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In addition to the question, please do correct anything below that I have incorrect.

I understand how the test statistics for a t-test and WRS/MWU test are calculated. In a t-test, $t=\frac{\bar x_{1}-\bar x_{2}}{SE}$, where the formula for $SE$ comes from a normal approximation, which is justified by the central limit theorem.

Even though the central limit theorem is used to justify generating a sampling distribution using a normal approximation, I understand from other posts on this site that the CLT isn't always guaranteed to generate a normal sampling distribution from non-normal date. It is my understanding that this is why normality is still a standard assumption for a t-test.

On the other hand, the calculation of $W$ or $U$ in a WRS/MWU is obviously a little more involved but does not involve any invocation of a distribution up to this point. Thus, why this test is considered distribution free.

However, once you try to calculate a p-value from a WRS/MWU test, I am confused as to how it remains different from a t-test in it's reliance on normality.

Just looking at wikipedia, the process for generating a p-value from $U$ simply involves generating a Z-score from the standard form $Z = \frac{U - m_{u}}{\sigma_{u}}$, where $m_{u} = \frac{n_{1}n_{2}}{2}$ and $\sigma_{u} = \sqrt{\frac{n_{1}n_{2}(n_{1} + n_{2}+ 1)}{12}}$, the latter two equations being derived from a normal approximation, when the sample sizes are "large enough." This sounds a lot like the CLT to me.

At this point, the steps for obtaining a p value seem the same between a t-test and a WRS/MWU test:

  1. The t-score in the t-test and z-score in WRS/MWU are both obtained using sampling distributions calculated from normal approximations
  2. These scores are then used on sampling distributions (t-distribution vs. normal distribution) to obtain the probability of an equal or more extreme score given the null hypothesis

Therefore, given that:

  1. Both these test utilize normal approximations to obtain p-values
  2. That the justification for both of these approximations seem to come from the CLT
  3. The reason that normality is an assumption for the t-test is because it is still possible for the CLT to be affected by non-normal data

Why is normality not also an assumption for p-values calculated from a WRS/MWU test?

Thanks for your help!

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  • $\begingroup$ In its original form, the t-test does not require a normal approximation. Its distribution is derived from the fact that the variables are exactly normal. If that is not the case then it only holds asymptotically, because of the CLT as you say. $\endgroup$ – JohnK May 27 '15 at 17:57
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First, let us address the problem with the link you posted. Assume there is a distribution with finite mean ($\mu$), variance ($\sigma^{2}$ and skew (third central moment = $\rho$). The Berry-Essen theorem states that

$|F_{N}(x) - \Phi(x)| \leq \frac{0.8 \rho}{\sigma^{3} \sqrt{N}}$

where $F_{N}(x)$ is the CDF obtained after a standard CLT-type scaling of a sum. Notice that for a fixed mean and variance, the larger the skew, the slower the convergence to normality. The author in the previous account uses loose parsing when he says it 'fails'. What he means is normality like behavior is not achieved. (Heavy tailed distributions have high skew)

For the Mann-Whitney test, the reason for non-normality is as $N \rightarrow \infty$, $\sigma_{U}$ and $\mu_{U}$ blow up. Finiteness is required for the CLT. When this condition is violated, you can end up converging in distribution to something else (see the section titled 'Relation to the law of large numbers' in http://en.wikipedia.org/wiki/Central_limit_theorem)

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  • $\begingroup$ $\rho$ in Berry-Esseen is not the third central moment* (nor strictly speaking is third central moment the same as skewness; since skewness is unit free but third central moment is in the original variable's units cubed). Look carefully at the definition of $\rho$ here: en.wikipedia.org/wiki/Berry-Esseen_theorem $\endgroup$ – Glen_b -Reinstate Monica May 25 '17 at 5:08

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