Calculate theoretical test errors and supplied-tail boundaries

Use the completed test foundation

Begin with hypotheses chosen before the data, a null reference and a rejection rule. Here calculate theoretical probabilities from iid normal observations with known positive variance. Unknown variance or non-normal populations need separate approximation arguments

Condition on the specified alternative

Type II error means not rejecting the fixed rule when the specified alternative mean μ₁ is true. First find the boundary under μ₀. Then standardise that same boundary under μ₁. Do not recalculate a new decision boundary centred on μ₁

Work a left-sided theoretical error

Take μ₀=12, n=18 and known σ²=18, so SE=1. Reject if Z₀≤−2, equivalently x̄≤10. At true μ₁=9, non-rejection is x̄>10 and z₁=(10−9)/1=1. Thus β=1−Φ(1)=0.1587 and power=0.8413 to four decimals. Under the null, α=Φ(−2)=0.0228

A left-sided test has null mean 12 and standard error 1. It rejects when Z₀≤−2. Give the exact rejection boundary for the sample mean

A left-sided test rejects when the sample mean is at most 10. Its standard error is 1. Give β when the true mean is 9, to four decimal places

What changes between the table rows?

Work the right-sided mirror

With μ₀=12 and SE=1, reject if Z₀≥2, so x̄≥14. If μ₁=15, non-rejection is x̄<14 and z₁=−1. Then β=Φ(−1)=0.1587. For a left-sided test use the upper non-rejection tail; for a right-sided test use the lower non-rejection tail

A right-sided test has rejection boundary 14 and standard error 1. Standardise this boundary under μ₁=15

Treat each supplied tail literally

If the null probabilities below L and above U are pL and pR, the total rejection probability is pL+pR. Use L=μ₀+SE×Φ⁻¹(pL) and U=μ₀+SE×Φ⁻¹(1−pR). Do not divide a probability already stated for one tail by two

Work a supplied two-tail allocation

For μ₀=6, n=27 and known σ²=12, SE=2/3. Prescribe 0.02 below L and 0.03 above U. Then α=0.05, L≈4.631 and U≈7.254 to three decimals; the null central probability is 0.95. The unequal tails explain the asymmetry. Use unrounded quantiles before rounding the final boundaries

A rule rejects with null probability 0.02 below its lower boundary and 0.03 above its upper boundary. Give its total rejection probability

A rule has total null rejection probability 0.05. Give its null probability of non-rejection

A rule assigns 1% to each tail. What is its total null rejection probability?

Explain the conditioning and rule, ungraded

Derive β for both a lower-boundary and an upper-boundary rejection rule. State which distribution supplies the probability, why its boundary is unchanged and what assumption makes equality irrelevant. Explain why a counting table estimates a probability rather than replacing this calculation

Model: compute the non-rejection event under μ₁

For reject x̄≤L, β=Prμ₁(x̄>L)=1−Φ((L−μ₁)/SE). For reject x̄≥U, β=Prμ₁(x̄<U)=Φ((U−μ₁)/SE). Continuity gives zero probability to equality. The procedure fixes L or U before the alternative generating distribution is used. Finite trial counts are random estimates of these model probabilities

Audit an unequal-tail design, ungraded

Derive L and U from separately supplied pL and pR. Explain why α is their sum and why unequal tails are not the standard symmetric equal-tail allocation. Identify the assumptions needed before treating these as exact error rates

Model: invert the two cumulative areas

The cumulative area at L is pL; at U it is 1−pR. Transform their standard-normal quantiles using the null mean and known SE. The two rejected events are disjoint, giving α=pL+pR and central probability 1−α. Iid normal observations and known variance give the exact sample-mean model. A different test design or estimated variance requires its own justification