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from what I understand the assumption of normality (that must be assumed if one wants to use parametric tests) refers to the sampling distribution of the mean. It does not imply that the distriubtion is normal in the population itself nor a in a given sample. With the CLT we can assume that with a large enough sample this assumption is met:

given random and independent samples of N observations each, the distribution of sample means approaches normality as the size of N increases, regardless of the shape of the population distribution

Therefore (from my understanding):

If we have a large sample ( I know the N > 30 is debatable, but say the N = 500 simple random sample) OR the variable of interest form population that we are sampling from is known to be normal, then the sampling distribution has to be normal and therefore we can use parametric tests. For example I stumbled across this in one of the scientific journals:

With large enough sample sizes (> 30 or 40), the violation of the normality assumption should not cause major problems (4); this implies that we can use parametric procedures even when the data are not normally distributed (8). If we have samples consisting of hundreds of observations, we can ignore the distribution of the data (3). According to the central limit theorem, (a) if the sample data are approximately normal then the sampling distribution too will be normal; (b) in large samples (> 30 or 40), the sampling distribution tends to be normal, regardless of the shape of the data (2, 8); and (c) means of random samples from any distribution will themselves have normal distribution (3). Link to the article

Going back to my question, if the normality assumption DOES NOT imply that the variable of interest in a sample is normaly distributed (only that the sampling distriubtion of the mean is normal). why do people plot histograms to see if the data in a sample is normally distributed? How is this relevant for the assumption of normality?

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    $\begingroup$ For any fixed sample size, there exists a population distribution (infinitely many, in fact) for which the CLT does not provide a reasonable approximation. For an extreme example, see the Cauchy distribution. $\endgroup$
    – knrumsey
    Jul 2, 2020 at 0:34
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    $\begingroup$ Using 'parametric' as a synonym for 'normal-based' is unfortunate. For example, there are several tests involving exponential distributions that are 'parametric'. // Unfortunately, there are online 'journals' nowadays in which publication requires payment of a fee and serious refereeing is not much of a consideration. One must check the reputation of a journal before taking its content seriously. $\endgroup$
    – BruceET
    Jul 2, 2020 at 0:46
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    $\begingroup$ In order for a 1-sample t-test to be valid, the t stat must have a Student's t dist'n. That requires (1) $\bar X$ in numerator to be normal, (2) $S$ in denom to be properly related to chi-sq dist'n, and (3) $\bar X$ and $S$ to be stochastically independent. So it's not enough to just quote CLT to show (1) is 'nearly' OK. See discussion on this pg for more on this topic. $\endgroup$
    – BruceET
    Jul 2, 2020 at 0:51
  • $\begingroup$ @BruceET thank you for the advice. What your saying is probably true. However in the Social Sciences parametric is used as a synonym for normal-based, since the tests we use always require this assumption. $\endgroup$
    – Bruno
    Jul 2, 2020 at 0:54
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    $\begingroup$ I probably should have clarified that I wasn't disputing what you said; I think we agree. Indeed the Cauchy doesn't have a mean for the sample mean to converge to. $\endgroup$
    – Glen_b
    Nov 13, 2021 at 9:51

2 Answers 2

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Some of the trouble is that the central limit theorem is that the convergence is a limit, not realized in finite sample sizes...

...except that it is realized in finite sample sizes if the data have a normal distribution. In fact, the distribution of the sample mean is normal if the sample size is just $1$.

Therefore, if you have normal data, you immediately have a normal sampling distribution.

Aother reason is that we sometimes do inference about parameters other than means. Another parameter we might want to test is the variance. Let's do a simulation to test the equality of variances for two equal exponential distributions.

library(ggplot2)
set.seed(2021)
B <- 1000
N <- 300
mean_ps <- var_ps <- rep(NA, B)
for (i in 1:B){
  
  x <- rexp(N, 1)
  y <- rexp(N, 1)
  mean_ps[i] <- t.test(x, y)$p.value
  var_ps[i] <- var.test(x, y)$p.value
}
s <- seq(0, 1, 0.001)
d1 <- data.frame(
  p = s,
  Quantile = ecdf(mean_ps)(s),
  Test = "Mean"
)
d2 <- data.frame(
  p = s,
  Quantile = ecdf(var_ps)(s),
  Test = "Variance"
)
d <- rbind(d1, d2)
ggplot(d, aes(x = p, y = Quantile, col = Test)) +
  geom_point() +
  geom_abline(slope = 1, intercept = 0) + 
  theme_bw()

enter image description here

The F-test of variance equality rejects the true null hypothesis much more often that the t-test of mean equality ($35\%$ rejections at $\alpha = 5\%$), which behaves approximately correctly (uniform distribution of p-values).

Finally, it's easy to check the data. To check the distribution of sample means, we need multiple sample means. There are ways of approximating such a distribution (e.g., bootstrap), but we don't have that distribution "for free" when we collect the data.

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To quote a source:

Said another way, CLT is a statistical theory stating that given a sufficiently large sample size from a population with a finite level of variance, the mean of all samples from the same population will be approximately equal to the mean of the population. Furthermore, all the samples will follow an approximate normal distribution pattern, with all variances being approximately equal to the variance of the population, divided by each sample's size.

So, per your question: "why do people plot histograms to see if the data in a sample is normally distributed?", one possible answer is that we are actually checking for "an approximate normal distribution pattern", per the source above.

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    $\begingroup$ Alright, so we plot a histogram to see if the distrubution strongly deviates from a normal distribution, right? But again, how is this connected to the assumption of normality? As I mentioned, this assumption reffers to the sampling distribution, not the distribution in a sample - just because the distribution within sample is not normal, does not mean that the assumption is violated. I’m not arguing for or agains anything, just trying to understand. $\endgroup$
    – Bruno
    Jul 2, 2020 at 13:16
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    $\begingroup$ This answer confuses the distribution of the data with the sampling distribution of a mean. $\endgroup$
    – whuber
    Jul 3, 2020 at 14:23

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