[R] Power Analysis

[Albyn Jones]

First, note that you are doing two separate power calculations,
one with n=2 and sd = 1.19, the other with n=3 and sd = 4.35.
I will assume this was on purpose.  Now...

> power.t.test(n = 2, delta = 13.5, sd = 1.19, sig.level = 0.05)

     Two-sample t test power calculation 

              n = 2
          delta = 13.5
             sd = 1.19
      sig.level = 0.05
          power = 0.9982097
    alternative = two.sided

Now, with n=2, the power is already .99.  With n=1, there are zero df.
So, what n corresponds to a power of .8?  

> power.t.test(n = 1.6305, delta = 13.5, sd = 1.19, sig.level = 0.05)

     Two-sample t test power calculation 

              n = 1.6305
          delta = 13.5
             sd = 1.19
      sig.level = 0.05
          power = 0.8003734
    alternative = two.sided

It looks like 1.63 subjects will do the job :-)

Finally, look at the power.t.test function, there is a line that explains
your error message:

 else if (is.null(n)) 
        n <- uniroot(function(n) eval(p.body) - power, c(2, 1e+07))$root

power.t.test() is making the sensible assumption that we only care about 
sample sizes of at least n = 2....

albyn

On Mon, Apr 18, 2011 at 02:31:19PM -0700, Schatzi wrote:
> I am trying to do a power analysis to get the number of replicas per
> treatment.
> 
> If I try to get the power it works just fine:
> setn=c(2,3)
> sdx=c(1.19,4.35)
> power.t.test(n = setn, delta = 13.5, sd = sdx, sig.level = 0.05,power =
> NULL)
> 
> If I go the other way to obtain the "n" I have problems.
> sdx=c(1.19,4.35)
> pow=c(.8,.8)
> power.t.test(n = NULL, delta = 13.5, sd = sdx, sig.level = 0.05, power =
> 0.8)
> 
> Is there any way to do this? Thank you.
> 
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-- 
Albyn Jones
Reed College
jones at reed.edu