psych (version 2.4.6.26)

cortest: Chi square tests of whether a single matrix is an identity matrix, or a pair of matrices are equal.

Description

Steiger (1980) pointed out that the sum of the squared elements of a correlation matrix, or the Fisher z score equivalents, is distributed as chi square under the null hypothesis that the values are zero (i.e., elements of the identity matrix). This is particularly useful for examining whether correlations in a single matrix differ from zero or for comparing two matrices. Jennrich (1970) also examined tests of differences between matrices.

Usage

cortest(R1,R2=NULL,n1=NULL,n2 = NULL, fisher = TRUE,cor=TRUE,method="pearson",
           use ="pairwise") #same as cortest.normal this does the steiger test
cortest.normal(R1, R2 = NULL, n1 = NULL, n2 = NULL, fisher = TRUE)      #the steiger test
cortest.jennrich(R1,R2,n1=NULL, n2=NULL)  #the Jennrich test
cortest.mat(R1,R2=NULL,n1=NULL,n2 = NULL) #an alternative test

Value

chi2

The chi square statistic

df

Degrees of freedom for the Chi Square

prob

The probability of observing the Chi Square under the null hypothesis.

Arguments

R1

A correlation matrix. (If R1 is not rectangular, and cor=TRUE, the correlations are found).

R2

A correlation matrix. If R2 is not rectangular, and cor=TRUE, the correlations are found. If R2 is NULL, then the test is just whether R1 is an identity matrix.

n1

Sample size of R1

n2

Sample size of R2

fisher

Fisher z transform the correlations?

cor

By default, if the input matrices are not symmetric, they are converted to correlation matrices. That is, they are treated as if they were the raw data. If cor=FALSE, then the input matrices are taken to be correlation matrices.

method

Which type of correlation to find ("pearson", "spearman","kendall")

use

How to handle missing data (defaults to pairwise)

Author

William Revelle

Details

There are several ways to test if a matrix is the identity matrix. The most well known is the chi square test of Bartlett (1951) and Box (1949). A very straightforward test, discussed by Steiger (1980) is to find the sum of the squared correlations or the sum of the squared Fisher transformed correlations. Under the null hypothesis that all the correlations are equal, this sum is distributed as chi square. This is implemented in cortest and cortest.normal

Yet another test, is the Jennrich(1970) test of the equality of two matrices. This compares the differences between two matrices to the averages of two matrices using a chi square test. This is implemented in cortest.jennrich.

Yet another option cortest.mat is to compare the two matrices using an approach analogous to that used in evaluating the adequacy of a factor model. In factor analysis, the maximum likelihood fit statistic is
\(f = log(trace ((FF'+U2)^{-1} R) - log(|(FF'+U2)^{-1} R|) - n.items\).

This in turn is converted to a chi square

\(\chi^2 = (n.obs - 1 - (2 * p + 5)/6 - (2 * factors)/3)) * f \) (see fa.)

That is, the model (M = FF' + U2) is compared to the original correlation matrix (R) by a function of \(M^{-1} R\). By analogy, in the case of two matrices, A and B, cortest.mat finds the chi squares associated with \(A^{-1}B\) and \(A B^{-1}\). The sum of these two \(\chi^2\) will also be a \(\chi^2\) but with twice the degrees of freedom.

References

Steiger, James H. (1980) Testing pattern hypotheses on correlation matrices: alternative statistics and some empirical results. Multivariate Behavioral Research, 15, 335-352.

Jennrich, Robert I. (1970) An Asymptotic \(\chi^2\) Test for the Equality of Two Correlation Matrices. Journal of the American Statistical Association, 65, 904-912.

See Also

cortest.bartlett corr.test

Examples

set.seed(42)
x <- matrix(rnorm(1000),ncol=10)
cortest.normal(x)  #just test if this matrix is an identity
#now create two correlation matrices that should be equal
x <- sim.congeneric(loads =c(.9,.8,.7,.6,.5),N=1000,short=FALSE)
y <- sim.congeneric(loads =c(.9,.8,.7,.6,.5),N=1000,short=FALSE)

cortest(x$r,y$r,n1=1000,n2=1000) #The Steiger test
cortest.jennrich(x$r,y$r,n1=100,n2=1000) # The Jennrich test

cortest.mat(x$r,y$r,n1=1000,n2=1000)   #twice the degrees of freedom as the Jennrich

#create a new matrix that is different
z <- sim.congeneric(loads=c(.8,.8,.7,.7, .6), N= 1000, short=FALSE)
cortest(x$r,z$r,n1=1000)  #these should be different