Table 1

Statistical table.

FigureData structureType of testStatistical valuep value
2A von Frey pooledTwo factors
(group and time point)
Two-way ANOVAF(2,44) = 41.0<0.001
2B withdrawal score pooledOne factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,44) = 123.3<0.001
2C von Frey treatmentOne factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(3,34) = 37.95<0.001
2D withdrawal scoreOne factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(3,34) = 52.9<0.001
3A Shannon baselineOne factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,15) = 2.7450.096
3A Shannon post-treatmentOne factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,15) = 87.15<0.001
3A Chao1 baselineOne factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,15) = 1.520.251
3A Chao1 posttreatmentOne factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,15) = 6.873<0.008
Bacteroidetes post-treatmentOne factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,15) = 8,844<0.003
Firmicutes post-treatmentOne factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,15) = 27.43<0.001
3F caecum weightOne factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(4,42) = 52.74<0.001
4 (BLA)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,26) = 22.3<0.001
4 (PAG)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,26) = 16.9<0.001
4 (LC)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,26) = 25.2<0.001
4 (CeA)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,26) = 73.97<0.001
4 (PVT)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,26) = 23<0.001
4 (AI)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,26) = 5.92<0.001
4 (BNST)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,26) = 55.7<0.001
4 (LHb)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,26) = 37.3<0.001
5 (BLA)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,24) = 18.8<0.001
5 (PAG)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,24) = 32.6<0.001
5 (LC)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,24) = 6.79<0.005
5 (CeA)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,24) = 5.13<0.014
5 (PVT)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,24) = 1.18<0.325
5 (AI)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,24) = 1.18<0.324
5 (BNST)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,24) = 2.24<0.129
5 (LHb)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,24) = 2.71<0.87
6 (BLA)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,26) = 2.45<0.106
6 (PAG)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,26) = 17.6<0.001
6 (LC)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,26) = 17.7<0.001
6 (CeA)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,26) = 17.9<0.001
6 (PVT)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,26) = 24.1<0.001
6 (AI)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,26) = 3.87<0.034
6 (BNST)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,26) = 21.6<0.001
6 (LHb)One factor (treatment)One-way ANOVA
Student–Newman–Keuls post hoc test
F(2,26) = 38.1<0.001