Using t test to compare z-scores from RNA expression data

Question

[Camilo Andres]

Hi! I need to compare the expression of ~220 genes between two groups (n = 62 and n = 777). I have a dataframe with Z-scores for the expression of those genes (they were actually part of the same group, but I split them based on some criteria, but the z-scores were calculated using them as a group). I have no access to other kind of data like count data. Is it right to use a t-test or a moderated t test to compare them? would be better to use a wilcox test?

Thanks in advance

Answer 1

[i.sudbery]

With those sorts of n numbers, you should start by looking at some distributional plots. Do the Z scores (or log Z scores) look even vaguely normally distributed? If so, then a moderated t-test, like that in limma might be applicable. However, with numbers that large, a wilcox is also likely to be well powered.

[Camilo Andres]

Thanks, this is what I'll do.

Answer 2

[Trivas]

I'm going to say you can't perform a statistical test on these data. A Z-score is a dimensionless metric to illustrate the difference between a sample from the mean. Since you have 777 samples in one group, they are all going to skew the mean in their favor.

[Camilo Andres]

Since you have 777 samples in one group, they are all going to skew the mean in their favor.

But I think thats true for every population you analyse. Given that they are taken from the same universe of samples, gene expressions should be near mean in both groups, if I found a difference between my two groups that would be because there is an actual difference between them, wouldn’t it be?

[i.sudbery]

I agree, as long as both samples are normalised the same way, it shouldn't change the relative distribution of samples in units of standard deviation.