0.5* std(Y(i))] (note that i have scaled the Y matrix as well) k is

In my last e-mails, I have asked for help regarding 1. 'defining functions inside loops' 2. 'integrating functions / vector arithmetics' 3. 'vectors out of lists?' 4. 'numerical integration' Since some of these topics seemed to be relevant (I'm guessing by the # of replies I got), I'm posting a modified section of my code. Any thoughts on improvements would

Basically I need to use the following data to calculate a squared error for each Sample based on the expected Survival for the zone. Basically, this code has Live/Dead for each sample, and I need to calculate the square error based on the Expected Mean (ie, Survival). The code looks up the expectation for each zone and applies for each sample in the zone using a loop: Data1 <-

Hello, I'm doing some linear regression: >lm<-lm(osas~alp,data) >summary(lm) However, the Rsquared in the output of summary() is not the same as the "standard" Rsquared calculated by spreadsheets, and outlined in statistical guidebooks, being SSR/SSTO. The output says "multiple Rsquared", but it is no multiple regression... What's the difference? Thanks,

Hi, I see a lot of folks verify the regression identity SST = SSE + SSR numerically, but I cannot seem to find a proof. I wonder if any folks on this list could guide me to a mathematical proof of this fact. Thanks. David. -- View this message in context: http://r.789695.n4.nabble.com/Regression-Identity-tp4636829.html Sent from the R help mailing list archive at Nabble.com.

Number of subjects = 25 Mean of Sample = 77 Standard Deviation (s) = 12 sem = 2.4 df = 24 The claim is that population mean is less than 80 * > 80 So our H0 (null hupotheis) is * > 80 > qt(.95,24) [1] 1.710882 > qt(0.05, 24) [1] -1.710882 tstat = -1.25 on t24 falls between 1.711 (.95,24) and *1.711 (.005,24) How Could I sketch t curve for the above data where my * would be at the

Hi all, I appreciate the help this list has given me before. I have a question which has been perplexing me. I have been working on doing a Bayesian calculating inserting studies sequentially after using a non-informative prior to get a meta-analysis type result. I created a function using three iterations of this, my code is below. I insert prior mean and precision (I add precision manually