7.
Product and Process Comparisons
7.4. Comparisons based on data from more than two processes 7.4.2. Are the means equal?


Sums of Squares help us compute the variance estimates displayed in ANOVA Tables  The sums of squares SST
and SSE previously computed for the oneway ANOVA are used
to form two mean squares, one for treatments and the second for
error. These mean squares are denoted by MST and MSE
respectively. These are typically displayed in a tabular form, known as
an ANOVA Table. The ANOVA table also shows the statistics used to
test hypotheses about the population means.
When the null hypothesis of equal means is true, the two mean sum of squares estimate the same quantity (error variance), and should be about of equal magnitude. In other words, their ratio should be close to 1. If the null hypothesis is false, MST should be larger than MSE. The mean squares are formed by dividing the sum of squares by the associated degrees of freedom. Let N = Sn_{i}. Then,
the degrees of freedom for treatment, DFT = k  1 and the degrees
of freedom for error, DFE = N_{ } k
MSE = SSE / DFE 

The Ftest 
The test statistic, used in testing the equality of treatment means is: F = MST / MSE. The critical value is the table value of the F distribution, based on the chosen a level and the degrees of freedom DFT and DFE. The calculations are displayed in an ANOVA table, as follows:
The word "source" stands for source of variation. Some authors prefer to use "among" and "within" instead of "treatments" and "error" respectively. 

A numerical example 
The data below resulted from measuring the difference in resistance resulting from subjecting identical resistors to three different temperatures for a period of 24 hours. The sample size of each group was 5. In the language of Design of Experiments we have an experiment in which each of three treatments was replicated 5 times.
The resulting ANOVA table is 

Example ANOVA table 
The test statistic is the F value of 9.59. Using an a of .05, we have that F_{.05; 2, 12} = 3.89. Since the test statistic is much larger than the critical value, we reject the null hypothesis of equal population means and conclude that there is a (statistically) significant difference among the population means. The pvalue for 9.59 is .00325, so the test statistic is significant at that level. The populations here are resistor readings while operating under the three different temperatures. What we do not know at this point is which of the three means are different from the others, and by how much. There are several techniques we might use to further analyze the differences. These are:
