Using the ANOVA table listed below, what should our conclusion be? ANOVA Source of Vartation Between Groups Within Groups Total S.S 3.631344 1.103017 4.734361 MS 2 1.815672 15 0.073534 17 F P-value 24.69145 0.0000179548 3.682320344 F crit our means are not significantly different from one another. We Since Ferit = 3.68, we fail to reject the null hypothesi must now complete Tukey's HSD to determine if there are mean differences between specific group pairs. Since F= 24.69, we reject the null hypothesis; our means are not significantly different from one another so we do not need to complete Tukey's HSD. Since p < .05, we reject the null hypothesis; our means are significantly different from one another. We must now complete Tukey's HSD to determine which means differ significantly from one another. Since p>.05, we fail to reject the null hypothesis; our means are significantly different from one another. We must now complete Tukey's HSD to determine which means differ significantly from one another.