Question

5. We have two independent samples of n observations X1,Xy,.. . ,x, and Y. Ya, . … We want to test the hypothesis Ho : μ,-μυ versus the alternative Hi : μ, μν. (a) First, assume that the null hypothesis Ho is true and find the MLE for μ (b) Then plug this estimate into the log likelihood along with the MLE μ'.. and 1,-j) to calculate the LRT statistic. (e) Is this likelihood ratio test equivalent to the test that rejects the null hypothesis when (t -g)2 > e? Justify your answer.

Thank you so much!

Answer 1

$(a)~The~likelihood~function,~L(\mu)=\frac{1}{(10\pi)^{n/2}(14\pi)^{n/2}}\exp\left(-\frac{1}{10}\sum_{i=1}^n(x_i-\mu)^2-\frac{1}{14}\sum_{i=1}^n(y_i-\mu)^2\right)\\ \log L(\mu)=l(\mu)=-\frac{n}{2}\log 140-n\log \pi-\frac{1}{10}\sum_{i=1}^n(x_i-\mu)^2-\frac{1}{14}\sum_{i=1}^n(y_i-\mu)^2\\ \frac{\mathrm{d} l(\mu)}{\mathrm{d} \mu}=\frac{1}{5}\sum_{i=1}^n(x_i-\mu)&plus;\frac{1}{7}\sum_{i=1}^n(y_i-\mu)=0~or,~\hat{\mu}=\frac{7\bar{x}&plus;5\bar{y}}{12}\\ (b)~\sup_{\mu=\mu_0}L(\mu)=\frac{1}{(10\pi)^{n/2}(14\pi)^{n/2}}\exp\left(-\frac{1}{10}\sum_{i=1}^n(x_i-\hat{\mu})^2-\frac{1}{14}\sum_{i=1}^n(y_i-\hat{\mu})^2\right)\\ \sup_{\mu}L(\mu_x,\mu_y)=\frac{1}{(10\pi)^{n/2}(14\pi)^{n/2}}\exp\left(-\frac{1}{10}\sum_{i=1}^n(x_i-\hat{\mu}_x)^2-\frac{1}{14}\sum_{i=1}^n(y_i-\hat{\mu}_y)^2\right),\\ ~where~\hat{\mu}_x=\bar{x},~\hat{\mu}_y=\bar{y}\\$$Likelihood~ratio=\Lambda=\frac{\sup_{\mu=\mu_0}L(\mu)}{\sup_{\mu_x,\mu_y}L(\mu_x,\mu_y)}=\exp\left(-\frac{1}{10}\sum_{i=1}^n(2x_i-\hat{\mu}-\hat{\mu}_x)(\hat{\mu}_x-\hat{\mu})-\frac{1}{14} \sum_{i=1}^n(2y_i-\hat{\mu}-\hat{\mu}_y)(\hat{\mu}_y-\hat{\mu})\right )\\ =\exp\left(-\frac{5}{288}(\bar{x}-\bar{y})^2-\frac{7}{288}(\bar{x}-\bar{y})^2 \right )=\exp\left(-\frac{1}{24}(\bar{x}-\bar{y})^2 \right )\\ (c)~We~reject~H_0~if\\ \Lambda<k~or~\exp\left(-\frac{1}{24}(\bar{x}-\bar{y})^2 \right )<k~or,~(\bar{x}-\bar{y})^2>c\\ where~c~is~obtained~from~size~condition.$