Table 3.

Power calculations of the Cumulative Geographic Residual Test for adjusted and unadjusted area-level and individual-level covariate analyses when spatial clustering exists independent of covariate and outcome relationship (c=1).

		Area-level Spatial Cluster				Individual-level Spatial Cluster
Type of Adjustment:		None	Area-Level			None	Individual-Level	None	Area-Level
	β	Pow	Pow	Sens	Accuracy	Pow	Pow	Pow	Pow
Independence γ = 0	−2	0.191	0.698	0.884	0.466	0.944	1.000	0.941	1.000
	−1	0.369	0.705	0.899	0.457	1.000	1.000	1.000	1.000
	0	0.710	0.701	0.906	0.459	1.000	1.000	1.000	1.000
	1	0.403	0.646	0.912	0.466	0.999	1.000	0.998	1.000
	2	0.173	0.663	0.908	0.461	0.936	1.000	0.943	1.000
Moderate Dependence γ = 0.5	−2	0.064	0.668	0.893	0.463	0.148	1.000	0.146	1.000
	−1	0.185	0.652	0.891	0.470	0.963	1.000	0.957	1.000
	0	0.687	0.688	0.890	0.460	1.000	1.000	1.000	1.000
	1	0.649	0.650	0.893	0.464	1.000	1.000	1.000	1.000
	2	0.433	0.673	0.888	0.462	1.000	1.000	1.000	1.000
Strong Dependence γ = 1.0	−2	0.042	0.578	0.858	0.468	0.053	1.000	0.055	0.898
	−1	0.074	0.610	0.879	0.465	0.318	1.000	0.325	0.887
	0	0.703	0.603	0.871	0.473	1.000	1.000	1.000	0.894
	1	0.863	0.618	0.877	0.458	1.000	1.000	1.000	0.909
	2	0.741	0.591	0.875	0.454	1.000	1.000	1.000	0.882

Model Framework:

Area-level Spatial Cluster: $X_i\mid Z_i^{\ast }\sim N(\gamma Z_i^{\ast },1)$, Y_i|X_i, $Z_i^{\ast }\sim N(c\sqrt{2}{Z}_i^{\ast }+\beta X_i,1)$, and w_i = 1 (i = 1,…,100)

Individual-level Spatial Cluster: $X_{ij}\mid Z_i^{\ast }\sim N(\gamma Z_i^{\ast },1)$, Y_ij|X_ij, $Z_i^{\ast }\sim N(c\sqrt{2}{Z}_i^{\ast }+\beta X_{ij},1)$ (i = 1, …, 100) (j = 1, …, 10)

Note: Area-level adjustment uses outcome ${\overline{Y}}_i={\sum }_{j=1}^{10}{Y}_{ij}/10$, area-level covariate ${\overline{X}}_i={\sum }_{j=1}^{10}{X}_{ij}/10$, and w_i = 10.

$\text{Pow}=\frac{1}{1000}{\displaystyle \sum _{j=1}^{1000}}I(P_j<0.05)$

$\text{Sens}=\frac{1}{1000}{\displaystyle \sum _{j=1}^{1000}}I(P_j<0.05)I(\{{\mathit{CritReg}}_j\cap Z^{\ast }\}\ne \varnothing )$

$\text{accuracy}=\frac{1}{1000}{\displaystyle \sum _{j=1}^{1000}}I(P_j<0.05)\frac{1}{\mid {\mathit{CritReg}}_j\mid }{\sum }_{k=1}^{\mid {\mathit{CritReg}}_j\mid }I(\{{\mathit{CritReg}}_{jk}\in Z^{\ast }\}\ne \varnothing )$