Power Analysis

It seems to me, with deltas this large (relative to the SD), that a significance test is a moot point!

David Cross
d.cross at tcu.edu
www.davidcross.us




On Apr 18, 2011, at 5:14 PM, Albyn Jones wrote:

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