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

7.4.3.8.

Models and calculations for the two-way ANOVA

Basic Layout
The balanced two-way factorial 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 a balanced factorial. In this case, the \(abr\) data points \(\{y_{ijk}\}\) can be shown pictorially as follows: $$ \begin{array}{cccc} & & \mbox{Factor } B \\ \mbox{Factor } A & 1 & 2 & \ldots & b \\ 1 & y_{111}, \, y_{112}, \, \ldots, \, y_{11r} & y_{121}, \, y_{122}, \, \ldots, \, y_{12r} & \ldots & y_{1b1}, \, y_{1b2}, \, \ldots, \, y_{1br} \\ 2 & y_{211}, \, y_{212}, \, \ldots, \, y_{21r} & y_{221}, \, y_{222}, \, \ldots, \, y_{22r} & \ldots & y_{2b1}, \, y_{2b2}, \, \ldots, \, y_{2br} \\ \vdots & \vdots & \vdots & \ldots & \vdots \\ a & y_{a11}, \, y_{a12}, \, \ldots, \, y_{a1r} & y_{a21}, \, y_{a22}, \, \ldots, \, y_{a2r} & \ldots & y_{ab1}, \, y_{ab2}, \, \ldots, \, y_{abr} \\ \end{array} $$
How to obtain sums of squares for the balanced factorial layout 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 \(A\), \(i = 1, \, \ldots, \, a\). The \(A_i\) are the row sums.

  • Let \(B_j\) be the sum of all observations of level \(j\) of factor \(B\), \(j = 1, \, \ldots, b\). 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 obtain (omitting derivations): $$ \begin{eqnarray} \mbox{CM } & = & \frac{(\mbox{Sum of all observations})^2}{rab} \\ & & \\ SS_{total} & = & \sum (\mbox{each observation})^2 - CM \\ & & \\ SS(A) & = & \frac{\sum_{i=1}^a A_i^2}{rb} - CM \\ & & \\ SS(B) & = & \frac{\sum_{j=1}^b B_j^2}{ra} - CM \\ & & \\ SS(AB) & = & \frac{\sum_{i=1}^a \sum_{j=1}^b (AB)_{ij}^2}{r} - CM - SS(A) - SS(B) \\ & & \\ SSE & = & SS_{total} - SS(A) - SS(B) - SS(AB) \end{eqnarray} $$

These expressions are used to calculate the ANOVA table entries for the (fixed effects) two-way ANOVA.

Two-Way ANOVA Example:
Data 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:


  Materials (\(B\))
LABS (\(A\)) 1 2 3

  4.1 3.1 3.5
1 3.9 2.8 3.2
  4.3 3.3 3.6

  2.7 1.9 2.7
2 3.1 2.2 2.3
  2.6 2.3 2.5

Row and column sums The preliminary part of the analysis yields a table of row and column sums.


Material (\(B\))

Lab (\(A\)) 1 2 3 Total (\(A_i\))

1 12.3 9.2 10.3 31.8
2 8.4 6.4 7.5 22.3

Total (\(B_j\)) 20.7 15.6 17.8 54.1
ANOVA table From this table we generate the ANOVA table.


Source SS df MS F p-value

A 5.0139 1 5.0139 100.28 0
B 2.1811 2 1.0906 21.81 .0001
AB 0.1344 2 0.0672 1.34 .298
Error 0.6000 12 0.0500    

Total (Corr) 7.9294 17      
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