[R] Difference betweeen cor.test() and formula everyone says to use

[JLucke at ria.buffalo.edu]

The distribution of the statistic $ndf * r^2 / (1-r^2)$ with  the true 
value $\rho = zero$ follows an $F(1,ndf)$ distribution.
So the t-test is the correct test for $\rho=0$. 
Fisher's z is an asymptotically normal  transformation for any value of 
$\rho$. 
Thus  Fisher's z is better for testing $\rho= \rho_0 $ or $\rho_1 = 
\rho_2$.
The two statistics will not be equivalent at $\rho=0$ because the 
statistics are based on different assumptions.




Jeremy Miles <jeremy.miles at gmail.com> 
Sent by: r-help-bounces at r-project.org
10/16/2014 07:32 PM

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Subject
[R] Difference betweeen cor.test() and formula everyone says to use






I'm trying to understand how cor.test() is calculating the p-value of
a correlation. It gives a p-value based on t, but every text I've ever
seen gives the calculation based on z.

For example:
> data(cars)
> with(cars[1:10, ], cor.test(speed, dist))

Pearson's product-moment correlation

data:  speed and dist
t = 2.3893, df = 8, p-value = 0.04391
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
 0.02641348 0.90658582
sample estimates:
      cor
0.6453079

But when I use the regular formula:
> r <- cor(cars[1:10, ])[1, 2]
> r.z <- fisherz(r)
> se <- se <- 1/sqrt(10 - 3)
> z <- r.z / se
> (1 - pnorm(z))*2
[1] 0.04237039

My p-value is different.  The help file for cor.test doesn't (seem to)
have any reference to this, and I can see in the source code that it
is doing something different. I'm just not sure what.

Thanks,

Jeremy

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