# Likelihood and Confidence Intervals

#### hellostudent

##### New Member
Hi, I'm trying to understand how LR-based Confidence intervals work. More specifically, to try to understand the topic, I was doing this exercise:

I hadn't particular problems with the first two parts, but I started to struggle while doing this third part. If I understood, I've to compute a confidence interval by using the likelihood ratios. So I've to find the values that satisfy the inequality:
where . Some of my classmates solved the problem in this way:

So they subtracted, from the likelihood function with respect to a generic parameter p, the likelihood of the sampling estimator ( ), obviously putting this subtraction greater or equal to -1.92
The aim is to find a value of p such that its likelihood is equal or greater than the log-lik on the estimator minus 1.92. But so, how can I obtain, from this computation, a limited interval like the one in the photo (p belonging to [0.183,0.51]. In particular, I don't understand how it is possible to arrive, from the inequality to that interval. How is it possible to estimate the parameter for which the difference between the log-lik of p and the log-lik of is the biggest one?
I'm sorry if my question could appear a bit confused, it may be connected to the fact that I haven't such a big knowledge about chi-squared distribution (I only know that a chi-squared random variable refers to the sum of some squared standard normal variables). So I don't know, in general, how to use the chi-squared distribution to build confidence intervals.