Test vs control group - how to increase the accuracy of my estimate? I would like to find out if there is a way to reduce the statistical noise from my estimate of incremental value added from a certain customer treatment. Assume we have a group of clients for which we want to increase our revenue. We schedule some treatment (e.g. an ad or a special offer) for part of the customers ('test group') and don't provide any special treatment for the remaining customers ('control group'). The clients are assigned to test and control groups randomly.
One way to calculate the value added from such action is:
$y=N_T*(\overline{x}_T-\overline{x}_C)$,
where $N_T$ is the number of clients in the test group, $\overline{x}_T$ and $\overline{x}_C$ are the mean revenues per client in test and control group respectively and $y$ is the incremental value added from the treatment.
Assuming $y$ is significantly greater than 0 at some $\alpha$, is there a way to reduce the uncertainty around estimate $y$?
 A: Comment continued: On the efficiency of balanced designs. 
In the following example, we have a total of a thousand
subjects available and hope to detect a difference of of 2 units between population means of two groups, each with standard deviation $\sigma = 10.$
Using a balanced two-sample design with sample sizes $n_1 = n_2 = 500,$ the power of a test
at significance level 5% is about 89%. 
By contrast, the power for an unbalanced
design with $n_1 = 100, n_2 = 900$ is only about 47%. That's worse power than with a balanced design using only 200 subjects in each group (about 
51%).
m = 10^5;  mu1 = 100;  mu2 = 102;  sg = 10

# balanced
n1 = 500;  n2 = 500
set.seed(707)  # for reprocibility 
pv.bal = replicate(m, t.test(rnorm(n1,mu1,sg), rnorm(n2,mu2,sg), var.eq=T)$p.val)
mean(pv.bal <= .05)
[1] 0.886       # aprx power

# unbalanced
n1 = 100; n2 = 900
pv.ub = replicate(m, t.test(rnorm(n1,mu1,sg), rnorm(n2,mu2,sg), var.eq=T)$p.val)
mean(pv.ub <= .05)
[1] 0.4736     # aprx power

# Smaller balanced
n1 = 200;  n2 = 200
set.seed(707)
pv.sm = replicate(m, t.test(rnorm(n1,mu1,sg), rnorm(n2,mu2,sg), var.eq=T)$p.val)
mean(pv.bal <= .05)
[1] 0.5127      # aprx power

Note: Various statistical softwares have procedures for finding
power. Also, some online power calculators are reliable. 
The theory
is not difficult: Under the alternative that the discrepancy $|\mu_1-\mu_2| = \delta,$ the $T$-statistic has a non-central t distribution
with $\nu= n_1 + n_2 - 2$ degrees of freedom, and non-centrality
parameter $\lambda = \frac{\delta}{\sigma\sqrt{1/n_1 + 1/n_2}}.$
Thus the power of the two sided, pooled t test at the 5% level, in the unbalanced case $n_1=100, n_2=900,$ is computed in R as 47.4%.
nu = 998; t.c = qt(.975, nu)
del = 2; sg = 10; lam =  del/(sg*sqrt(1/100 + 1/900))
1 - diff(pt(c(-t.c, t.c), nu, lam))
[1] 0.4743743

