3.4.1, 3.4.2.4 - Spearman's rank and significance

Association and inference

A scatter graph reveals a paired relationship; a coefficient summarises it, while an inferential test asks how compatible the observed relationship is with a stated null hypothesis under suitable assumptions.

$r_s$ lies from −1 to +1. +1 means identical rank order; −1 means completely reversed order. A value near zero indicates little monotonic association, not necessarily no relationship: a U-shaped pattern can have weak rank correlation despite a clear curved relationship. Inspect the graph and data before choosing the statistic.

The paired unit must be clear. Distance and water temperature from each of six distinct sites form six pairs; mixing one site's distance with another's temperature destroys the relationship. Ranked ordinal or numerical data can be suitable. The test assumes independent paired units; readings repeated at one site or strongly related neighbouring units may not supply independent evidence. Selection and measurement affect generalisation even when calculation is correct.

Descriptive $r_s$ summarises these pairs. Statistical inference compares it with what might occur under a null model, enabling a limited statement about evidence against no association. Neither a strong coefficient nor a small p-value establishes a cause: altitude, shade or time could confound an apparent geographical gradient. Relational enquiry therefore combines a clear question, paired measurements, graph inspection, calculation and contextual judgement.

Ranking and calculation without ties

Rank both variables in the same direction, keep each site's pair on the same row, and calculate the rank difference before applying the no-tie shortcut.

In the original illustration, ascending distance and temperature ranks are:

$\sum d^2=2$, so $r_s=1-6(2)/[6(6^2-1)]=1-12/210=0.942857\ldots$, or +0.943 to three decimal places. The strong positive rank association is imperfect because B and C swap order. Reversing the ranking direction for both variables leaves the answer unchanged; reversing just one changes its sign and misrepresents the original association.

Check every pair, $n$, the squared differences and the final interval −1 to +1. A negative difference still produces a positive squared difference. Keep unrounded working until the end. If both sets of raw values are ranked smallest =1, an increase in one with a decrease in the other should give a negative coefficient, providing a useful sign check.

Testing significance in context

A significance decision requires a stated hypothesis, level and appropriate reference distribution, rather than calling every non-zero coefficient significant.