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 random sample x₁, x₂, …, x₁₆ from a N(μ, 16) population. We want to test the simple hypotheses   H₀: μ = 10  against  H₁: μ = 15 at a significance level of α = 0.05 using the most powerful test (by the Neyman–Pearson lemma). Step 1. Set Up the Likelihood Ratio Since the sample is from a normal distribution with known variance, the joint density under μ is   L(μ) = ∏[1/√(2π·16)] exp[–(xi – μ)²/(2·16)]     = constant × exp[–1/32 ∑(xi – μ)²]. The likelihood ratio comparing μ = 15 to μ = 10 is   Λ(x) = L(15) / L(10)     = exp{ –1/32 [∑(xi – 15)² – ∑(xi – 10)²] }. With some algebra (or by noticing that the test will be based on a monotone function of the sample mean) one can show that the most powerful test is of the form: reject H₀ when the sample mean x̄ is sufficiently large. Step 2. Find the Rejection Region Under H₀, x̄ is normally distributed:   x̄ ~ N(10, σ²/n) = N(10, 16/16) = N(10, 1). Since we desire a test of size α = 0.05 and larger values of x̄ make H₁ more plausible, we choose the critical region to be   C = { x̄ > c }. We need to choose c so that   P(x̄ > c | H₀) = 0.05. Under H₀, standardizing gives   P(x̄ > c) = P(Z > c – 10)  (where Z ~ N(0,1)). Thus, we require   P(Z > c – 10) = 0.05. The 95th percentile of the standard normal distribution is approximately 1.645, so we set   c – 10 = 1.645  ⇒  c = 11.645. Thus, the most powerful test is:   Reject H₀ if x̄ > 11.645. Step 3. Compute the Power The power of the test is the probability of correctly rejecting H₀ when H₁ is true (i.e., when μ = 15). Under H₁, the sample mean is distributed as   x̄ ~ N(15, 16/16) = N(15, 1). Thus, the power is   Power = P(x̄ > 11.645 | μ = 15)     = P(Z > (11.645 – 15))    (where Z ~ N(0,1))     = P(Z > –3.355). Since for a standard normal variable P(Z > –a) = 1 – Φ(–a) = Φ(a), we have   Power = Φ(3.355). Looking up the value (or using a calculator), Φ(3.355) is very close to 1 (approximately 0.9996). Step 4. Conclusion The most powerful test at the 0.05 significance level is: reject H₀ (i.e., conclude μ is 15 rather than 10) whenever the sample mean x̄ > 11.645. The power of this test, when μ is truly 15, is approximately 0.9996. Thus, the solution is: • Critical region: Reject H₀ if x̄ > 11.645. • Power of the test (at μ = 15): Approximately 99.96%.