Critical region of likelihood ratio test

Question

This is problem # 5 from RSS's 2014 Graduate Diploma Module 2:

$$\mathbb{P}(X_j=k) = \begin{cases} (1-p)^3 & k=0\\ 3p(1-p) & k=1\\ p^3 & k=2\\ 0 & \text{otherwise} \end{cases}$$

$$Y_k = \sum_{i=1}^n I(X_i=k)\quad\text{for }\, k=0,1,2$$

It is straightforward to get an MLE estimate of $p$: $$\hat{p} = \frac{Y_1+3Y_2}{3Y_0+2Y_1+3Y_2}$$

Required to test: $H_0: p=p_0 \quad \text{versus}\quad H_1: p\neq p_0$.

I am stuck in the following part:

Using an asymptotic result, find the critical region of this test if its size is to be approximately 0.05

Attempt

Likelihood ratio(LR) test: $P(LR < k) = \alpha = 0.05$

\begin{align*} LR &= \frac{L(p|\mathbf{X})|_{p=p_0}}{L(p|\mathbf{X})|_{p=\hat{p}}}\\ &= \frac{(1-p_0)^{3Y_0}(3p_0(1-p_0))^{Y_1}(p_0^3)^{Y_2}}{(1-\hat{p})^{3Y_0}(3\hat{p}(1-\hat{p}))^{Y_1}(\hat{p}^3)^{Y_2}}\\ &= \left(\frac{p_0}{\hat{p}}\right)^{Y_1+3Y_2} \left(\frac{1-p_0}{1-\hat{p}}\right)^{Y_1+3Y_0} \end{align*}

EDIT 1:

Using Wilk's theorem: $-2\log(LR) \stackrel{D}\longrightarrow \chi^2_1(0.95)$ in distribution.

Thus critical region is

$$C = \left\{\mathbf{Y}: (Y_1+3Y_2) \log{\frac{p_0}{\hat{p}}} + (Y_1+3Y_0) \log\left(\frac{1-p_0}{1-\hat{p}}\right) \geq \frac{-\chi^2_1(0.95)}{2} \right\}$$

Is this further reducable since the final subpart of the question asks:

Show that in the case $Y_0 = Y_2$ this confidence interval can be written as $$p(1-p) \geq 0.25 \exp{\left(\frac{-1.92}{Y_1+3Y_2}\right)}$$

Answer 1

$\chi^2_1(0.95) = 3.841$

$C = \{\mathbf{Y}: (Y_1+3Y_2) \log{\frac{p_0}{\hat{p}}} + (Y_1+3Y_0) \log(\frac{1-p_0}{1-\hat{p}}) \geq \frac{-\chi^2_1(0.95)}{2} \}$

When $Y_0 =Y_2$, $\hat{p} = 1/2$

Thus, $\log(4p_0(1-p_0)) \geq \frac{-3.841}{2(Y_1+3Y_2)} $ $\implies p_0(1-p_0) \geq 0.25 \exp{(\frac{-1.92}{Y_1+3Y_2})}$