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
>
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>
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