[R] how to determine power in my analysis?
Kristi Glover
kristi.glover at hotmail.com
Sat Nov 8 23:56:05 CET 2014
Dear Dennis, I really appreciated for your and Bert's help. I read the paper and it seems that once the study is completed, power calculations do not inform us in any way as to the conclusions of the present study. But I am really now confused whether we can't improve the research design for future or next year monitoring based on the present results. I would really be grateful for your suggestions and insights. Can't we take reference from the present study for improving future sampling?Thanks KG
> Date: Sat, 8 Nov 2014 13:36:35 -0800
> Subject: Re: [R] how to determine power in my analysis?
> From: djmuser at gmail.com
> To: kristi.glover at hotmail.com
> CC: gunter.berton at gene.com; r-help at stat.math.ethz.ch
>
> Hi Kristi:
>
> I think this paper elucidates the problem Bert mentioned. A thorough
> and careful reading of the last two sections should clarify what
> post-hoc power is and is not.
>
> http://www.stat.uiowa.edu/files/stat/techrep/tr378.pdf
>
> Dennis
>
> On Sat, Nov 8, 2014 at 11:25 AM, Kristi Glover
> <kristi.glover at hotmail.com> wrote:
> > Hi Bert, Thanks for the message. So far I know we can test whether my sample size in my analysis is enough or not. It is also post hoc property. For example, we can calculate standard deviations, error variance etc in the data sets, and then we can use them to determine whether the sample size was enough or not with certain level of alpha and power. we can do it is some of the statistical programs, but I was not aware in R. thanks KG
> >
> >> Date: Sat, 8 Nov 2014 10:55:56 -0800
> >> Subject: Re: [R] how to determine power in my analysis?
> >> From: gunter.berton at gene.com
> >> To: kristi.glover at hotmail.com
> >> CC: r-help at stat.math.ethz.ch
> >>
> >> Kristi:
> >>
> >> Power is a prespecified property of the design, not a post hoc
> >> property of the analysis (SAS procedures notwithstanding). So you're a
> >> day late and a dollar short.
> >>
> >> I suggest you consult with a local statistician about such matters, as
> >> you appear to be out of your depth.
> >>
> >> Cheers,
> >> Bert
> >>
> >> Bert Gunter
> >> Genentech Nonclinical Biostatistics
> >> (650) 467-7374
> >>
> >> "Data is not information. Information is not knowledge. And knowledge
> >> is certainly not wisdom."
> >> Clifford Stoll
> >>
> >>
> >>
> >>
> >> On Sat, Nov 8, 2014 at 3:49 AM, Kristi Glover <kristi.glover at hotmail.com> wrote:
> >> > Hi R Users,
> >> > I was trying to determine whether I have enough samples and power in my analysis. Would you mind to provide some hints?. I found a several packages for power analysis but did not find any example data. I have two sites and each site has 4 groups. I wanted to test whether there was an effect of restoration activities and sites on the observed value. I used a two way factorial ANOVA and now I wanted to test the power of the analysis (whether the sample sizes are enough for the analysis? what are the alpha and power in the analysis using this data set? if it is not enough, how much samples should be collected for alpha 0.05 and power=0.8 and 0.9 for the analysis (two way factorial analysis).
> >> > The example data:data<-structure(list(observedValue = c(0.08, 0.53, 0.14, 0.66, 0.37, 0.88, 0.84, 0.46, 0.3, 0.61, 0.75, 0.82, 0.67, 0.37, 0.95, 0.73, 0.74, 0.69, 0.06, 0.97, 0.97, 0.07, 0.75, 0.68, 0.53, 0.72, 0.34, 0.12, 0.49, 0.77, 0.45, 0.07, 0.97, 0.34, 0.68, 0.48, 0.65, 0.7, 0.57, 0.66, 0.4, 0.29, 0.88, 0.36, 0.68, 0.32, 0.8, 0, 0.11, 0.48, 0.85, 0.94, 0.12, 0.12, 0, 0.89, 0.66, 0.2, 0.57, 0.09, 0.27, 0.81, 0.53, 0.09, 0.5, 0.41, 0.89, 0.47, 0.39, 0.85, 0.71, 0.89, 0.01, 0.71, 0.42, 0.72, 0.62, 0.3, 0.56, 0.99, 0.97, 0.03, 0.09, 0.27, 0.27, 0.94, 0.23, 0.97, 0.81, 0.95), condition = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L), .Label = c("good!
> > ", "!
> >> > medium", "poor", "verygood"), class = "factor"), areas = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("Restored", "unrestored"), class = "factor")), .Names = c("observedValue", "condition", "areas"), class = "data.frame", row.names = c(NA, -90L))
> >> > test= aov(observedValue~condition*areas,data=data)summary(test)
> >> > power of the analysis?
> >> > thanks for your help.
> >> > Sincerely, KG
> >> >
> >> > [[alternative HTML version deleted]]
> >> >
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> >
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