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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.

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Your approach is correct. You have the right expression for the MLE.

$N$ 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.

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  • $\begingroup$ Thanks, just the fact that I was on the right track was really helpful. I managed to produce the plot and so on and get what seems to be the right answer. $\endgroup$ – David Pfeiffer Apr 22 at 23:24

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