Table 1.

True positive rate (TPR), false positive rate (FPR) and false discovery rate (FDR) for the proposed method of assessing whether a chemical is active. A perfect classifier has a TPR of 1 and an FPR/FDR of 0. Shown is the mean (SD) across simulations

		J = 0	J = 5	J = 10	J = 15	J = 20
TPR	K = 1	0.72 (0.08)	0.66 (0.08)	0.64 (0.07)	0.58 (0.06)	0.52 (0.09)
	K = 3	0.98 (0.03)	0.97 (0.03)	0.97 (0.03)	0.95 (0.04)	0.93 (0.05)
	K = 5	1.00 (0.01)	1.00 (0.01)	1.00 (0.01)	0.99 (0.01)	0.98 (0.02)
FPR	K = 1	0.00 (0.00)	0.00 (0.00)	0.00 (0.01)	0.00 (0.00)	0.00 (0.00)
	K = 3	0.00 (0.01)	0.00 (0.01)	0.00 (0.01)	0.00 (0.01)	0.00 (0.01)
	K = 5	0.00 (0.01)	0.00 (0.01)	0.00 (0.01)	0.00 (0.01)	0.00 (0.01)
FDR	K = 1	0.00 (0.01)	0.00 (0.01)	0.00 (0.01)	0.00 (0.00)	0.00 (0.01)
	K = 3	0.00 (0.01)	0.00 (0.01)	0.00 (0.01)	0.00 (0.01)	0.00 (0.01)
	K = 5	0.00 (0.01)	0.00 (0.01)	0.00 (0.01)	0.00 (0.01)	0.00 (0.01)