Table 4 p-values of the one-sided paired t-test for the hypotheses $$\mathbb {E}\left( \rho _{\text {NetSparse}}-\rho _{\text {Method}}\right) =0$$

Base$$3.21\times 10^{-2}$$$$3.98\times 10^{-2}$$$$4.52\times 10^{-3}$$$$1.92\times 10^{-4}$$$$1.67\times 10^{-3}$$
$$S_{10}$$$$2.05\times 10^{-8}$$$$6.96\times 10^{-6}$$$$7.45\times 10^{-9}$$$$9.36\times 10^{-6}$$$$4.48\times 10^{-9}$$
$$S_{100}$$$$4.63\times 10^{-4}$$$$3.64\times 10^{-3}$$$$2.18\times 10^{-4}$$$$2.79\times 10^{-5}$$$$1.52\times 10^{-4}$$
$$S_{1000}$$$$5.00\times 10^{-1}$$$$3.48\times 10^{-3}$$$$1.33\times 10^{-1}$$$$1.29\times 10^{-3}$$$$9.67\times 10^{-2}$$
$$D_{\text {medium}}$$$$5.75\times 10^{-3}$$$$1.47\times 10^{-1}$$$$5.00\times 10^{-1}$$$$1.19\times 10^{-1}$$$$2.10\times 10^{-1}$$
$$D_{\text {extreme}}$$$$2.09\times 10^{-2}$$$$1.72\times 10^{-1}$$$$-1.85\times 10^{-6}$$$$-7.57\times 10^{-6}$$$$-5.45\times 10^{-6}$$
$$E_A$$$$5.75\times 10^{-3}$$$$1.95\times 10^{-1}$$$$1.50\times 10^{-3}$$$$2.39\times 10^{-4}$$$$1.14\times 10^{-2}$$
$$E_C$$$$8.92\times 10^{-2}$$$$9.61\times 10^{-3}$$$$1.98\times 10^{-1}$$$$1.04\times 10^{-1}$$$$1.60\times 10^{-1}$$
$$E_I$$$$7.92\times 10^{-3}$$$$1.67\times 10^{-3}$$$$4.13\times 10^{-4}$$$$8.27\times 10^{-5}$$$$5.27\times 10^{-4}$$
1. Each row corresponds to a scenario, as summarized in TableĀ 1. The cells containing negative values had p close to 1 and therefore we chose to put $$-(1-p)$$ in those cell instead. The minus serves to clearly identify those cases while the $$(1-p)$$ represents the p-value for superiority of GBLUP-AD and BayesB-AD over NetSparse