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. 2019 Sep 4;39(36):7049–7060. doi: 10.1523/JNEUROSCI.2499-18.2019

Table 1.

Statistical analysis of behavioral and biochemistry data of animals

Factor F value p value
A. First stage transition between trained groups of animals (one-way ANOVA, α* = 0.05); Fig. 2A
    AV session F(4,31) = 2.323 p = 0.0768
    AQ session F(2,21) = 1.832 p = 0.1847
    RT Session Unpaired Student's t test p = 0.2761
B. Quantification absolute values of BCN ECM-fraction (one-way ANOVA, α* = 0.05); Fig. 3C,D
    145 kDa BCN ACx F(5,41) = 3.342 p = 0.0127
    55 kDa BCN ACx F(5,41) = 3.434 p = 0.0111
    145 kDa BCN CA F(5,41) = 0.3497 p = 0.8795
    55 kDa BCN CA F(5,41) = 1.062 p = 0.3954
C. Quantification absolute values of TNR ECM-fraction (one-way ANOVA, α* = 0.05); Fig. 4C,D
    Full-length TNR ACx F(5,41) = 2.085 p = 0.0868
    Full-length TNR CA F(5,41) = 1.818 p = 0.1306
D. Quantification absolute values of BCN cellular fraction (one-way ANOVA, α* = 0.05); Fig. 3C,D
    145 kDa BCN ACx F(5,41) = 1.758 p = 0.1433
    55 kDa BCN ACx F(5,41) = 15.79 p < 0.0001
    145 kDa BCN CA F(5,41) = 1.19 p = 0.3309
    55 kDa BCN CA F(5,41) = 5.221 p = 0.0008
E. Quantification absolute values of TNR cellular fraction (one way ANOVA, α* = 0.05); Fig. 4C,D
    Full-length TNR ACx F(5,41) = 1.668 p = 0,1640
    Full-length TNR CA F(5,41) = 1.422 p = 0.2367
F. Quantification total levels of BCN (one-way ANOVA, α* = 0.05); Fig. 7A,B
    Sum of 55kDa BCN ACx F(5,41) = 14.6 p < 0.0001
    Sum of 145kDa BCN ACx F(5,41) = 5.567 p = 0.0005
    Sum of 55kDa BCN CA F(5,41) = 6.377 p = 0.0002
    Sum of 145kDa BCN CA F(5,41) = 1.632 p = 0.1733
G. Quantification total levels of TNR (one-way ANOVA, α* = 0.05); Fig. 7A,B
    Sum of full-length TNR ACx F(5,41) = 2.025 p = 0.0952
    Sum of full-length TNR CA F(5,41) = 3.773 p = 0.0067

All tests were based on significance level of α* = 0.05; for ANOVA post hoc analysis Tukey's tests of multiple comparisons was used. Corresponding figures for tests are indicated.