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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

ScenarioGBLUPBayesBGBLUP-ADBayesB-ADGBLUP-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 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