Designs with Factors at Three Levels

Abstract

Sometimes, we wish to examine the impact of a factor at three levels rather than at two levels as discussed in previous chapters. For example, to determine the differences in quality among three suppliers, one would consider the factor “supplier” at three levels. However, for factors whose levels are measured on a numerical scale, there is a major and conceptually-different reason to use three levels: to be able to study not only the linear impact of the factor on the response (which is all that can be done when studying a factor that has only two levels), but also the nonlinear impact. The basic analysis-of-variance technique treats the levels of a factor as categorical, whether they actually are or not. One (although not the only) logical and useful way to orthogonally break down the sum of squares associated with a numerical factor is to decompose it into a linear effect and a quadratic effect (for a factor with three numerical levels), a linear effect, a quadratic effect, and a cubic effect (for a factor with four numerical levels), and so forth.

Appendix

Example 12.6 Selling Toys Example using SPSS

From the perspective of SPSS , the toy-store problem is a two-factor cross-classification with replication. The data were entered as shown in Table 12.13 and SPSS was told that the first and second columns represent the levels of the row (A) and column (B) factors , respectively, and the third column represents the dependent variable . Table 12.14 shows the SPSS output.

Table 12.13 SPSS input format for toy study

Table 12.14 SPSS ANOVA table for toy-store example

The breakdown of the main-effect sum of squares into linear and quadratic components in an augmented ANOVA table is not provided, but would have to be independently and separately generated. For this, we can use a syntax to specify the coefficients for factors A and B, as shown in Fig. 12.10, where L1 and L2 are the linear and quadratic components, respectively. This will generate the results shown in Tables 12.15 and 12.16.

Fig. 12.10

Steps for setting up the contrast coefficients in SPSS

Table 12.15 Decomposition of sum of squares for factor A in SPSS

Table 12.16 Decomposition of sum of squares for factor B in SPSS

Example 12.7 Selling Toys Example using R

In this example, we will use the fac.design() introduced in previous chapters. We will use −1, 0, and 1 to identify the levels, but we could have used “low,” “medium,” and “high” or any other classification.

  > design <- fac.design(nlevels=3, nfactors=2, factor.names=list+(A=c(-1,0,1), B=c(-1,0,1)), replications=2, randomize=FALSE) creating full factorial with 9 runs ..

Next, we create a vector with the responses and add it to the design table:

  > y <- c(47, 57, 70, 34, 80, 105, 60, 81, 88, 45, 43, 86, 46, 92, 99, 80, +67, 92) > design <- add.response(design=design, response=y) > design

	run.no	run.no.std.rp	A	B	Blocks	y
1	1	1.1	-1	-1	.1	47
2	2	2.1	0	-1	.1	57
3	3	3.1	1	-1	.1	70
4	4	4.1	-1	0	.1	34
5	5	5.1	0	0	.1	80
6	6	6.1	1	0	.1	105
7	7	7.1	-1	1	.1	60
8	8	8.1	0	1	.1	81
9	9	9.1	1	1	.1	88
10	10	1.2	-1	-1	.2	45
11	11	2.2	0	-1	.2	43
12	12	3.2	1	-1	.2	86
13	13	4.2	-1	0	.2	46
14	14	5.2	0	0	.2	92
15	15	6.2	1	0	.2	99
16	16	7.2	-1	1	.2	80
17	17	8.2	0	1	.2	67
18	18	9.2	1	1	.2	92

class=design, type= full factorial NOTE: columns run.no and run.no.std.rp are annotation, not part of the data frame

To obtain the ANOVA table:

> design.aov <- aov(y~A*B, data=design) > summary(design.aov)

	Df	Sum Sq	Mean Sq	F value	Pr(>F)
A	2	4336	2168.0	28.034	0.000136	***
B	2	1456	728.0	9.414	0.006222	**
A:B	4	1472	368.0	4.759	0.024406	*
Residuals	9	696	77.3

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

As in the procedure described in Chap. 5, we have to set up a matrix with the contrast coefficients that will be used to decompose the sum of squares.

> matrix <- matrix(c(-1,0,1,0.5,-1,0.5), nrow=3, ncol=2, +byrow=F) > matrix

	[,1]	[,2]
[1,]	-1	0.5
[2,]	0	-1.0
[3,]	1	0.5

> contrasts(design$A) <- matrix > contrasts(design$B) <- matrix

To determine if the coefficients have been correctly assigned, we use

> design.aov$contrasts $A

[,1]	[,2]
-1	0.5
0	-1.0
1	0.5

$B

[,1]	[,2]
-1	0.5
0	-1.0
1	0.5

We can run ANOVA again without the interaction term, assuming we are only interested in the linear and quadratic components. Otherwise, R will use the coefficients to break down SSQ_interaction, as shown below. The following command decomposes the sum of squares and is valid for both situations – with or without the interaction term.

> summary.aov (design.aov2, split=list(A=list(1, 2, 3), +B=list(1,2,3)))

	Df	Sum Sq	Mean Sq	F value	Pr(>F)
A	2	4336	2168	28.034	0.000136	***
A: C1	1	4332	4332	56.017	3.75e-05	***
A: C2	1	4	4	0.052	0.825172
A: C3	1
B	2	1456	728	9.414	0.006222	**
B: C1	1	1200	1200	15.517	0.003410	**
B: C2	1	256	256	3.310	0.102195
B: C3	1
A:B	4	1472	368	4.759	0.024406	*
A:B: C1.C1	1	72	72	0.931	0.359804
A:B: C2.C1	1	24	24	0.310	0.591051
A:B: C3.C1	1
A:B: C1.C2	1	864	864	11.172	0.008626	**
A:B: C2.C2	1	512	512	6.621	0.030036	*
A:B: C3.C2	1
A:B: C1.C3	1
A:B: C2.C3	1
A:B: C3.C3	1
Residuals	9	696	77

--- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1