Cattell’s salient similarity index (s) may be used to compare two solutions’ patterns of
loadings. Consider the following hypothetical loading matrices:
Variable
X1
X2
X3
X4
X5
SSL
Group 1
Factor2
.12
.75
.67
-.53
-.14
1.32
Factor1
.90
.63
.15
-.09
-.74
1.78
Factor3
.21
.34
.24
-.16
-.19
0.28
Factor1
.45
.65
.27
-.15
.95
1.62
Group 2
Factor2
.49
-.15
.80
.09
-.79
1.53
Factor3
-.70
.22
.04
.67
.12
1.00
Each loading is classified as Positively Salient (Cattell used a criterion of > .10, I’ll use a
higher cut, > .30), Negatively Salient (< -.30) or neither (HyperPlane). One then constructs a third
order square [PS, HP, NS] matrix comparing Group 1 with Group 2. I’ll abbreviate the contents of this
table using these cell indices:
Group2
PS
HP
NS
Group 1
PS
HP
NS
11
12
13
21
22
23
31
32
33
The loadings for X1 - F1 are PS for both groups, so it is counted in cell 11. Ditto for X2 - F1. X3 F1 is HP in both groups, so it is counted in cell 22. Ditto for X4 - F1. X5 - F1 is NS in Group 1 but PS in
Group 2, so it is counted in cell 13.
Thus, the table for comparing Factor 1 in Group 1 with Factor 1 in Group 2 with frequency
counts inserted in the cells looks like this:
Group2
PS
HP
NS
Group 1
PS
HP
NS
2
0
1
0
2
0
0
0
0
The 1 in the upper right corner reflects the difference in the two patterns with respect to X5.
Counts in the main diagonal, especially in the upper left and the lower right, indicate similarity of
structure; counts off the main diagonal, especially in the upper right or lower left, indicate dissimilarity.
Cattell’s s is computed from these counts this way:
s
11  33  13  31
(The numbers here are cell indices.)
11  33  13  31  .5(12  21  23  32)
For our data,
s
2  0  1 0
 .33
2  0  1  0  .5(0  0  0  0)
Cattell et al. (Educ. & Psych. Measurement, 1969, 29, 781-792) provide tables to convert s to
an approximate significance level, P, for testing the null hypothesis that the two factors being
compared (one from population 1, one from population 2) are not related to one another. [I have
these tables, in Tabachnick & Fidell, 1989, pages 717 & 718, if you need them.] These tables require
you to compute the percentage of hyperplane counts (60, 70, 80, or 90) and to have at least 10
variables (the table has rows for 10, 20, 30, 40, 50, 60, 80, & 100 variables). We have only 5
variables, and a hyperplane percentage of only 40%, so we can’t use the table. If we had 10
variables and a hyperplane percentage of 60%, P = .138 for s = .26 and P = .02 for s = .51. Under
those conditions our s of .33 would have a P of about .10, not low enough to reject the null hypothesis
(if alpha = .05) and conclude that the two factors are related (similar). In other words, we would be
left with the null hypothesis that Factor 1 is not the same in population 1 as population 2.
It is not always easy to decide which pairs of factors to compare. One does not always
compare Factor 1 in Group 1 with Factor 1 in Group 2, and 2 in 1 with 2 in 2, etc. Factor 1 in Group 1
may look more like Factor 2 in Group 2 than it does like Factor 1 in Group 2, so one would compare 1
in 1 with 2 in 2. Remember that factors are ordered from highest to lowest SSL, and sampling error
alone may cause inversions in the orders of factors with similar SSL’s. For our hypothetical data,
comparing 1 in 1 with 1 in 2 makes sense, since F1 has high loadings on X1 and X2 in both groups.
But what factor in Group 2 would we choose to compare with F2 in Group 1 - the structures are so
different that a simple eyeball test tells us that there is no factor in Group 2 similar to F2 in Group 1.