91Ó°ÊÓ

From the definition of the likelihood function for Bernoulli distribution, for \(n\) samples, the likelihood function is: \(L(\theta;X)=\theta^{\sum_{i=1}^{n} X_{i}}(1-\theta)^{n - \sum_{i=1}^{n} X_{i}}.\) The likelihood ratio test statistic is the ratio of the maximum likelihood under \(H_{0}\) to the maximum likelihood under \(H_{1}\), i.e. \(\lambda=\frac{L(\theta_{0};X)}{L(\hat{\theta};X)}\), where \(\hat{\theta}\) is MLE of \(\theta\) which equals to \(\bar{X}=Y/n\). Therefore, the likelihood ratio test statistic \(\lambda\) takes the form: \(\lambda=\left(\frac{\theta_{0}^{Y}(1-\theta_{0})^{n-Y}}{(\bar{X})^{Y}(1-\bar{X})^{n-Y}}\right)\).

Under the null hypothesis \(H_{0}: \theta=\theta_{0}\), each \(X_{i}\) follows a Bernoulli(\(\theta_{0}\)) distribution, which means \(Y=\sum_{i=1}^{n} X_{i}\) follows a \(binomial(n, \theta_{0})\) distribution.

To reject the null hypothesis at significance level \(\alpha=0.05\), you need to find the values \(c_{1}\) and \(c_{2}=100-c_{1}\) such that \(P(Y \leq c_{1} or Y \geq c_{2})=0.05\). You can use the Central Limit Theorem to approximate this probability. The Central Limit Theorem states that the variable \(Y\) follows a distribution that is roughly \(normal(n\theta_{0}, n\theta_{0}(1-\theta_{0}))\) when \(n\) is large. Therefore when \(n=100\) and \(\theta_{0}=0.5\), the normal distribution is \(normal(50, 25)\), and you can use the standard normal z-table to find \(z_{0.025}\) so that \(P(Z > z_{0.025}) = 0.025\). Then, use the approximation \(c_{1}= n \theta_0 - z_{0.025} \sqrt{n\theta_0 (1-\theta_0)}\) and \(c_{2}= n\theta_{0} + z_{0.025} \sqrt{n\theta_0 (1-\theta_0)}\).