Pregunta: I'm stuck in the green part, speaking algebraically, I don't know what else to do to get to what they ask me to demonstrate, if you help me and your answer is correct I'll upvote.

7. Sean $X_1,\dots X_n\stackrel{i.i.d}{\sim }$ Bernoulli $(\theta )$. Utilice el cociente de verosimilitudes generalizado (LRT) para probar que la región de rechazo asociada a la prueba de hipótesis $H_0:\theta ={\theta }_0\text{ vs }{H}_1:\theta \ne {\theta }_0$ está dada por: $C:=\left\{\left(X_1,\dots ,X_n\right)\mid {\sum }_{i=1}^n{X}_i\ge k_1\delta {\sum }_{i=1}^n{X}_i\le k_2\right\}$ Sea $x_1,\dots ,x_n$ m.a de una bernoulli $(\theta )$ La función de verosimilitud bajo $H_0$ es: $L\left({\theta }_0\right)={\prod }_{i=1}^{n}p\left(x_i;{\theta }_0\right)={\prod }_{i=1}^n{\theta }_0^{x_1}{\left(1-{\theta }_0\right)}^{1-x_1}$ De forma analoga la funcion de verosimilitud bajo $H_1$ es: $L(\theta )={\prod }_{i=1}^{n}P\left(x_i;\theta \right)={\prod }_{i=1}^n{\theta }^{x_1}(1-\theta {)}^{1-x_1}$ Por ejercicios anteriores sabemos que el estimador por maxima verosimilitud para $x_1,\dots ,x_n$ m.a Bernoulli $(\theta )$ $\hat{\theta }{m}_v=\bar{x}$ Y la razon de verosimilitud esta dada por: $\begin{array}{rl}\lambda (x)& =\frac{L\left({\theta }_0\right)}{L(\theta )}=\frac{L\left({\theta }_0\right)}{L(\bar{x})}\\ & =\frac{{\prod }_{i=1}^n{\theta }_0^{x_1}{\left(1-{\theta }_0\right)}^{1-x_1}}{{\prod }_{i=1}^n{\bar{x}}^{x_1}(1-\bar{x}{)}^{1-x_1}}=\prod _{i=1}^n{\left(\frac{{\theta }_0}{\bar{x}}\right)}^{x_i}{\left(\frac{1-{\theta }_0}{1-\bar{x}}\right)}^{1-x_i}\end{array}$ Aplicamos in: $\ln \left(\frac{L\left({\theta }_0\right)}{L(\theta )}\right)={\sum }_{i=1}^n{x}_i\ln \left(\frac{{\theta }_0}{\bar{x}}\right)+{\sum }_{i=1}^n\left(1-x_i\right)\ln \left(\frac{1-{\theta }_0}{1-\bar{x}}\right)$