$\newcommand{\szdp}[1]{\!\left(#1\right)} \newcommand{\szdb}[1]{\!\left[#1\right]}$ Problem Statement: Suppose that independent random samples of sizes $n_1$ and $n_2$ are to be selected from normal populations with means $\mu_1$ and $\mu_2,$ respectively, and common variance $\sigma^2.$ For testing $H_0:\mu_1=\mu_2$ versus $H_a:\mu_1-\mu_2>0$ ($\sigma^2$ unknown), show that the likelihood ratio test reduces to the two-sample $t$ test presented in Section 10.8.
Note: This is Exercise 10.94 from Mathematical Statistics with Applications, 5th. Ed., by Wackerly, Mendenhall, and Scheaffer.
My Work So Far: We have the likelihood as $$L(\mu_1, \mu_2,\sigma^2)= \szdp{\frac{1}{\sqrt{2\pi}}}^{\!\!(n_1+n_2)} \szdp{\frac{1}{\sigma^2}}^{\!\!(n_1+n_2)/2} \exp\szdb{-\frac{1}{2\sigma^2}\szdp{\sum_{i=1}^{n_1}(x_i-\mu_1)^2 +\sum_{i=1}^{n_2}(y_i-\mu_2)^2}}.$$ To compute $L\big(\hat\Omega_0\big),$ we need to find the MLE for $\sigma^2:$ \begin{align*} \hat\sigma^2&=\frac{1}{n_1+n_2}\szdp{\sum_{i=1}^{n_1}(x_i-\mu_1)^2 +\sum_{i=1}^{n_2}(y_i-\mu_2)^2}. \end{align*} This is the MLE for $\sigma^2$ regardless of what $\mu_1$ and $\mu_2$ are. Thus, under $H_0,$ we have that $$\hat\sigma_0^2=\frac{1}{n_1+n_2}\szdp{\sum_{i=1}^{n_1}(x_i-\mu_0)^2 +\sum_{i=1}^{n_2}(y_i-\mu_0)^2},$$ and the unrestricted case is $$\hat\sigma^2=\frac{1}{n_1+n_2}\szdp{\sum_{i=1}^{n_1}(x_i-\overline{x})^2 +\sum_{i=1}^{n_2}(y_i-\overline{y})^2}.$$ Under $H_0,\;\mu_1=\mu_2=\mu_0,$ so that \begin{align*} L\big(\hat\Omega_0\big) &=\szdp{\frac{1}{\sqrt{2\pi}}}^{\!\!(n_1+n_2)} \szdp{\frac{1}{\hat\sigma_0^2}}^{\!\!(n_1+n_2)/2} \exp\szdb{-\frac{n_1+n_2}{2}}\\ L\big(\hat\Omega\big) &=\szdp{\frac{1}{\sqrt{2\pi}}}^{\!\!(n_1+n_2)} \szdp{\frac{1}{\hat\sigma^2}}^{\!\!(n_1+n_2)/2} \exp\szdb{-\frac{n_1+n_2}{2}}, \end{align*} and the likelihood ratio is given by \begin{align*} \lambda &=\frac{L\big(\hat\Omega_0\big)}{L\big(\hat\Omega\big)}\\ &=\szdp{\frac{\hat\sigma^2}{\hat\sigma_0^2}}^{\!\!(n_1+n_2)/2}\\ &=\szdp{\frac{\displaystyle\sum_{i=1}^{n_1}(x_i-\overline{x})^2 +\sum_{i=1}^{n_2}(y_i-\overline{y})^2} {\displaystyle\sum_{i=1}^{n_1}(x_i-\mu_0)^2 +\sum_{i=1}^{n_2}(y_i-\mu_0)^2}}^{\!\!(n_1+n_2)/2}. \end{align*} It follows that the rejection region, $\lambda\le k,$ is equivalent to \begin{align*} \frac{\displaystyle\sum_{i=1}^{n_1}(x_i-\overline{x})^2 +\sum_{i=1}^{n_2}(y_i-\overline{y})^2} {\displaystyle\sum_{i=1}^{n_1}(x_i-\mu_0)^2 +\sum_{i=1}^{n_2}(y_i-\mu_0)^2}&<k^{2/(n_1+n_2)}=k'\\ \frac{\displaystyle\sum_{i=1}^{n_1}(x_i-\overline{x})^2 +\sum_{i=1}^{n_2}(y_i-\overline{y})^2} {\displaystyle\sum_{i=1}^{n_1}(x_i-\overline{x})^2+n_1(\overline{x}-\mu_0)^2 +\sum_{i=1}^{n_2}(y_i-\overline{y})^2+n_2(\overline{y}-\mu_0)^2}&<k'\\ \frac{1}{1+\dfrac{n_1(\overline{x}-\mu_0)^2+n_2(\overline{y}-\mu_0)^2} {\displaystyle\sum_{i=1}^{n_1}(x_i-\overline{x})^2 +\sum_{i=1}^{n_2}(y_i-\overline{y})^2}}&<k'\\ \frac{n_1(\overline{x}-\mu_0)^2+n_2(\overline{y}-\mu_0)^2} {\displaystyle\sum_{i=1}^{n_1}(x_i-\overline{x})^2 +\sum_{i=1}^{n_2}(y_i-\overline{y})^2}&>\frac{1}{k'}-1=k''\\ \frac{n_1(\overline{x}-\mu_0)^2+n_2(\overline{y}-\mu_0)^2} {\displaystyle(n_1-1)S_1^2+(n_2-1)S_2^2}&>k''\\ \frac{n_1(\overline{x}-\mu_0)^2+n_2(\overline{y}-\mu_0)^2} {\displaystyle\dfrac{(n_1-1)S_1^2+(n_2-1)S_2^2} {n_1+n_2-2}}&>k''(n_1+n_2-2)\\ \frac{n_1(\overline{x}-\mu_0)^2+n_2(\overline{y}-\mu_0)^2} {S_p^2}&>k''(n_1+n_2-2). \end{align*} Here $$S_p^2=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}.$$
My Question: The goal is to get this expression somehow to look like $$t=\frac{\overline{x}-\overline{y}}{S_p\sqrt{1/n_1+1/n_2}}>t_{\alpha}.$$ But I don't see how I can convert my expression, with the same sign for $\overline{x}$ and $\overline{y},$ to the desired formula with its opposite signs. What am I missing?
