# 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 ?

• I highly suspect it does not exactly follow a t-distribution, even though it may be approximated well by one. The least squares regression coefficient does exactly follow a t-distribution when the errors are normally distributed. Tests of the OLS parameter are consistent with tests of correlation coefficient, meaning they converge to the same answer as n gets big. Another assumption to consider is that the null must be true; when the null is not true, statistics normally distributed as T will have non central T distributions. Dec 27, 2017 at 18:43
• @AdamO: I see what you mean, but I am not sure it answers my question: As you say the $\hat{\beta}_1$ in a regression $y=\beta_0+\beta_1 x +\epsilon$ is linked to the correlation coefficient. Moreover, as you say, (if all necessary conditions are met) $t=\hat{\beta}_1/s_{\hat{\beta}_1}$ is t-distributed. So your reasoning may hold if this $t$ is exactly equal to the formula in my question. Is that the case ?
– user83346
Dec 29, 2017 at 4:37

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.

• This post shows the intermediate steps of the last equation. Feb 25, 2021 at 7:33

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.