Under H0 a daily signal is N(0, 1) and under H1 it is N(0.3, 1). With 100 iid observations and a 5% size constraint, construct the most powerful test, state its rejection region, and compute its power.

Under H0 a daily signal is N(0, 1) and under H1 it is N(0.3, 1). With 100 iid observations and a 5% size constraint, construct the most powerful test, state its rejection region, and compute its power.

Approach: Form the likelihood ratio for the two simple hypotheses, show it is monotone in a single statistic, then set the constant by the size constraint.

0.912. The Neyman-Pearson lemma says the most powerful test of one simple hypothesis against another rejects when the likelihood ratio exceeds a constant. Here log(L1/L0) = 0.3*sum(x_i) - 100*0.045, which is strictly increasing in the sample mean, so the rejection region is xbar > c. Under H0 the sample mean is N(0, 0.01), so size 0.05 fixes c = 1.645/sqrt(100) = 0.1645. Under H1 the sample mean is N(0.3, 0.01), so the power is P(Z > (0.1645 - 0.3)/0.1) = P(Z > -1.355) = 0.912. No other test of size 0.05 has higher power against this alternative. The region does not depend on the specific value 0.3, only on its sign, so the same test is uniformly most powerful against every mu > 0.

Follow-up: Why does no uniformly most powerful test exist for the two-sided alternative mu not equal to zero, and what is the standard replacement?

Key concepts: Neyman-Pearson lemma, likelihood ratio, rejection region, power.