9.4 Testing homogeneity
Proposition 9.4.1. Under homogeneity, with the \(\hat \beta _k\) approximately normal and the \(v_k\) treated as known, \[ Q = \sum _k w_k(\hat \beta _k - \hat \beta _W)^2 \] has approximately the \(\chi ^2_{K - 1}\) distribution.
Proof. Let \(z_k = \sqrt {w_k}(\hat \beta _k - \beta )\), independent standard normal, and \(u\) the unit vector with components \(\sqrt {w_k}/\sqrt {\sum _l w_l}\). Then \(\sqrt {w_k}(\hat \beta _k - \hat \beta _W) = z_k - u_k\,(u^\top z)\), since \(u^\top z = \sum _l w_l(\hat \beta _l - \beta )/\sqrt {\sum _l w_l} = \sqrt {\sum _l w_l}\,(\hat \beta _W - \beta )\). So \(Q = \|(I - uu^\top )z\|^2\), the squared length of the projection of a standard normal vector onto the \((K - 1)\)-dimensional subspace orthogonal to \(u\), which is \(\chi ^2_{K - 1}\). □
The test has low power when the strata are small, so a non-significant result is weak evidence of homogeneity. Its value is in flagging large, clear differences, which should then be reported as effect modification.
Example 9.4.2. In Example 9.1.1 the weights \(1/v_k\) are \(6.11\), \(12.86\) and \(10.63\), and the log odds ratios \(0.563\), \(0.693\) and \(0.901\). Then \[ \hat \beta _W = \frac {6.11 \times 0.563 + 12.86 \times 0.693 + 10.63 \times 0.901}{29.60} = 0.741 , \] so the pooled odds ratio is \(e^{0.741} = 2.10\), with \(\text {SE} = 1/\sqrt {29.60} = 0.184\) and \(95\%\) interval \((1.46,\ 3.01)\). The homogeneity statistic is \begin {align*} Q &= 6.11(0.563 - 0.741)^2 + 12.86(0.693 - 0.741)^2 \\ &\quad + 10.63(0.901 - 0.741)^2 = 0.49 \end {align*}
on \(2\) degrees of freedom, \(P = 0.78\). Nothing suggests that the odds ratio varies with age, and the age-adjusted estimate \(2.10\) replaces the crude \(3.37\). The test of no association, \(z = 0.741/0.184 = 4.03\), is highly significant.
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