Table 4.

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

Scenario	GBLUP	BayesB	GBLUP-AD	BayesB-AD	GBLUP-ADAA
Base	$3.21\times {10}^{-2}$	$3.98\times {10}^{-2}$	$4.52\times {10}^{-3}$	$1.92\times {10}^{-4}$	$1.67\times {10}^{-3}$
$S_{10}$	$2.05\times {10}^{-8}$	$6.96\times {10}^{-6}$	$7.45\times {10}^{-9}$	$9.36\times {10}^{-6}$	$4.48\times {10}^{-9}$
$S_{100}$	$4.63\times {10}^{-4}$	$3.64\times {10}^{-3}$	$2.18\times {10}^{-4}$	$2.79\times {10}^{-5}$	$1.52\times {10}^{-4}$
$S_{1000}$	$5.00\times {10}^{-1}$	$3.48\times {10}^{-3}$	$1.33\times {10}^{-1}$	$1.29\times {10}^{-3}$	$9.67\times {10}^{-2}$
$D_{\text{medium}}$	$5.75\times {10}^{-3}$	$1.47\times {10}^{-1}$	$5.00\times {10}^{-1}$	$1.19\times {10}^{-1}$	$2.10\times {10}^{-1}$
$D_{\text{extreme}}$	$2.09\times {10}^{-2}$	$1.72\times {10}^{-1}$	$-1.85\times {10}^{-6}$	$-7.57\times {10}^{-6}$	$-5.45\times {10}^{-6}$
$E_A$	$5.75\times {10}^{-3}$	$1.95\times {10}^{-1}$	$1.50\times {10}^{-3}$	$2.39\times {10}^{-4}$	$1.14\times {10}^{-2}$
$E_C$	$8.92\times {10}^{-2}$	$9.61\times {10}^{-3}$	$1.98\times {10}^{-1}$	$1.04\times {10}^{-1}$	$1.60\times {10}^{-1}$
$E_I$	$7.92\times {10}^{-3}$	$1.67\times {10}^{-3}$	$4.13\times {10}^{-4}$	$8.27\times {10}^{-5}$	$5.27\times {10}^{-4}$

Each row corresponds to a scenario, as summarized in Table 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