Let’s start with the application of the central limit theorem to champagne drinkers. First, there’s the distinction between “liver weights are normally distributed” and “mean of a sample of liver weights is normally distributed”. The latter is much better-justified, since we compute the mean by adding a bunch of (presumably independent) random variables together. And the latter is usually what we actually use in basic analysis of experimental data—e.g. to decide whether there’s a significant different between the champagne-drinking group and the non-champagne-drinking group. That does not require that liver weights themselves be normally distributed.
I think your statement in bold font is wrong. I think in cases such as champagne drinkers vs non-champagne-drinkers people are likely to use Student’s two-sample t-test or Welch’s two-sample unequal variances t-test. It assumes that in both groups, each sample is distributed normally, not that the means are distributed normally.
No, the student’s two-sample t-test does not require that individual samples are distributed uniformly. You certainly could derive it that way, but it’s not a necessary assumption. All it actually needs is normality of the group means via CLT—see e.g. here.
I think your statement in bold font is wrong. I think in cases such as champagne drinkers vs non-champagne-drinkers people are likely to use Student’s two-sample t-test or Welch’s two-sample unequal variances t-test. It assumes that in both groups, each sample is distributed normally, not that the means are distributed normally.
No, the student’s two-sample t-test does not require that individual samples are distributed uniformly. You certainly could derive it that way, but it’s not a necessary assumption. All it actually needs is normality of the group means via CLT—see e.g. here.