similar to: chi-Squared distribution

Dear R helpers: Thanks for the previous reply. I am using Friedman racing test. According the the book "Pratical Nonprametric Statistic" by WJ Conover, after computing the statistics, he suggested to use chi-squared or F distribution to accept or reject null hypothesis. After looking into the source code, I found that R uses chi-sqaured distribution as below: PVAL <-

Hi, Attention chi-squared distribution, unlike F distribution, has only df1 as parameter, not df1 and df2. So correct into: outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1, df2)) outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1)) ^^^^^^^^^^^^^^^^^^^^ Regards, Vito you wrote: Dear Rs: outer(1:3, 1:3, function(df1, df2) qf(0.95, df1, df2)) I compare this F

Dear Rs I have two vectors A and B where A is V1 d 1 0.000100000 1.123278 2 0.002186431 1.120448 3 0.004351214 1.106661 4 0.006515998 1.107713 5 0.008680781 1.107667 6 0.013010348 1.106353 7 0.019504698 1.104077 8 0.034658181 1.103202 9 0.051976447 1.103200 10 0.073624280 1.094825 11 0.093085682 1.085123 12 0.095250465 1.087325 13 0.132051782 1.086158 14 0.168853098

Dear Rs: BY having the following code: candidates<-expand.grid(e=c("nearest-neighbor","exaustive"), d=c(70,75,80,85,90,92,94,96,98,99), n=c(20,25,30,35,40)) results in : e d n 1 nearest-neighbor 70 20 2 exaustive 70 20 3 nearest-neighbor 75 20 4 exaustive 75 20 ................ 90 exaustive 90 40 91 nearest-neighbor 92 40 92 exaustive 92 40 93 nearest-neighbor 94 40 94

Dear Rs I have a single table with three columns in the following form: 1 100 150 1 45 32 1 99 100 2 150 33 2 22 87 2 71 31 .... .... 1000 64 32 1 100 150 1 45 32 1 99 100 2 22 89 2 31 44 2 88 11 .... .... 1200 64 32 1 100 150 1 45 32 1 99 100 2 150 33 2 22 87 2 71 31 ... ... 1100 31 34 Totally 1000+1200+1100 rows. Now, I need to group by first column and average then second and third column

Dear R helpers, I have a vector of 500 numbers and I need to plot them in such a way that the first 250 values should occupy 80% of the plot and the remaining ones should take 20%. More precisely, x axis ranges form 1:500 and the idea is to give the snapshot of the first 250 values. I tried "axis()" and log="x", but I am not getting the required output. Thanks

Full_Name: Jerry W. Lewis Version: 2.6.1 OS: Windows XP Professional Submission from: (NULL) (24.147.191.250) pchisq(0,0,ncp=lambda) returns 0 instead of exp(-lambda/2) pchisq(x,0,ncp=lambda) returns NaN instead of exp(-lambda/2)*(1 + SUM_{r=0}^infty ((lambda/2)^r / r!) pchisq(x, df + 2r)) qchisq(.7,0,ncp=1) returns 1.712252 instead of 0.701297103 qchisq(exp(-1/2),0,ncp=1) returns 1.238938

Hi, pchisq -> distribution function dchisq -> density function pval is the area under the curve, to calculte it you use distribution function which is the integral of density function. See: http://www.itl.nist.gov/div898/handbook/eda/section3/eda362.htm http://mathworld.wolfram.com/DistributionFunction.html f(x) density function F(x) distribution function =Pr(X<x)= integral(f(x))

Hi, How do we get the value of a chi square as we usually look up on the table on our text book? i.e. Chi-square(0.01, df=8), the text book table gives 20.090 > dchisq(0.01, df=8) [1] 1.036471e-08 > pchisq(0.01, df=8) [1] 2.593772e-11 > qchisq(0.01, df=8) [1] 1.646497 > nono of them give me 20.090 Thanks, cruz

Please could you tell me how to make code to make chi-square distribution table? Please help

Hi there, How accurate is the aproximation R makes to the Chi-Square distribution? For example, if I run: > qchisq(1/1000000,6) [1] 0.03650857 how accurate is 0.0365 compared to the theoretical percentile? What kind of approximations have been made in the software's algorithm? It woudl be useful to know since I am working with tiny percentiles such as one one-millionth and one

Hi The following is a database table named friends. +-----+------+------+-------+ | sid | id | fid | ftype | +-----+------+------+-------+ | 30 | 1 | 2 | F | | 31 | 1 | 3 | R | | 32 | 3 | 2 | F | | 33 | 3 | 4 | F | +-----+------+------+-------+ I want to update the ftype field based on id and fid. I want to achive the following. Update friends

Dear R experts I am looking for a package which gives me latin hyper cube samples from the grid of values produced from the command "expand.grid". Any pointers to this issue might be very useful. Basically, I am doing the following: > a<-(1:10) > b<-(20:30) > dataGrid<-expand.grid(a,b) Now, is there a way to use this "dataGrid" in the package

On Wed, Sep 13, 2017 at 1:03 PM, GiangCoi Mr <ltrgiang86 at gmail.com> wrote: > Hi all > Hi GiangCoi, The Good news is that now we have gluster-block [1] which will help you configure block storage using gluster very easy. gluster-block will take care of all the targetcli and tcmu-runner configuration for you, all you need as a pre-requisite is a gluster volume. And the sad part is

Hi, I want to determine the confidence interval on the sum of two sigma's. Is there an easy way to do this in R? I guess I have to use some sort of chisquare convolution algorithm??? Thanx, Roy -- The information contained in this communication and any atta...{{dropped}}

Full_Name: Viktor Witkovsky Version: 2.9.2 OS: Windows XP Submission from: (NULL) (78.98.89.227) Hello, I have found strange behavior of the function qchisq (the non-central qchisq is based on inversion of pchisq, which is further based on pgamma). The function gives wrong results without any warning. For example: qchisq(1e-12,1,8.94^2,lower.tail=FALSE) gives 255.1840972465858 (notice that

Hi, I want to compute the quantiles of Chi^2 distributions with different degrees of freedom like x<-cbind(0.005, 0.010, 0.025, 0.05, 0.1, 0.5, 0.9, 0.95, 0.975, 0.99, 0.995) df<-rbind(1:100) m<-qchisq(x,df) and hoped to get back a length(df) times length(x) matrix with the quantiles. Since this does not work, I use x<-c(0.005, 0.010, 0.025, 0.05, 0.1, 0.5, 0.9, 0.95, 0.975,

Full_Name: David Clayton Version: 1.8.1 OS: Linux Submission from: (NULL) (131.111.126.242) qchisq behaves very strangely when ncp is passed as zero (forcing internal qnchisq to be called) when first argument is small. Eg > qchisq(1-1e-6, 1, ncp=0, lower.tail=TRUE) qchisq(1-1e-6, 1, ncp=0, lower.tail=TRUE) [1] 1024 while, if ncp is unspecified, > qchisq(1-1e-6, 1) qchisq(1-1e-6, 1)

Full_Name: Drouilhet R?my Version: 1.8.1 OS: Linux Submission from: (NULL) (195.221.43.136) qchisq(1,10) works well but qchisq(1,10,ncp=0) does not work whereas ncp=0 is the default value of the function qchisq(1,10). (of course, 10 will be replaced by any integer value). Let us notice that this bug occurs only when applying probability one. (qchisq(seq(0,.9,.1),10,ncp=0) works very well).

Dear Rs I have a 3x3 multiple plot. I would like to have a overall legend in the outer right margin. From the help archive, I found that it can be done by setting par(xpd=NA). However, I couldn't find the correct values for x and y co-ordinates for the legend. Please find the code snippet below: par(mfrow=c(3,3), mar=c(4,4,0.9,0.5), oma=c(1,2,2,4),cex.main=1.1)