similar to: chi-Squared distribution in Friedman test

Dear Rs: outer(1:3, 1:3, function(df1, df2) qf(0.95, df1, df2)) I compare this F distribution results with the table, the answers were perfect. But I need to see for chi-sqaured distribution. When I employed the similar formula outer(1:3, 1:3, function(df1, df2) qchisq(0.95, df1, df2)) , I am getting unexpected results. I need to see the following values: p=0.750 ..... 1 1.323

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

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))

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

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 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

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

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)

Hi, I'm getting the following error when I'm trying to compile/link llvm with a backend I have. Running make on the /lib/M68K directory ends without any errors, and I have made all the required files and functions as outlined in http://llvm.org/releases/2.5/docs/WritingAnLLVMBackend.html I copied most of the files from Sparc backend and remove the content from functions mostly. Anyone has

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).

Hi all, I am having a trouble with this function I wrote ################################################### p26=function(x,alpha){ # dummy variable j=1 ci=matrix(ncol=2,nrow=3) while (j<4){ if (j==2) {x=x+c(-1,1)*0.5} ci[j,]= x+qnorm(1-alpha/2)^2/2+ c(-1,1)*qnorm(1-alpha/2)* sqrt(x+qnorm(1-alpha/2)^2/4) j=j+1 if (j==3) { # exact x=x-c(-1,1)*0.5

Hello, I want to use the quantile function so I read the doc but I don't understand with this > qchisq(seq(0.05,0.95,by=0.05),df=(length(don)-1)) [1] 62667.11 62795.62 62882.42 62951.47 63010.74 63064.00 63113.39 63160.27 63205.65 63250.33 63295.04 63340.48 63387.48 63437.03 63490.53 63550.14 63619.68 [18] 63707.24 63837.16 Can you help me please?

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

Hello, We have a video conference server using tcp and udp 3001 prot in internal, external user said that can''t connect to video server and held on 3001 fail, the following is file configuration, nat: 1.2.3.4 eth1:3 192.168.0.18 rule: video/ACCEPT net loc:192.168.0.18 marco.video: PARAM - - tcp 3000 PARAM - -