t-test for pearson correlation coeffcient Hypothesis testing for the correlation coefficient (at least one of them) is based on a t-distribution of the test statistic $t=\frac{r}{\sqrt{(1-r^2)/(N-2)}}$ where $r$ is the sample correlation coefficient and $N$ is the number of cases.  
Can someone tell me what assumptions are made in order for this $t$ to have a $t$-distribution ? I assume that bivariate normality is one of them ?
 A: The statistic you've mentioned $t=\frac{r}{\sqrt{(1-r^2)/(N-2)}}$ is essentially a significance of regression testing problem under the given setting :-
Let $(x_i, y_i)$ be a bivariate normal random sample of size $N$. Then we know that $E(Y|X=x)=\left(\mu_Y-\rho \dfrac{\sigma_Y}{\sigma_X} \mu_X\right)+\left(\rho \dfrac{\sigma_Y}{\sigma_X}\right)x = \beta_0 + \beta_1x$ 
Where $\beta_0 = \left(\mu_Y-\rho \dfrac{\sigma_Y}{\sigma_X} \mu_X\right)$ and $\beta_1 = \left(\rho \dfrac{\sigma_Y}{\sigma_X}\right)$
Now to test $H_0 : \rho = 0$ is equivalent to testing $H_0 : \beta_1 = 0$.
Let $\hat{\beta_1}$ be the OLS estimator of $\beta_1$ then under $H_0 : \beta_1 =0$ 
$t=\frac{\hat{\beta_1}}{s.e(\hat{\beta_1})}=\frac{r}{\sqrt{(1-r^2)/(N-2)}}$  follows a t-distribution with $N-2$ degrees of freedom.
Hence the only thing you need to ensure is that the samples are independent and follow Bivariate Normal Distribution. Note that checking for individual normality will not work as marginal normality doesn't imply joint normality.   
A: The assumptions that are required are as follows;


*

*independent observations; 

*the population correlation, ρ = 0;

*normality: the 2 variables involved are bivariately normally
distributed in the population. However, this is not needed for a
reasonable sample size -say, N ≥ 20 or so.

