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Well, okay, so perhaps the proof isn’t all that particularly enlightening, but perhaps if we take a look at a simple example, we’ll become more enlightened. of an exponential random variable is:for$x ≥ 0$. Under the hypothesis $H \colon \theta = 3$, the p. d.

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d. f. PrintCasella, George, and Roger L. We say that C is the most powerful size $\alpha$ test. or p.

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d. To differentiate the lemma from theories that have a name and a Greek letter (like Glasss Delta or Fleiss kappa), its sometimes written as Lemma (Neyman-Pearson). Need help with a homework or test question? With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. Is the hypothesis $H \colon \theta = 3$ a simple or a composite hypothesis?The p. is:Note that for the sake of ease, we drop the reference to the sample $X_1 , X_2 , \dots , X_n$ in using $L (\theta)$ as the notation for the likelihood function.

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Wiley StatsRef: Statistics Reference Online. 2nd ed. 05\) to test the simple null hypothesis $H_{0} \colon \mu = 10 $ against the simple alternative hypothesis $H_{A} \colon \mu = 15 $. A good hypothesis test will official statement both of these errors, which inherently involves a trade-off between them. Berger.

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Well, okay, so perhaps the proof isn’t all that particularly enlightening, but perhaps if we take a look at a simple example, we’ll become more enlightened. d. Find the test with the best critical region, that is, find the most powerful test, with significance level $\alpha = 0. The lemma tells us that the ratio of the likelihoods under the null and alternative must be less than some constant k:Simplifying, we get:And, simplifying yet more, we get:Now, taking the natural logarithm of both sides of the inequality, collecting like terms, and multiplying through by 32, we get:And, moving the constant term on the left-side of the inequality to the right-side, and dividing through by −160, we get:That is, the Neyman Pearson Lemma tells us that the rejection region for the most powerful test right here testing \(H_{0} \colon \mu = 10 $ against $H_{A} \colon \mu = 15 $, under the normal probability model, is of the form:where $k^*$ is selected so that the size of the critical region is \(\alpha = 0.

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The lemma basically tells us that good hypothesis tests are likelihood ratio tests. Thus, we know that\begin{align} (\phi(x) – \widetilde{\phi}(x))(f(x; \theta_1) – \kappa f(x; \theta_0)) \geq 0 \\ \implies \phi(x) f(x; \theta_1) – \widetilde{\phi}(x)f(x; \theta_1) – \kappa\phi(x)f(x; \theta_0) + \kappa\widetilde{\phi}(x)f(x; \theta_0) \geq 0 \\ \end{align}Integrating both sides (integration respects inequalities), we haveNotice that these terms correspond to the respective power functions: $\int\left[\phi(x) f(x; \theta_1)\right] = \beta(\theta_1)$, $\int\left[\phi(x) f(x; \theta_0)\right] = \beta(\theta_0)$, $\int\left[\widetilde{\phi}(x) f(x; \theta_1)\right] = \widetilde{\beta}(\theta_1)$, and $\int\left[\widetilde{\phi}(x) f(x; \theta_0)\right] = \widetilde{\beta}(\theta_0)$. Find the test with the best critical region, that is, find the most powerful test, with a sample size of $n = 16$ and a significance level \(\alpha = 0. f.

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d. Is the hypothesis $H \colon \mu = 12$ a simple or a composite hypothesis?The p. Under the hypothesis $H \colon \theta = 3$, the p.

Creative Commons Attribution NonCommercial License 4. f. The power of a hypothesis test is the probability that test correctly rejects the null hypothesis when the alternate hypothesis is true.

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), in which part of the class’s parameter space do you think the true parameter lives?Typically, the parameter space $\Theta$ is divided into two parts, $\Theta_0$ and $\Theta_1$, which are the parameter values corresponding to the null and alternative hypotheses, respectively. Suppose we have a random sample $X_1 , X_2 , \dots , X_n$ from a probability distribution with parameter $\theta$. The simple accept/reject framework implies that there exist two types of errors: false positives (falsely reject) and false negatives (falsely accept). That said, how can we be sure that the T-test for a mean $\mu$ is the “most powerful” test we could use? Is there instead a K-test or a V-test or you-name-the-letter-of-the-alphabet-test that would provide us with more power? A very important result, known as the Neyman Pearson Lemma, will reassure us that each of the tests we learned in Section 7 is the most powerful test for testing statistical hypotheses about the parameter under the assumed probability distribution. .