cortest.bartlett {psych} | R Documentation |
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.
cortest.bartlett(R, n = NULL,diag=TRUE)
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. |
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.
chisq |
Assymptotically chisquare |
p.value |
Of chi square |
df |
The degrees of freedom |
William Revelle
Bartlett, M. S., (1951), The Effect of Standardization on a chi square Approximation in Factor Analysis, Biometrika, 38, 337-344.
cortest.mat
, cortest.normal
, cortest.jennrich
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)