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


The balanced
2way factorial layout
How to obtain sums of squares for the balanced factorial layout
2way ANOVA Example 
Basic Layout
Factor A has 1, 2, ..., a levels. Factor B has 1, 2, ..., b levels. There are ab treatment combinations (or cells) in a complete factorial layout. Assume that each treatment cell has r independent obsevations (known as replications). When each cell has the same number of replications, the design is balanced factorial. In this case, the abr data points {y_{ijk}}can be shown pictorially as follows:
Next, we will calculate the sums of squares needed for the ANOVA table. Let A_{i }be the sum of all observations of level i of factor A, i = 1, ... ,a. The A_{i} are the row sums. Let B_{j} be the sum of all observations of level j of factor B, j = 1, ...,b. The B_{j} are the column sums. Let (AB)_{ij }be the sum of all observations of level i of A and level j of B. These are cell sums. Let r be the number of replicates in the experiment, that is: the number of times each factorial treatment combination appears in the experiment. Then the total number of observations for each level of factor A is rb and the total number of observations for each level of factor B is ra and the total number of observations for each interaction is r. Finally, the total number of observations n in the experiment is abr. With the help of these expressions we arrive (omitting derivations) at These expressions are used to calculate the ANOVA table entries for the (fixed effects) 2way ANOVA. Example: An evaluation of a new coating applied to 3 different materials was conducted at 2 different laboratories. Each laboratory tested 3 samples from each of the treated materials. The results are given in the next table: The preliminary part of the analysis yields a table of row and column
sums
From this table we generate the ANOVA table. 