Re: [R] Power calculation for detecting linear trend
Erik, I haven't seen an answer to your question, so I'll try to answer it. The problem is that you switched the degrees of freedom. You had: 1 - pf(qf(.95, Vl, 1, ncp = 0), Vl, 1, ncp = Dl) [1] 0.05472242 But it should be: 1 - pf(qf(.95, 1, Vl, ncp = 0), 1, Vl, ncp = Dl) [1] 0.532651 Cheers, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Reseach Institute for Nature and Forest Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 [EMAIL PROTECTED] www.inbo.be Do not put your faith in what statistics say until you have carefully considered what they do not say. ~William W. Watt A statistical analysis, properly conducted, is a delicate dissection of uncertainties, a surgery of suppositions. ~M.J.Moroney -Oorspronkelijk bericht- Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Namens Meesters, Erik Verzonden: woensdag 7 maart 2007 15:50 Aan: r-help@stat.math.ethz.ch Onderwerp: [R] Power calculation for detecting linear trend Dear people, I've a problem in doing a power calculation. In Fryer and Nicholson (1993), ICES J. mar. Sci. 50: 161-168 page 164 an example is given with the following characteristics T=5, points in time R=5, replicates Var.within=0.1 q=10, a 10% increase per year The degrees of freedom for the test are calculated as Vl=T*R-2=23 and the non-centrality parameter Dl=4.54. Using this they get a power of 0.53, but the result that I'm getting is 0.05472242. I've tried this several ways in R, but I'm not able to come up with the same number. Am I doing something wrong in the calculation of the power? Here's my code: T-5 R-5 sigmasq-0.1 q-10 Vl-(T*R)-2 Dl-(R*(T-1)*T*(T+1)/(12*sigmasq))*(log(1+(q/100)))^2 #Dl result is still similar power.1-1-pf(qf(.95,(T*R-2),1,ncp=0),(T*R-2),1,ncp=Dl) Thank you for any suggestions/help. I'm using R2.4.1, on windowsXP. Erik Meesters [[alternative HTML version deleted]] __ 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] Power calculation for detecting linear trend
Dear people, I've a problem in doing a power calculation. In Fryer and Nicholson (1993), ICES J. mar. Sci. 50: 161-168 page 164 an example is given with the following characteristics T=5, points in time R=5, replicates Var.within=0.1 q=10, a 10% increase per year The degrees of freedom for the test are calculated as Vl=T*R-2=23 and the non-centrality parameter Dl=4.54. Using this they get a power of 0.53, but the result that I'm getting is 0.05472242. I've tried this several ways in R, but I'm not able to come up with the same number. Am I doing something wrong in the calculation of the power? Here's my code: T-5 R-5 sigmasq-0.1 q-10 Vl-(T*R)-2 Dl-(R*(T-1)*T*(T+1)/(12*sigmasq))*(log(1+(q/100)))^2 #Dl result is still similar power.1-1-pf(qf(.95,(T*R-2),1,ncp=0),(T*R-2),1,ncp=Dl) Thank you for any suggestions/help. I'm using R2.4.1, on windowsXP. Erik Meesters [[alternative HTML version deleted]] __ 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.
Re: [R] Power calculation for detecting linear trend
Meesters, Erik wrote: Dear people, I've a problem in doing a power calculation. In Fryer and Nicholson (1993), ICES J. mar. Sci. 50: 161-168 page 164 an example is given with the following characteristics T=5, points in time R=5, replicates Var.within=0.1 q=10, a 10% increase per year The degrees of freedom for the test are calculated as Vl=T*R-2=23 and the non-centrality parameter Dl=4.54. Using this they get a power of 0.53, but the result that I'm getting is 0.05472242. I've tried this several ways in R, but I'm not able to come up with the same number. Am I doing something wrong in the calculation of the power? Here's my code: T-5 R-5 sigmasq-0.1 q-10 Vl-(T*R)-2 Dl-(R*(T-1)*T*(T+1)/(12*sigmasq))*(log(1+(q/100)))^2 #Dl result is still similar power.1-1-pf(qf(.95,(T*R-2),1,ncp=0),(T*R-2),1,ncp=Dl) Thank you for any suggestions/help. I think your DF are upside-down: power.1-1-pf(qf(.95,1,(T*R-2),ncp=0),1,(T*R-2),ncp=Dl) power.1 [1] 0.532651 -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ 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.
