Estimation of critical values in Spearman rank correlation According to Wikipedia, to evaluate significance of the Spearman rank correlation, you can use:
$$t = r \sqrt{\frac{n-2}{1-r^2}}$$
but I don't understand how to use this or how this generates the values as shown here:
http://web.anglia.ac.uk/numbers/biostatistics/spearman/local_folder/critical_values.html
Nothing I try matches and online explanations are scarce.
 A: There's no great mystery here.


*

*The distribution of the Spearman correlation is discrete. (The set of ranks of $n$ values are discrete, so the Spearman correlation will necessarily be discrete.)

*The relevant Wikipedia page lists several approximations for the distribution of the Spearman correlation coefficient; I think the t-approximation is the second listed. The resulting p-values are not exact in small samples (but are reasonably close considering it's a continuous approximation of a discrete distribution - see the plot).

*Since the exact small sample distribution (under the null) for the Spearman correlation can be computed, exact small sample tables exist. The table you link to appears to be a table of critical correlations (yielding type I error no greater than the listed significance levels) derived from the exact distribution.

A continuity correction would probably help here.
Edit: actually, it looks like a suitable continuity correction helps quite a lot overall (however it's slightly less accurate for small portions of the range of $r_s$, so while most of the time it improves the approximation, sometimes it doesn't).
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How to use the t-statistic that Wikipedia gives: you can get one- or two-tailed p-values from the cdf of the $t$ with $n-2$ d.f. just in the same way as you would for any t-statistic with given degrees of freedom.
You can get critical values from the inverse CDF of the $t$. As noted above, these will be approximate, but the approximation kicks in pretty quickly. You don't need very large $n$ before it's suitable for most purposes.
A: Here's how reconcile your table. For instance, look at the df=15 and $\alpha=0.05$, you see the critical value 0.446. Plug this thing into the formula, and look at the value of cumulative distribution function of Student t distribution with df=15-2. Here's the code in R:
> pt(0.446*sqrt(13/(1-0.446^2)),13)
[1] 0.9521751

You see that the right tail is about 5%, so, for two tailed test you get 10% in the tails - exactly what you asked with $\alpha=0.1$.
The formula in Wikipedia tell you that the expression is from Student t with n-2 degrees of freedom. Btw, this is how MATLAB calculates Spearman's critical value when there's a tie. Otherwise, it uses AS89 algo, to which I referred you in your other thread, but you didn't seem to like it :)
