I am working on comparisons between two time points 2013 and 2019, I have 180 subjects. I have a numeric "score" from 2013 and 2019. I calculated the average score at both time points and would like to assess significance. I've run a paired T-test on this, however when I check the normality assumption it is not there. So I moved onto using the Wilcoxon test for non parametric data. I suppose my main question is, reading a few things online it appears the T-test tests if the average score from 2013 is significantly different from the 2019 average. Does the Wilcoxon test do the same? A few of the things I've read seem to indicate it compares median?
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The term "Wilcoxon test" tends to be reserved for an unpaired test: Wilcoxon Mann-Whitney U. What exactly do you want to do? Do you indeed have pairing (such as before and after on the same subjects)? Is your lack of normality on the pairwise differences? // The Wilcoxon Mann-Whitney U test has a funky null hypothesis. There are advocates of using Wilcoxon as a test of means, however, when normality conditions are not met. – Dave Aug 11 '21 at 18:54
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Yes it is before and after (the same survey administered to the same 180 people once in 2013 and once in 2019), I have been using Wilcoxon signed-rank test, at least I thought I was with: wilcox.test(before1, after1, paired = TRUE). To assess the normality I conducted a Shapiro Wilks test on the differences (2013 scores-2019 scores) – Jamie Imada Aug 11 '21 at 19:00
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[Normality testing with a test like Shapiro-Wilk is less helpful than one would hope.](https://stats.stackexchange.com/questions/2492/is-normality-testing-essentially-useless) – Dave Aug 11 '21 at 19:03
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This was an interesting thread, thanks Dave. I'm wondering though, what is the appropriate path forward? – Jamie Imada Aug 11 '21 at 19:17
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The normality issue is delicate, and I have described it on here a few times. I want to take the stance that we should default to the Wilcoxon test, as it outperforms the t-test in some situations where t-test assumptions (particularly normality) are violated and performs almost as well when t-test assumptions are met. However, colleagues and clients like to look under the hood of my work and are more comfortable using the t-test method that they know. Given that the t-test is quite robust under many circumstances, I relent in my pursuit of nonparametrics and rely on convergence theorems. – Dave Aug 11 '21 at 21:52
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1Something else to keep in mind is that, if the normality assumptions are violated, differences in means might not even be interesting anymore or at least not all that is interesting. Consider $exp(1)$ and $N(1, 1)$. Those have the same mean and even the same variance, so testing either *should not* result in rejections of such equality, yet those are rather different distributions. – Dave Aug 11 '21 at 21:54
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I think what @Dave was getting at is that testing for normality once and moving on isn't a great approach. There's a reason there isn't one gold standard "perfect" test for the assumptions of least squares-based statistics. Have you tried just looking at the data? Does the lack of normality seem reasonable? Additionally, I hate to say it because I wouldn't want to insult anyone's intelligence. I find that many people misinterpret the results of tests like Shapiro-Wilk and Levene's. – Kat Aug 11 '21 at 21:58
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@Kat You are extremely right that people misinterpret the p-values resulting from Shapiro-Wilk (deciding if the data are normal or if nonparametric approaches should be used) and Levene's test of variance (deciding if Welch's t-test should be used). On the latter of those, I take the stance that we should default to Welch's t-test, and R software agrees with me, as `t.test` assumes unequal variance without a user override. – Dave Aug 11 '21 at 22:01
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@Dave I don't think that it is the case that "Wilcoxon test" is reserved for the rank sum test. It's perhaps more common in the stats literature but in a number of application areas seems to more often refer to the signed rank test. (It's usually best to be explicit though, because it's definitely ambiguous either way) – Glen_b Aug 12 '21 at 03:07
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The Signed Rank test compares pseudomedians rather than medians. See https://stats.stackexchange.com/questions/406984/pseudo-median-from-wilcoxon-signed-rank-test-larger-than-sample-median/407051#407051 for a discussion of a case that makes a difference. If the assumption (required under the null, but not necessarily under the alternative) of the symmetry of pair differences, it would be a test of median difference, but even when that assumption under the null is true, you don't know that it holds in the population (you don't know the null holds, that's what the test is for), so the estimate.. – Glen_b Aug 12 '21 at 04:17
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... of the shift is best presented as what it is, which is not a median of the differences (nor a difference of medians!). Also see the discussion in the later part of this answer: https://stats.stackexchange.com/questions/214472/why-is-my-spearmans-rho-result-contradicting-my-wilcoxon-matched-pairs/214478#214478 – Glen_b Aug 12 '21 at 04:18
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Wilcoxon rank test tells you whether there is a location shift between two samples. Wilcoxon signed-rank test tests is for one sample only and tests the null hypothesis $\mu = \mu_0$ for $\mu_0$ of your choice
So, if you wanted to run Wilcoxon rank test on your sample then you can see if the samples are shifted positively or negatively

Kaan Yolsever
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