# Likelihood interval for binomial counts

I have an assignment question regarding a "likelihood interval" that I don't really understand.

The question asks to consider counts of $$X_i$$, with $$i\in \{1,...,N\}$$, modelled as independent binomial variates (typo?) with constant known $$n$$ and mean constant unknown $$p$$.

Firstly I found the MLE of $$p$$ to be (not sure about this):$$\hat{p}=\frac{\sum_{i=1}^N x_i}{Nn}$$ We are then told to determine a $$10\%$$ likelihood interval based on this data where $$n=10$$ (why does $$i$$ start at $$0$$ now??):

$$\begin{matrix} i & 0 & 1 & 2 & 3 & 4 & 5 \\ X_i & 2 & 0 & 1 & 5 & 3 & 3 \\ \end{matrix}$$

and also calculate a $$10\%$$ likelihood interval based on a single observation of $$X=2$$ with $$n=10$$.

My attempt for the first part: According to our course notes, a $$c\%$$ likelihood interval for $$p$$ is given by $$\bigg{\{} p:\frac{L(p;x)}{L(\hat{p};x)} \ge \frac{c}{100}\bigg{\}}$$

so with the the information given and taking $$N=6$$ (not sure due to the first count being for $$i=0$$ if this should be $$N=5$$ or not), I substituted everything into $$\displaystyle \frac{L(p;x)}{L(\hat{p};x)}$$ and obtained

$$\frac{L(p;x)}{L(\hat{p};x)}=(1.433678665*10^{14})*p^{14}(1-p)^{46}$$

so the $$10\%$$ likelihood interval becomes

$$\bigg{\{} p:(1.433678665*10^{14})*p^{14}(1-p)^{46} \ge 0.1\bigg{\}}$$

First of all, I feel like it's very unlikely that this is correct. Second of all, this isn't an interval in a form like $$(x,y)$$ and I don't know how to get it to be.

Does anyone have any insight? There is almost no information on the internet about likelihood intervals so I can't really make sense of this. Let me know if something is excluded or doesn't make sense and I will provide more details of my working. Also I should add that I would like to do this without the use of software if possible.

$$N$$ is indeed $$6$$.
For the confidence interval, plot the expression $$(1.43*10^{14})*p^{14}*(1-p)^{46}$$ as a function of $$p.$$ You will see that it is concentrated around the MLE value which is $$\dfrac{14}{60}=0.2333$$ in this case.
From the plot you will get an idea of the confidence interval. You can solve for the exact $$(a,b)$$ for the confidence interval using a root finder software.