Table 3.

One-way ANOVA test on the classification performance of nine tractography algorithms for three diagnostic tasks when using the raw matrices as features.

Diagnostic task	Degrees of freedom	F	Sig.
AD vs. NC	Between groups	8	1.111	0.358
	Within groups	171
AD vs. MCI	Between groups	8	1.348	0.223
	Within groups	171
MCI vs. NC	Between groups	8	1.945	0.056
	Within groups	171

The “F” column presents computed F score and the “Sig.” column gives the p-value. Results with Sig. value < 0.05 are treated as nominally significant, so no differences were detectable. “Between Groups” represents sum of the squared deviations from the mean between groups, which captures variability between each group. “Within Groups” represents sum of the squared deviations from the mean within groups, which captures variability within each group. We have nine tractography algorithms, so the number of degrees of freedom for the Between Groups comparison is 9-1 = 8. And since we have 20 splits for each algorithm, the number of degrees of freedom for the Within Groups comparison is 20x9-9 = 171. Since α = 0.05 and the number of degrees of freedom = (8,171), we accept H₀ if F_8,171 ≤ 1.9929. All our three F-values (1.111, 1.348, and 1.945) are all less than 1.9929, so we accept our H₀. In other words, there are no significant group differences in these nine tractography algorithm-derived networks using the raw matrices as features.