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I have two groups of data (N > 30 in each group). I want to know if the two groups differ from each other. So I thought about making an independent t-test. But then I remembered that the groups must be normally distributed in order to perform a t-test. So I performed a Shapiro test and it appears that both groups are not normally distributed.

My question is: Am I still allowed to perform a t-test knowing that my groups are not normally distributed and that I have N > 30 in each group?

Thank you for your time.

Edit: When I do a QQ-plot on R with package ggpubr, both groups look like this (group 1 first, then group 2): This is the QQ-plot for the first group.

enter image description here

It seems that there are some flat "levels" in both cases.

Here are the histograms (group 1 first, then group 2): enter image description here

enter image description here

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  • $\begingroup$ Formal testing of normality is less helpful than one might hope. $\endgroup$ – Dave Jun 18 at 13:55
  • $\begingroup$ Seeing the QQ plots, are the histograms trimodal? $\endgroup$ – Dave Jun 18 at 14:53
  • $\begingroup$ @Dave No I don't think the histograms are trimodal. I also added them. I checked the link you posted above. According to what I could read, it really depends on how your data look like, right? In my case, I think my data are quite far away from "almost normally distributed", meaning that t-test is a no-go? $\endgroup$ – arkadryyx Jun 18 at 15:03
  • $\begingroup$ Kontroll seems to have a mode around 800, another around 120, and another around 150; Versuch appears to have a model around 100, a model around 120, and a model around 160. // Perhaps the t-test is a no-go. Some people argue that we should default to the Wilcoxon Mann-Whitney U test, wilcox.test in R, so we do not have to address issues like this. $\endgroup$ – Dave Jun 18 at 15:15
  • $\begingroup$ The shapes of those histograms make me wonder if a focus on the means using a t-test is going to capture all (or any) of the possibly interesting features of the groups. $\endgroup$ – Michael Lew Jun 18 at 21:01
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Here are fictitious data, sampled using R, somewhat similar to your data.

set.seed(1234)
x1 = c(rnorm(50,100, 2), rnorm(140, 125, 5), rnorm(40,155,3))
x2 = c(rnorm(50,105, 2), rnorm(150, 125, 5), rnorm(40,156,2))
summary(x1); length(x1); sd(x1)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  95.31  118.30  124.70  125.01  131.94  163.76 
[1] 230
[1] 18.12978
summary(x2); length(x2); sd(x2)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  98.53  117.40  124.34  125.83  130.64  160.12 
[1] 240
[1] 16.01616

par(mfrow=c(1,2))
 hist(x1, br=5, col="skyblue2")
 hist(x2, br=5, col="skyblue2")
par(mfrow=c(1,1))

enter image description here

x = c(x1,x2)
g = c(rep(1,230),rep(2,240))
boxplot(x~g, notch=T, col="skyblue2", horizontal=T)

Notches in the sides of boxplots are nonparametric CIs roughly calibrated so that overlapping CIs suggest no significant difference in location.

enter image description here

Shapes of the samples are similar, so a nonparametric 2-sample Wilcoxon rank sum test should test for a significant difference in location. The P-value $0.75 > 0.05 = 5\%$ indicates no significant difference.

wilcox.test(x~g)

        Wilcoxon rank sum test 
        with continuity correction

data:  x by g
W = 27134, p-value = 0.7518
alternative hypothesis: 
 true location shift is not equal to 0

If these were the actual data, I would stop here. Of course, your data may show different results.

For the reasons you mention, I would not trust a 2-sample t test to give reliable results, but for such large sample sizes its P-value may be roughly correct. Again, no indication of a significant difference.

t.test(x~g, var.eq=T)  # pooled t test

        Two Sample t-test

data:  x by g
t = -0.51847, df = 468, p-value = 0.6044
alternative hypothesis: 
 true difference in means is not equal to 0
95 percent confidence interval:
 -3.914830  2.280273
sample estimates:
mean in group 1 mean in group 2 
       125.0149        125.8322 

Even though I would not trust a pooled t statistic to have Student's t distribution in these circumstances, the t statistic does seem a reasonable way to measure a difference between means of the two samples.

A permutation test using the pooled t statistic as metric, does not find a significant difference, its P-value is about the same as for the pooled t test. No significant difference.

set.seed(2021)
t.obs = t.test(x~g, var.eq=T)$stat
t.prm = replicate(10^5, t.test(x~sample(g), var.eq=T)$stat)
mean(abs(t.prm) > abs(t.obs))
[1] 0.6094  # P-value of simulated permutation test

hdr = "Simulated Permutation Dist'n"
hist(t.prm, prob=T, col="skyblue2", main=hdr)
abline(v = c(t.obs,-t.obs), col="red", lwd=2)

enter image description here

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