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Hi All, I am trying to reproduce this using R instead. [image: Full-size image (38 K)] I tried using the following code *SChla <- read.csv("SM_Chla_data.csv")* *Atlantis <- SChla[16:66,]* *head(Atlantis)* * * Seamount Station Depth Pico Nano Micro Total_Ch dbar Latitude Longitud 16 Atlantis 1217 Surface 0.0639 0.1560 0.0398 0.2597 2.082 -32.71450 57.29733 17 Atlantis 1217 Shallow 0.0305 0.1250 0.0740 0.2295 24.524 -32.71450 57.29733 18 Atlantis 1217 Deep 0.1660 0.3560 0.0734 0.5954 49.573 -32.71450 57.29733 19 Atlantis 1217 Fm...

# Peter B. Mandeville kindly offered to send me code for Hotelling's T^2 # Test. Unfortunately there seems to be no route to his machine. # So i'm trying to reach him via the Mailing List. ------------------------------------------------------------------------ Sir, this morning i recieved your message about the availability of the code for Hotelling's Test. I hurried to find out

...Fluorescence Uncalibrated Maximum observed depth: 36 m QUAL has one digit for each of pressure, temp., sal. and fluor. QUAL=1:Uncal., QUAL=2:OK, QUAL=6:Interp., QUAL=9:No data DEPTH CTDPRS CTDTMP CTDSAL RAWFLU NUMB. QUAL M DBAR IPTS-68 PSS-78 OBS. ******* ******* ******* ******* 1 1.0 2999 2 2.0 5.9793 35.1629 .107 17 2221 3 3.0 5.9797 35.1631 .101 17 2221 4 4.0 5.9809 35.1631 .118 12 2221 5 5.1 5.9811 35.1629 .1...

By accident, I left out the lines defining n1 and n2. Here it is as a function. Peter B. hotelling <- function(d1,d2){ k <- ncol(d1) n1 <- nrow(d1) n2 <- nrow(d2) xbar1 <- apply(d1,2,mean) xbar2 <- apply(d2,2,mean) dbar <- xbar2-xbar1 v <- ((n1-1)*var(d1)+(n2-1)*var(d2))/(n1+n2-2) t2 <- n1*n2*dbar%*%solve(v)%*%dbar/(n1+n2) f <- (n1+n2-k-1)*t2/((n1+n2-2)*k) cat("F:",f,"\n") cat("PROBABILITY:",1-pf(f,k,n1+n2-k-1),"\n") } > source("c:/r/...

Hello there, I have two data sets with 14 variables each, and wish to do the test for equality of their covariance matrices and mean vectors. Normally these tests should be done by chi square test (box provided) and Hotelling's T square test respectively. Which R functions could do this kind of test? I just find some functions could do for one dimension, but no for multidimension. Some one

...per.d<-numeric(iter) #set.seed(987654321) #begining of simulation loop######## for (i in 1:iter){ r.NB<-rnbinom(n, mu = mu, size = size) y<-sort(r.NB) iter.num<-i print(y) print(iter.num) #empirical estimates or sample moments xbar<-mean(y) variance<-(sum((y-xbar)^2))/length(y) dbar<-variance/xbar #sample estimate of proportion of zeros and zero inflation index pbar<-length(y[y==0])/length(y) ### Simplified function ############################################# NegBin<-function(th){ mu<-th[1] d<-th[2] n<-length(y) arg1<-n*mean(y)*ifelse(mu >= 0, log(...

...effort and more importantly speed up builds. Build acceleration tools commonly use the build options, and among them macros, to determine if the output from a build is equivalent or not with a later build of the same compilation unit. The commands ... $ clang -DFOO example.c .. and ... $ clang -DBAR example.c ... produce 100% identical output, since neither macro FOO nor BAR affect the compilation unit. However, the build acceleration tools that do not run the preprocessor will not detect that the input to the compilation stage is identical and so will treat the output as unequal and run the...

I would like to test the hypothesis that the difference between pairs, for several variables, is zero. This is easily done separately for each variable with: lm(Y ~ rep(0, nrow(Y))) where Y is a matrix whose columns are the differences for each variable between pair members. However, I would like to get an overall probability across all variables from a Wilks or Pillai-Bartlett statistic as in

...we cannot make a conclusion by comparing the absolute mean value of each sample. So I turn to a means of hypothesis testing as I post in the title. Now my strategy is using permutation test. Like this: permutation.test.bias=function(x,y,mc=1000){ n1=length(x) n2=length(y) n=n1+n2 xy=c(x,y) dbar=abs(abs(mean(x))-abs(mean(y))) z=c() for(i in 1:mc){ mn=sample(1:n,n1) z[i]=abs(mean(xy[mn]))-abs(mean(xy[-mn])) } p.value=sum(abs(z)>dbar)/mc p.value } Although it seems plausible, this function doesn't work > n=1000 > x=rnorm(n)+10 > y=rnorm(n)-10 > permutation.test.b...

R 1.8.0 on Windows XP Professional. A huge THANK YOU to the R Team for this marvelous software. I am making lattice plots of oceanographic data. The usual layout does not conform to plotting conventions that marine scientists use when depth is the independent variable. Under those conventions, plots are made with the origin at the upper left, depth on the vertical axis (increasing as it

...ated as part of the input string. This means we don't have to worry about complexities like escaping "[[" etc. In the (unlikely) event that somebody's YAML needs to include the literal "[[FOO]]", they simply should not also use -DFOO - use a different name instead, e.g. -DBAR. On Mon, 3 Feb 2020 at 22:04, Fangrui Song <maskray at google.com<mailto:maskray at google.com>> wrote: I am adding -D k=v to yaml2obj, similar to clang -D. This makes it easy to generate {32-bit,64-bit} x {big-endian,little-endian} tests. --- !ELF FileHeader: Class: ELFC...

I am adding -D k=v to yaml2obj, similar to clang -D. This makes it easy to generate {32-bit,64-bit} x {big-endian,little-endian} tests. --- !ELF FileHeader: Class: ELFCLASS[[BITS]] Data: ELFDATA2[[ENCODE]] Type: ET_DYN Machine: EM_X86_64 # RUN: yaml2obj -D BITS=32 -D ENCODE=LSB %s -o %t.32le # RUN: yaml2obj -D BITS=32 -D ENCODE=MSB %s -o %t.32le # RUN: yaml2obj

Hi all, I have a user login form which the browser keeps populating because of its form fill feature. So when anyone goes to log in it shows the previous login''s username and password. How can I stop the browser from populating a text field? Thanks in advance. Cheers, Diego --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to

Hello everybody, I use the sweetpotato database included in R package: data(sweetpotato) This dataset contains two variables: yield(continous variable) and virus(factor variable). Due to Levene test is significant I cannot assume homogeneity of variances and I apply Welch test in R instead of one-way ANOVA followed by Tukey posthoc. Nevertheless, the problems come from when I apply posthoc