Table 1.

Statistical analyses for the data in Figures 15

Data structureType of testStatisticp-value
Figure 1ANormal distributionThree-way RM ANOVA
 TimeF(3.347, 147.3) = 283.8p < 0.001
 (Female vs male)F(1, 44) = 2.288p = 0.14
 (WT vs BCL)F(1, 44) = 17.15p < 0.001
 Time × (female vs male)F(5, 220) = 1.081p = 0.37
 Time × (WT vs BCL)F(5, 220) = 3.895p = 0.002
 (Female vs male) × (WT vs BCL)F(1, 44) = 2.454p = 0.12
 Time × (female vs male) × (WT vs BCL)F(5, 220) = 1.172p = 0.32
Figure 1BNormal distributionThree-way RM ANOVA
 TimeF(4.139, 182.1) = 4.579p = 0.001
 (Female vs male)F(1, 44) = 0.08293p = 0.77
 (WT vs BCL)F(1, 44) = 1.423p = 0.24
 Time × (female vs male)F(5, 220) = 0.8589p = 0.51
 Time × (WT vs BCL)F(5, 220) = 0.6759p = 0.64
 (Female vs male) × (WT vs BCL)F(1, 44) = 0.1692p = 0.68
 Time × (female vs male) × (WT vs BCL)F(5, 220) = 0.1891p = 0.97
Figure 1CNormal distributionThree-way RM ANOVA
 TimeF(5, 220) = 4.380p < 0.001
 (Female vs male)F(1, 44) = 0.01810p = 0.89
 (WT vs BCL)F(1, 44) = 0.05027p = 0.82
 Time × (female vs male)F(5, 220) = 1.053p = 0.39
 Time × (WT vs BCL)F(5, 220) = 0.1713p = 0.97
 (Female vs male) × (WT vs BCL)F(1, 44) = 2.059p = 0.16
 Time × (female vs male) × (WT vs BCL)F(5, 220) = 1.059p = 0.38
Figure 1D,ENormal distributionThree-way RM ANOVA
 TimeF(4, 176) = 55.66p < 0.001
 (Female vs male)F(1, 44) = 0.1896p = 0.67
 (WT vs BCL)F(1, 44) = 37.92p < 0.001
 Time × (female vs male)F(4, 176) = 1.207p = 0.31
 Time × (WT vs BCL)F(4, 176) = 0.2075p = 0.93
 (Female vs male) × (WT vs BCL)F(1, 44) = 8.123p = 0.007
 Time × (female vs male) × (WT vs BCL)F(4, 176) = 0.2403p = 0.92
Figure 1FNormal distributionOne-way ANOVAF(3, 76) = 3.746p = 0.01
Figure 2ANormal distributionThree-way RM ANOVA
 SexF(1, 90) = 18.70p < 0.001
 (WT vs BCL2)F(1, 90) = 1.734p = 0.191
 (Sham vs SNI)F(1, 90) = 0.07338p = 0.787
 Sex × (WT vs BCL2)F(1, 90) = 7.799p = 0.006
 Sex × (sham vs SNI)F(1, 90) = 0.6937p = 0.407
 (WT vs BCL2) × (sham vs SNI)F(1, 90) = 0.2768p = 0.600
 Sex × (WT vs BCL2) × (sham vs SNI)F(1, 90) = 0.01009p = 0.920
Figure 2BNormal distributionThree-way RM ANOVA
 SexF(1, 96) = 1.457p = 0.230
 (WT vs BCL2)F(1, 96) = 3.684p = 0.058
 (Sham vs SNI)F(1, 96) = 135.6p < 0.001
 Sex × (WT vs BCL2)F(1, 96) = 5.630p = 0.020
 Sex × (sham vs SNI)F(1, 96) = 0.03694p = 0.848
 (WT vs BCL2) × (sham vs SNI)F(1, 96) = 1.294p = 0.258
 Sex × (WT vs BCL2) × (sham vs SNI)F(1, 96) = 0.4683p = 0.495
Figure 3ANormal distributionThree-way ANOVA
 SexF(1, 79) = 3.007p = 0.087
 (WT vs BCL2)F(1, 79) = 8.180p = 0.005
 (Sham vs SNI)F(1, 79) = 23.96p < 0.001
 Sex × (WT vs BCL2)F(1, 79) = 1.719p = 0.194
 Sex × (sham vs SNI)F(1, 79) = 0.04524p = 0.832
 (WT vs BCL2) × (sham vs SNI)F(1, 79) = 0.1233p = 0.726
 Sex × (WT vs BCL2) × (sham vs SNI)F(1, 79) = 2.325p = 0.131
Figure 3BNormal distributionThree-way ANOVA
 SexF(1, 79) = 206.0p < 0.001
 (WT vs BCL2)F(1, 79) = 33.22p < 0.001
 (Sham vs SNI)F(1, 79) = 2.709p = 0.104
 Sex × (WT vs BCL2)F(1, 79) = 12.23p < 0.001
 Sex × (sham vs SNI)F(1, 79) = 6.929p = 0.010
 (WT vs BCL2) × (sham vs SNI)F(1, 79) = 2.733p = 0.102
 Sex × (WT vs BCL2) × (sham vs SNI)F(1, 79) = 0.2839p = 0.596
Figure 4ANormal distributionThree-way ANOVA
 SexF(1, 81) = 4.726p = 0.033
 (WT vs BCL2)F(1, 81) = 0.4593p = 0.500
 (Sham vs SNI)F(1, 81) = 11.41p = 0.001
 Sex × (WT vs BCL2)F(1, 81) = 0.9078p = 0.344
 Sex × (sham vs SNI)F(1, 81) = 0.6091p = 0.437
 (WT vs BCL2) × (sham vs SNI)F(1, 81) = 6.308p = 0.014
 Sex × (WT vs BCL2) × (sham vs SNI)F(1, 81) = 0.1998p = 0.656
Figure 4BNormal distributionThree-way ANOVA
 SexF(1, 81) = 15.31p < 0.001
 (WT vs BCL2)F(1, 81) = 0.4593p = 0.500
 (Sham vs SNI)F(1, 81) = 0.4492p = 0.505
 Sex × (WT vs BCL2)F(1, 81) = 0.6998p = 0.405
 Sex × (sham vs SNI)F(1, 81) = 1.336p = 0.251
 (WT vs BCL2) × (sham vs SNI)F(1, 81) = 0.5748p = 0.451
 Sex × (WT vs BCL2) × (sham vs SNI)F(1, 81) = 0.03412p = 0.854
Figure 5Normal distributionThree-way ANOVA
 SexF(1, 35) = 4.067p = 0.051
 (WT vs BCL2)F(1, 35) = 0.06186p = 0.805
 (Sham vs SNI)F(1, 35) = 0.3581p = 0.553
 Sex × (WT vs BCL2)F(1, 35) = 3.708p = 0.062
 Sex × (sham vs SNI)F(1, 35) = 3.467p = 0.071
 (WT vs BCL2) × (sham vs SNI)F(1, 35) = 0.1879p = 0.667
 Sex × (WT vs BCL2) × (sham vs SNI)F(1, 35) = 4.557p = 0.040