# Why t-test of correlation coefficient can't be used for testing non-zero?

t-test of a correlation coefficient is used for testing the sample correlation against population correlation of zero. To test an assumed value of the population coefficient other than zero, we should use z-test for a correlation coefficient.

Question:

1. What makes the tests different such that the t-test can only test for zero? Isn't the z-test a special case of the t-test (infinite degree of freedom)?

2. Since the t-test can't be used for tested for non-zero correlation. Why can't we always use the z-test? What's the point of the t-test?

EDITED for reference:

100 Statistical Tests by Gopal K.Janji

• I have not heard this before, do you have a source? (I may just be ignorant!) Feb 17, 2017 at 5:21
• @GeoMatt22 I added a reference, please look at Limitations. Feb 17, 2017 at 5:27
• These might be helpful: en.wikipedia.org/wiki/… and en.wikipedia.org/wiki/… . I am assuming the latter has been done in Test 13 your text is referring to? Feb 17, 2017 at 7:02
• Possible duplicate of When is Fisher's z-transform appropriate? Nov 1, 2018 at 12:50

Short answer: You don't use a t-test of the depicted form when $$\rho\neq 0$$ simply because the statistic $$\frac{r\sqrt{n-2}}{\sqrt{1-r^2}}$$ has a $$t_{n-2}$$ distribution when $$\rho= 0$$ (and the other assumptions hold) and it doesn't have a $$t_{n-2}$$ distribution otherwise.
There's several different z-tests for a correlation coefficient (whether $$\rho$$ is $$0$$ or not), but they're approximate, large sample tests. Some approximations are better than others. It may be that you can also make a good t approximation, but I don't recall seeing a derivation of any that would give you a suitable $$\nu$$.