I have some data, belonging to paired groups $A,B$. From each group I get a non-negative statistic $d^A,d^B$ which is averaged on all samples in group. My interest is in the ratio $\frac{d^A}{d^B}$, i.e. how far are these statistics from each other (straight subtraction makes no sense). My equivalence hypotheses (inspired by the 80/125 rule of bioequivalence) are: $$H_0:\frac{E[d^A]}{E[d^B]}>\epsilon\text{ or }\frac{E[d^A]}{E[d^B]}<\epsilon^{-1},\quad H_1:\epsilon^{-1}\le \frac{E[d^A]}{E[d^B]}\le\epsilon$$

The null hypothesis can be rewritten as $H_0:E[d^A]-\epsilon E[d^B]>0 \text{ or }E[d^A]-\epsilon^{-1} E[d^B]<0$ which makes things a bit easier: While we don't know the actual (or theoretical) distribution of the ratio, we can relate to the term $\bar{d}_{U}=\frac{1}{N}(\sum_i d^A_i-\epsilon\sum_i d^B_i)$ as the sample mean, calculate the sample variance $s^2_{U}=Var(d^A_i-\epsilon d^B_i)$ and then the statistic $t_U=\frac{\bar{d}_{U}-0}{\sqrt{s^2_{U}}}$ has $t$ distribution and we can test whether or not $t_U \le t_{\alpha,N-1}$.

In the same manner we define $\bar{d}_{L}=\frac{1}{N}(\sum_i d^A_i-\epsilon^{-1}\sum_i d^B_i),s^2_{L}=Var(d^A_i-\epsilon^{-1} d^B_i)$ and $t_L=\frac{\bar{d}_{L}-0}{\sqrt{s^2_{L}}}$, in order to test whether or not $t_L\ge t_{1-\alpha,N-1}$. Finally, we reject $H_0$ if $t_U \le t_{\alpha,N-1}~\text{ and }~t_L\ge t_{1-\alpha,N-1}$.

All of the above is nice, but now we want to make it a one-sided test. That is, our updated hypotheses are

$$H_0:\left|\log\left(\frac{E[d^A]}{E[d^B]}\right)\right|>\epsilon,\quad H_1:\left|\log\left(\frac{E[d^A]}{E[d^B]}\right)\right|\le\epsilon$$

We can obviously write $|\log(E[d^A])-\log(E[d^B])|-\log(\epsilon)>0$ but I'm not really sure how does this help me in formulating a test, what are the sample statistics I get here or anything else. As the sum and log operations are not commutative, I'm not really sure how to progress here of is it a smart move at all. It is not at all like the case of $t$-testing a log-transformed data (unlike here and here), because I only transform my statistic (and not the whole data).

Would love to hear ideas and thoughts.



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.