Population contains two independent parts: Group A & Group B with size $N_A$ nad $N_B$.

$N = N_A + N_B$

Now sample from Group A and Group B separately. Sample size $n_A$, $n_B$.

$n = n_A + n_B$

To Estimate Group A mean and Group B mean, it's simply the sample mean in sample A and sample B. But to estimate the population (A&B) mean: I come up with two ways:

  1. combine sample A and sample B together, calculates its mean $x$, which is $(x_A n_A + x_B n_B)/(n_A + n_B)$
  2. calculates sample mean respectively, $x_A, x_B$, then estimate for population mean is $(x_A N_A + x_B N_B)/(N_A + N_B)$

Obvious, one of them has to be wrong (maybe both?). But I don't know why. I also have the problem when it comes to estimation of population variance.

  • 2
    $\begingroup$ if you use $x_i$ and $y_i$ to stand for your samples and $\bar{X}$, $\bar{Y}$ to stand for means then your question might be earsier to be understood, I think. $\endgroup$
    – Deep North
    Jun 17, 2015 at 6:45

1 Answer 1


Consider $N_A = N_B$, $n_A \gg n_B$. Then option 1's estimate is dominated by the estimate from the sample of $A$, whereas option 2's is equally weighted.

If you think $A$ and $B$ have the same distribution but are sampled separately for some other reason, then option 1 will be better; in this example, option 2 is giving equal weight to a good estimate (because $n_A$ is large) and a bad estimate (because $n_B$ is small).

If you think $A$ and $B$ are distinct subpopulations with possibly different mean values, then option 2 is better; it takes your knowledge about group membership and sizes into account, which option 1 ignores.


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