7.
Product and Process Comparisons
7.4. Comparisons based on data from more than two processes


Fixed and Random Factors and Components of Variance  
A fixed level of a factor or variable means that the levels in the experiment are the only ones we are interested in  In the previous example, the levels of the factor temperature were considered as fixed; that is, the three temperatures were the only ones that we were interested in (this may sound somewhat unlikely, but let us accept it without opposition). The model employed for fixed levels is called a fixed model. When the levels of a factor are random, such as operators, days, lots or batches, where the levels in the experiment might have been chosen at random from a large number of possible levels, the model is called a random model, and inferences are to be extended to all levels of the population.  
Random levels are chosen at random from a large or infinite set of levels  In a random model the experimenter is often interested in estimating components of variance. Let us run an example that analyzes and interprets a component of variance or random model.  
Components of Variance Example for Random Factors  
Data for the example 
A company supplies a customer with a larger number of batches of raw
materials. The customer makes three sample determinations from each
of five randomly selected batches to control the quality of the
incoming material. The model is
$$ Y_{ij} = \mu + \tau_i + \epsilon_{ij} \, , $$
and the \(k\)
levels (e.g., the batches) are chosen at random
from a population with variance \(\sigma_\tau\).
The data are shown below.


ANOVA table for example 
A oneway ANOVA is performed on the data with the following results:


Interpretation of the ANOVA table 
The computations that produce the SS are the same for both the fixed
and the random effects model. For the random model, however, the
treatment sum of squares, SST, is an estimate of \(\{\sigma_\epsilon^2 + 3 \sigma_\tau^2\}\).
This is shown in the EMS (Expected Mean Squares) column of the ANOVA
table.
The test statistic from the ANOVA table is F = 36.94 / 1.80 = 20.5. If we had chosen an \(\alpha\) value of 0.01, then the F value from the table in Chapter 1 for a df of 4 in the numerator and 10 in the denominator is 5.99. 

Method of moments  Since the test statistic is larger than the critical value, we reject the hypothesis of equal means. Since these batches were chosen via a random selection process, it may be of interest to find out how much of the variance in the experiment might be attributed to batch diferences and how much to random error. In order to answer these questions, we can use the EMS column. The estimate of \(\sigma_\epsilon^2\) is 1.80 and the computed treatment mean square of 36.94 is an estimate of \(\sigma_\epsilon^2 + 3 \sigma_\tau^2\). Setting the MS values equal to the EMS values (this is called the Method of Moments), we obtain $$ s_\epsilon^2 = 1.80 \,\, \mbox{ and } \,\, s_\epsilon^2 + 3 s_\tau^2 = 36.94 \, , $$ where we use \(s^2\) since these are estimators of the corresponding \(\sigma^2\) values.  
Computation of the components of variance  Solving these expressions $$ s_\tau^2 = \frac{36.94  1.80}{3} = 11.71 \, . $$ The total variance can be estimated as $$ s_{\mbox{total}}^2 = s_\tau^2 + s_\epsilon^2 = 11.71 + 1.80 = 13.51 \, . $$  
Interpretation  In terms of percentages, we see that 11.71/13.51 = 86.7 percent of the total variance is attributable to batch differences and 13.3 percent to error variability within the batches. 