Critical Values of r

Three points

Three students’ hours studied and scores are (2, 3), (5, 6) and (8, 8). Their correlation coefficient is r = 0.993, very close to 1.

That says little. Almost any three points lie close to some straight line, so a high r from three points can easily come from two quantities that are not related at all. The smaller the sample, the larger r must be before it counts as evidence.

The three points (2, 3), (5, 6) and (8, 8). They lie almost on one line, and r = 0.993.

The hypotheses

The sample’s r estimates $\rho$ρ, the correlation of the whole population the pairs come from. The null hypothesis is $H_0$H₀: $\rho =0$ρ = 0, no straight-line correlation in the population.

The alternative depends on the question asked before the data are seen. $H_1$H₁: $\rho \ne 0$ρ ≠ 0 asks whether there is any correlation, a two-tailed test. $H_1$H₁: $\rho >0$ρ > 0 asks whether there is positive correlation, and $H_1$H₁: $\rho <0$ρ < 0 negative correlation, each a one-tailed test.

The table of critical values

For each sample size n, the table gives the critical value: how large r must be to reject $H_0$H₀. At the 5 percent level, two-tailed, it is 0.878 for n = 5, 0.632 for n = 10 and 0.444 for n = 20. For n = 3 it is 0.997, so the three students’ r = 0.993 is not enough to reject $H_0$H₀.

For a one-tailed test at 5 percent, all of the 5 percent is in one tail, so the critical values are lower: 0.805 for n = 5, 0.549 for n = 10 and 0.378 for n = 20.

The values come from the t distribution. If $H_0$H₀ is true and both measurements are normal, $t=\frac{r\sqrt{n-2}}{\sqrt{1-r^2}}$t = r√(n − 2)/√(1 − r²) has n − 2 degrees of freedom. For n = 10, the two-tailed 5 percent value of t on 8 degrees of freedom is 2.306, and $r=\frac{t}{\sqrt{t^2+n-2}}=\frac{2.306}{\sqrt{5.318+8}}=0.632$r = t/√(t² + n − 2) = 2.306/√(5.318 + 8) = 0.632.

Ten pairs

Ten pairs give r = 0.81, and the question is whether there is any correlation: $H_1$H₁: $\rho \ne 0$ρ ≠ 0. The critical value for n = 10, two-tailed at 5 percent, is 0.632.

0.81 > 0.632, so reject $H_0$H₀. There is evidence at the 5 percent level of correlation between the two measurements in the population. The p-value is 0.0045.

The critical value for n = 10, two-tailed at 5 percent, is at 0.632. The sample’s r = 0.81 is beyond it.

Falling short

Another ten pairs give r = 0.41. 0.41 < 0.632, so do not reject $H_0$H₀. The p-value is 0.239: with no correlation at all in the population, ten pairs give an r at least this far from 0 about 24 times in 100.

That does not show $\rho =0$ρ = 0. Ten pairs are simply too few to tell a correlation of about 0.4 from chance. A larger sample might.

For a negative r, the two-tailed test compares its size: r = −0.81 with n = 10 is also beyond 0.632, so $H_0$H₀ is rejected. For a one-tailed test with $H_1$H₁: $\rho <0$ρ < 0, r must be below −0.549.

What the test assumes

The critical values assume both measurements come from a bivariate normal distribution. On a scatter plot that looks like an even, oval cloud: no bend, no fan shape, and no outliers.

Six pairs of heights and reaches, (1, 2), (2, 4), (4, 4), (5, 7), (7, 7) and (8, 9), form such a cloud. With x̄ = 4.5 and ȳ = 5.5, Sxx = 37.5, Syy = 33.5 and Sxy = 33.5, so $r=\frac{33.5}{\sqrt{37.5\times 33.5}}=\frac{33.5}{35.44}=0.945$r = 33.5/√(37.5 × 33.5) = 33.5/35.44 = 0.945. For n = 6, two-tailed at 5 percent, the critical value is 0.811, and 0.945 > 0.811: reject $H_0$H₀.

One outlier can create or hide a correlation on its own, and a curve can give a small r with a strong relationship. In either case the table’s critical value does not apply.

The six points (1, 2), (2, 4), (4, 4), (5, 7), (7, 7) and (8, 9): an even band with no bend and no outlier, and r = 0.945.

Significant is not the same as strong

The critical value falls as n grows. With n = 20, r = 0.45 is significant two-tailed at 5 percent, since 0.45 > 0.444, yet it is only a moderate correlation. A significant r says the correlation is unlikely to be chance, not that it is strong.

The usual mistakes

Using the critical value for the wrong n. The table is read at the number of pairs: ten pairs, n = 10.

Using the two-tailed value when the question named a direction in advance, or the one-tailed value when it did not. For n = 10 at 5 percent they are 0.549 and 0.632.

Saying $H_0$H₀ is proved when r falls short. Not rejecting $H_0$H₀ means only that the sample was not enough evidence of a correlation.

Reading a significant r as a strong one. Significance depends on n as well as on r.

Two researchers, one r

In the application below, ten tomato plants give r = 0.600. One researcher predicted positive correlation in advance and the other asked only whether there is any correlation, so they compare the same r with 0.5494, one-tailed, and 0.6319, two-tailed, both for n = 10 at 5 percent.