Re: [R] [newbie] mesuring adequation of normal distribution with avariable
I am not sure what you mean with adequation, but maybe ?shapiro.test is what you are looking for? BW, Marco -Oorspronkelijk bericht- Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Namens [EMAIL PROTECTED] Verzonden: zondag 1 april 2007 10:55 Aan: r-help@stat.math.ethz.ch Onderwerp: [R] [newbie] mesuring adequation of normal distribution with avariable Hello, I'm looking for a way of mesuring the adequation of a given variable with the normal distribution. Does R provide a standard test for this purpose, or is there a statistical methodology? Best, Sylvain Loiseau __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] only need the p-value
Dear R users, I am calculating several cox proportional hazard models after each other (I know this is unusual, but I am just exploring the data). For the purpose of multiple testing correction I need to construct an array of these p-values. However since the output is not an array in itself, I cannot find a way to obtain the p-value only. attach(tms) goal-rep(0.7*FREQUENC[1:13],6) event- Surv(TIJD,FREQUENCgoal) results-coxph(event~ TYPETREA) summary(results) Call: coxph(formula = event ~ TYPETREA) n=76 (2 observations deleted due to missing) coef exp(coef) se(coef) z p TYPETREAnon-guided -0.826 0.4380.484 -1.71 0.088 exp(coef) exp(-coef) lower .95 upper .95 TYPETREAnon-guided 0.438 2.28 0.169 1.13 Rsquare= 0.041 (max possible= 0.829 ) Likelihood ratio test= 3.2 on 1 df, p=0.0737 Wald test= 2.91 on 1 df, p=0.088 Score (logrank) test = 3.08 on 1 df, p=0.0794 Does anyone now how to extract the p-value? Many thanks!, Marco __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] paste? 'cmd /c c:\\pheno\\whap --file c:\\pheno\\smri --alt 1'
Dear R users,
This command works (calling a programm -called whap- with file specifiers etc.):
system('cmd /c c:\\pheno\\whap --file c:\\pheno\\smri --alt 1 --perm 500',
intern=TRUE)
Now I need to call it from a loop to replace the 1 by different number,
however I get lost using the quotes:
I tried numerous versions of:
i-1
system(paste(c('cmd /c c:\\pheno\\whap --file c:\\pheno\\smri --alt, i,
--perm 500', sep= )), intern=TRUE)
However no luck! I would be gratefull for any help.
Thanks,
Marco
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[R] help a newbie with a loop
Hi,
I am new in R and stumbled on a problem my (more experienced) friends
can not help with with. Why isnt this code working?
The function is working, also with the loop and the graph appears,
only when I build another loop around it (for different values of p) ,
R stays in a loop?
Can't it take more then 2 loops in one program?
powerb-function(x,sp2,a,b,b1,m)
{ sx-(sum(x^2)-(sum(x)^2)/length(x))/length(x)
n0-ceilingqnorm(1-(a/2))+qnorm(1-b))/b1)^2)*(((m+1)/m)*sp2/sx))
repeat
{
n1-ceilingqt(1-(a/2),n0-4)+qt(1-b,n0-4))/b1)^2)*(((m+1)/m)*sp2/sx))
if(n0==n1) break
n0-n1
}
return(c(sx,n1))
}
x-rnorm(1000,0,1)
x-x[order(x)]
res-matrix(0,1000,2)
#use the function and plot for different values of ind and p
for ( p in c(0.05,0.10,0.15,0.20,0.25,0.30,0.40,0.50))
{ risk-p*(2-p)
nonrisk-(1-p)^2
m-nonrisk/risk
for (ind in 1:500)
{res[ind,]-powerb(x[c(1:(500-ind),(500+ind):1000)],4,0.05,0.20,0.1,m)}
plot(res[,1],res[,2],type=p,ylab=n,xlab=var(x),main=b=0.1,power=0
.80,alpha=0.05,dominant met p=0.25)}
I would appreciate the help,
Marco
MPM Boks, MD PhD,
Department of Psychiatry, B01.206
University Medical Centre Utrecht,
PO box 85500, 3508 GA Utrecht
The Netherlands.
Tel: +31 30 2506370
Fax: +31 30 2505509
Email: [EMAIL PROTECTED] mailto:[EMAIL PROTECTED]
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