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

## Models and calculations for the two-way ANOVA

The balanced 2-way factorial layout

How to obtain sums of squares for the balanced factorial layout

2-way 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 {yijk}can be shown pictorially as follows:

Next, we will calculate the sums of squares needed for the ANOVA table.

Let Ai  be the sum of all observations of level i of factor A, i = 1, ... ,a. The Ai are the row sums.

Let Bj be the sum of all observations of level j of factor B,  j = 1, ...,b. The Bj 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) 2-way 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.