**Here’s a quick note on something that I often find in discussions on tests, even though it treats “power”, which is a capacity-of-test notion, as if it were a fit-with-data notion…..**

1.** Take a one-sided Normal test T+: with n iid samples:**

**H _{0}: µ ≤ _{ }0 against H_{1}: µ > _{ }0**

σ = 10, n = 100, σ/√n =σ** _{x}**= 1, α = .025.

So the test would reject H_{0} iff Z > c_{.025} =1.96. (1.96. is the “cut-off”.)

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**Simple rules for alternatives against which T+ has high power:**

- If we add σ
_{x }(here 1) to the cut-off (here, 1.96) we are at an alternative value for µ that test T+ has .84 power to detect. - If we add 3σ
to the cut-off we are at an alternative value for µ that test T+ has ~ .999 power to detect. This value, which we can write as µ_{x }^{.}^{999}= 4.96

Let the observed outcome just reach the cut-off to reject the null,z_{0 }= 1.96.

If we were to form a “likelihood ratio” of μ = 4.96 compared to μ_{0} = 0 using

[Power(T+, 4.96)]/α,

it would be 40. (.999/.025).

It is absurd to say the alternative 4.96 is supported 40 times as much as the null, understanding support as likelihood or comparative likelihood. (The data 1.96 are even closer to 0 than to 4.96). The same point can be made with less extreme cases.) What is commonly done next is to assign priors of .5 to the two hypotheses, yielding

Pr(H_{0} |z_{0}) = 1/ (1 + 40) = .024, so Pr(H_{1} |z_{0}) = .976.

Such an inference is highly unwarranted and would almost always be wrong. Continue reading