3.
Production
Process Characterization
3.2. Assumptions / Prerequisites 3.2.3. Analysis of Variance Models (ANOVA)


Description  When we have two factors with at least two levels and one or more observations at each level, we say we have a twoway layout. We say that the twoway layout is crossed when every level of Factor A occurs with every level of Factor B. With this kind of layout we can estimate the effect of each factor (Main Effects) as well as any interaction between the factors.  
Model 
If we assume that we have K observations at each combination
of I levels of Factor A and J levels of Factor B, then we can model the
twoway layout with an equation of the form:
This equation just says that the kth data value for the jth level of Factor B and the ith level of Factor A is the sum of five components: the common value (grand mean), the level effect for Factor A, the level effect for Factor B, the interaction effect, and the residual. Note that (ab) does not mean multiplication; rather that there is interaction between the two factors. 

Estimation  Like the oneway case, the estimation for the twoway layout can be done either by calculating the variance components or by using CLM techniques.  
Click here for the value splitting example 
For the twoway ANOVA, we display the data in a twodimensional
table with the levels of Factor A in columns and the levels of Factor B
in rows. The replicate observations fill each cell. We can sweep out the
common value, the row effects, the column effects, the interaction effects
and the residuals using valuesplitting
techniques. Sums of squares can be calculated and summarized
in an ANOVA table as shown below.
\( \bar{y}_{i..} = \frac{1}{JK} \sum_{j=1}^{J}{\sum_{k=1}^{K}{y_{ijk}}} \) \( \bar{y}_{.j.} = \frac{1}{IK} \sum_{i=1}^{I}{\sum_{k=1}^{K}{y_{ijk}}} \) \( \bar{y}_{ij.} = \frac{1}{K} \sum_{k=1}^{K}{y_{ijk}} \) \( \bar{y}_{...} = \frac{1}{IJK} \sum_{i=1}^{I}{\sum_{j=1}^{J}{\sum_{k=1}^{K}{y_{ijk}}}} \) The row labeled, "Corr. Total", in the ANOVA table contains the corrected total sum of squares and the associated degrees of freedom (DoF). We can use CLM techniques to do the estimation. We still have the problem that the model is saturated and no unique solution exists. We overcome this problem by applying the constraints to the model that the two main effects and interaction effects each sum to zero. 

Testing  Like testing in the oneway case, we are testing that two main effects and the interaction are zero. Again we just form a ratio of each main effect mean square and the interaction mean square to the residual mean square. If the assumptions stated below are true then those ratios follow an F distribution and the test is performed by comparing the F_{0} ratios to values in an F table with the appropriate degrees of freedom and confidence level.  
Assumptions  For estimation purposes, we assume the data can be adequately modeled as described in the model above. It is assumed that the random component can be modeled with a Gaussian distribution with fixed location and spread.  
Uses  The twoway crossed ANOVA is useful when we want to compare the effect of multiple levels of two factors and we can combine every level of one factor with every level of the other factor. If we have multiple observations at each level, then we can also estimate the effects of interaction between the two factors.  
Example 
Let's extend the oneway machining
example by assuming that we want to test if there are any differences in
pin diameters due to different types of coolant. We still have five different
machines making the same part and we take five samples from each machine
for each coolant type to obtain the following data:


Analyze  For analysis details see the crossed twoway value splitting example. We can summarize the analysis results in an ANOVA table as follows:  


Test  By dividing the mean square for machine by the mean square for residuals we obtain an F_{0} value of 8.8 which is greater than the critical value of 2.61 based on 4 and 40 degrees of freedom and a 0.05 significance level. Likewise the F_{0} values for Coolant and Interaction, obtained by dividing their mean squares by the residual mean square, are less than their respective critical values of 4.08 and 2.61 (0.05 significance level).  
Conclusion  From the ANOVA table we can conclude that machine is the most important factor and is statistically significant. Coolant is not significant and neither is the interaction. These results would lead us to believe that some toolmatching efforts would be useful for improving this process. 