Simulations by Razali et al (2011) showed that the Shapiro-Wilk test of normality provided the most power (at a fixed significance level) when compared to Anderson-Darling, Kolmogorov-Smirnov, etc. My question is this:

For data $X_1, \cdots, X_n$, suppose I want to test \begin{align*} H_0: X_i \sim F \qquad \text{vs} \qquad H_1: X_i \not\sim F \end{align*} for some continuous distribution $F$. This is equivalent to \begin{align*} H_0: \Phi^{-1}(F(X_i)) \sim N(0, 1) \qquad \text{vs} \qquad H_1: \Phi^{-1}(F(X_i)) \not\sim N(0, 1) \end{align*} through probability integral transforms, where $\Phi$ is the standard normal CDF. Since Shapiro-Wilk does well in terms of power, this procedure would give us a rather powerful test. Why isn't this idea more prevalent in the literature?

Razali, Nornadiah; Wah, Yap Bee (2011). "Power comparisons of Shapiro–Wilk, Kolmogorov–Smirnov, Lilliefors and Anderson–Darling tests"

  • 1
    $\begingroup$ Some clarification would help and might even lead directly to an answer: what, specifically is the alternative hypothesis implied by "most powerful"? $\endgroup$
    – whuber
    Commented Oct 11, 2018 at 18:10
  • $\begingroup$ Yes, I should clarify that tests of goodness of fit for distributions will rarely yield a uniformly most powerful test. For normality, at least, the Shapiro-Wilk appears to be the best for the select distributions used in Razali et al (2011) and when compared against the other tests listed above. When we perform goodness of fits for distributions other than normal, why can't we extend the Shapiro-Wilk framework to there? $\endgroup$
    – Tom Chen
    Commented Oct 11, 2018 at 18:40
  • 2
    $\begingroup$ Razali et al seem to have missed a lot of the goodness of fit literature, (or perhaps have simply chosen to essentially repeat earlier studies without citing them, but I'll assume that's not the case). The comparisons there seem to have been done (over a wider range of alternatives) decades earlier than their paper -- to the same conclusion (that the Shapiro Wilk is often fairly powerful against most of the interesting alternatives, in many cases having the highest power against typical alternatives people might find of interest). ... ctd $\endgroup$
    – Glen_b
    Commented Oct 11, 2018 at 21:44
  • $\begingroup$ ctd... However, typically more powerful tests have been published in the since the Shapiro-Wilk test; the Chen-Shapiro test (1995) is one example that often outperforms the Shapiro-Wilk on typical sorts of alternatives. $\endgroup$
    – Glen_b
    Commented Oct 11, 2018 at 21:44

1 Answer 1


The Shapiro-Wilk test is for testing normality with unspecified $\mu$ and $\sigma$.

In proposing to transform to uniformity and then standard normal, you're assmuing that $F$ is completely specified (no unknown parameters), so that $F(X)$ is standard uniform; and then that transformation to normality will produce only a standard normal (fully specified distribution).


  • If $F$ is completely known, the Shapiro-Wilk test will be wasting power on treating-as-unspecified parameters in the normal that are actually known; it may be better in that case to use a test specifically for a fully specified distribution such as the Anderson-Darling. (It might still do quite well, but I wouldn't automatically expect it would do quite as well as in the situation it was designed for and which the power comparison you mention was based on.)

  • If some parameters of $F$ are unknown, estimating those parameters of $F$ and computing $\hat{F}(X_i)$ will produce a conservative (and therefore lower-power than if it were completely specified) test. (This is akin to using a Kolmogorov-Smirnov test when you should be using a Lilliefors test, and with similar lowering of the power.)

  • The sorts of alternatives you may be interested in for $F$ may not be "typical" alternatives that the Shapiro-Wilk does particularly well on. There are alternatives where it's not best or even second best and it may be that for some choices of $F$ that the interesting alternatives might be focused on what would be a narrow subset of cases at the normal (e.g. perhaps this might be the case when testing the Pareto, where in some applications extreme-upper-tail differences might be of substantial interest, but would correspond to only a small subset of typical alternatives of interest when looking at testing normality)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.