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Background: We have RNA-seq data (a method to quantify the expression of genes in a cell) from different cell types. These data were normalized using a method called TMM and then log2-transformed followed by Z-transformation.

We then plotted the Z-scores for two groups of genes as shown in the plot below for two cell types.

We aim to establish that the difference in expression between the two cell types is greater for genes in group 1 than for those in group 2.

Using t.test or wilcox.test on both groups results in highly significant p-values < 0.00000000000000022 so comparing these two p-values is not too informative.

Would there be a more elaborate way to establish this difference?

enter image description here

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  • $\begingroup$ You want to put all types/groups into a single model, not use two separate tests and try to compare them. If you have a two-way anova, for the way your data are come out when plotted, a significant interaction would indicate a greater "difference in difference" for one group. $\endgroup$ – Sal Mangiafico Sep 10 '19 at 19:32
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I think for what your want to test you shouldn't apply test individually. In individual tests you are only checking whether mean for different cell types is different for a given group. You can try the following:

Let your data be $\{x_{ij}\}$, where $i$ represents cell type and $j$ is group.

Now define $y_j = x_{1j} - x_{2j}, \forall j=1,2$.

Now you test the hypothesis:

$H_0: \mu_1>\mu_2$

$H_a: \mu_1 \leq \mu_2$

where $\mu_j \equiv E(Y_j)$.

Note that unlike the usual null that $\mu_1=\mu_2$, your null is different so your test will be one sided only. Whatever software you are using should show these options.

Side note: Looking at the box plot of your data, it seems that your data may not be normally distributed, which is an assumption of t-test. You can check for normality of $y_j$ (say using JB test). In any case, wilcoxon test will do fine if your distribution is at least symmetric, if not normal.

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