ANOVA F-test results in p | anova p value
Isupposeyoumusthave$r=15$replicationsineachofyour$g=4$groups.Accordingtothenullhypothesis(allfourgrouppopulationmeansequal),theF-statistichasdistribution$mathsf{F}(3,56).$Thecriticalvaluesfortestingatthe5%and1%levelsare2.796and4.152,respectively,ascomputedinRstatisticalsoftware.(Theideaistocut5%or1%oftheareafromtheuppertailofthedistribution$mathsf{F}(3,56).$InR,qfistheinverseCDForquantilefunctionofF;maybeyoucanfigureouthowtogetsimilarresultsfromExcel.)SowithobservedFof62.911,youarefaraboveth...
I suppose you must have $r = 15$ replications in each of your $g = 4$ groups. According to the null hypothesis (all four group population means equal), the F-statistic has distribution $mathsf{F}(3, 56).$ The critical values for testing at the 5% and 1% levels are 2.796 and 4.152, respectively, as computed in R statistical software. (The idea is to cut 5% or 1% of the area from the upper tail of the distribution $mathsf{F}(3, 56).$ In R, qf is the inverse CDF or quantile function of F; maybe you can figure out how to get similar results from Excel.) So with observed F of 62.911, you are far above the critical value for any reasonable significance level.
qf(.95, 3, 56); qf(.99, 3, 56) ## 2.769431 # 5% crit val ## 4.151941 # 1%If the observed $F$ is 62.911, then the P-value is the probability $P(F > 62.911) $ $= 1 - P(F le 62.911) approx 0,$ where $F sim mathsf{F}(3,56).$ The P-value is the probability of a more extreme result than what you observed; here,...