5.6 Tests
A sample coefficient \(r_s\) estimates the population coefficient \(\rho _s\), and the question is whether an observed \(r_s\) is far enough from zero to rule out \(\rho _s=0\). The critical values are tabulated for small \(n\) and approximated for large \(n\).
- (i).
- For \(H_0:\rho _s\leq 0\) against \(H_a:\rho _s>0\), reject \(H_0\) if \(r_s\) exceeds the critical value at level \(\alpha \).
- (ii).
- For \(H_0:\rho _s\geq 0\) against \(H_a:\rho _s<0\), reject \(H_0\) if \(r_s<0\) and \(\left |r_s\right |\) exceeds the critical value at level \(\alpha \).
- (iii).
- For \(H_0:\rho _s=0\) against \(H_a:\rho _s\neq 0\), reject \(H_0\) if \(\left |r_s\right |\) exceeds the critical value at level \(\alpha /2\).
The halving belongs to the two-sided test alone. A one-sided test spends its whole \(\alpha \) in one tail, and using \(\alpha /2\) there makes the test conservative without cause.
For moderate \(n\), roughly \(20<n<40\), \[t=r_s\sqrt {\dfrac {n-2}{1-r_s^{2}}}\ \thicksim \ t_{n-2},\] referred to one- or two-tailed critical values of the \(t\) distribution on \(n-2\) degrees of freedom. An equivalent form sometimes quoted is \[F=\dfrac {1+\left |r_s\right |}{1-\left |r_s\right |}\ \thicksim \ F_{(n-2,\,n-2)} .\] For \(n\geq 40\) the simpler large-sample result \[Z=r_s\sqrt {n-1}\ \thicksim \ N(0,1)\] is adequate.
Example 5.5. Two agronomists independently rank five maize varieties for drought tolerance:
| variety | A | B | C | D | E |
| \(R(X_i)\), first agronomist | 1 | 2 | 3 | 4 | 5 |
| \(R(Y_i)\), second agronomist | 2 | 3 | 1 | 5 | 4 |
| \(d_i=R(X_i)-R(Y_i)\) | \(-1\) | \(-1\) | \(2\) | \(-1\) | \(1\) |
| \(d_i^{2}\) | 1 | 1 | 4 | 1 | 1 |
Test at the \(5\%\) level whether the two agronomists agree, against the one-sided alternative \(H_a:\rho _s>0\).
Coefficient. Here \(n=5\), \(\sum d_i^{2}=8\) and \(n^{3}-n=125-5=120\), so \[r_s=1-\dfrac {6\sum d_i^{2}}{n^{3}-n}=1-\dfrac {6\times 8}{120} =1-\dfrac {48}{120}=0.6 .\]
Critical value. For \(n=5\) the one-tailed \(5\%\) critical value is \(r_{s(0.05,5)}=0.900\).
Decision. Since \(0.6<0.900\) we fail to reject \(H_0\). There is no evidence at the \(5\%\) level that the agronomists agree more than chance would produce.
Note 5.6. A coefficient of \(0.6\) sounds substantial, and it is not significant. The reason is \(n=5\): there are only \(5!=120\) possible rankings, and enumerating them gives the exact null distribution directly. It shows \[P\left (r_s\geq 0.6\right )=\dfrac {21}{120}=0.175,\] so a value of \(0.6\) or better arises by chance in about one ranking in six. Only \(r_s\geq 0.9\) is rare enough, with \(P=5/120=0.0417\), and \(r_s=1\) — perfect agreement — has \(P=1/120=0.0083\).
This is worth remembering whenever a correlation is quoted from a handful of observations. With \(n=5\) nothing short of near-perfect agreement can reach the \(5\%\) level, so a non-significant result carries almost no information. The sample was too small to answer the question, which is a different finding from the two agronomists disagreeing.
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