Ononot — Given a sample of size n from a distribution f(x|\theta), th

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Inferential Statisticshard

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Given a sample of size $n$ from a distribution $f(x|\theta)$ , the Score function $U(\theta)$ and the Fisher Information $I(\theta)$ are used to test $H_0: \theta = \theta_0$ . If we define the Score test statistic as $S = U(\theta_0)^2 / I(\theta_0)$ , what is the fundamental advantage of this test over the Wald test in terms of the likelihood function?