I am not statistician so I am sorry if my question formulation sounds confusing.

The ultimate goal is to comapre two methods (A and B). Each method creates 24 numbers (so, 24 pairs). Apart from the method A and B, each of these numbers are linked to the same object. Something like this:

  Protein|structure|Method A  |Method B
  MDM2     a1       23          19
  MDM2     a2       25          8
  MDM2     a3       45          37
  P53      b1       19          18 
  P53      b2       ...         ...
  P53      b3
  XXX      c1
  XXX      c2
  XXX      c3
  YYY      d1
  YYY      d2
  YYY      d3

Ultimate goal is to compare method A to method B and determine whether one method has statistically significant difference to the other.

However, in my case comparison should be considered as pairwise within the Protein (MDM2 for example). So, I should not compare it as a1A to a1B; a2A to a2B.... It should more likely be pairwise within a protein: a1A to a1a2a3B; a2A to a1a2a3B; a3A a1a2a3B. Then, b1A to b1b2b3B.

So ultimately, statistical methods tells me whether method A is statistically different than method B across entire sample. So that, at the end, if I "shuffle" the numbers within each method, but within Protein, I get same result.

Should I just create new, pairwise sample and do simple 2-sample t-test?


  • $\begingroup$ Hey @Pitouille, thanks for your time reading my question. My field is computational chemistry/drug design. I am not sure whether two-way ANOVA is the best way to go, I only have one comparison between method A and method B, however, important thing here is to eliminate dependency of the "Protein (aaa;bbb;ccc)". So that if I shuffle the numbers within the method A itself and do the same with method B, I would always get the same statistics? Bcs currently I am comparing a1A to a1B; but in reality it can also be a1A to a2B; a1A to A3B (pairwise). $\endgroup$
    – sergio
    Sep 8, 2021 at 7:34
  • $\begingroup$ So instead of doing pairwise comparison, I am thinking whether two-way ANOVA can help elimination of dependency of "Protein" aaa;bbb;ccc;ddd;eee;fff;ggg. $\endgroup$
    – sergio
    Sep 8, 2021 at 7:35
  • $\begingroup$ Sorry... I deleted my comment because I am not sure to understand your data...! So, this is not paired data... you want to analyze the 2 methods and the protein does not actually matter... is that right? $\endgroup$
    – Pitouille
    Sep 8, 2021 at 7:39
  • $\begingroup$ @Pitouille This is exactly what I want :) Sorry if my formulation was not clear enough $\endgroup$
    – sergio
    Sep 8, 2021 at 7:48
  • $\begingroup$ Is it continous data that you are collecting? $\endgroup$
    – Pitouille
    Sep 8, 2021 at 7:49

1 Answer 1


Based on the information you provided, it seems that you want to compare the means of two populations (continuous data, sample size n=24) and determine whether they differ. Therefore, your hypothesis could be stated as below:

$H_0:$ The means for the 2 populations are equal

$H_a:$ The means for the 2 populations are not equal

If you want to compare both method independently of the protein, a 2-sample t-test will do the work assuming that t-test requirements are met (random sample, independent groups...). In R, using t.test, you can specify if it concerns paired test, equal or unequal variance...

Now if you want to compare methods within each subgroup (a, b, c), t-test will no longer work, so ANOVA could be an option. However, you might face a challenge since I do not think your data will meet the minimum requirement in terms of sample size (my understanding is that your original sample size is 24... so, I supposed it will at least less than 8 since you mentioned a minimum of 3 groups).

  • $\begingroup$ Thanks a lot for for your suggestions and taking a time to propose ways based on your knowledge and experience. I am thinking of doing something like: taking the mean of a1a2a3 for each protein (with providing standard deviation) and doing 2-sample t-test because it is a random sample and two groups are independent. Thanks a lot! $\endgroup$
    – sergio
    Sep 8, 2021 at 11:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.