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Power calculation for detecting linear trend
3 messages · Meesters, Erik, Peter Dalgaard, ONKELINX, Thierry
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 ?ster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk) FAX: (+45) 35327907
6 days later
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 Thierry.Onkelinx op inbo.be 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: r-help-bounces op stat.math.ethz.ch
[mailto:r-help-bounces op stat.math.ethz.ch] Namens Meesters, Erik
Verzonden: woensdag 7 maart 2007 15:50
Aan: r-help op 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
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