psych (version 2.4.6.26)

cortest.bartlett: Bartlett's test that a correlation matrix is an identity matrix

Description

Bartlett (1951) proposed that -ln(det(R)*(N-1 - (2p+5)/6) was distributed as chi square if R were an identity matrix. A useful test that residuals correlations are all zero. Contrast to the Kaiser-Meyer-Olkin test.

Usage

cortest.bartlett(R, n = NULL,diag=TRUE)

Value

chisq

Assymptotically chisquare

p.value

Of chi square

df

The degrees of freedom

Arguments

R

A correlation matrix. (If R is not square, correlations are found and a warning is issued.

n

Sample size (if not specified, 100 is assumed).

diag

Will replace the diagonal of the matrix with 1s to make it a correlation matrix.

Author

William Revelle

Details

More useful for pedagogical purposes than actual applications. The Bartlett test is asymptotically chi square distributed.

Note that if applied to residuals from factor analysis (fa) or principal components analysis (principal) that the diagonal must be replaced with 1s. This is done automatically if diag=TRUE. (See examples.)

An Alternative way of testing whether a correlation matrix is factorable (i.e., the correlations differ from 0) is the Kaiser-Meyer-Olkin KMO test of factorial adequacy.

References

Bartlett, M. S., (1951), The Effect of Standardization on a chi square Approximation in Factor Analysis, Biometrika, 38, 337-344.

See Also

cortest.mat, cortest.normal, cortest.jennrich

Examples

set.seed(42)   
x <- matrix(rnorm(1000),ncol=10)
r <- cor(x)
cortest.bartlett(r)      #random data don't differ from an identity matrix
#data(bfi)
cortest.bartlett(bfi[1:200,1:10])    #not an identity matrix
f3 <- fa(Thurstone,3)
f3r <- f3$resid
cortest.bartlett(f3r,n=213,diag=FALSE)  #incorrect

cortest.bartlett(f3r,n=213,diag=TRUE)  #correct (by default)