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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