Comparing a subgroup mean with its containing sample mean (${\overline{X}}_A-\overline{X}=(1-w)({\overline{X}}_A-{\overline{X}}_B)$)

Partition a sample into two disjoint groups $A,B$ with fixed positive sizes $n_A,n_B$, and let $w=n_A/(n_A+n_B)$. The overall sample mean is the weighted meanSubtracting gives the displayed identity. The subgroup and overall sample means share observations, so they are not independent. If the two disjoint groups are sampled independently, thenThus treating the subgroup and the combined sample as two independent samples uses the wrong standard error. Equality of the subgroup and overall population means is equivalent to equality of subgroup and complement means when both groups have positive weight, but the sampling distribution must respect the shared observations.