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I have two groups : Infection group (n=26) and healthy control group (n=127). The aim is to see if there is a difference in the mean of T cells absolute count* values between the two groups.

Ho= the infection group’s T cells absolute count mean is no different than the healthy group’s.

Originally the welch’s t test seemed like the straight forward answer to conduct my hypothesis testing. However after running the shapiro wilk test, the healthy group (n=127) turns out to be not normally distributed, which violates the t test’s assumption of the groups being normally distributed. On the other hand the data Infection group (n=26) had a normal distribution according to the shapiro wilk test. Now switching to a non parametric alternative like Mann whitney u doesn’t work either, since both data needs to be not normally distributed to be valid for this test.

In this case which test do you recommend ?

Considering the difference in sample size I wonder if bootstraping for welch’s t test would be any helpful in ignoring the difference in distribution ?

PS: despite the violation, I did run the welch’s t test on spss, with a 95% confidence interval the results were as follow :

t value = -13.733

Alpha(2tailed)= 0.0000 (4.77 e-25)

Mean difference = -4.88

Standard error difference = 36.67

Running welch’s t test with Bootstrapping based on 1000 sample gave pretty close results but with an alpha of 0.001, would this result be reliable for interpretation?

* T cells absolute counts is a continuous numerical data that can range from 0 to 3000 or even more in some rare cases

Mounmoun
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    [Formal normality testing is less helpful than you seem to think.](https://stats.stackexchange.com/questions/2492/is-normality-testing-essentially-useless) // Wilcoxon does not require both distributions to be non-normal (or even either group to be non-normal), though there is a convenient interpretation when both have the same shape and just a location shift. // What do you want to know about your data? – Dave Jul 15 '21 at 00:08
  • @Dave thank you for your involvement in this question, as I mentioned before I want to know if the means of the infection group is statistically different than the control group’s. I am trying to prove that the means of the two samples are not equal. // Isn’t mann whitney test the wilcoxon test for independent samples t test? According to this website https://www.statisticshowto.com/mann-whitney-u-test/ and several other sources, the test does require both distributions to be non-normal – Mounmoun Jul 15 '21 at 00:34
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    What does “statistically different” mean to you? // Those sources are wrong. There’s a school of thought—to which a respected member on here and founding chairman of biostatistics at Vanderbilt, Frank Harrell, belongs—saying that we should default to Wilcoxon (Mann-Whitney U) because it’s almost as good as the t-test when the distributions are normal and is quite superior when they are not. // Without testing with Shapiro-Will, your domain knowledge lets you *know* that the distributions are not normal, since they cannot take values below zero. A normal distribution can take any real value. – Dave Jul 15 '21 at 00:43
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    My stance on that school of thought is that I want to be in it but have found clients/customers to be much more comfortable using the t-test, and the t-test is (often) quite robust to deviations from normality. Further, if we have dramatically different distributions, we might be wise to consider what other differences there are (variance, multimodality, etc), and I don’t want the analysis to stop with a Wilcoxon test. – Dave Jul 15 '21 at 00:45
  • Can you show us a plot (two histograms with many bins)? – kjetil b halvorsen Jul 15 '21 at 01:33
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    What matters is not so much what the samples suggest, but what you believe would be the case when the null is true; a perfectly reasonable sequence of alternatives may well exist that would lead to different shapes as the means become more different, without impacting the significance level at all. – Glen_b Jul 15 '21 at 07:01

1 Answers1

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Maybe a specific example with fictitious data will help you see how results of the two-sample Wilcoxon test can be interpreted when two populations (hence samples) have somewhat different shapes.

Data, sampled in R:

set.seed(2021)
x1 = rgamma(127, 6, .12)       # right-skewed
summary(x1); length(x1); sd(x1)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  13.57   33.10   43.47   47.63   56.38  122.00 
[1] 127      # sample size
[1] 21.6026  # sample SD

x2 = rnorm(26, 50, 18)         # normal
summary(x2); length(x2); sd(x2)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  28.47   46.21   54.97   57.83   66.94   98.00 
[1] 26
[1] 18.11204

x = c(x1,x2);  g = c(rep(1,127),rep(2,26))
boxplot(x~g, horizontal=T, notch=T, 
  col=c("skyblue2","wheat"),varwidth=T)

The two samples have different sample medians (also means) and somewhat different shapes. The notches in the the sides of the boxes in the boxplots below represent nonparametric confidence intervals for the medians. They are calibrated so tha non-overlapping intervals suggest a difference in two medians.

enter image description here

Wilcoxon rank sum test:

A formal nonparametric two sample Wilcoxon rank sum test finds a difference between the populations from which the samples were drawn: the P-value is $0.008 < 0.05 = 5\%.$

wilcox.test(x1,x2)

        Wilcoxon rank sum test 
        with continuity correction

data:  x1 and x2
W = 1103, p-value = 0.007822
alternative hypothesis: 
  true location shift is not equal to 0

Because the shapes of the two distributions are somewhat different, one should not say just that the difference is a difference in location (different medians).

Stochastic dominance:

From empirical CDF (ECDF) plots we can see that the smaller sample (brown) stochastically dominates the larger one. (That is, the smaller sample tends to have larger values than the larger one.) The brown ECDF plot is shifted to the right (hence plots below) the blue ECDF plot.

plot(ecdf(x1), col="blue")
 lines(ecdf(x2), col="brown")

enter image description here

Note: A 2-sample Kolmogorov-Smirnov test also finds a difference between the two populations. It's $D$-statistic is the maximum vertical difference between the two ECDF plots.

ks.test(x1,x2)

        Two-sample Kolmogorov-Smirnov test

data:  x1 and x2
D = 0.35373, p-value = 0.00641
alternative hypothesis: two-sided
BruceET
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