Suppose \( x_{1}, x_{2}, x_{2}, \ldots x_{n} \) is a random sample from a normal population with mean \( \mu \) and variance 16 . Find the test with the best critical region, that is, find the most powerful test, with a sample size of \( \mathrm{n}=16 \) and a significance level \( \propto=0.05 \) to test the simple null \( H_{\mathbf{a}}: \theta=10 \) hypothesis against the simple alternative hypothesis \( H_{1}: \theta=15 \), Also find the power of the test.

We are given a sample from N(µ, 16) with n = 16. We wish to test   H₀: µ = 10  versus  H₁: µ = 15 at a significance level of α = 0.05. Step 1. Identify the Test Statistic Since the sample comes from a normal population with a known variance, the sample mean X̄ is normally distributed. Under any given µ, we have   X̄ ~ N(µ, σ²/n)  =  N(µ, 16/16) = N(µ, 1). Step 2. Obtain the Most Powerful Test (Neyman–Pearson Lemma) For testing two simple hypotheses, the Neyman–Pearson lemma tells us that the most powerful test rejects H₀ for large values of the likelihood ratio. Given the problem and the alternatives, the decision will be based on X̄. Intuitively, since H₁ has a larger mean than H₀, we reject H₀ when X̄ is large. Step 3. Determine the Critical Value We want the test to have a size of 0.05. Therefore, under H₀ (µ = 10) we find the value c such that   P(X̄ > c | µ = 10) = 0.05. Since X̄ ~ N(10, 1) under H₀, we standardize:   P(X̄ > c) = P(Z > c−10) = 0.05, where Z ~ N(0, 1). From the standard normal table, the 95th percentile is approximately 1.645. Therefore,   c − 10 = 1.645  ⟹  c = 11.645. Thus, the most powerful test rejects H₀ if X̄ > 11.645. Step 4. Compute the Power of the Test The power of the test is the probability of rejecting H₀ when H₁ is true (i.e., when µ = 15). Under H₁,   X̄ ~ N(15, 1). Then, the power is   Power = P(X̄ > 11.645 | µ = 15). Standardizing under H₁:   P(X̄ > 11.645) = P(Z > 11.645 − 15) = P(Z > −3.355), where Z ~ N(0, 1). Since P(Z > −3.355) = 1 − P(Z ≤ −3.355) and using symmetry P(Z ≤ −3.355) = P(Z ≥ 3.355), we note that P(Z ≥ 3.355) is very small. Alternatively, because the standard normal is symmetric, we have   P(Z > −3.355) = Φ(3.355). Using standard normal tables or a calculator, Φ(3.355) is approximately 0.9996. Thus, the power is roughly 99.96%. Step 5. Final Answer The most powerful test at the 0.05 significance level is:   Reject H₀ if X̄ > 11.645. The power of this test when µ = 15 is approximately 0.9996 (99.96%).