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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

Short Answer

Expert verified
The best critical region for testing the hypothesis \(H_{0}: \theta=1\) versus \(H_{1}: \theta=2\) is \(C=\left\{\left(x_{1}, x_{2}, \ldots, x_{n}\right): c \leq \prod_{i=1}^{n} x_{i}\right\}\)

Step by step solution

01

Set up the likelihood ratio

The likelihood ratio for the hypothesis \(H_{0}: \theta=1\) versus \(H_{1}: \theta=2\) is given by:\[ \Lambda = \frac{L(1)}{L(2)} \]In these expressions, \(L(1)\) and \(L(2)\) denote the likelihoods of the data under the respective hypotheses. Each one is the product of the probability density function \(f(x; \theta)\) evaluated at each datum under \(\theta=1\) and \(\theta=2\), respectively.
02

Compute likelihoods

Now, compute the likelihoods as follows:\[ L(1) = \prod_{i=1}^{n} f(x_i; 1) = \prod_{i=1}^{n} x_i^{1-1} = 1 \]\[ L(2) = \prod_{i=1}^{n} f(x_i; 2) = \prod_{i=1}^{n} 2 x_i^{2-1} = 2^n \prod_{i=1}^{n} x_i \]The likelihood ratio thus evaluates to \(\Lambda = \frac{1}{2^n \prod_{i=1}^{n} x_i}\).
03

Show that the likelihood ratio is a monotonic function

The likelihood ratio \(\Lambda\) is a decreasing function of the product \(\prod_{i=1}^{n} x_i\). It follows that the most powerful test rejects \(H_0\) for low values of \(\Lambda\), i.e., large values of \(\prod_{i=1}^{n} x_i\), and accepts \(H_0\) for high values of \(\Lambda\), i.e., small values of \(\prod_{i=1}^{n} x_i\). Therefore, the best critical region is of the form \(C = \{\left(x_{1}, x_{2}, \ldots, x_{n}\right): c \leq \prod_{i=1}^{n} x_{i}\}\), where the threshold \(c\) is chosen to satisfy the desired significance level.

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Key Concepts

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

Statistical Hypothesis Testing
Statistical hypothesis testing is a method used by statisticians to make decisions about the properties of populations based on sample data. At its core, the method involves proposing two competing hypotheses: the null hypothesis, denoted as \(H_0\), which represents a default position, and an alternative hypothesis, \(H_1\), which we seek to demonstrate evidence for.

In a typical hypothesis test, the goal is to determine whether there is sufficient evidence from the sample data to reject \(H_0\) in favor of \(H_1\). This decision is made by calculating a test statistic from the sample data and determining whether the statistic falls into a critical region or not. The critical region represents a set of values for which the null hypothesis is considered to be unlikely and is therefore rejected.

To prevent incorrect conclusions, statisticians set a significance level \(\alpha\), often at 0.05 or 5%, which directly corresponds to the probability of rejecting the null hypothesis when it is actually true (a type I error). The lower the significance level, the more stringent and conservative the test.
Probability Density Function
The probability density function (PDF), denoted as \(f(x; \theta)\), describes how the probability is distributed over the values of a continuous random variable. For any given value of \(x\), the PDF gives the relative likelihood of that value occurring within a small interval around \(x\). It is important to note that the function itself is not a probability, but an integral of the PDF over a range of values yields a probability for the variable to lie within that range.

In the context of the example exercise, the PDF is defined as \(f(x ; \theta)=\theta x^{\theta-1}\) for \(0 < x < 1\), which highlights that the likelihood of observing a particular sample can be calculated as a function of \(\theta\) and values of \(x_{i}\). By establishing this relationship, the PDF serves as the fundamental building block for computing likelihoods and conducting hypothesis tests.
Critical Region
The critical region, also called the rejection region, is a key concept in hypothesis testing. It represents the set of all possible outcomes of a test statistic that would lead to the rejection of the null hypothesis \(H_0\). The boundaries of this region are defined and determined based on the desired significance level \(\alpha\) and the distribution of the test statistic under the null hypothesis.

In the step-by-step solution, the critical region is expressed as \(C = \{\left(x_{1}, x_{2}, \ldots, x_{n}\right): c \leq \prod_{i=1}^{n} x_{i}\}\), showing the best set of outcomes for which we reject \(H_0\) when comparing two values of \(\theta\). Essentially, if the calculated test statistic falls within this critical region, there is sufficient evidence to conclude that the sample data is significantly different from what would be expected under the null hypothesis, thus supporting the alternative hypothesis \(H_1\). It's imperative for the significance level and resultant region to be established prior to conducting the test to avoid bias.

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Most popular questions from this chapter

Suppose \(X_{1}, \ldots, X_{n}\) is a random sample on \(X\) which has a \(N\left(\mu, \sigma_{0}^{2}\right)\) distribution, where \(\sigma_{0}^{2}\) is known. Consider the two-sided hypotheses $$ H_{0}: \mu=0 \text { versus } H_{1}: \mu \neq 0 $$ Show that the test based on the critical region \(C=\left\\{|\bar{X}|>\sqrt{\sigma_{0}^{2} / n} z_{\alpha / 2}\right\\}\) is an unbiased level \(\alpha\) test.

Let \(X\) have the pdf \(f(x ; \theta)=\theta^{x}(1-\theta)^{1-x}, x=0,1\), zero elsewhere. We test \(H_{0}: \theta=\frac{1}{2}\) against \(H_{1}: \theta<\frac{1}{2}\) by taking a random sample \(X_{1}, X_{2}, \ldots, X_{5}\) of size \(n=5\) and rejecting \(H_{0}\) if \(Y=\sum_{1}^{h} X_{i}\) is observed to be less than or equal to a constant \(c\). (a) Show that this is a uniformly most powerful test. (b) Find the significance level when \(c=1\). (c) Find the significance level when \(c=0\). (d) By using a randomized test, as discussed in Example \(4.6 .4\), modify the tests given in parts (b) and (c) to find a test with significance level \(\alpha=\frac{2}{32}\).

Let \(X_{1}, \ldots, X_{n}\) and \(Y_{1}, \ldots, Y_{m}\) follow the location model $$ \begin{aligned} X_{i} &=\theta_{1}+Z_{i}, \quad i=1, \ldots, n \\ Y_{i} &=\theta_{2}+Z_{n+i}, \quad i=1, \ldots, m, \end{aligned} $$ where \(Z_{1}, \ldots, Z_{n+m}\) are iid random variables with common pdf \(f(z) .\) Assume that \(E\left(Z_{i}\right)=0\) and \(\operatorname{Var}\left(Z_{i}\right)=\theta_{3}<\infty\) (a) Show that \(E\left(X_{i}\right)=\theta_{1}, E\left(Y_{i}\right)=\theta_{2}\), and \(\operatorname{Var}\left(X_{i}\right)=\operatorname{Var}\left(Y_{i}\right)=\theta_{3}\). (b) Consider the hypotheses of Example 8.3.1, i.e., $$ H_{0}: \theta_{1}=\theta_{2} \text { versus } H_{1}: \theta_{1} \neq \theta_{2} \text { . } $$ Show that under \(H_{0}\), the test statistic \(T\) given in expression \((8.3 .4)\) has a limiting \(N(0,1)\) distribution. (c) Using part (b), determine the corresponding large sample test (decision rule) of \(H_{0}\) versus \(H_{1}\). (This shows that the test in Example \(8.3 .1\) is asymptotically correct.)

Consider a distribution having a pmf of the form \(f(x ; \theta)=\theta^{x}(1-\theta)^{1-x}, x=\) 0,1, zero elsewhere. Let \(H_{0}: \theta=\frac{1}{20}\) and \(H_{1}: \theta>\frac{1}{20} .\) Use the Central Limit Theorem to determine the sample size \(n\) of a random sample so that a uniformly most powerful test of \(H_{0}\) against \(H_{1}\) has a power function \(\gamma(\theta)\), with approximately \(\gamma\left(\frac{1}{20}\right)=0.05\) and \(\gamma\left(\frac{1}{10}\right)=0.90\)

Let \(X_{1}, X_{2}, \ldots, X_{20}\) be a random sample of size 20 from a distribution that is \(N(\theta, 5)\). Let \(L(\theta)\) represent the joint pdf of \(X_{1}, X_{2}, \ldots, X_{20}\). The problem is to test. \(H_{0}: \theta=1\) against \(H_{1}: \theta=0 .\) Thus \(\Omega=\\{\theta: \theta=0,1\\}\). (a) Show that \(L(1) / L(0) \leq k\) is equivalent to \(\bar{x} \leq c\). (b) Find \(c\) so that the significance level is \(\alpha=0.05 .\) Compute the power of this test if \(H_{1}\) is true. (c) If the loss function is such that \(\mathcal{L}(1,1)=\mathcal{L}(0,0)=0\) and \(\mathcal{L}(1,0)=\mathcal{L}(0,1)>0\), find the minimax test. Evaluate the power function of this test at the points \(\theta=1\) and \(\theta=0 .\)

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