By following the distribution I provided here on stackmathematics
Q. A biased die favours the number 2. It's rolled 4 times and 2 comes up twice. It's rolled again 4 times and 2 comes up once. Calculate the Likelihood and find the MLE.
The likelihood is: $L(\theta;3) = \binom{8}{3}\theta^3(1-\theta)^5$, whereas we have the local maximum at $\theta = \frac{3}{8}$.
I want to verify that, approximately $(0.0718, 0.6026)$ is a 2-likelihood region for $\theta$ by making a plot to indicate this.
From what I understand, the 2-likelihood region is given as the following:
The region $R_k = \{\theta \in \mathbb{R}: L(\theta;x) \ge e^{-k}L(\hat{\theta};x)\}$ for the k-unit likelihood region for parameters $\theta$ based on data $x$.
We know that $x=3$ and $\theta = \frac{3}{8}$, then we apply the following:
$$\binom{8}{3}\theta^3(1-\theta)^5 \ge e^{-2}L\left(\frac{3}{8};3\right)\approx0.035$$
So I think the next step is to plug $0.0718, 0.6026$ into $\theta$ to compare the inequality.
Something like:
$$\binom{8}{3}(0.0718)^3(1-0.0718)^5 \approx 0.014$$
However, $0.014 < 0.035$ and so $0.0718$ does not fall in the 2-likelihood region?
If this is the case, is there an easier way to check for values that fall within the confidence interval of the likelihood region on R?