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Let $X$ be a r.v which is Weibull distributed with $k=4$ and $\lambda>0$. I'm looking at the following test:

\begin{align*} H_0:\lambda &= 1\\ H_1: \lambda&\not = 1. \end{align*}

I have already done the calculations for the actual likelihood ratio, it is \begin{align*} \lambda_n(\tilde{x}) &= \dfrac{L(1,\tilde{x})}{L(\hat{\lambda},\tilde{x}) }\\ &=\left(\dfrac{1}{n}\sum_{i\leq n}x_i^4\right)^n\exp\left(\sum_{i\leq n}x_i^4+n\right)\\ &=A^ne^{-n(A-1)}, \end{align*} where $A =A(\tilde{x}) = \frac1n\sum_{i\leq n}x^4_i$. We define the critical region as $R = \left\{ \tilde{x} \mid \lambda_n(\tilde{x}) \leq \lambda_0 \right\}$, with $\lambda_0$ such that $$P(\Lambda_n(\tilde{X})\leq \lambda_0) = 0.05,$$ and we reject the null hypothesis iff $\tilde{x} \in R$.

I know how to calculate the critical region if I know the correct value for $\lambda_0$. But I don't know how to find that value, since it seems I need to know how $\Lambda_n$ is distributed, which I don't.

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  • $\begingroup$ What does "$\Lambda_n(\tilde X)$" refer to? Presumably it's a test statistic, so you can compute its distribution under the null hypothesis and thereby solve for $\lambda_0.$ Why, then, don't you know its distribution? $\endgroup$
    – whuber
    Aug 9 '20 at 14:56
  • $\begingroup$ @whuber $\Lambda(\tilde{X}_n)= \lambda(\tilde{X}_n)$ is the likelihood ratio for the likelihood ratio test. Finding the distribution of this r.v is precisely what I don't know how to do. Also, it has to be exact, so I can't use any asymptotic results. $\endgroup$ Aug 9 '20 at 16:00

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