Testing for zero correlationEdexcel International A Level Further Maths: Revision notes

A sample correlation coefficient comes from only some of the population, so even if there is no correlation in the population, a sample can show a non-zero value by chance. A hypothesis test decides whether the sample value is far enough from 0 to be convincing evidence of correlation in the population. The population correlation coefficient is written $\rho$ (product moment) or $\rho_s$ (Spearman). The hypotheses are always about the population parameter, never about the sample value $r$ or $r_s$.

$H_0$: $\rho=0$ (no correlation). Two-tailed: $H_1$: $\rho\neq0$, when the question asks whether there is any correlation. One-tailed: $H_1$: $\rho>0$ (positive correlation) or $H_1$: $\rho<0$ (negative correlation), when the question states a direction in advance. The direction comes from the wording of the question, not from the sign of the sample value. Use the same pattern for Spearman's, with $\rho_s$.

Use it when the data are random pairs from a bivariate Normal distribution, so that the relationship is linear. Method: find $r$, find the critical value for the sample size $n$ and the significance level from the tables, and compare. Two-tailed: reject $H_0$ if $|r|\ge$ critical value. One-tailed positive: reject if $r\ge$ critical value. One-tailed negative: reject if $r\le-$ critical value. Example: $n=10$, $r=0.62$, two-tailed 5% critical value $0.6319$. Since $0.62<0.6319$, do not reject $H_0$: there is insufficient evidence of correlation.

Use it when the data are ranks, or when the relationship is monotonic but not necessarily linear, or when the data are not bivariate Normal. Rank the data, find $r_s=1-\frac{6\sum d^2}{n(n^2-1)}$, and compare with the tabulated critical value for $n$, using the same rules as above. For $n=8$, the 5% critical values are $0.6429$ (one-tailed) and $0.7381$ (two-tailed). For $n=10$, they are $0.5636$ and $0.6485$. Worked example: $n=8$, $\sum d^2=26$ so $r_s=0.690$. For $H_1$: $\rho_s>0$ at 5% the critical value is $0.6429$, so reject $H_0$. With a two-tailed test, $0.690<0.7381$ and $H_0$ is not rejected.

If the sample value is in the critical region, reject $H_0$: 'there is sufficient evidence of (positive/negative) correlation between ... and ...'. If it is not, do not reject $H_0$: 'there is insufficient evidence of correlation between ... and ...'. Never say that this proves there is no correlation: the sample may be too small to detect one. A significant result shows evidence of correlation, not that one variable causes the other. Larger samples give more reliable decisions and a smaller critical value.