similar to: Chi-Square test

Hello everyone, I am using the following program to get the p-value of some numbers (column 'LR' of the data.dat file). I want to write the 1st and 2nd column of the output file (data.out) as an integer while the program change them to double. Could anybody please tell me how I can write the code which writes the values of the first two columns as integer? Thanks library ('MASS')

Hello everyone, I am using the following program to get the p-value of some numbers (column 'LR' of the data.dat file). I want to write the 1st and 2nd column of the output file (data.out) as an integer while the program change them. Could anybody please tell me how I can write the code which writes the values of the first two columns as integer? For your convenience, I have attached the

Hello all, I have this simple barplot code: ifn <- "id.dat" dat <- read.table(ifn) ofn <- "id.png" bitmap(ofn, type = "png256", width = 30, height = 30, pointsize = 30, bg = "white",res=50) par(mar=c(5, 5, 3, 2),lwd=5) par(cex.main=1.6,cex.lab=1.6,cex.axis=1.6) names(dat)<-c("NumberOfPeople","Average")

Hello, I have used the following data to draw my barplot: BL LR Q 36.35 1.00 1.92 36.91 4.00 0.00 25.70 6.00 0.00 34.38 3.00 1.92 05.32 0.50 0.00 BL<-c(36.35, 36.91, 25.70, 34.38, 05.32) LR<-c(1.00, 4.00, 6.00, 3.00, 0.50) Q<-<(1.92, 0.00, 0.00, 1.92, 0.00) barplot(dt$LR, main='LR Value', col='orange', border='black',

Hello All: If I do have: x = (2, 4, 5, 5, 6, 4, 5, 2, 1) y = (9, 11.5, 12.5, 13, 14, 19, 20, 21, 22) I wanted to find a simple function in R which calculates the averages of X and Y for every 4 unit increase of Y. The results should look like: x = (4, 6, 3) #where (2+ 4+ 5+ 5)/ 4 = 4 and ... y = (12, 14, 20.5) #where (9+11.5+12.5+13)/ 4 = 12 and ... Thanks, Mohsen

Dear Lister, Is there an interface in R with SQL server that allows me to append records to table in the DB? Might I do that using RODBC? Thanks a lot. -- WenSui Liu A lousy statistician who happens to know a little programming (http://spaces.msn.com/statcompute/blog)

Right now, I am using SAS and S+/R. As a new year resolution, I am planning to learn a scripting language. from statisticians' point of view, which scripting language is worth to learn, perl, python, or any other recommendation? (Most likely, I will be learning it in windows.) Since I am not in research, I will prefer one widely used in industry and related to statistical work. if you

Dear all, I want to design a histogram and I need to have the frequency at certain points. For example I have the following 2 columns: *X Y* 0.1 25 0.4 22 0.45 11 0.55 21 I want the chart to have 4 columns. First column is from 0.0-0.1 (on X) and frequency is 25. Next colum is wider and form 0.1-0.4 with 22 frequency. Next column is narrow with 11 frequency and the last column is

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

Hi to all I found an formular of an ** ***p-Value Calculator for the Chi-Square test* *http://www.danielsoper.com/statcalc/calc11.aspx* *with the formula* *http://www.danielsoper.com/statkb/topic11.aspx* *what's the gamma function of this formula in r?* *df=5* *ch2=25.50878* *the following code does not give the result <0.001 for the values above * *p=

if i want to plot the chi-square distribution with a different degree of freedom how can i plot it in the graph?Sometimes i plot the histogram and cut it in a lot of piece.It's distribution like a chi-square.So i want to plot the chi-square with a different degree of freedom to compare it . -- View this message in context:

I'm wondering if R has functions to return pvalues with given x-squares and df. I googled for it but couldn't seem to find one. Appreciate any helps here. An example: df=4, x<- c(33.69, 32.96, 30.90) which are the statistic for chi-square, I'd like to get the corresponding pvalues for each values in x. Thanks. -Karen [[alternative HTML version deleted]]

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

Full_Name: Alan Bain Version: 2.4.0 OS: XP Submission from: (NULL) (155.140.122.227) Code for pnchisq contains following if (tSml) { if (x> f+ theta+ 5*sqrt(2*(f+2*theta) ))){ return 1.; /* better than 0 --- but definately FIXME */ } } This needs to check which tail has been requested; it is only correct if the default lower_tail=1 has been requested; for upper tail should return 0

Hi, This is more of a general question than a pure R one, but I hope that is OK. I want to combine one-tailed independent p-values using the weighted version of fisher's inverse chi-square method. The unweighted version is pretty straightforward to implement. If x is a vector with p-values, then I guess that this will do for the unweighted version: statistic <- -2*sum(log(x)) comb.p <-

Hello All: I am having the following data file named "data.dat" Number of people Number of pets Age of trees ifn <- "data.dat" dat <- read.table(ifn) colnames(dat)<-c("Variables") I want R to read all these in a string but when I ask R to read these, it gives me error because there more than one word in each line. Any comments to solve this problem?

Hi R-experts, I have data from four conditions of an experiment. I tried to create four subsets of the data with read.table, for example, read.table("Experiment.csv",subset=(condition=="1")) . I found a similar post in the archive, but the answer to that post was no. Any new ideas about reading subsets of data with read.table? Thanks! Feng [[alternative HTML version

Hello All: Can anybody tell me what is the problem with my program please. I have an error message as appears below. My program is: ifn <- "Jul08_09.LM" data <- read.table(ifn) ofn <- "Jul.png" bitmap(ofn, type = "png256", width = 30, height = 30, pointsize = 30, bg = "white",res=50) par(mar=c(5, 5, 3, 2),lwd=5)

Dear R-users, is there a function for the Chi-square Goodness of Fit Test in R (compare function chisq.gof in S/S-PLUS). Thanks for help. Matthias Seitzinger -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the

Hi There, There are 300 boy students and 100 girl students in a class. One interesting question is whether boy is smarter than girl or not. first given the exam with a difficulty level 1, the number of the student who got A is below 31 for boy, 10 for girl. Then we increase the difficulty level of the exam to level 2, the number of the student who got A is below 32 for boy, 10 for girl. We