I know the approach using student t-test to derive p-value from spearman rho coefficient. I also saw permutation test for this operation, but I am somehow confused and prefer to stick to the z-score formula $frac{x-mu}sigma$.

In other words, how can I convert Spearman rho coefficient to the normal approximation z statistic with formula z-score formula $frac{x-mu}sigma$?

Link to related unanswered question: Spearman rho statistical significance value (z)

**Contents**hide

#### Best Answer

Let $R^x_1, R^x_2,ldots, R^x_{n_x}$ be the ranks of $x$ and $R^y_1, R^y_2,ldots,R^y_{n_y}$ be the ranks of $y$. Spearman correlation is determined as begin{equation} r_s = 1 – frac{6D}{n^3-n} end{equation} where begin{equation} D = sum_{i=1}^n d_i^2 = sum_{i=1}^n (R^x_{(i)}-R^y_{(i)})^2 end{equation}

Significance testing of $r_s$ is based on the standard normal distribution. Using the expected value of $D$ as begin{equation} E(D) = frac{n^3-n}{6} = frac{n(n-1)(n+1)}{6} end{equation} begin{equation} V(D) = frac{n^2(n+1)^2(n-1)}{36} end{equation} begin{equation} Z = frac{D – E(D)}{sqrt{V(D)}} Longrightarrow frac{x-mu}{sigma} end{equation}

### Similar Posts:

- Solved – Interpretation of Spearman’s rank correlation coefficient – beyond its significance
- Solved – Why is a Pearson correlation of ranks valid despite normality assumption
- Solved – Why is a Pearson correlation of ranks valid despite normality assumption
- Solved – Can Spearman’s correlation be run on z scores
- Solved – Problem with Spearman correlation in case of many ties