Question

The test statistic for testing the significance of Spearman's rank correlation coefficient \((r_s)\) is :

• A. \(t = \dfrac{r_s\sqrt{n-2}}{\sqrt{1-r_s^2}}\)
• B. \(t = \dfrac{r_s\sqrt{n-1}}{\sqrt{1-r_s^2}}\)
• C. \(t = \dfrac{\sqrt{1-r_s^2}}{r_s}\sqrt{n-1}\)
• D. \(t = \dfrac{\sqrt{1-r_s^2}}{r_s}\sqrt{n-2}\)

Hint:

The test statistic for Spearman's \(r_s\) has the same form as the one used for testing Pearson's correlation coefficient, with \((n-2)\) degrees of freedom.

Answer 1

The Correct Option is A

Step 1: State the hypothesis. Let \(r_s\) be the sample Spearman rank correlation coefficient computed from \(n\) paired ranks. To test \(H_0: \rho_s = 0\) (no association between the ranks) against \(H_1: \rho_s \neq 0\), a test statistic is needed whose distribution is known under \(H_0\).

Step 2: Use the analogy with the ordinary correlation coefficient. Just as Pearson's \(r\) is tested using \(t = \dfrac{r\sqrt{n-2}}{\sqrt{1-r^2}}\) with \((n-2)\) degrees of freedom, for reasonably large \(n\) (usually \(n > 10\)) the same form applies to \(r_s\), since \(r_s\) behaves like an ordinary correlation coefficient computed on the ranks. So \[t = \frac{r_s\sqrt{n-2}}{\sqrt{1-r_s^2}}.\]

Step 3: Identify the degrees of freedom. Two quantities (analogous to the two regression constants in the linear fit between ranks) are estimated from the \(n\) paired ranks before computing \(r_s\), so the statistic follows Student's \(t\) distribution with \((n-2)\) degrees of freedom under \(H_0\).

Step 4: Apply the decision rule. Compute \(|t|\) from the sample and compare with the tabulated \(t\) value at the chosen level of significance with \((n-2)\) df. If the computed value exceeds the table value, \(H_0\) is rejected and \(r_s\) is declared significant.

Step 5: Match with the options. This is exactly option (A). \[\boxed{t = \dfrac{r_s\sqrt{n-2}}{\sqrt{1-r_s^2}}}\]