Consider the situation where there are 2 positive distributions with unknown expected values $\mu_1$ and $\mu_2$,

We have samples $S_1 = \{x_1, x_2, ..., x_n\}$ and $S_2 = \{y_1, y_2, ..., y_n\}$.

We would like to use a statistical test under the null hypothesis: $\mu_1 = k \times \mu_2$.

Do you know what kind of test should we use?

  • $\begingroup$ I assume that you looking for a test with a specific value of $k$ in mind. Are there any assumptions about the underlying distributions, or about the variances of $S_1$ and $S_2$? $\endgroup$ – EdM Oct 19 '19 at 21:24
  • $\begingroup$ O don't have an assumption about the distributions ir variances. Do you know a test in this framework that uses an assumption that is not mentioned above? $\endgroup$ – Eduardo Cesar N. Coutinho Oct 19 '19 at 21:46
  • $\begingroup$ I notice you have the same sample size for both samples. Is this because there's some pairing or are the samples independent? $\endgroup$ – Glen_b Oct 20 '19 at 1:32

If your null is that the distributions are the same i.e. that $F_1(x) = F_2(x k)$ when you could perhaps do a permutation test: multiply the observations in sample 2 by $k$ ($x_{2i}^*=kx_{2i},\,i=1,2,...,n$ and then test whether they have the same mean, because under the null, the set of $kx_{2i}$ and $x_{1j}$ would be exchangeable.

If you are additionally interested specifically in a pure scale shift alternative (and want an easily-obtained interval for it) you might divide the first sample through by $k$ (or multiply the second by $k$), take logs and look at a location shift (whether a permutation test or a rank based test like the Wilcoxon-Mann-Whitney), before backtransforming the interval

If you have a suitable parametric assumption, using that information will generally be advantageous (power-wise), if the assumption is true or very nearly true.

Note that failure to reject doesn't mean the multiplier is $k$. You can't verify that; you might consider whether an equivalence test comes nearer to what you need in that case.


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.