Which test should I use to compare a normal distribution sample with skewed distribution sample? 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
 A: 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.

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")


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

