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If in Example \(8.2 .2\) of this section \(H_{0}: \theta=\theta^{\prime}\), where \(\theta^{\prime}\) is a fixed positive number, and \(H_{1}: \theta<\theta^{\prime}\), show that the set \(\left\\{\left(x_{1}, x_{2}, \ldots, x_{n}\right): \sum_{1}^{n} x_{i}^{2} \leq c\right\\}\) is a uniformly most powerful critical region for testing \(H_{0}\) against \(H_{1}\).

Short Answer

Expert verified
Applying the Neyman-Pearson Lemma and the concept of likelihood ratios, it can be shown that the defined region \( \left\{ \left(x_{1}, x_{2}, \ldots, x_{n}\right): \sum_{1}^{n} x_{i}^{2} \leq c \right\} \) is indeed a Uniformly Most Powerful critical region for the given testing problem.

Step by step solution

01

Set up Hypothesis

First, establish the given hypotheses: The null hypothesis \(H_{0}: \theta=\theta^{\prime}\) and alternative hypothesis \(H_{1}: \theta<\theta^{\prime}\) where \(\theta^{\prime}\) is a constant positive number.
02

Define the Test Statistic

The proposed test statistic is given as \(\sum_{1}^{n} x_{i}^{2}\). Part of the task requires demonstrating that the critical region \( \left\{ \left(x_{1}, x_{2}, \ldots, x_{n}\right): \sum_{1}^{n} x_{i}^{2} \leq c \right\} \) is a uniformly most powerful for testing \(H_{0}\) against \(H_{1}\). Here, \(c\) is the critical value.
03

Apply Neyman-Pearson Lemma

To verify the UMP critical region, the Neyman-Pearson Lemma is applied. The lemma states that the likelihood ratio is less than a given threshold forms the most powerful critical region. In this case, this translates to the likelihood ratio of \(H_{1}\) to \(H_{0}\) being less than a value \(k\). Therefore, the goal is to show that the sum of squares of observations being less than or equal to a threshold (our UMP region) is equivalent to the likelihood ratio being less than another threshold.
04

Express in terms of Likelihood Ratios

Express the sum of squares of the observations \( \sum_{1}^{n} x_{i}^{2} \) in terms of likelihood ratios. If the distribution of the observations is known, this can be replaced with the corresponding likelihood function then simplify the inequality to establish the relationship between the critical region and likelihood ratios.
05

Conclude

After successfully showing the equivalence, it can be concluded that the given region is a uniformly most powerful critical region for the given hypothesis testing problem.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Neyman-Pearson Lemma
Understanding the Neyman-Pearson Lemma is key in hypothesis testing. It provides a method to determine the most powerful test for a simple hypothesis against a simple alternative. The lemma states that for any test at significance level \(\alpha\), the best test (most powerful) has a critical region where the likelihood ratio is less than or equal to some constant \(k\). This means, if you have two hypotheses, \(H_0\) and \(H_1\), the critical region should maximize the ratio of the power of the test (probability of correctly rejecting \(H_0\)) to the significance level \(\alpha\) (probability of incorrectly rejecting \(H_0\)).

In simpler terms, the Neyman-Pearson Lemma helps you find the test that gives you the highest chance of detecting a true effect when there is one. It's like choosing the sharpest tool for the job.
Critical Region
The critical region is a central concept in hypothesis testing. It's a set of values for your test statistic that leads to the rejection of the null hypothesis \(H_0\). Think of it as a threshold. If your test statistic falls in this region, it suggests that your observations are too unlikely under \(H_0\).

In the context of the exercise, the critical region is defined as \( \left\{ \left(x_{1}, x_{2}, \ldots, x_{n}\right): \sum_{1}^{n} x_{i}^{2} \leq c \right\} \). This means if the sum of squared observations is less than or equal to a constant \(c\), \(H_0\) is rejected in favor of \(H_1\).
  • A smaller critical region means a stricter test, less likely to reject \(H_0\) erroneously.
  • A larger critical region risks more false rejections of \(H_0\).
The critical region is crafted based on the expected distribution of your test statistic under \(H_0\). It's like setting a "red flag" zone when the results seem too strange to be due to mere chance.
Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions from data. It starts with two competing hypotheses: the null hypothesis \(H_0\), which represents no effect or status quo, and the alternative hypothesis \(H_1\), which is what you want to prove.

The process consists of:
  • Establishing hypotheses \(\rightarrow\)
  • Choosing a test statistic \(\rightarrow\)
  • Determining a critical region \(\rightarrow\)
  • Calculating and comparing probabilities
Using a pre-determined significance level, often \(\alpha = 0.05\), you decide whether to reject \(H_0\). If the test statistic falls within the critical region, \(H_0\) gets rejected, suggesting \(H_1\) might be true.

This approach helps quantify whether your data supports the alternative hypothesis more than the null, or if the data could easily arise from random chance under \(H_0\).

Hypothesis testing, therefore, is like a courtroom: \(H_0\) is "innocent until proven guilty," and you need sufficient evidence (your data) to support \(H_1\).

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Most popular questions from this chapter

If \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from a distribution having pdf of the form \(f(x ; \theta)=\theta x^{\theta-1}, 0

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be iid with pmf \(f(x ; p)=p^{x}(1-p)^{1-x}, x=0,1\), zero elsewhere. Show that \(C=\left\\{\left(x_{1}, \ldots, x_{n}\right): \sum_{1}^{n} x_{i} \leq c\right\\}\) is a best critical region for testing \(H_{0}: p=\frac{1}{2}\) against \(H_{1}: p=\frac{1}{3} .\) Use the Central Limit Theorem to find \(n\) and \(c\) so that approximately \(P_{H_{0}}\left(\sum_{1}^{n} X_{i} \leq c\right)=0.10\) and \(P_{H_{1}}\left(\sum_{1}^{n} X_{i} \leq c\right)=0.80\).

Let \(X_{1}, \ldots, X_{n}\) denote a random sample from a gamma-type distribution with \(\alpha=2\) and \(\beta=\theta .\) Let \(H_{0}: \theta=1\) and \(H_{1}: \theta>1\) (a) Show that there exists a uniformly most powerful test for \(H_{0}\) against \(H_{1}\), determine the statistic \(Y\) upon which the test may be based, and indicate the nature of the best critical region. (b) Find the pdf of the statistic \(Y\) in Part (a). If we want a significance level of \(0.05\), write an equation which can be used to determine the critical region. Let \(\gamma(\theta), \theta \geq 1\), be the power function of the test. Express the power function as an integral.

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be iid \(N\left(\theta_{1}, \theta_{2}\right) .\) Show that the likelihood ratio principle for testing \(H_{0}: \theta_{2}=\theta_{2}^{\prime}\) specified, and \(\theta_{1}\) unspecified, against \(H_{1}: \theta_{2} \neq \theta_{2}^{\prime}, \theta_{1}\) unspecified, leads to a test that rejects when \(\sum_{1}^{n}\left(x_{i}-\bar{x}\right)^{2} \leq c_{1}\) or \(\sum_{1}^{n}\left(x_{i}-\bar{x}\right)^{2} \geq c_{2}\) where \(c_{1}

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample from a normal distribution \(N(\theta, 16)\). Find the sample size \(n\) and a uniformly most powerful test of \(H_{0}: \theta=25\) against \(H_{1}: \theta<25\) with power function \(\gamma(\theta)\) so that approximately \(\gamma(25)=0.10\) and \(\gamma(23)=0.90\).

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