psych (version 2.2.9)

# 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.

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

Run this code
``````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(psychTools::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)

``````

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