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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 \(P_{H_{1}}\left(\sum_{1}^{n} X_{i} \leq c\right)=0.80\).

Short Answer

Expert verified
First, it's proven that C is the best critical region. The Central Limit Theorem is then invoked to bring in Normal approximation. Finally, by using the standard Normal table and the given probabilities under both null and alternate hypothesis, we solve the equations to find the required n and c respective to those probabilities.

Step by step solution

01

Check critical region

We first prove that C is the best critical region. For this, it must satisfy both Neyman–Pearson conditions: The ratio \(\frac{f(x ; p_{1})}{f(x ; p_{0})}\) must be monotone increasing and the randomized test must have the form: Reject \(H_{0}\) if \(\frac{f(x ; p_{1})}{f(x ; p_{0})} > k\), do not reject \(H_{0}\) if \(\frac{f(x ; p_{1})}{f(x ; p_{0})} < k\), and random rejection when \(\frac{f(x ; p_{1})}{f(x ; p_{0})} = k\). Here, \(p_{1}=1/3\) and \(p_{0}=1/2\) and x is the sum of all random variables, \(x_{i}\). From the given pmf, we know that the likelihood ratio decreases as x increases. Hence, the first condition is satisfied.
02

Implement Central Limit Theorem

According to the Central Limit Theorem (CLT), the distribution of the sum (or average) of a large number of independent, identically distributed variables approaches the Normal distribution, regardless of the shape of the original distribution. Because the \(X_i\) are independent and identically distributed and because \(n\) is large, the CLT can be applied. The mean and variance under \(H_{0}\) and \(H_{1}\) can be calculated using \(E(X)=np\) and \(Var(X)=np(1-p)\).
03

Find n and c

Let Z denote the standardized test statistic. Our task is to find n and c so that \(P_{H_{0}}(\frac{\sum_{1}^{n} X_{i} - n*p_{0}}{\sqrt{n*p_{0}(1-p_{0})}} \leq Z) = 0.10\) and \(P_{H_{1}}(\frac{\sum_{1}^{n} X_{i} - n*p_{1}}{\sqrt{n*p_{1}(1-p_{1})}} \leq Z) = 0.80\). Using the standard Normal table, we can solve these equations for n and c. The equation under \(H_{0}\) gives Z-value corresponding to the area 0.10, and the equation under \(H_{1}\) gives Z-value corresponding to the area 0.80. The system of these two equations will yield the required n and c.

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental concept in statistics. It states that, for a large number of independent and identically distributed random variables, the distribution of their sum tends towards a normal distribution. This happens regardless of the original distribution's shape.

Why is this important? It allows us to make inferences about population parameters using sample statistics. In this problem, we have random variables following a Bernoulli distribution, but if our sample size is large enough, we can approximate the sum of these variables with a normal distribution. This makes calculations more manageable and enables the use of z-scores.

The CLT is crucial because it makes hypothesis testing feasible with large samples. For example, if the sample size \(n\) is big, the sum \(\sum_{i=1}^{n} X_i\) under hypothesis \(H_0\) can be approximated as a normal distribution with mean \(n \times \frac{1}{2}\) and variance \(n \times \frac{1}{2} \times \left(1 - \frac{1}{2}\right)\). This simplifies finding probabilities and determining critical values.
Neyman–Pearson Lemma
The Neyman–Pearson Lemma is a cornerstone of hypothesis testing. It provides a way to find the most powerful test (one that maximizes the probability of rejecting a false null hypothesis) for a given size \(\alpha\).

In this problem, we're dealing with hypotheses \(H_0: p=\frac{1}{2}\) and \(H_1: p=\frac{1}{3}\). The lemma states that the likelihood ratio test, expressed as \(\frac{f(x; p_1)}{f(x; p_0)}\), identifies the best critical region for rejecting \(H_0\).

The critical region \(C\) is defined such that you reject \(H_0\) when the likelihood ratio \(> k\). Here, we need to verify that the ratio is monotone, which is true because it decreases as the sum \(x\) increases. This ensures that the region \(C\) provides the highest power of the test within the set constraints.
Likelihood Ratio Test
The Likelihood Ratio Test (LRT) is a method for comparing two statistical models based on their likelihoods. For two competing hypotheses, it evaluates which hypothesis is more likely given the observed data.

To use LRT, we calculate the likelihoods under both \(H_0\) and \(H_1\). The test statistic is the ratio of these likelihoods: \(\frac{L(H_1)}{L(H_0)}\). In our specific problem, this is expressed using the probability mass function for Bernoulli variables.

A decision rule is then established based on this ratio. We reject \(H_0\) in favor of \(H_1\) if the ratio is higher than a preset threshold \(k\). Our model assumes a specific parametric form, where parameters like \(p\) change under different hypotheses. The LRT helps us decide if the observed data significantly favors one hypothesis over the other.

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Most popular questions from this chapter

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 \(H_{1}: \theta_{1}=\theta_{1}^{\prime \prime}=1, \theta_{2}=\theta_{2}^{\prime \prime}=4\).

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 \(\Omega\).

Let \(X_{1}, X_{2}, \ldots, X_{10}\) denote a random sample of size 10 from a Poisson distribution with mean \(\theta .\) Show that the critical region \(C\) defined by \(\sum_{1}^{10} x_{i} \geq 3\) is a best critical region for testing \(H_{0}: \theta=0.1\) against \(H_{1}: \theta=0.5 .\) Determine, for this test, the significance level \(\alpha\) and the power at \(\theta=0.5\).

Let the independent random variables \(Y\) and \(Z\) be \(N\left(\mu_{1}, 1\right)\) and \(N\left(\mu_{2}, 1\right)\), respectively. Let \(\theta=\mu_{1}-\mu_{2}\). Let us observe independent observations from each distribution, say \(Y_{1}, Y_{2}, \ldots\) and \(Z_{1}, Z_{2}, \ldots\) To test sequentially the hypothesis \(H_{0}: \theta=0\) against \(H_{1}: \theta=\frac{1}{2}\), use the sequence \(X_{i}=Y_{i}-Z_{i}, i=1,2, \ldots\) If \(\alpha_{a}=\beta_{a}=0.05\), show that the test can be based upon \(\bar{X}=\bar{Y}-\bar{Z} .\) Find \(c_{0}(n)\) and \(c_{1}(n)\)

Let \(X\) have the pmf \(f(x ; \theta)=\theta^{x}(1-\theta)^{1-x}, x=0,1\), zero elsewhere. \(W_{\text {e }}\) test the simple hypothesis \(H_{0}: \theta=\frac{1}{4}\) against the alternative composite hypothesis \(H_{1}: \theta<\frac{1}{4}\) by taking a random sample of size 10 and rejecting \(H_{0}: \theta=\frac{1}{4}\) if aulul only if the observed values \(x_{1}, x_{2}, \ldots, x_{10}\) of the sample observations are such that \(\sum_{1}^{10} x_{i} \leq 1 .\) Find the power function \(\gamma(\theta), 0<\theta \leq \frac{1}{4}\), of this test.

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