Ho: mu(1) = mu(2) = mu(3) = .... = mu(k)

Ha: at least one of the means mu(k) is different from the rest

An F statistic is computed which compares the ratio of the variance between the groups to the variance within the groups.

If this ratio is "large enough" (i.e., significantly larger than 1.0), then we have evidence to support rejection of Ho.

==========================================

Refer to the attached image for the data set and the ANOVA Table. This is a 1-way ANOVA set up to compare the means of 3 groups of 10 data points each. Our alpha level for the test is set at .05.

F = MSb/MSw --> mean-square between / mean-square within

MSb = SSb/df(b) --> sum of squares between / dof between

MSw = SSw/df(w) --> sum of squares within / dof within

df(b) = #groups - 1 = 2

df(w) = (#groups * sample size) - #groups = (3*10)-3 = 27

SSb = summation of [ n * (group mean - grand mean)^2 ]

= 10*(20.89-19.76)^2 + 10*(20.45-19.76)^2 + 10*(17.94-19.76)^2

= 50.732

SSw = summation of [ (n-1) * group variance ]

= (9*11.03) + (9*13.446) + (9*2.371)

= 241.626

then:

MSb = 50.732/2 = 25.366

MSw = 241.626/27 = 8.949

then:

F = 25.366/8.949 = 2.834

F(crit) = 3.354

Therefore we do not have sufficient evidence to reject Ho. If we look at the p-value for the computed F ratio, we see a probability of 0.076.

In other words, if Ho is in fact true, then there is a 7.6% chance of getting the F ratio that we computed, or larger. However, it is not quite "large enough" or "unlikely" enough to reject Ho.