A correlation is not significant or not on its own
Ask whether r = 0.6 "is significant" and the honest answer is: at what n? The p-value for a Pearson correlation rides on both numbers, because the test statistic does: t = r√(n − 2) ÷ √(1 − r²), judged against a t distribution with n − 2 degrees of freedom. At n = 20, r = 0.6 gives t = 3.182 and p = 0.0052, clearly significant. The same r at n = 8 gives t = 1.837, p = 0.116, nothing established. The correlation did not change; the evidence behind it did. That is why the tool asks for the pair (r, n) and nothing else, and why reporting r without n is reporting half a result.
Read the t formula as a signal-to-noise ratio. The numerator grows r with √(n−2): more pairs, more signal. The denominator √(1 − r²) shrinks as the relationship tightens, charging less noise. Small samples of weak correlations lose on both: n = 12, r = 0.3 gives t = 0.994, p = 0.344, while n = 102, same r, gives t = 3.145, p = 0.002. Significance is a property of the dataset, not of the coefficient alone.
The confidence interval tells the rest
The calculator adds the 95% interval for ρ built with Fisher's z transform, because r's own sampling distribution is skewed near ±1 and the plain plus-minus formula misbehaves there. Fisher's transform stretches r onto a scale where the standard error is simply 1/√(n − 3); compute the interval there, then squash it back. For r = 0.6 at n = 20 the interval runs roughly 0.21 to 0.83: significant, yes, but compatible with anything from weak to strong. The p-value answered yes/no; the interval answers how well you actually know ρ, and it is usually the more honest number.
Testing a correlation you selected because it looked big. If you scanned 40 variable pairs and tested the largest r, its p-value is meaningless: selection inflated it before the test began. Test pre-chosen pairs, or treat screened correlations as hypotheses for fresh data.
Frequently asked questions
What sample size makes r = 0.5 significant?
Around n = 16 at the 0.05 two-tailed level: t = 0.5 x sqrt(14)/sqrt(0.75) = 2.16, p = 0.048. Smaller samples leave even a strong-looking 0.5 unproven.
Can I get a one-tailed p-value here?
Halve the two-tailed value when the observed r is in the predicted direction. The tool reports the standard two-tailed default.
Does a significant r prove the variables are related linearly in the population?
It rejects zero linear correlation. Curved relationships can hide behind small r values, and a significant r never establishes causation on its own.