Question
When doing AOV tests over two treatments, i.e., H0: u1 = u2 = ... = ut, for our data y_ij, there's an assumption of constant
When doing AOV tests over two treatments, i.e., H0: u1 = u2 = ... = ut, for our data y_ij, there's an assumption of constant variance and that the errors are iid normal. In the case, there's a violation, and we often need to perform a transform on y_ij, i.e., x_ij = g(y_ij), which will ideally give us constant variance (or more constant).
We can now then do AOV tests, but H0 is now (note the prime) H0': u1_x = u2_x = ...= ut_x, which is in the transformed data space and not in the original space.
Question is:
1) Under what conditions are H0' equivalent to H0?
2) Further, is there any relationship between the taylor series approximation of 'g' and the equivalency of H0' and H0?
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