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)

  • $\begingroup$ Please let me know in comment if I am missing something. $\endgroup$ – Woeitg Feb 12 '17 at 15:20
  • $\begingroup$ Link to related unanswered question: stats.stackexchange.com/questions/217422/… $\endgroup$ – Woeitg Feb 12 '17 at 15:23
  • $\begingroup$ I don't think you can use $\frac{x-\mu}{\sigma}$ to convert Spearman's $\rho$ to z-score here. You need Fisher transformation as the linked wikipedia page on the question in your comment shows. Check this question and Nick Cox's answer: stats.stackexchange.com/questions/59714/… . I don't think it is hard to use Fisher transformation to obtain z-score. $\endgroup$ – T.E.G. - Reinstate Monica Feb 12 '17 at 15:46
  • $\begingroup$ The problem is I get different p-value using fisher transformation and t-distribution. For example in the wikipedia xample, rho=-0.1757 and p-value=0.62. Using Fisher transformation. I get F(r)=Arctan(-0.1757)=-0.1775 and z-score=SQRT((10-3)/1.06)*-0.1775=-0.456. Using normal distribution table, pvalue==2*(1-NORMSDIST(ABS(-0.456)))=0.648 which is different from wikipedia example $\endgroup$ – Woeitg Feb 12 '17 at 16:19
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    $\begingroup$ I see and I am sorry if I misled you, but don't get confused with Glen_b's answer. That question is about comparing two separate correlations ("running separate Spearman's analyses for each group"). I sent it to show that using Fisher transformation for Spearman's $\rho$ is admissible. $\endgroup$ – T.E.G. - Reinstate Monica Feb 12 '17 at 16:54

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}

  • $\begingroup$ you're welcome. It is the closed form approximation that works, so testing is not really needed, unless you mean "debug your code." $\endgroup$ – JoleT Feb 12 '17 at 17:22

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