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Retrospective study involving three groups of patients. Group 1 are patient's diagnosed with sepsis. Group 2 are patient's diagnosed with a localized infection. Group 3 are "healthy" controls. I'm looking at parameters called cell population data available on a haematology analyser. For example, the CPD parameter LY-X is normally distributed in group 2 and 3 but doesn't show a normal distribution in group 3. I used Shapiro Wilks in SPSS to assess for normality. I did this to find out if I need a parametric or non parametric test. I've also tried to transform my data as logs but this hasn't changed the outcome. I don't know how to compare the three groups of patients given two are normally distributed and one is not. It would be easier if all three groups were consistent. I hope this makes sense. I have no statistics background and my university will not provide any help.

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    $\begingroup$ Welcome to Cross Validated! What does the SW test report back? How does that compare with your visual assessment of the data, such as with a histogram, kernel density plot, or normal quantile-quantile plot? $\endgroup$
    – Dave
    Commented May 23 at 21:49
  • $\begingroup$ I wonder whether Steven Frank's How to read probability distributions as statements about process is relevant? Is the difference between the distributions an artifact of the way the data were gathered, or is it trying to you something? $\endgroup$ Commented May 24 at 8:25
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    $\begingroup$ Are you sure that you have asked the right question? What if the effect of infection on the cell population data is a change of distribution shape rather than (or as well as) a change in mean? In that case you would be seemingly ignoring interesting information in trying to do a test of means or any other index of population centrality. (As implied by @SimonCrase's comment.) You might want to be asking blood cell biologists about the distributions rather than statisticians for a potentially unhelpful test. $\endgroup$ Commented May 28 at 2:17

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First, regression and ANOVA do not require normal data. They make an assumption of normal errors, but even this is often not that vital. So, the fact that one group is not normal and the others are is irrelevant.

Second, don't use SW to test for normality. Use quantile normal plots.

Third, if you do find that the residuals are markedly non-normal, you can use robust regression or quantile regression.

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Chiming in on @Peter Flom's answer, you do NOT need each group to be normally distributed. It is the residuals (for ANOVA) or errors (regression) which need to be normally distributed. And normality of the groups is neither a neccesary, nor a sufficient condition for normality of the residuals.
So feel free to run an ANOVA, or a regression. And then analyze the residuals for normality; you may have a good surprise.
Even if a formal test for normality fails (such tests are not fool-proof; they often fail for small sample sizes, and will almost always fail for very large sample sizes. And they are negatively affected by ties -i.e. measurements with the same value- which is in practice a common situation, due to the limited resolution of measurement equipment). As long as the residuals appear unimodal, with not too large a skew, the F test will be quite robust. You could even run a series of 2-sample t-tests; t-tests are also quite robust to departures from normality, provided that you use the Welch variant (to avoid issues with non-equality of the variances). And of course use a proper multiple comparison correction. Large sample sizes also help here (because of the CLT). If really a parametric test is not warranted, you could use a Kruskall-Wallis (K-W) non-parametric test. But you need to be careful about what it tests for: it is not about equality means, nor medians (as often reported: but that needs very strong assumptions of symmetry, or "exact same" distribution, for all the groups, which evidently your data does not comply with). It is instead about stochastic superiority (comparing 2 groups A & B, $P(X_A>X_B)>.5$), and for K-W, among all the groups compared there exists at least one pair (i,j) such that $P(X_i>X_j)>.5$.
Last, but not least, you can use Bootstrapping. See here for example (R package) or here on CV.

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  • $\begingroup$ Very nice answer ! :) (+1) obviously :) $\endgroup$
    – Joe King
    Commented May 24 at 11:47

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