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I'm reading PDQ Statistics[1], I have a small base of statistics and I can't figure out this statement from the book.

They assure in the book that

It seems that most people who have taken a stats course forget this basic idea when they start to worry about when to use parametric statistics such as t tests. Although it is true that parametric statistics hang on the idea of a normal distribution, all we need is a normal distribution of the means, not of the original data.

Second point I find this sentence a bit fuzzy

The basic difference is that, in parametric statistics, the DV (note: dependent variable) is some measured quantity (a ratio or interval level variable), so it makes sense to calculate means and SDs. With nonparametric stats, the DV is usually either a count or ranking (such as of dead bodies, which happen often, or cures, which happen rarely) for which it makes no sense to calculate means (eg,“The average religion of Americans is 2.67”).

I always thought as also stated here that:

A parametric test is a test in which you assume as working hypothesis an underlying distribution for your data, while a non-parametric test is a test done without assuming any particular distribution.

Is the book a true simplification or it's totally wrong? Am I not catching something?

[1]: Norman, G.R. and D.L. Streiner (2003)
PDQ Statistics, Volume 1, 3rd Ed,
People's Medical Publishing House, Shelton, CT
(p17, p28)

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Although it is true that parametric statistics hang on the idea of a normal distribution,

Unless the author has previously redefined "parametric statistics" to exclude most parametric statistics, this statement is untrue. Parametric statistics do not in general "hang on the idea of a normal distribution" (though a lot of the commonly used tests do base their inference on a normal assumption they don't constitute more than a tiny fraction of the different things that are done and those are an even tinier fraction of what could be done with parametric statistics).

If I compare the means of two samples that I assume to be drawn from exponential distributions or compute a confidence interval for the ratio, I will be doing a parametric statistical procedure but it won't relate to normality.

all we need is a normal distribution of the means, not of the original data.

This is also untrue. The t-statistic consists not only of a numerator, but also of a denominator, and the t-statistic has a t-distribution because of the way the ratio of the numerator to denominator is distributed (under the assumption of normal individual observations, the t-distrbution for the statistic follows).

Applying the CLT to the numerator does not give a t-statistic a t-distribution.

In the limit as $n\to\infty$, the use of two theorems (along with some other assumptions) can get you a statistic with an asymptotic normal distribution, but this is not the claim being made; there's also the issue that under non-normal sampling, in some situations the sample size has to be very, very large indeed before that normal approximation is useful.

With nonparametric stats, the DV is usually either a count or ranking

Actually, a lot of the corresponding nonparametric tests start with an assumption of continuity in order to derive the permutation distribution. If that's not the case, the calculations can still be done in principle but they can be complicated; a number of popular software packages by default either rely on asymptotic approximation or even ignore the occurrence of ties.

I would say that most commonly count data are analyzed with methods more directly suited to counts (including chi-square tests, Poisson or binomial regression) rather than using nonparametric tests. (However some books seem to count chi-square tests as nonparametric; maybe that's the intent there. One can perhaps make some argument for that in the case of chi-square goodness-of-fit tests)

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