\[ T = \frac{\Delta}{\sqrt{Var(\Delta)}} \]

Let’s expand on the previous relationship:

\[ T = \frac{ \bar{X}_t - \bar{X_c}}{\sqrt{Var(\bar{X}_t) + Var(\bar{X}_c) + 2COV(\bar{X}_t, \bar{X}_c)}} \] Since the two samples are independent, the covariance of the means is zero. The variance of the mean is:

\[ Var(X_t) = Var(\frac{X_1+X_2+X_{n_t}}{n_t}) = \frac{1}{n_t^2} \bigg[ Var(X_1) + Var(X_2) + ... \bigg] = \frac{n_t}{n_t^2} s_t^2 = \frac{s_t^2}{n_t} \]

where we assumed that all samples are independent and come from the same distribution, with empirical variance \(s_t^2\). Similarly, we can obtain the same result for the control group. Hence, our t-test can be written as:

\[ T = \frac{ \bar{X}_t - \bar{X_c}}{\sqrt{\frac{s_t^2}{n_t}+ \frac{s_c^2}{n_c}}} \]