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

cortest.bartlett {psych} | R Documentation

## 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)

### 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

### Author(s)

William Revelle

### 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)

[Package psych version 1.7.8 Index]