A drug trial is conducted identically in two different groups that are essentially similar at baseline. Can I obtain a joint mean and a joint SE?

For example, Group A has placebo group of n=224 mean of 37.6 SE of 3.3 Group B has n=236 mean of 30.9 SE of 3.0 What is their combined mean and SE and how do you do it?


1 Answer 1


Yes you can. Essentially you have two sample sizes, means, and standard deviations. This is enough information to compute the pooled mean and variance.

The mean will be the weighted average of the individual means. Call the combined group $X$ and let $N=N_A+N_B$. Essentially you need the sum of observations which can you write in terms of $\mu_A$ and $\mu_B$. $$ \mu_X = \frac{1}{N}\sum_i x_i = \frac{1}{N_A+N_B}(N_A\mu_A+N_B\mu_B) $$ To compute the sample variance you do the exact same procedure, this time you need the sum of squared observations which you can write in terms of $s_A^2$ and $s_B^2$.

For $A$ (and similarly for $B$), $$ \sum_i a_i^2 = (N_A-1)s_A^2+N_A\mu_A^2. $$ Then we are ready to compute: $$ \begin{align*} s_X^2&=\frac{1}{N - 1}\left(\sum_i x_i^2 - N\mu_X^2\right)\\ &=\frac{1}{N - 1}\left((N_A - 1)s_A^2 + N_A\mu_A^2 + (N_B - 1)s_B^2 + N_B\mu_B^2 - N\mu_X^2\right) \end{align*} $$

  • $\begingroup$ +1. A similar answer to a generalization of this question (to any number of subgroups and arbitrary invertible functions of any moments) is given at stats.stackexchange.com/questions/51622: the method continues to be compute the sums (of relevant powers) and then reduce those to moments. $\endgroup$
    – whuber
    Commented Apr 7, 2017 at 21:15

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