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This question already has an answer here:

Are non-parametric methods preferable to parametric methods since the former do not force the model to have a parametric structure?

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marked as duplicate by Silverfish, usεr11852, gung, John, conjugateprior Dec 12 '15 at 4:27

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In some cases a parametric test confers more power when certain assumptions are met. As an example, if one were to compare exponential survival analysis, to weibull survival analysis, to Cox proportion hazards, the relative efficiency of the tests of the relative hazard parameter is 1:2:3 meaning that when data are actually exponentially distributed, the parametric assumptions are 3 times as efficient as those of the Cox model. Of course, even with machine precision engineering, few things in life are actually exponentially distributed in their survival times (except radioactive decay, interestingly).

The question of "what happens when assumptions fail" is important. What happens is much, much worse than simply a loss of power. Usually, the test becomes inconsistent in that the test will reject the null hypothesis is circumstances where it is true. The results, then, become very hard to interpret. Envisioning parametric methods as "enforcing" a structure does help one to see why this is an issue.

For that reason, I would agree that any statistical methods which rely on numerous assumptions, testable or otherwise, are generally not a good thing to rely upon. Assumption free tests can be a bit different from non-parametric, though, and some non-parametric tests are absolute power sinks, requiring thousands upon thousands of observations to make simple inference.

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  • $\begingroup$ That's very interesting, do you have some references, where I can read up more details regarding the efficiency of Cox vs. parametric proportional hazards models? I actually had trouble finding something that specific. $\endgroup$ – Björn Apr 7 '16 at 19:16
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For example, if a population is normal distributed, then using a two-sample t-test (a parametric tool) is more efficient than using a Mann-Whitney test (a non-parametric tool).

By efficient I mean that you make better use of the information contained in the data, in the sense that the variance of the parametric estimator is smaller than the variance of the non-parametric estimator; of course, the parametric method makes a more restrictive assumption about the nature of the population than does the non-parametric method.

For given sample sizes, the power of the parametric method is greater than the power of the non-parametric method; this reflects that you make better use of the information contained in the data with a parametric method, if you have been able to correctly identify the underlying distribution of the data.

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    $\begingroup$ Generally speaking, the t-test is always more efficient than a Mann-Whitney. Here, giving efficiency it's proper meaning, I am referring to the power of the test. The t-test is root-n efficient whereas Mann-Whitney is root-root-n efficient. $\endgroup$ – AdamO Dec 11 '15 at 23:48
  • $\begingroup$ @AdamO Can you support that comment with a link or a derivation? (I don't think it can be correct -- I think if your claim were true, the ARE of the Mann Whitney would have to be 0, since that looks at what happens when $n\to\infty$. In fact, aside from a $\sqrt{n}$ factor for both scaling the size of the location-shift, the power curves of the two remain in almost the same relative position as $n$ changes.) $\endgroup$ – Glen_b Dec 11 '15 at 23:54
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    $\begingroup$ @Glen_b Using the asymptotic results, $\sqrt{n} \left( \bar{X} - \mu \right) $ has a limiting normal distribution and $n^{1/4} \left( \mathbb{F}^{-1}(.5) - F^{-1}(.5) \right)$ has a limiting normal distribution. The rank test is a test of medians when the distribution is symmetric, hence also a test of means. Some notes discussing this at greater detail here, stat.washington.edu/~jaw/COURSES/580s/581/LECTNOTES/ch6c.pdf $\endgroup$ – AdamO Dec 12 '15 at 0:03
  • $\begingroup$ @AdamO the relevant calculation of relative efficiency would seem to be exercise 4.6 on page 47 of your reference (part (ii) for the case of the normal). While no answer is given there (though it's simple enough and can be found in many places), you can see that $12\sigma^2(\int f^2(x)dx)^2$ is a non-zero constant. If I recall correctly it comes out to be $3/\pi$ $\endgroup$ – Glen_b Dec 12 '15 at 0:58
  • $\begingroup$ While that's an asymptotic result, clearly there can't be different powers of $n$ in the finite-sample efficiency -- if there were, the ratio of the efficiencies would involve $n$ (and therefore the one with the smaller power of $n$ would have ARE 0, not $3/\pi$). $\endgroup$ – Glen_b Dec 12 '15 at 1:08

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