# cortest.bartlett: Bartlett's test that a correlation matrix is an identity... In psych: Procedures for Psychological, Psychometric, and Personality Research

## 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

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

## 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.

## 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.

## Value

 `chisq` Assymptotically chisquare `p.value ` Of chi square `df` The degrees of freedom

William Revelle

## References

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`
 ``` 1 2 3 4 5 6 7 8 9 10 11``` ```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) ```