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Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) denote a random sample from a population having a Poisson distribution with mean \(\lambda_{1} .\) Let \(X_{1}, X_{2}, \ldots, X_{m}\) denote an independent random sample from a population having a Poisson distribution with mean \(\lambda_{2}\). Derive the most powerful test for testing \(H_{0}: \lambda_{1}=\lambda_{2}=2\) versus \(H_{a}: \lambda_{1}=1 / 2, \lambda_{2}=3\)

Short Answer

Expert verified
Reject \( H_0 \) if \( \sum Y_i - 3 \sum X_j \) is greater than a critical value.

Step by step solution

01

Define the Hypotheses

We need to test the null hypothesis \( H_0: \lambda_1 = \lambda_2 = 2 \) (the means of the two Poisson distributions are equal to 2) against the alternative hypothesis \( H_a: \lambda_1 = \frac{1}{2}, \lambda_2 = 3 \) (mean of the first distribution is 1/2 and mean of the second distribution is 3).
02

Formulate the Likelihoods

For the null hypothesis \( H_0 \), the likelihood function is formed by the joint distribution:\[ L_0 = \prod_{i=1}^n \frac{e^{-2} (2)^{Y_i}}{Y_i!} \prod_{j=1}^m \frac{e^{-2} (2)^{X_j}}{X_j!} \]And for the alternative hypothesis \( H_a \), the likelihood is:\[ L_a = \prod_{i=1}^n \frac{e^{-\frac{1}{2}} (\frac{1}{2})^{Y_i}}{Y_i!} \prod_{j=1}^m \frac{e^{-3} (3)^{X_j}}{X_j!} \]
03

Calculate the Likelihood Ratio

The likelihood ratio \( \Lambda \) is given by the ratio of \( L_0 \) to \( L_a \):\[\Lambda = \frac{L_0}{L_a} = \frac{\prod_{i=1}^n e^{-2} (2)^{Y_i} \prod_{j=1}^m e^{-2} (2)^{X_j}}{\prod_{i=1}^n e^{-\frac{1}{2}} (\frac{1}{2})^{Y_i} \prod_{j=1}^m e^{-3} (3)^{X_j}}\]
04

Simplify the Likelihood Ratio

Simplify the expression for \( \Lambda \):\[\Lambda = e^{-2n - 2m + \frac{n}{2} + 3m} \cdot \frac{2^{\sum Y_i + \sum X_j}}{(\frac{1}{2})^{\sum Y_i} 3^{\sum X_j}} = e^{-\frac{3n}{2} + m} \cdot 2^{\sum Y_i + \sum X_j} \cdot \left( 2^{\sum Y_i} \right) \cdot 3^{-\sum X_j}\]
05

Determine the Test Criterion

The most powerful test criterion is based on rejecting \( H_0 \) for values of the statistic \( T = \sum_{i=1}^n Y_i - 3 \sum_{j=1}^m X_j \) greater than a critical value that is determined by the significance level \( \alpha \).
06

Conclusion

The most powerful test involves calculating the statistic \( T \) and rejecting \( H_0 \) when \( T \) exceeds a critical value determined using the condition \( \Lambda < c \) for some constant \( c \), based on the significance level.

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Key Concepts

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

Poisson distribution
The Poisson distribution is a significant concept in statistics, particularly for modeling count data. This distribution is especially useful for predicting the number of events occurring within a fixed interval of time or space when these events happen with a known constant mean rate and independently of each other. For instance, if you want to measure the number of cars passing through a toll booth in an hour, the Poisson distribution can be a suitable model.

The key parameter in a Poisson distribution is \( \lambda \), which represents the average rate of occurrence. If events occur with an average of \( \lambda \) per unit time, we describe this situation using a Poisson distribution with parameter \( \lambda \). In the context of our exercise, we deal with samples drawn from populations with known means \( \lambda_1 \) and \( \lambda_2 \), where each has its own distribution characterized by these means.
  • Mean and Variance: Both are equal to \( \lambda \) for a Poisson distribution.
  • Probability Mass Function: The probability of observing \( y \) events is given by \( P(Y = y) = \frac{e^{-\lambda} \lambda^{y}}{y!} \).
likelihood ratio test
The Likelihood Ratio Test (LRT) is a statistical method employed to compare the likelihoods of two hypotheses. It evaluates whether the observed data are more likely under one hypothesis than the other. This test is particularly useful when testing complex models in which parameters are estimated by maximizing a likelihood function.

In the exercise, for testing between the null hypothesis (\( H_0 \)) and the alternative hypothesis (\( H_a \)), the likelihood ratio \( \Lambda \) is calculated. The likelihoods under both hypotheses are compared, and a ratio is formed:
  • Null Hypothesis Likelihood (\( L_0 \)): Based on the assumption that both \( \lambda_1 \) and \( \lambda_2 \) are 2.
  • Alternative Hypothesis Likelihood (\( L_a \)): Based on the mean values \( \lambda_1 = \frac{1}{2} \) and \( \lambda_2 = 3 \).
The test statistic \( \Lambda \) is the ratio \( \frac{L_0}{L_a} \). If this ratio is notably small, it suggests that the observed data are more supportive of the alternative hypothesis than the null hypothesis, prompting a rejection of \( H_0 \).
most powerful test
The quest for the most powerful test revolves around identifying a statistical test that can most effectively distinguish between two hypotheses. In simple terms, it's about maximizing the probability of correctly rejecting a false null hypothesis, known as maximizing the power of the test.

For a test to be considered most powerful, it must achieve the highest power for a given significance level \( \alpha \). This means the test effectively minimizes the Type II error (failing to reject a false null hypothesis) at a given level of Type I error (incorrectly rejecting a true null hypothesis).

With Poisson samples, as in the exercise, devising the most powerful test involves using the likelihood ratio criterion. The test statistic \( T = \sum_{i=1}^n Y_i - 3 \sum_{j=1}^m X_j \) is used to decide whether to reject \( H_0 \). If \( T \) exceeds a certain critical value indicated by the LRT, \( H_0 \) is rejected. This critical value is determined based on the desired significance level, ensuring that the test is both powerful and reliable.
null and alternative hypotheses
Formulating null and alternative hypotheses is a cornerstone of hypothesis testing. These hypotheses are essentially educated guesses about population parameters, forming the basis for statistical testing.

In the context of the exercise:
  • Null Hypothesis (\( H_0 \)): Assumes that both Poisson distributions have the same mean: \( \lambda_1 = \lambda_2 = 2 \). This is the statement we aim to test against.
  • Alternative Hypothesis (\( H_a \)): Posits that the distributions have different means: \( \lambda_1 = \frac{1}{2} \) and \( \lambda_2 = 3 \). This suggests that the observed samples arise from populations with these differing parameters.
By setting up these hypotheses, we define our test's structure and direction. The null hypothesis serves as a default assumption, while the alternative hypothesis represents the change we suspect. Statistical testing then revolves around deciding whether to reject \( H_0 \) in favor of \( H_a \), based on the computed test statistic and significance level.

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Most popular questions from this chapter

What assumptions are made when a Student's \(t\) test is employed to test a hypothesis involving a population mean?

The output voltage for an electric circuit is specified to be \(130 .\) A sample of 40 independent readings on the voltage for this circuit gave a sample mean 128.6 and standard deviation \(2.1 .\) Test the hypothesis that the average output voltage is 130 against the alternative that it is less than \(130 .\) Use a test with level. \(05 .\)

An experimenter has prepared a drug dosage level that she claims will induce sleep for \(80 \%\) of people suffering from insomnia. After examining the dosage, we feel that her claims regarding the effectiveness of the dosage are inflated. In an attempt to disprove her claim, we administer her prescribed dosage to 20 insomniacs and we observe \(Y\), the number for whom the drug dose induces sleep. We wish to test the hypothesis \(H_{0}: p=.8\) versus the alternative, \(H_{a}: p<.8 .\) Assume that the rejection region \(\\{y \leq 12\\}\) is used. a. In terms of this problem, what is a type I error? b. Find \(\alpha\) c. In terms of this problem, what is a type II error? d. Find \(\beta\) when \(p=.6\) e. Find \(\beta\) when \(p=.4\)

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

Suppose that \(Y_{1}, Y_{2}, \ldots, Y_{n}\) denote a random sample from the probability density function given by $$f\left(y | \theta_{1}, \theta_{2}\right)=\left\\{\begin{array}{ll} \left(\frac{1}{\theta_{1}}\right) e^{-\left(y-\theta_{2}\right) / \theta_{1}}, & y > \theta_{2} \\ 0, & \text { elsewhere } \end{array}\right.$$ Find the likelihood ratio test for testing \(H_{0}: \theta_{1}=\theta_{1,0}\) versus \(H_{a}: \theta_{1} > \theta_{1,0},\) with \(\theta_{2}\) unknown.

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