Two-Factor Full Factorial Design with Replications

[Jain, Ch. 22]

We now add to Chapter 21:

• additive term for interactions among factors A and B in model equation
• effect of error through replication

Model

where

• is effect of interaction between factor A at level j and factor B at level i
• is experimental error in observation at factor A level k, factor B level i, and replication k
• Each row or column sum of interactions is zero:

  

  

• The sum of errors for any particular choice of factors is zero:

  

Computing Effects, Interactions, and Errors

Example

Consider the number bytes required to code five workloads (I,J,K,L,M) on each of four computers (W,X,Y,Z).

First, should we use additive or multiplicative model?

• Range , or one order of magnitude.
• Physically, one processor might increase the code size by 50% over another, or might decrease size by one half. Factors 50% or 1/2 are multipliers.

Thus use multiplicative model: take of code sizes as data.

See Table 22.2

Computing Effects

Table for computation of effects has average of replicas for factor levels i and j in cell i,j.

See TABLE 22.3

As before, effects and are obtained by

1. summing row and columns,
2. dividing by number of addends to obtain means, and
3. subtracting from row and column means (mean of all abr observations).

Formally:

Where did above formulas come from?

• Sum model equation for all i,j,k and solve for .
• Sum model equation for all i,k and solve for .
• Sum model equation for all j,k and solve for .
• Sum model equation for all k and solve for .

(See Ch. 20 (one-factor design) notes for example.)

Computing Interactions

By last formula above, interactions for (i,j)th cell are computed by subtracting from cell mean .

See TABLE 22.4

Note that row and column sums are zero.

Computing Errors

Given i and j, error in kth replication is, as usual, the difference between kth observation and mean of replications:

As usual, we only need to compute errors to visually test assumptions.

Allocation of Variation

How much of the variation is due to:

• factor A?
• factor B?
• interaction of factors A and B?
• experimental error?

As in earlier designs, square both sides of model equation (crossterms are again zero):

All terms except SSE are easy to compute. Therefore SSE can be found from the other terms:

For our example,

Therefore both the computer and the workload alone explain about 96% of the variation, while the factor interactions appears insignificant. Also, the error is virtually zero!

Analysis of Variance with ANOVA Table

Simply generalize last design (two factors, no replication). Add SSAB term, and thus a third F-test for AB significance.

Recall:

and similarly for MS0, MSA, MSB, and MSE. Degrees of freedom for SSY, SS0, SSA, SSB are same as last design; SSAB and SSE differ:

SSE has ab(r-1) degrees of freedom because the r errors add to zero; thus only r-1 are required.

MSA, MSB are same as last design; MSAB and MSE differ:

Finally, computed F statistics for A and B are same as last design; statistic for MSAB is new:

Table Arrangement for Computation

Differences from last design (two factor, no replication):

• Add row for component AB.
• Change DF column for e to ab(r-1).

 
Table 1: Based on Jain Table 22.5

See TABLE 22.6

Conclusion:

factors A, B, and AB interaction are all significant!

Confidence Intervals for Effects and Interactions

Compared to last design (two factors, no replication), formulas for variance based on are almost identical: just divide by number of replicas, r. Also add row for .

and, as usual, .

See TABLES 22.8, 22.9 for confidence intervals in our example.

Notes:

• Can use normal rather than student-t distribution because degress of freedom are large (ab(r-1)=40).
• Confidence intervals are very narrow.

Visual validation of assumptions

Validation that errors are independent of response AND that variance of errors is constant w.r.t. response variable:

See Fig. 22.1

Validation that errors are normally distributed:

See Fig. 22.2