Question

Let \(X_{1}, X_{2}, \ldots, X_{10}\) be a random sample from a distribution that is \(N\left(\theta_{1}, \theta_{2}\right)\). Find a best test of the simple hypothesis \(H_{0}: \theta_{1}=\theta_{1}^{\prime}=0, \theta_{2}=\theta_{2}^{\prime}=1\) against the alternative simple hypothesis \(H_{1}: \theta_{1}=\theta_{1}^{\prime \prime}=1, \theta_{2}=\theta_{2}^{\prime \prime}=4\).

Short answer

The best test of the stated simple hypothesis against the alternative hypothesis is to reject the null hypothesis H_0 if the test statistic \(\frac{4}{5}\sum_{i=1}^{10}X_i\) is less than \(\ln2\). This indicates that the data is more likely under the alternative hypothesis.

Step-by-step solution

Setting Up the Likelihood Ratio Test

Under the null hypothesis \(H_{0}:\), the likelihood function is: \[ L_{H_0}(X_1, X_2, ..., X_{10}) = \frac{1}{(2\pi)^{5}} e^{-\frac{1}{2}\sum_{i=1}^{10} X_i^2 } \] Under the alternative hypothesis \(H_{1}:\), the likelihood function is: \[ L_{H_1}(X_1, X_2, ..., X_{10})=\frac{1}{(8\pi)^5} e^{-\frac{1}{8}\sum_{i=1}^{10}(X_i-1)^2} \] The likelihood ratio test statistic is then given by: \[ \lambda=\frac{L_{H_0}(X_1, ..., X_{10})}{L_{H_1}(X_1, ..., X_{10})} \]

Simplifying the Test Statistic

Simplify the test statistic \(\lambda\). We note that the likelihood ratio \(\lambda\) will be small if the data is 'in favor of' \(H_1\). Also, note that the decision rule for testing will reject \(H_0\) if \(\lambda<k\) for some constant K. We simplify to get: \[ \lambda=\left(\frac{1}{4}\right)^5 e^{ -\frac{1}{8}\sum_{i=1}^{10}((X_i-1)^2- X_i^2)} \] Thus our decision rule is to reject \(H_0\) if \(\lambda<k\) or equivalently: \[ \sum_{i=1}^{10}((X_i-1)^2- X_i^2)<8 \ln4-10 \ln4.\]

Finding the Critical Region

The above inequality will provide us the critical region. After simplification, we get: \[ \sum_{i=1}^{10}X_i^2 - 20\sum_{i=1}^{10}X_i+20 <8 \ln4-10\ln4 \] which simplifies to: \[ \frac{4}{5}\sum_{i=1}^{10}X_i < \ln2 \] So, we can conclude that our critical region is \(C=\{\; x \in R_{10} \;|\; \frac{4}{5}\sum_{i=1}^{10}X_i < \ln2 \}\)