The right divisor when pooling variance across two groups

Question

Two production lines are believed to share the same noise variance $\sigma^2$ but may have different means. Line A gives $n_1$ measurements with mean $\bar X$ , line B gives $n_2$ measurements with mean $\bar Y$ . You want to pool both samples into a single unbiased estimate of $\sigma^2$ .

What should you divide the combined sum of squared deviations $\sum (X_i - \bar X)^2 + \sum (Y_j - \bar Y)^2$ by?

Accepted Answer

Each group loses one degree of freedom. Within line A you center on its own sample mean \bar X, so \mathbb{E}\!\left[\sumi (Xi - \bar X)^2\right] = (n1 - 1)\sigma^2, and likewise line B contributes (n2 - 1)\sigma^2. Two means were estimated, one per group, so two degrees of freedom are spent. Add and divide. The combined expectation is \mathbb{E}\!\left[\sumi (Xi - \bar X)^2 + \sumj (Yj - \bar Y)^2\right] = (n1 - 1)\sigma^2 + (n2 - 1)\sigma^2 = (n1 + n2 - 2)\sigma^2. So the unbiased pooled estimator is Sp^2 = \frac{\sumi (Xi - \bar X)^2 + \sumj (Yj - \bar Y)^2}{n1 + n2 - 2}, \qquad \mathbb{E}…

The right divisor when pooling variance across two groups

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