The Question:
Let $x_1,\dots,x_n$ be a large sample taken from a distribution with density function $f(x;\theta)$ where $\theta$ is a scalar parameter. Let $\ell (\theta)$ be the log-likelihood of the distribution, and $\hat \theta$ be the MLE of $\theta$.
How does the solution(s) for $\theta$ in
$$\ell (\theta) = \ell (\hat \theta) - 1.92$$
relate to an approximate $95$% confidence interval for $\theta$?
Note: If $Z \sim \chi_1^2$, then $\Bbb P(Z≤3.84)=0.95$.
My Attempt:
I see that $3.84$ is twice of $1.92$, so it might be related to some two-tailed intervals?
By the CLT, we have that
\begin{align} & \sqrt{\Bbb E \bigg(-\frac{d^2 \ell}{d \theta^2} \bigg)}\big(\hat \theta - \theta \big) \approx N(0,1) \\ \implies & \Bbb E \bigg(-\frac{d^2 \ell}{d \theta^2} \bigg) \, \big(\hat \theta - \theta \big)^2 \approx \chi_1^2 \end{align}
and then I am stuck. Any hints?
Firstly, you need to get the Fisher information $ I(\theta) = \Bbb E \left(-\frac{d^2 \ell}{d \theta^2} \right)$. According to the CLT, we should have $\hat{\theta} - \theta \sim N\left(0, \frac{1}{nI(\theta)}\right)$.
Then, you can get the 95% confidence interval easily, right?