Description Usage Arguments Details Value Author(s) References See Also Examples

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.

1 | ```
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`

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(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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