Find a 10% likelihood interval

Question

The function is:

(n choose x)$[(1-y)^{k}]^{x}[1-(1-y)^{k}]^{n-x}$

Suppose n = 100, k = 10, x = 89

I found the maximum likelihood of y-hat to be 0.0116

Now I need to find a 10% likelihood interval. This means that I need 90% confidence(I think).

So the interval is:

[y-hat - 1.645*sqrt(1/n * (y-hat(1 - y-hat))) , y-hat + 1.645*sqrt(1/n * (y-hat(1 - y-hat)))]

When I plug in the values, I get [-0.006014, 0.029217]

The correct answer is [0.0056, 0.0207]

What am I doing wrong?

Answer 1

If we let $\theta = (1-y)^k$ then this is a binomial inference problem. Your MLE is correct, but the approximation methods you are using are not valid for this problem. Note that $y=.016 \implies \theta \approx 0.89$, which is rather skewed and hence a sample size of 100 is likely not going to cut it for using normal approximations (which is what you are trying to do). You will need to numerically solve this:

1. First, find the un-normalized likelihood at the MLE: $(0.89)^{89}(0.11)^{11} = 8.93\times 10^{-16}$
2. Now, you need to solve $\theta^{89}(1-\theta)^{11} = 0.1 \times 8.93\times 10^{-16} = 8.93\times 10^{-17}$
3. Converting to log-space, we need to solve: $89\ln(\theta)+11\ln(1-\theta) + 17\ln(10)-\ln(8.93)=0$
4. Using "goal seek" in excel, I get $\theta_1 = 0.945148$ and $\theta_2=0.811309$
5. We can simply reverse-transform from $\theta \rightarrow y: y=1-\theta^{1/10}$ to get $[0.0056,0.0207]$