7.
Product and Process Comparisons
7.4. Comparisons based on data from more than two processes 7.4.3. 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.  
Ratio of MST and MSE  When the null hypothesis of equal means is true, the two mean squares estimate the same quantity (error variance), and should be of approximately equal magnitude. In other words, their ratio should be close to 1. If the null hypothesis is false, MST should be larger than MSE.  
Divide sum of squares by degrees of freedom to obtain mean squares 
The mean squares are formed by dividing the sum of squares by the
associated degrees of freedom.
Let N = n_{i}. Then, the degrees of freedom for treatment, DFT = k  1, and the degrees of freedom for error, DFE = N_{ } k. The corresponding mean squares are:
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 tabular value of the F distribution, based on the chosen level and the degrees of freedom DFT and DFE. The calculations are displayed in an ANOVA table, as follows: 

ANOVA table 
The word "source" stands for source of variation. Some authors prefer to use "between" and "within" instead of "treatments" and "error", respectively. 

ANOVA Table Example  
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 


Interpretation of the ANOVA table  The test statistic is the F value of 9.59. Using an of .05, we have that F_{.05; 2, 12} = 3.89 (see the F distribution table in Chapter 1). 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.  
Techniques for further analysis 
The populations here are resistor readings while operating under the
three different temperatures. What we do not know at this
point is whether the three means are all different or which of the
three means is different from the other two, and by how much.
There are several techniques we might use to further analyze the differences. These are: 