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 '15 at 6:45

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.