Q1: You can, but you probably should not. You should instead try to find out why the data are heteroscedastic and to find a more appropriate way to analyse the data (are you possibly trying to compare apples to peaches? would some transformation make sense? should you better assume some other distribution? ...)

Q2: No, not really.

Your last sentence is unclear, as there is no "Mann-Whitney t.test". Which test ((Welch-)t or U) is the best depends also on what hypothesis you actually want to test. The U-test might not test the hypothesis you have in mind, particularly not when the data are heteroscedastic.

Muhammad Salman Munir Malik, you have told us that n1 = 60 and n2 = 77. Can you also tell us the means and SDs for the two groups? Can you show us a pair of histograms, one for each group? And can you tell us what the dependent variable is? Thanks.

Jochen Wilhelm Thanks. First I would learn and try to transform the data.

Q2: Sorry It's Mann-Whitney U test (not t test). I want to test the hypothesis that the concentrations of my test variable are reduced in my target population (age group >65) as compared to the control (age group

Thanks Muhammad Salman Munir Malik. Do you have the actual ages of the subjects? Or do you only know which age group they belong to? If you do know the actual ages, why would you not estimate a regression model that treats age as a quantitative variable? I suggest that you generate a scatter-plot with X = age and Y = your concentration variable (or maybe log-concentration*) and post it here.

* I am not a lab scientist, so analysis of concentrations is not an area or expertise for me. But AFAIK, concentrations are often log-transformed. Jochen Wilhelm and other readers are better able to advise you about that than I am.

PS- Here is an image of the PowerPoint slide Muhammad Salman Munir Malik uploaded.

If you want to transform the data, the square root transformation works reasonably well, or you might try a Box-Cox transformation.

But given your sample size, and the distribution of your data, the result from a t-test won't change much. This is because the distribution of the means of the underlying populations would follow a normal distributions with these sample sizes.

You don't need equal sample sizes for a Wilcoxon-Mann-Whitney test. (And in this case, the results will be similar to that from a t-test anyway.)

Below I'll paste some R code with an approximation of your data, and some transformations with the t-tests. I know it's difficult to read on this platform, but if you use R, you can try it.

P.S. It goes without saying that you shouldn't try various transformations and conduct tests and choose which one you like based on the results. The point here is just to demonstrate that t-tests on moderately skewed samples with reasonably large sample sizes don't change much from the case if the data are transformed so that the samples are more normal in distribution.

########################

if(!require(tidyr)){install.packages("tidyr")}

if(!require(rcompanion)){install.packages("rcompanion")}

if(!require(lattice)){install.packages("lattice")}

Data = read.table(header=TRUE, sep=",", text="

Group, Y, Count

A, 0.9430882770366171, 1.1538461538461497

A, 1.9393982401362484, 1.1538461538461497

A, 2.948481407891002, 8.076923076923073

A, 4.031162564372897, 24.890109890109883

A, 5.011353959693443, 16.15384615384615

A, 5.935586553667733, 7.087912087912084

A, 6.98359758322858, 5.109890109890106

A, 7.990856007461172, 11.04395604395604

A, 8.97804225295, 6.0989010989010985

A, 9.96705324196099, 2.142857142857139

A, 10.967012692104943, 4.120879120879117

A, 11.957848424638094, 1.1538461538461497

B, 1.9999999999999996, 1.1578947368421062

B, 2.9999999999999996, 6.105263157894736

B, 4, 15.157894736842104

B, 5, 7.157894736842106

B, 6, 9.157894736842104

B, 7, 4.210526315789476

B, 8, 6.210526315789474

B, 8.947368421052634, 7.052631578947368

B, 10.000000000000002, 5.2631578947368425

B, 11, 3.157894736842106

")

Data$Group = factor(Data$Group)

Data$Y = round(Data$Y)

Data$Count = round(Data$Count)

library(tidyr)

Long = uncount(Data, Count)

rownames(Long) = seq(1:nrow(Long))

Long$logY = log(Long$Y)

Long$rootY = sqrt(Long$Y)

library(rcompanion)

Long$intY = blom(Long$Y)

library(lattice)

histogram(~ factor(Y) | Group, data = Long, layout = c(1,2))

histogram(~ Y | Group, data = Long, layout = c(1,2))

qqnorm(Long$Y); qqline(Long$Y, col="red")

histogram(~ logY | Group, data = Long, layout = c(1,2))

histogram(~ rootY | Group, data = Long, layout = c(1,2))

qqnorm(Long$rootY); qqline(Long$rootY, col="red")

histogram(~ intY | Group, data = Long, layout = c(1,2))

qqnorm(Long$intY); qqline(Long$intY, col="red")

t.test(Y ~ Group, data=Long)

t.test(rootY ~ Group, data=Long)

t.test(intY ~ Group, data=Long)

wilcox.test(Y ~ Group, data=Long)

Sal Mangiafico, did you notice that the groups are age groups with a cut point at 65? If the actual ages are available, a model that treats age as a quantitative explanatory variable will (almost certainly) be better than a t-test.

The Welch's test should replace the Student's t-test if the variances of the respective two groups being compared are different. What are the variances of each of your groups?

Use the Mann-Whitney if your data are fail tests of normality. If you transform the data and then retest for normality then by all means use students t-test if the transformed data appear to be normally distributed.

Either way estimate the variance of each group and Welch is the variance are significantly different between groups as the the assumption of the other tests is that variances are similar.

Patrice Showers Corneli, I have several comments on what you posted.

1) Muhammad Salman Munir Malik has two age groups. If he has subjects' actual ages, he should (almost certainly) treat age as a quantitative explanatory variable, not as a dichotomy.

2) How sensitive t-tests (and ANOVA models) are to heterogeneity of variance depends very much on how similar the two sample sizes are. If the sample sizes are equal (or nearly so), these tests are extremely robust to variance heterogeneity. You might remember George Box's famous statement about rowing boats and ocean liners--see the attached image. See also:

Article Of rowing boats, ocean liners and tests of the ANOVA homogen...

3) Basing choices about which model to use on statistical tests of normality (or variance heterogeneity) are generally ill-advised. This issue has been discussed many times on RG. I believe you follow Jochen Wilhelm on RG. He as stated very clearly many times in multiple threads that testing for normality is ill-advised. See also my short conference presentation on testing for normality.

Presentation Silly or Pointless Things People Do When Analyzing Data: 1. ...

You may view things differently, of course. But given my view of these issues, I could not let your post go unchallenged. ;-)