What hypothesis test does R use by default for Pearson's correlation coefficient when using cor.test()?

Question: What hypothesis test does R use by default for Pearson's correlation coefficient when using cor.test()?

6 weeks ago by

SpacemanSpiffo • 30

SpacemanSpiffo • 30 wrote:

I have two vectors, X and Y

I perform a correlation test using them

cor.test(x, y)

From this, I get a p value and a correlation value, among other things.

In a manuscript, I report my results like so:

Pearson’s r(283) = 0.943, p < 2.2e-16

I have been asked to define which statistical test is being used. It was my understanding that Pearson's correlation test was itself the test. Is this incorrect?

I have read that a student's T test can be used to calculate the p value for the hypothesis test, which I believe is testing the hypothesis that the true correlation is 0.

Does the R language use a T-test for the p value by default in cor.test() ? Or is it correct to say that Pearson's is itself the test?

statistics R • 194 views

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modified 6 weeks ago by i.sudbery ♦ 11k • written 6 weeks ago by SpacemanSpiffo • 30

6 weeks ago by

i.sudbery ♦ 11k

Sheffield, UK

i.sudbery ♦ 11k wrote:

The null hypothesis being tested is that there is no correlation i.e. h0: rho=0

Here is what the r documentation has to say about the caculatoin of the p-value:

the test statistic is based on Pearson's product moment correlation coefficient cor(x, y) and follows a t distribution with length(x)-2 degrees of freedom if the samples follow independent normal distributions. If there are at least 4 complete pairs of observation, an asymptotic confidence interval is given based on Fisher's Z transform.

Wikipedia has this to say about the approximation of the distributino of rho to the t-distribution:

For pairs from an uncorrelated bivariate normal distribution, the sampling distribution of a certain function of Pearson's correlation coefficient follows Student's t-distribution with degrees of freedom n − 2. Specifically, if the underlying variables are white and have a bivariate normal distribution, the variable

t=r*sqrt ((n-2)/(1-r^2))

has a student's t-distribution in the null case (zero correlation).[20] This holds approximately in case of non-normal observed values if sample sizes are large enough.[21]

ADD COMMENT • link written 6 weeks ago by i.sudbery ♦ 11k

Thank you. Based on this, is it correct to say that a Student's t-test is being peformed?

ADD REPLY • link written 6 weeks ago by SpacemanSpiffo • 30

Its a test based on the t-statistic, but I probably wouldn't go as far as saying a t-test was being performed. Both in its original usage (https://www.york.ac.uk/depts/maths/histstat/student.pdf) and in common usage the term "t-test" implies a test for the difference between the means of two populations. That is not what is being don't here, even if the underlying distribution is the same.

ADD REPLY • link written 5 weeks ago by i.sudbery ♦ 11k

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