# Posts Tagged With: homework problems

## Answer to the Homework & a New Exercise

Debunking the “power paradox” allegation from my previous post. The authors consider a one-tailed Z test of the hypothesis H0: μ ≤ 0 versus H1: 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 σx = 2, the cut-off for rejection would be 0 + 1.65(2) = 3.30.

So, in the second case (σx = 2) the probability of erroneously accepting H0, 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 p1 < p2[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] p1= .081 and p2 = .242.

Categories: Statistics |