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

From: Bayesian neural networks with variable selection for prediction of genotypic values

Scenario

GBLUP

BayesB

GBLUP-AD

BayesB-AD

GBLUP-ADAA

Base

\(3.21\times 10^{-2}\)

\(3.98\times 10^{-2}\)

\(4.52\times 10^{-3}\)

\(1.92\times 10^{-4}\)

\(1.67\times 10^{-3}\)

\(S_{10}\)

\(2.05\times 10^{-8}\)

\(6.96\times 10^{-6}\)

\(7.45\times 10^{-9}\)

\(9.36\times 10^{-6}\)

\(4.48\times 10^{-9}\)

\(S_{100}\)

\(4.63\times 10^{-4}\)

\(3.64\times 10^{-3}\)

\(2.18\times 10^{-4}\)

\(2.79\times 10^{-5}\)

\(1.52\times 10^{-4}\)

\(S_{1000}\)

\(5.00\times 10^{-1}\)

\(3.48\times 10^{-3}\)

\(1.33\times 10^{-1}\)

\(1.29\times 10^{-3}\)

\(9.67\times 10^{-2}\)

\(D_{\text {medium}}\)

\(5.75\times 10^{-3}\)

\(1.47\times 10^{-1}\)

\(5.00\times 10^{-1}\)

\(1.19\times 10^{-1}\)

\(2.10\times 10^{-1}\)

\(D_{\text {extreme}}\)

\(2.09\times 10^{-2}\)

\(1.72\times 10^{-1}\)

\(-1.85\times 10^{-6}\)

\(-7.57\times 10^{-6}\)

\(-5.45\times 10^{-6}\)

\(E_A\)

\(5.75\times 10^{-3}\)

\(1.95\times 10^{-1}\)

\(1.50\times 10^{-3}\)

\(2.39\times 10^{-4}\)

\(1.14\times 10^{-2}\)

\(E_C\)

\(8.92\times 10^{-2}\)

\(9.61\times 10^{-3}\)

\(1.98\times 10^{-1}\)

\(1.04\times 10^{-1}\)

\(1.60\times 10^{-1}\)

\(E_I\)

\(7.92\times 10^{-3}\)

\(1.67\times 10^{-3}\)

\(4.13\times 10^{-4}\)

\(8.27\times 10^{-5}\)

\(5.27\times 10^{-4}\)

  1. 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