Formulated Question
In a two-sample bootstrap procedure for testing the difference in means, why is it insufficient to subtract the group-specific mean from each bootstrapped observation (centering each group around zero)? This approach makes the common mean equal to zero for both groups. Instead, why does one need to adjust the bootstrap samples using the pooled overall average? Mathematically:
Let $$ X = \{x_1, x_2, \dots, x_n\} $$ and $$ Y = \{y_1, y_2, \dots, y_m\} $$ be two independent samples, with:
- Sample means: $$ \bar{x} = \frac{1}{n} \sum_{i=1}^n x_i, \quad \bar{y} = \frac{1}{m} \sum_{j=1}^m y_j $$
- Observed difference in means: $$ \Delta_{\text{obs}} = \bar{x} - \bar{y} $$
The centered bootstrap samples are obtained by subtracting the respective group means:
- For group ( X ): $$ x_i^* = x_i - \bar{x}, \quad \text{for } i = 1, 2, \dots, n $$
- For group ( Y ): $$ y_j^* = y_j - \bar{y}, \quad \text{for } j = 1, 2, \dots, m $$ After centering, the means of both groups are zero: $$ \bar{x}^* = 0, \quad \bar{y}^* = 0 $$ This ensures that the bootstrap samples simulate a scenario where the group-specific means are equal.
However, some approaches suggest that instead of centering the data around the group-specific means $$( \bar{x}, \bar{y} )$$, the bootstrap samples should be adjusted by adding the pooled mean: $$ \bar{z} = \frac{n \cdot \bar{x} + m \cdot \bar{y}}{n + m} $$ The adjusted bootstrap samples become:
- For group ( X ): $$ x_i^* = x_i - \bar{x} + \bar{z} $$
- For group ( Y ): $$ y_j^* = y_j - \bar{y} + \bar{z} $$
Question: Why is this additional adjustment to the pooled mean $$( \bar{z} )$$ necessary? If the centered bootstrap samples already have group-specific means of zero, doesn’t this ensure that the null hypothesis of equal means is satisfied? Why is it not sufficient to center around zero?
