Combining two group means into one efficient estimate

Question

Two independent survey batches measure the same population mean $\mu$ with the same per-response variance $\sigma^2$ . Batch 1 has $n_1 = 30$ responses with mean $\bar X_1$ ; batch 2 has $n_2 = 10$ responses with mean $\bar X_2$ . You combine the two batch means as $w\bar X_1 + (1-w)\bar X_2$ .

Find the weight $w$ that minimizes the variance, and identify the resulting estimator.

Accepted Answer

Variances of the batch means. Each batch mean has variance inversely proportional to its size: \operatorname{Var}(\bar X1) = \frac{\sigma^2}{n1}, \qquad \operatorname{Var}(\bar X2) = \frac{\sigma^2}{n2}. Apply inverse-variance weighting. The optimal weight is proportional to the inverse variance, which here is proportional to the sample size: w \propto \frac{1}{\operatorname{Var}(\bar X1)} = \frac{n1}{\sigma^2} \propto n1. Normalizing, w^\star = \frac{n1}{n1 + n2} = \frac{30}{40} = 0.75, so batch 1 gets weight 0.75 and batch 2 gets 0.25. Recognize the result. The optimal blend is w^\star\bar …

Combining two group means into one efficient estimate

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