Is a correlation significant? The t-test of correlation

What does the t-test of a correlation test?

It tests whether the true population correlation, $\rho$, differs from zero, given only the sample correlation coefficient $r$ you measured. A sample of paired returns will almost never show exactly zero correlation, even for unrelated series. The question is whether the $r$ you see is too far from zero to be luck.

This is a parametric test of correlation : it assumes both variables are approximately normally distributed. It is the test of the Pearson correlation coefficient , the ordinary correlation $r$. The usual hypotheses are two-tailed.

$$H_0: \rho = 0 \qquad H_A: \rho \neq 0$$

How do you compute the t-statistic?

Scale the correlation by the square root of the degrees of freedom and divide by the square root of what the correlation leaves unexplained.

$$t = \frac{r\sqrt{n-2}}{\sqrt{1-r^2}}, \qquad df = n - 2$$

The t-statistic for a correlation follows a t-distribution with $n-2$ degrees of freedom . If $|t|$ is larger than the critical value, reject $H_0$ and call the correlation statistically significant .

Questions also turn the formula around: they give $t$ and $n$ and ask for $r$, or give $r$ and $t$ and ask how many observations were used.

Why does sample size matter so much?

Because $n$ sits inside the numerator. Keep $r$ fixed and every extra observation makes $t$ larger, so a weak correlation that is noise in a small sample becomes significant in a big one.

The larger the sample, the smaller the correlation needed to reject $H_0: \rho = 0$. In very large data sets almost any correlation passes.

Compute t from r and nH₀: ρ = 0, df = n − 2

• |t| greater than the critical value Memorize a large t is far from zero
  • Reject H₀ the correlation is statistically significant
• |t| at or below the critical value Memorize the sample could easily be noise
  • Fail to reject H₀ not significant, which is not the same as proving ρ = 0

Exam tip Failing to reject is not proof that the correlation is zero. It only means this sample is not strong enough evidence. Never write “accept $H_0$”.

How do you read a correlation matrix?

Use it

Fund selectors test every pair in a correlation matrix . Take four funds with 48 monthly returns each. The two-tailed 5% critical value for 46 degrees of freedom is 2.013.

Pair	r	t	Significant at 5%?
A–B	0.62	5.36	yes
A–C	0.41	3.05	yes
A–D	0.18	1.24	no
B–C	0.35	2.53	yes
B–D	0.09	0.61	no
C–D	0.27	1.90	no

Three pairs are significant. The pairs whose correlation is not significantly different from zero (A–D, B–D, C–D) are the ones that offer the clearest diversification, though not significant is not the same as zero.

There is a shortcut. With one critical value $t_c$ and a fixed $n$, the smallest significant correlation is $t_c \div \sqrt{t_c^2 + n - 2}$. Here that is 2.013 ÷ √(2.013² + 46) = 0.285, so any |r| above about 0.285 is significant, and no calculation per pair is needed.

Go deeper Statistical significance is not economic significance

A significant correlation can still be too weak to build a trade or a hedge on, as the huge-sample example shows. Conversely, a large correlation in a tiny sample can fail the test. Report both the size of $r$ and the evidence behind it. And remember the test says nothing about cause: two series can move together because of a third factor or by coincidence.

Go deeper Link to simple regression

If you regress one variable on the other with a single explanatory variable, the t-statistic of the slope coefficient is exactly the same number as the t-statistic for the correlation, with the same $n-2$ degrees of freedom. Testing “the slope is zero” and “the correlation is zero” are the same hypothesis seen from two sides. You meet the regression version in Learning Module 10, simple linear regression.