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.ber...@gene.com
> To: kristi.glo...@hotmail.com
> CC: r-h...@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.glo...@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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> > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
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