For some reason that I can’t quite remember, I thought it would be a good idea to investigate the properties of Lawley’s test for the equality of correlations in a correlation matrix. So the null hypothesis of this test kind of looks like this:

I have absolutely no idea why I thought this was a good idea. I’m thinking it may have been back in the days where I was not as proficient in Structural Equation Modelling (SEM) so I’d dwell in my mathematical training to accomplish things that can be trivially set up as SEMs. Following on my poorly-documented code and meager notes, it seems like I thought I could tweak the test to see whether or not the assumption of Parallel Tests or Tau-Equivalent Tests could be evaluated. You know, to make sure people would know whether their Cronbach’s alpha was a true estimate of reliability or just a lower bound. Little did I know at that point that people just use alpha whether the assumptions behind it hold or not. The truth is nobody really cares.

The one thing I remember is that I couldn’t find any R package that would run the test. So I ended up coding it myself. I’m not sure if this would be of any use to anyone, but I’m hoping this may save some time to someone out there who may need it for whatever reason.

So the R function you’d have to declare is:

lawley <- function(data) {

```
```R <- cor(data)

m <- cor(data)

r <- nrow(m)

c <- ncol(m)

u <- c

z <- lower.tri(c,diag = FALSE)

u[z] <- 0

l <- c

z1 <- upper.tri(c,diag = FALSE)

l[z1] <- 0

p <- r

n <- nrow(data)

mcp <- 0*1:ncol(m)

for (i in 1:ncol(m))

{

mcp[i] <- (sum(m[,i])-1)/(ncol(m)-1)

} # vector of average values of non-diagonal elements of each row

lR <- R

z2 <- upper.tri(R,diag = FALSE)

lR[z2] <- 0

mc <- (2/(p*(p-1))) * sum(R-lR)

TR <- R

z3 <- upper.tri(R,diag = TRUE)

TR[z3] <-0

A <- (TR - mc)^2

A[z3] <-0

A <- sum(A)

B <- sum((mcp-mc)^2)

C <- ((p-1)^2*(1-(1-mc)^2))/(p-((p-2)*(1-mc)^2))

X2 <- (((n-1)/((1-mc)^2))*(A-C*B))

v <- ((p+1)*(p-2))/2

P <- 1-pchisq(X2,v,ncp = 0, lower.tail = TRUE, log.p = FALSE)

result <- rbind(c(X2, v, P))

dimnames(result) <- list(c("X2"), c("statistic", "df", "p.value"))

`print(result)`

}

And the way you use it is very simple. Let’s try it with a mock dataset we’ll generate in the R package `lavaan`

. Here we’re generating data from a One Factor model with equal loadings of 0.3 and equal error variances:

library(lavaan)

set.seed(123)

```
```pop = 'f1 =~ 0.3*x1 + 0.3*x2 + 0.3*x3 + 0.3*x4 + 0.3*x5'

dat = simulateData(model=pop, sample.nobs=500)

`lawley(dat)`

statistic df p.value

X2 7.840194 9 0.5503277

So… yeah. Non-significant p-value (as per glorious alpha of .05) tells us that the null hypothesis is true and the population correlation matrix indeed has equal elements all around.

Let’s change one loading and see what happens:

library(lavaan)

set.seed(123)

```
```pop = 'f1 =~ 0.4*x1 + 0.3*x2 + 0.3*x3 + 0.3*x4 + 0.3*x5'

dat = simulateData(model=pop, sample.nobs=500)

`lawley(dat)`

statistic df p.value

X2 18.78748 9 0.02706175

So the null hypothesis is now rejected because one of the loadings is different so there is one element of the covariance matrix that is not equal to every other element.

I literally have no clue why Lawley thought having such a test was a good idea. But then again I was investigating it a few years ago so maybe *I* thought the test was a good idea in the first place.

Anyhoo, I hope this helps someone if they need it.

]]>Let me show you why.

Say we have the very simple scenario of calculating power for the easy-cheesy *t-test* of the Pearson correlation coefficient. We are going to be extra indulgent with ourselves and claim the population effect size is (so a LARGE effect size à la Cohen). If you plug in the usual specifications in G*Power (Type I error rate of .05, desired power of 0.8, population effect size of against the null of ) this is what we get:

So your sample size should be 26. Just for kicks an giggles, I simulated the power curve for this exact scenario and marked with a line where the 80% power would be located.

Same answer as with G*Power, somewhere a little over *n=25*. Pretty straightforward, right? Well… sure… if you’re comfortable with **the assumption that your data is bivariate normal.** Both in the R simulation I made and in G*Power, the software assumes that your data looks like this:

For even more kicking and giggling, let’s assume your data is NOT normal (which, as we know, is the more common case). In this particular instance, both variables are -distributed with 1 degree of freedom (quite skewed). Each variable looks like this:

And their joint density (e.g. if you do a scatterplot), looks like that:

But here’s the catch… because of how I simulated them (through a Gaussian copula if you’re wondering), they both have the *same* population effect size of 0.5. What does the power curve look like in this case? It looks like this:

So that for the same large population effect size, you need a little over **TWICE** the sample size to obtain the same 80%.

You see where I’m going with this? Where’s the my-data-is-not-normal option in G*Power? Or my data-has-missing-values? Or my-data-has-measurement error? Or my data has all of those at once? Sure, I realize that this is a little bit of an extreme case because the sample size is not terribly large, the non-normality is severe and by the time *n=100*, the malicious influence of the non-normality has been “washed away” so to speak. The power curves look more and more similar as the sample size grows larger and larger. But it is still a reminder that every time I see people report their power analyses through G*Power my mind immediately goes to… “is this really power or a lower/upper bound to power?” And, moreover… if you go ahead, do your analyses and find your magic *p-value* under .05, you’re probably going to feel even *more* confident that your results are the real deal, right? I mean, you did your due diligence, you’re aware of the issues and you tried to address them the best way you could. And that’s exactly what kills me. Sometimes your best is just not good enough.

Solutions? Well… I dunno. Unless someone makes computer simulations mandatory in research methods classes, the only other option I have is usually to close my eyes and hope for the best.

]]>