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Correlation Significance Test

Two paired variables, is their correlation real, or sampling weather?

Test a correlation

Paired X and Y values, in the same order.

Quick answer: t = r√(n−2)÷√(1−r²). Example: r = 0.9953 on 5 pairs gives t(3) = 17.73, p = 0.0004. Rule of thumb: |r| above about 2÷√n is in significant territory.

Is this correlation distinguishable from zero?

A sample correlation of 0.4 can be decisive evidence with n = 200 and pure noise with n = 8. The significance test converts r into a t statistic on n − 2 degrees of freedom, asking whether the population correlation ρ could plausibly be 0. The calculator also reports the Fisher-z confidence interval for ρ, the more informative half of the answer.

Correlation test statistict = r√(n − 2) ÷ √(1 − r²),   df = n − 2
Worked example

Five students: study hours 2, 3, 4, 5, 6 against marks 50, 55, 60, 68, 72. r = 0.9953, nearly a straight line. t = 17.73 on 3 df, p = 0.0004. Even with only five pairs, a correlation this tight is not a coincidence of sampling. The 95% interval for ρ (about 0.943 to 0.9996 by Fisher's method) stays sky-high, small n, but an unmissable relationship.

What significance does not rescue

Watch out

Quoting r² = 0.99 as "99% accurate predictions". r² is the share of variance linearly shared between the two variables in this sample, not a forecast guarantee, and it inflates fast on tiny samples.

Frequently asked questions

How large must r be to matter with my n?

Roughly, |r| above 2÷√n is in significant territory: 0.63 at n = 10, 0.28 at n = 50, 0.20 at n = 100. The exact fence is what this page computes.

Does the test assume normality?

The t conversion assumes the pair is roughly bivariate normal: in practice, no extreme outliers and no strong curvature. For a ranks-based alternative, use Spearman's coefficient.