I have a biological sample and I test its response to two different stimuli $A_1$ and $A_2$. I repeat the measurements multiple times in a pseudo random order: e.g. $A_1$ $A_1$ $A_2$ $A_1$ $A_2$ $A_2$ ...

Now the sample undergoes a treatment and the two responses are measured again, now labeled with $B$: $B_1$ $B_1$ $B_2$ $B_1$ $B_2$ $B_2$ ...

I need to check whether the treatment has changed the relative responses, in other words, if $A_1/A_2$ is different from $B_1/B_2$.

Therefore want to compare the two quotient distributions $A_1/A_2$ to $B_1/B_2$.

First of all I need to obtain the quotient distributions. As there are multiple $A_1$ and $A_2$ values which don't come paired, my guess is that the best estimation would be to calculate the quotient of all possible pairings of $A_1$ and $A_2$. The distribution of these $A_1/A_2$ pairs is roughly a log-normal distribution, but the values aren't independent, as every $A_1$ value is used multiple times, as I pair it with every $A_2$ value.

Now I have to compare the distributions of $A_1/A_2$ with $B_1/B_2$. But as they are neither Gaussian, nor are they independent or paired, I haven't found a proper way of comparing them in a statistically sound way.

  • 1
    $\begingroup$ Working on logarithmic scale would make things much easier here. For instance, you could compare $\log A_1- \log A_2$ against $\log B_1- \log B_2$ with the Mann-Whitney or the t-test for paired data. Otherwise it seems hard. $\endgroup$ – utobi Dec 1 '16 at 14:42
  • $\begingroup$ Hmm, but why can I apply Mann-Whitney to the logarithmic data? The quotient or in the logarithmic case difference distributions are not independent, because to sample the A_1/A_2 distribution, I use all possible combinations of A_1 and A_2. So each A_1 value is represented multiple times. Same applies for A_2, B_1 and B_2. $\endgroup$ – Randrian Dec 1 '16 at 23:15
  • 1
    $\begingroup$ The paired Mann-Whitney takes care of the dependency. Applying it as I suggested would tell you if the diffs are significantly different. Another thing you could do is permutation testing. $\endgroup$ – utobi Dec 2 '16 at 6:25
  • $\begingroup$ But I don't have pairs of data. I don't understand how i should apply a paired test on that. Neither pairs of A_1 and A_2 belong directly together (as a pair) nor do A_1/A_2 to B_1/B_2 belong together as a pair. $\endgroup$ – Randrian Dec 2 '16 at 11:06
  • $\begingroup$ What exactly is the difference between $A_1$ and $A_2$? Furthermore, do you have N probes on which you obtain N measurements of something (what is it?) or you are measuring N times on a single probe? $\endgroup$ – utobi Dec 2 '16 at 13:14

It seems to me that you can calculate four quantities: M1 = mean(log(A1)); M2 = mean(log(A2)); M3 = mean(log(B1)); and M4 = mean(log(B2)).

Then you can calculate: (M1 - M2) - (M3 - M4) and propagate uncertainty in the usual way.

Does this give you the inference you want?

  • $\begingroup$ I am not sure how this will give me a decision if the two quotient distributions are different. Your approach results in just one value with a uncertainty. $\endgroup$ – Randrian Dec 10 '16 at 7:40
  • $\begingroup$ Right, and the value is nonzero if the quotient distributions are different, no? If you had a way to pair off observations A1 and A2, then you would just take the actual quotients, and compare them. Instead, you just want to know whether, on average, a randomly selected observation from population A1 divided by a randomly selected observation from population A2. So you ask is mean(log(A1)-log(A2))) different from mean(log(B1)-log(B2))? But mean(log(A1) - log(A2)) = mean(log(A1)) - mean(log(A2)). Which is what I suggested you calculate. $\endgroup$ – Jacob Socolar Dec 12 '16 at 0:31
  • $\begingroup$ Yes I need to compare them, but as the measurements have some randomness involved, I can't just check whether they have the same value, because even if the distributions are the same, the result won't be the exact same value, therefore I need some statistical tool. $\endgroup$ – Randrian Dec 12 '16 at 11:27
  • $\begingroup$ As the bounty time is running up and you are the only one who has tried to provide an answer, I give the bounty to you. But I still don't consider the question as solved. $\endgroup$ – Randrian Dec 12 '16 at 11:31
  • $\begingroup$ The point here isn't to literally compute the sample means and do the subtraction, the point is to compute estimates of the population means, and do the subtraction with the appropriate error propagation. Does that make sense? $\endgroup$ – Jacob Socolar Dec 12 '16 at 16:02

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.