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Suppose that \(Y_{1}, Y_{2}, \ldots, Y_{n}\) denote a random sample from a population having an exponential distribution with mean \(\theta\) a. Derive the most powerful test for \(H_{0}: \theta=\theta_{0}\) against \(H_{a}: \theta=\theta_{a},\) where \(\theta_{a}<\theta_{0}\) b. Is the test derived in part (a) uniformly most powerful for testing \(H_{0}: \theta=\theta_{0}\) against \(H_{a}: \theta<\theta_{0} ?\)

Short Answer

Expert verified
a. Use \( \sum Y_i < k' \) as the critical region. b. Yes, the test is UMP for \( \theta < \theta_0 \).

Step by step solution

01

Understand the Hypotheses and Distribution

We want to test the null hypothesis \( H_0: \theta = \theta_0 \) against the alternative hypothesis \( H_a: \theta = \theta_a \) where \( \theta_a < \theta_0 \). The random sample \( Y_1, Y_2, \ldots, Y_n \) is from an exponential distribution with mean \( \theta \). The probability density function (pdf) for an exponential distribution is given by \( f(y; \theta) = \frac{1}{\theta} e^{-y/\theta} \) for \( y > 0 \).
02

Determine the Likelihood Function

The likelihood function given the sample \( Y_1, Y_2, \ldots, Y_n \) for the exponential distribution is \( L(\theta) = \prod_{i=1}^n \frac{1}{\theta}e^{-Y_i/\theta} = \frac{1}{\theta^n} e^{-\sum Y_i / \theta} \).
03

Apply the Neyman-Pearson Lemma

The Neyman-Pearson Lemma is used for deriving the most powerful test for simple hypotheses. For it, we need \( \frac{L(\theta_a)}{L(\theta_0)} \), which simplifies to \( \left(\frac{\theta_0}{\theta_a}\right)^n e^{(1/\theta_0 - 1/\theta_a) \sum Y_i} \). To reject \( H_0 \), the value of \( \frac{L(\theta_a)}{L(\theta_0)} \) should be greater than \( k \) for some constant \( k \).
04

Simplify the Test Statistic

The test statistic simplifies as \( \sum Y_i < k' \) for some constant \( k' \), because if \( \theta_a < \theta_0 \), \( (1/\theta_0 - 1/\theta_a) > 0 \). This leads to the critical region for our test to be based on the sum of the sample values, \( \sum Y_i \).
05

Analyze Uniformly Most Powerful Test (UMPT)

To determine if a test is uniformly most powerful, it must be most powerful for all possible alternatives \( \theta < \theta_0 \), not just a specific \( \theta_a \). Exponential distributions have the monotone likelihood ratio property, which guarantees that the test based on \( \sum Y_i \) is UMP for one-sided alternatives.

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Key Concepts

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

Exponential Distribution
The exponential distribution is a continuous probability distribution that is used to model the time between events in a Poisson process. If you think of random events occurring continuously and independently, like requests to a server or radioactive decay, this distribution can represent the time until the next event. The probability density function (pdf) for an exponential distribution is given by \[ f(y; \theta) = \frac{1}{\theta} e^{-y/\theta} \]where \( y > 0 \) and \( \theta \) is the mean of the distribution. Here, \( \theta \) determines both the scale and spread of the distribution. This makes the exponential distribution flexible for modeling real-world phenomena where the rate of occurrence of an event is constant over time. Breaking down the pdf, the term \( \frac{1}{\theta} \) scales the function depending on the mean, and \( e^{-y/\theta} \) represents the decay probability, giving it the characteristic 'memoryless' property.
Neyman-Pearson Lemma
The Neyman-Pearson Lemma is a fundamental result in statistical hypothesis testing. It provides a method to construct the most powerful test for simple hypotheses between two hypotheses. In simpler terms, when testing a null hypothesis \( H_0 \) against an alternative \( H_a \), the lemma gives a criterion to maximize the test's power, i.e., the probability of correctly rejecting \( H_0 \) when \( H_a \) is true. According to the lemma, to achieve the most powerful test, you should use the likelihood ratio \( \frac{L(\theta_a)}{L(\theta_0)} \) and reject \( H_0 \) if this ratio is greater than some constant \( k \). This approach ensures that the test is optimal in detecting differences hypothesized by the alternative hypothesis while keeping the probability of incorrectly rejecting \( H_0 \) at a predefined significance level.
Uniformly Most Powerful Test
A uniformly most powerful test (UMPT) is a hypothesis test that remains the most powerful for all values in the alternative hypothesis space, not just for a specific parameter. In the context of the exponential distribution and the given hypotheses where \( \theta_a < \theta_0 \), the test derived by Neyman-Pearson remains most powerful for all \( \theta < \theta_0 \), due to certain properties of the exponential distribution. Specifically, the exponential distribution exhibits the monotone likelihood ratio property, which facilitates the test's power to remain the strongest uniformly across all possible alternatives within the test's scope. Thus, the test based on the statistic \( \sum Y_i \) for alternatives \( \theta < \theta_0 \) ensures strength and simplicity.
Likelihood Ratio
The likelihood ratio is a crucial tool in statistical hypothesis testing. It compares the likelihoods of two competing hypotheses by constructing a ratio of their probabilities given the observed data. For our exponential distribution problem, the ratio is\[ \frac{L(\theta_a)}{L(\theta_0)} = \left(\frac{\theta_0}{\theta_a}\right)^n e^{(1/\theta_0 - 1/\theta_a) \sum Y_i} \].This ratio becomes the test statistic used in Neyman-Pearson testing. By calculating this value based on your data, it allows you to decide between the null and alternative hypotheses. The likelihood ratio, through its structure, accounts for how likely the data would be under each hypothesis, guiding where the evidence leans more comfortably. The critical part of applying it effectively is determining the threshold \( k \), which defines whether \( H_0 \) is rejected.
Critical Region
The critical region in hypothesis testing is the set of all outcomes that lead to the rejection of the null hypothesis \( H_0 \). It is determined based on the chosen significance level \( \alpha \), representing the probability of falsely rejecting \( H_0 \). In the case of the exponential distribution and the test derived, we utilize the statistic \( \sum Y_i \) and set our critical region such that \( \sum Y_i < k' \), where \( k' \) is determined by the required significance level and the hypothesized parameters. The critical region effectively partitions the sample space into areas where the null hypothesis is rejected or not, providing a clear decision rule for hypothesis testers. The precise choice of \( k' \) ensures that our test maintains the necessary trade-off between sensitivity (detecting when the alternative is true) and specificity (not mistakenly rejecting \( H_0 \)).

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Most popular questions from this chapter

What conditions must be met for the \(Z\) test to be used to test a hypothesis concerning a population mean \(\mu ?\)

Show that a likelihood ratio test depends on the data only through the value of a sufficient statistic. [Hint: Use the factorization criterion.]

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Suppose that independent random samples of sizes \(n_{1}\) and \(n_{2}\) are to be selected from normal populations with means \(\mu_{1}\) and \(\mu_{2},\) respectively, and common variance \(\sigma^{2} .\) For testing \(H_{0}: \mu_{1}=\mu_{2}\) versus \(H_{a}: \mu_{1}-\mu_{2}>0\left(\sigma^{2} \text { unknown }\right)\), show that the likelihood ratio test reduces to the two ample \(t\) test presented in Section \(10.8 .\)

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