Can I do a t-test Power Analysis for Unequal Size Groups which Produces 2 Different Minimum n's? It's usually straightforward to do a Power Analysis to compute the minimum sample size, especially in R which is my preferred statistical computing environment.
However, I am being asked to conduct a Power Analysis that's a little different than anything I've done or which I can find reference to online. I'm wondering if what I'm being asked for is even possible/valid.
The project basically has two unequal groups of states and the hypothesis is that these two groups are significantly different in terms of an outcome variable (which is the duration of phone calls to customers). The "control" group consists of 40 states and produced about 2,500 observations. The "test" group has about 10 states and 500 observations.
Initially, I found group means + pooled standard deviation, which I used to calculate an Effect Size. Then I used a package called pwr in R and found that I needed a minimum sample size of about 135 observations per group, given .05 significance and .8 power. 
However, they rejected my answer because they want one group to be much bigger than the other like it is now, and they are expecting either two different minimum numbers of observations per group or a minimum % of the population in terms of numbers of states or observations that have to go into their "test" group.
I see Power Analyses for two sample t-tests (i.e. the R function pwr.t2n.test), but I'd have to specify at least one of the sample sizes whereas they want me to tell them the minimal sample size for both groups (either as numbers or percentages) and this function doesn't reflect the differences in standard deviations for the two groups.
Is this possible or do I just tell them that's not how it works (i.e. the best I can do is tell them that given one of the sample sizes and a pooled standard deviation the second group has to be at least a certain size)?
 A: You can do sample size calculations for unequal sample sizes.
For example, you can decide the n's are in some ratio (such as in proportion to the populations perhaps).
It's then possible to do power calculations (at the least you can simulate to obtain the power under any particular set of circumstances, whether or not you are able to do the algebra).
The problem is that it's relatively inefficient at finding differences compared to the same total number of observations at equal sample sizes. 
Imagine you had a total sample of $n=n_1 + n_2$, with equal variance in the population and close to equal sample variance, and that your choice was between a 50-50 split and a 90-10 split ($n_1 = 0.5n$ vs $n_1=0.9n$).
The two-sample t-statistic is:
$t = \frac{\bar {X}_1 - \bar{X}_2}{s_{\text{pooled}} \cdot \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$
The impact of the sample size is in the term $1/{\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$.
If you have the 50-50 split it's like having a 40% smaller standard deviation; at a given $n_1+n_2$ you can pick up a substantially smaller effect with the even split.
If the combined sample size is not an effective constraint, this calculation may pointless however. It matters in cases where every observation carries the same marginal cost, which is not always relevant.
A: First off, why are you assuming equal variances in the two groups?  Please don't say, "Because it's convenient."  I seriously doubt that the group variances are equal, although in the case of equal sample sizes that isn't crucial.  Your degrees of freedom will be off, but you know you have at least 130, so who cares?  There are much bigger questions to address.
If you are going to permit (or require) unequal group sample sizes, the problem will not have a unique solution.  There are two unknowns ($n_1$ and $n_2$ and only one constraint (the power must be at least $\phi$.)  I don't think the problem can be solved without an additional constraint.  There are two obvious possibilities.  The first is to fix one of the sample sizes (e.g., the sponsors want at least 300 observations from Group I).  The other is to fix the ratio (e.g., because Group I is ten times the count of Group II, we want $n_1 = 10\, n_2$).  Now proceed with your power analysis.
