Orthogonality, Orthogonal Decomposition, and Their Role in Modern Experimental Design

Abstract

In Chap. 2, we saw how to investigate whether or not one factor influences some dependent variable. Our approach was to partition the total sum of squares (TSS), the variability in the original data, into two components – the sum of squares between columns (SSB_c), attributable to the factor under study, and the sum of squares within a column (SSW), the variability not explained by the factor under study, and instead explained by “everything else.” Finally, these quantities were combined with the appropriate degrees of freedom in order to assess statistical significance. We were able to accept or reject the null hypothesis that all column means are equal (or, correspondingly, reject or accept that the factor under study has an impact on the response). In Chap. 4, we discussed multiple-comparison techniques for asking more detailed questions about the factor under study; for example, if not all column means are equal, how do they differ? We now present a more sophisticated, flexible, and potent way to analyze (or “decompose”) the impact of a factor on the response, not limited to pairwise comparisons.

Appendix

Example 5.7 Drug Comparison Using SPSS

We use the drug comparison example to illustrate the use of SPSS when using orthogonal contrasts . Data from Table 5.5 are repeated in Table 5.15 for convenience.

Table 5.15 Column means for drug-comparison study

Running the ANOVA using SPSS yields the same basic result as we have seen previously, shown in Table 5.16. Once again, we conclude that our result is significant ( p < .01), so not all “treatments” are the same with respect to efficacy.

Table 5.16 One-way ANOVA table for drug-comparison study in SPSS

We now set out to see how they differ. Similar to what we have done, we will break down the differences into three components: (a) “real” drugs versus placebo ; (b) aspirin 1 versus aspirin 2; and (c) aspirin drugs versus non-aspirin. To use SPSS to enter the contrast values, we click on Contrasts... under Analyze > Compare Means > One-Way ANOVA and enter the three sets (rows) of coefficients in the orthonormal matrix we have built previously (click Previous or Next to change contrasts , and Continue once the three contrasts have been included), as shown in Fig. 5.2; the output in Table 5.17 verifies that these were the coefficients used.

Fig. 5.2

Steps for including contrasts in SPSS

Table 5.17 Orthonormal matrix in SPSS

Unfortunately, SPSS does not provide the augmented ANOVA table in one fell swoop. What it does provide, in the notation of this chapter, are the values of the Z’s and of the t _calc’s, which are the respective square roots of the F _calc’s in the augmented ANOVA table, as well as the p-values. It does this under the banner of an analysis that uses the mean-square error in the original ANOVA table as the basis of the standard error for all contrasts (as opposed to using a variance estimate based only on the particular columns, weighted according to the contrast’s coefficients). The SPSS results are in Table 5.18. Note that the “Value of Contrast” column in Table 5.18 consists of the same set of values as the Z’s in Table 5.8 and the numbers in the “t” column are each the square root of the corresponding F _calc value in Table 5.9, the augmented ANOVA table (some numbers are off a tiny bit due to rounding).

Table 5.18 Contrast tests in SPSS

Example 5.8 Drug Comparison Using R

We will use the same drug comparison example to demonstrate how the analysis is done in R. After importing the data, we have to access the order in which the levels of the independent variable (V1) are organized.

> drug <- read.csv(file.path("/Users/documents", "ex5.8.csv"), +header=F) > levels(drug$V1) [1] "Aspirin1" "Aspirin2" "NonA" "Placebo"

We will notice that the levels have been organized alphabetically and this has to be taken into account when setting up the contrast matrix. In the command below, the data are entered per row (byrow=T) using the contrast values in the orthonormal matrix we discussed previously. Note that the order of the values has changed to reflect the order of the levels in R.

> matrix <- matrix(c(0.2887, 0.2887, 0.2887, -0.866, -0.7071, +0.7071, 0, 0, -0.4082, -0.4082, 0.8165, 0), nrow=3, ncol=4, +byrow=T) > matrix

	[,1]	[,2]	[,3]	[,4]
[1,]	0.2887	0.2887	0.2887	-0.866
[2,]	-0.7071	0.7071	0.0000	0.000
[3,]	-0.4082	-0.4082	0.8165	0.000

R gives an error message if we try to use this matrix in further analysis. This is because the software package deals with contrasts when the values are organized in columns, rather than rows. For this reason, we have to transpose the contrast matrix, as follows:

> matrix_t <- t(matrix) > matrix_t

	[,1]	[,2]	[,3]
[1,]	0.2887	-0.7071	-0.4082
[2,]	0.2887	0.7071	-0.4082
[3,]	0.2887	0.0000	0.8165
[4,]	-0.8660	0.0000	0.0000

Alternatively, it is possible to create the transposed matrix, which would eliminate one step in the programming – we just have to remember to set the correct number of columns and row and byrow=F. Next, we set the contrast :

> contrasts(drug$V1) <- matrix_t > drug.aov <- aov(V2~V1, data=drug)

# This command is used to verify if the contrast matrix has been correctly assigned to the levels.

> drug.aov$contrasts $V1

	[,1]	[,2]	[,3]
Aspirin1	0.2887	-0.7071	-0.4082
Aspirin2	0.2887	0.7071	-0.4082
NonA	0.2887	0.0000	0.8165
Placeo	-0.8660	0.0000	0.0000

This will also provide the augmented ANOVA table, which can be viewed using the summary.aov() function. Note that in this command we specify how the SSB_c will be split, addressing our three questions regarding the differences between the treatments.

> summary.aov(drug.aov, split=list(V1=list("P vs. P'"=1, "A1 vs. +A2"=2, "A vs. Non-A"=3)))

	Df	Sum Sq	Mean Sq	F value	Pr(>F)
V1	3	112.00	37.33	7.467	0.000802	***
V1: P vs. P ′	1	42.67	42.67	8.533	0.006820	**
V1: A1 vs. A2	1	4.00	4.00	0.800	0.378718
V1: A vs. Non-A	1	65.33	65.33	13.067	0.001168	**
Residuals	28	140.00	5.00

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