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This was asked to me by a colleague. It seems to be an easy one but I wasn't able to find the derivation online. I've added my own attempt below.

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We have $H$ strata, each with sampled mean $\bar Y_{h}$, taken from population of size $N_h$ and error $\epsilon_h$. $\bar Y, N, \epsilon$ are the corresponding quantities for the whole sample.

\begin{align} \bar Y &= \sum_{h=1}^H {N_h \over N}\bar Y_h\\ \Rightarrow V(\bar Y)&= \sum_{h=1}^H {N_h^2 \over N^2} V(\bar Y_h)\\ \end{align}

The error for any confidence interval $\%$ is proportional to the standard error, $\epsilon = k\sqrt{V(\bar Y)}$. Also, we said all $\epsilon_h$ are the same. It is reasonable that the confidence level is also fixed:

\begin{align} V(\bar Y) &= \sum_{h=1}^H {N_h^2 \over N^2} {\epsilon_h^2 \over k_h^2}\\ V(\bar Y) &= {\epsilon_H^2 \over k_H^2N^2} \sum_{h=1}^H N_h^2\\ \Rightarrow \epsilon &= k{\epsilon_H \over k_H N} \sqrt{\sum_{h=1}^H N_h^2} \end{align}

If we again assume the same confidence level is desired for the global mean, we end up with

$$\epsilon = {\epsilon_H \over N} \sqrt{\sum_{h=1}^H N_h^2}$$

Since the sum of squares is equal or less than the square of the sum, the error for the global mean $\bar Y$ is guaranteed not to exceed the error at each stratum, and it diminishes with the number of strata used.

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