Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample from a normal distribution \(N(\theta, 16)\). Find the sample size \(n\) and a uniformly most powerful test of \(H_{0}: \theta=25\) against \(H_{1}: \theta<25\) with power function \(\gamma(\theta)\) so that approximately \(\gamma(25)=0.10\) and \(\gamma(23)=0.90\).

Short answer

The solution involves four major steps. The first one is understanding the power function used in hypothesis testing. The second step is the formulation of the likelihood ratio test using the Neyman-Pearson lemma. The third step is calculating the critical region by rearranging the inequality that determines the likelihood ratio test boundary. The final step involves using the second condition of the power function to find out the sample size needed. Calculations will depend on the specific information provided in the question.

Step-by-step solution

Understand the power function

The power function measures the power of a test to reject the null hypothesis when the actual parameter value is \( \theta \). It is given by \( \gamma(\theta)=P[Reject\ H_{0}|\theta] \). Therefore, the value of this function indicates us the performance of the testing procedure.

Formulate the likelihood ratio test

The likelihood ratio test is obtained by comparing the likelihood of the observed sample under the null hypothesis and under the alternative hypothesis. Since we are dealing with a normal distribution, the likelihood function is \( L(\theta) = \prod_{i=1}^{n} (1/(4\sqrt{\pi})) e^{-1/32 (x_{i} - \theta)^{2}} \). So, the likelihood ratio is \( \Lambda = L(25)/L(\theta) \). Reject \( H_{0} \) if \( \Lambda>K \) for some constant \( K \). To find \( K \), use the condition that \( \gamma(25)=0.1 \) meaning that \( P[Reject\ H_{0}|\theta=25]=0.1 \).

Compute the critical region

Solving the inequality \( \Lambda>K \) gives us the critical region of the test. Here, we rearrange the terms to get an equivalent inequality in terms of the sample mean \( \bar{X} \). This partition the space of \( \bar{X} \) into rejection and non-rejection regions.

Use the second power function condition

With the power function \( \gamma(23)=0.9 \), meaning that \( P[Reject\ H_{0}|\theta=23]=0.9 \), find the sample size \( n \) that satisfies this condition.