2
$\begingroup$

I am trying to construct a LRT to test the hypothesis $H_0: p \ge p_0$ and $H_1: p < p_0$ where $\alpha = .1$ and $p_0 = .6$ and give a critical region.

Attempt:

$\lambda(x) = \frac{"restricted" MLE}{"unrestricted" MLE} = \frac{L(\hat{\theta_0}|X)}{L(\hat{\theta}|X)}$

I am looking for $P(X \in \Re) \le \alpha$ and $P(\lambda(x) \le c) \le \alpha$ where $C \in (0,1)$

$Reject: \{ x: \lambda(x) \le c \}$

My confusion, is with this little information, how can I calculate $\frac{L(\hat{\theta_0}|X)}{L(\hat{\theta}|X)}$.

$\endgroup$
1
  • 1
    $\begingroup$ Without any kind of a definition for what $p$ even is? $\endgroup$
    – Glen_b
    Dec 14, 2013 at 2:15

2 Answers 2

3
$\begingroup$

The question is underdetermined, so it cannot be answered. As @Glen_b suggests, you need a defintion of $p$, and you also need a probability model i.e. a definition of $L(\theta|X)$.

$\endgroup$
0
0
$\begingroup$

I think it’s just that you first find $\theta\in \Theta_0$ that maximize the likelihood function, say it’s $\hat \theta_0$, and then MLE of $\theta$, which is denoted by $\hat \theta$. Then calculate the ratio of $f(x|\hat \theta_0)$ and $f(x|\theta)$.

$\endgroup$

This site is temporarily in read-only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .