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Show that the likelihood ratio principle leads to the same test when testing a simple hypothesis \(H_{0}\) against an alternative simple hypothesis \(H_{1}\), as that given by the Neyman- Pearson theorem. Note that there are only two points in \(\Omega\).

Short Answer

Expert verified
Both the Neyman-Pearson theorem and the likelihood ratio principle are fundamentally based on calculating a test statistic for making a decision about the hypothesis. In both the tests, the hypothesis is rejected if the calculated test statistic (or the likelihood ratio) is greater than a certain critical value. Hence, the Neyman-Pearson theorem and likelihood ratio principle yield the same test for the simple hypothesis testing.

Step by step solution

01

Understanding of Likelihood Ratio Principle

The likelihood ratio for a given set of observations is the ratio of the likelihoods under two different hypotheses. It is a fundamental tool for statistical inference. The likelihood ratio test is a statistical test used for comparing the goodness of fit of two statistical models - the null model against an alternative model.
02

Understanding of Neyman-Pearson Theorem

The Neyman-Pearson (NP) fundamental lemma is a result in testing hypothesis that says a test with size \(\alpha\) that has maximum power among all tests of size \(\alpha\) is a likelihood-ratio test. In other words, it is a method for testing a simple hypothesis vs another simple hypothesis that maximizes the power of the test.
03

Comparing the two principles

When using the Neyman-Pearson theorem for hypothesis testing, we utilize the test statistic, set up the critical region, then the test is done. The acceptance or rejection of the hypothesis is based on whether the test statistic lies in the critical region. On the other hand, the likelihood ratio principle forms a ratio of probabilities under the two hypotheses, and the decision on acceptance or rejection of the hypothesis is based on whether the likelihood ratio exceeds a certain threshold. From these definitions, it can be inferred that both the Neyman-Pearson theorem and the likelihood ratio principle are essentially methods that utilize a test statistic to make a decision about a hypothesis.
04

Demonstrating the Equivalence

The Neyman-Pearson theorem and likelihood ratio principle are equivalent for testing a simple hypothesis, if the data set consists of n independent and identically distributed random variables. In the Neyman-Pearson test, the test statistic is the likelihood ratio statistic, and we reject \(H_0\) if likelihood ratio statistic is greater than a certain threshold. Similarly, in the likelihood ratio test, we reject \(H_0\) if the calculated likelihood ratio is greater than the critical value calculated using a desired significance level. Hence, the two tests are equivalent for simple hypothesis test.

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Most popular questions from this chapter

Suppose that a manufacturing process makes about \(3 \%\) defective items, which is considered satisfactory for this particular product. The managers would like to decrease this to about \(1 \%\) and clearly want to guard against a substantial increase, say to \(5 \%\). To monitor the process, periodically \(n=100\) items are taken and the number \(X\) of defectives counted. Assume that \(X\) is \(b(n=100, p=\theta)\). Based on a sequence \(X_{1}, X_{2}, \ldots, X_{m}, \ldots\), determine a sequential probability ratio test that tests \(H_{0}: \theta=0.01\) against \(H_{1}: \theta=0.05 .\) (Note that \(\theta=0.03\), the present level, is in between these two values.) Write this test in the form $$ h_{0}>\sum_{i=1}^{m}\left(x_{i}-n d\right)>h_{1} $$ and determine \(d, h_{0}\), and \(h_{1}\) if \(\alpha_{a}=\beta_{a}=0.02\).

Let \(X_{1}, X_{2}, \ldots, X_{10}\) be a random sample from a distribution that is \(N\left(\theta_{1}, \theta_{2}\right)\). Find a best test of the simple hypothesis \(H_{0}: \theta_{1}=\theta_{1}^{\prime}=0, \theta_{2}=\theta_{2}^{\prime}=1\) against the alternative simple hypothesis \(H_{1}: \theta_{1}=\theta_{1}^{\prime \prime}=1, \theta_{2}=\theta_{2}^{\prime \prime}=4\).

Consider a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from a distribution with pdf \(f(x ; \theta)=\theta(1-x)^{\theta-1}, 00 .\) (a) Find the form of the uniformly most powerful test of \(H_{0}: \theta=1\) against \(H_{1}: \theta>1\) (b) What is the likelihood ratio \(\Lambda\) for testing \(H_{0}: \theta=1\) against \(H_{1}: \theta \neq 1 ?\)

Let \(X\) be a random variable with pdf \(f_{X}(x)=\left(2 b_{X}\right)^{-1} \exp \left\\{-|x| / b_{X}\right\\}\), for \(-\infty0\). First, show that the variance of \(X\) is \(\sigma_{X}^{2}=2 b_{X}^{2} .\) Next, let \(Y\), independent of \(X\), have pdf \(f_{Y}(y)=\left(2 b_{Y}\right)^{-1} \exp \left\\{-|y| / b_{Y}\right\\}\), for \(-\infty0\). Consider the hypotheses $$ H_{0}: \sigma_{X}^{2}=\sigma_{Y}^{2} \text { versus } H_{1}: \sigma_{X}^{2}>\sigma_{Y}^{2} $$ To illustrate Remark \(8.3 .2\) for testing these hypotheses, consider the following data set (data are also in the file exercise8316.rda). Sample 1 represents the values of a sample drawn on \(X\) with \(b_{X}=1\), while Sample 2 represents the values of a sample drawn on \(Y\) with \(b_{Y}=1\). Hence, in this case \(H_{0}\) is true. $$ \begin{array}{|c|rrrr|} \hline \text { Sample } & -0.389 & -2.177 & 0.813 & -0.001 \\ 1 & -0.110 & -0.709 & 0.456 & 0.135 \\ \hline \text { Sample } & 0.763 & -0.570 & -2.565 & -1.733 \\ 1 & 0.403 & 0.778 & -0.115 & \\ \hline \text { Sample } & -1.067 & -0.577 & 0.361 & -0.680 \\ 2 & -0.634 & -0.996 & -0.181 & 0.239 \\ \hline \text { Sample } & -0.775 & -1.421 & -0.818 & 0.328 \\ 2 & 0.213 & 1.425 & -0.165 & \\ \hline \end{array} $$ (a) Obtain comparison boxplots of these two samples. Comparison boxplots consist of boxplots of both samples drawn on the same scale. Based on these plots, in particular the interquartile ranges, what do you conclude about \(H_{0}\) ? (b) Obtain the \(F\) -test (for a one-sided hypothesis) as discussed in Remark 8.3.2 at level \(\alpha=0.10\). What is your conclusion? (c) The test in part (b) is not exact. Why?

The effect that a certain drug (Drug A) has on increasing blood pressure is a major concern. It is thought that a modification of the drug (Drug B) will lessen the increase in blood pressure. Let \(\mu_{A}\) and \(\mu_{B}\) be the true mean increases in blood pressure due to Drug \(\mathrm{A}\) and \(\mathrm{B}\), respectively. The hypotheses of interest are \(H_{0}: \mu_{A}=\mu_{B}=0\) versus \(H_{1}: \mu_{A}>\mu_{B}=0 .\) The two-sample \(t\) -test statistic discussed in Example \(8.3 .3\) is to be used to conduct the analysis. The nominal level is set at \(\alpha=0.05\) For the experimental design assume that the sample sizes are the same; i.e., \(m=n .\) Also, based on data from Drug \(A, \sigma=30\) seems to be a reasonable selection for the common standard deviation. Determine the common sample size, so that the difference in means \(\mu_{A}-\mu_{B}=12\) has an \(80 \%\) detection rate. Suppose when the experiment is over, due to patients dropping out, the sample sizes for Drugs \(A\) and \(B\) are respectively \(n=72\) and \(m=68 .\) What was the actual power of the experiment to detect the difference of \(12 ?\)

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