Debunking the “power paradox” allegation from my previous post. The authors consider a one-tailed Z test of the hypothesis *H*_{0}: μ ≤ 0 versus *H*_{1}: μ > 0: our Test T+. The observed sample mean is = 1.4 and in the first case _{σx} = 1, and in the second case _{σx} = 2.

First case: The power against μ = 3.29 is high, .95 (i.e. P(Z >* *1.645; μ=3.29) =1-φ(-1.645) = .95), and thus the DDS assessor would take the result as a good indication that μ < 3.29.

Second case: For σ** _{x}** = 2, the cut-off for rejection would be 0 + 1.65(2) = 3.30.

So, in the second case (σ_{x} = 2) the probability of erroneously accepting *H*_{0}, even if μ were as high as 3.29, is .5! (i.e. P(Z ≤* *1.645; μ=3.29) = φ(1.645-(3.29/2)) ~.5.) Although p_{1} < p_{2}[i] the justifiable upper bound in the first test is *smaller* (closer to 0) than in the second! Hence, the DDS assessment is entirely in keeping with the appropriate use of error probabilities in interpreting tests. There is no conflict with p-value reasoning.

NEW PROBLEM

The DDS power analyst always takes the worst cast of just missing the cut-off for rejection. Compare instead

SEV(μ < 3.29) for the first test, and SEV(μ < 3.29) for the second (using the actual outcomes as SEV requires).

[i] p_{1}= .081 and p_{2} = .242.