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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.

Short Answer

Expert verified
The UMP test is based on the statistic \(Y=\Sigma X_{i}\) using the critical region \(Y < c\), where \(c\) is determined by \(P(Y<c | \theta = 1) = 0.05\). The power function of the test is given by the integral \(\gamma(\theta) = \int_0^c \frac{\theta^{2n} y^{2n-1} e^{-\theta y}}{(2n-1)!} dy\).

Step by step solution

01

Writing down the likelihood ratio

The likelihood ratio for testing \(H_{0}\) against \(H_{1}\) is \(\frac{L(\theta=1)}{L(\theta>1)} = \frac{(\frac{1}{\theta})^{2n} e^{-\frac{\Sigma X_{i}}{\theta}}}{(\frac{1}{\theta})^{2n} e^{-\Sigma X_{i}}}= e^{\Sigma X_{i} - n\theta}\). The likelihood ratio is a function of \(\Sigma X_{i}\) so the statistic \(Y\) on which the UMP test can be based is \(Y=\Sigma X_{i}\).
02

Finding the critical region and UMP test

Given that we are testing \(H_{0}:\theta=1\) against \(H_{1}:\theta>1\), and \(\theta=1\) in \(H_{0}\) we can determine that the UMP test rejects \(H_{0}\) if \(Y=\Sigma X_{i} < c\). This is because for \(\theta > 1\), the likelihood ratio increases with \( Y=\Sigma X_{i}\), so it is more likely for \(H_{1}\) to be true when \( Y=\Sigma X_{i}\) is small. Therefore, the critical region is of the form \(Y < c\). This means that when the sum of the random sample is less than a certain threshold \(c\), we reject the null hypothesis.
03

Finding the pdf of \(Y\)

The random variable \(Y=\Sigma X_{i}\) follows a gamma distribution with parameters \(\alpha = 2n\) and \(\beta = \theta\). The pdf of this distribution is given by \( f_{Y}(y) = \frac{\theta^{2n} y^{2n-1} e^{-\theta y}}{(2n-1)!}\) for \(y > 0\).
04

Finding the critical value using the significance level

The significance level is 0.05, which means we reject \(H_{0}\) if the probability of a type I error is 0.05. This corresponds to the probability of \(Y < c\) under the null hypothesis. So the equation to determine the critical region \(c\) is \(P(Y < c | \theta = 1) = 0.05\). This equation can be solved numerically to find \(c\).
05

Expressing the power function as an integral

The power function is the probability of rejecting \(H_{0}\) given that \(H_{1}\) is true. It is given by \(\gamma(\theta) = P(Y 1)\) = \(\int_0^c \frac{\theta^{2n} y^{2n-1} e^{-\theta y}}{(2n-1)!} dy\).

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Key Concepts

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

Likelihood Ratio Test
The likelihood ratio test is a statistical tool used to test two competing hypotheses, often referred to as the null hypothesis (\(H_0\)) and an alternative hypothesis (\(H_1\)). In this context, we are interested in testing \(H_0: \theta=1\) against \(H_1: \theta>1\) within a gamma distribution.

To perform a likelihood ratio test, we construct a ratio of likelihood functions for the two hypotheses. In our given exercise, the likelihood ratio involves comparing the likelihood of the sample under \(\theta=1\) and \(\theta>1\). This results in the expression \(e^{\Sigma X_{i} - n\theta}\), where \(Y=\Sigma X_{i}\) serves as our test statistic. The test statistic \(Y\) is a sufficient and complete statistic for the scale parameter \(\theta\) under the gamma distribution assumption.

The nature of the best critical region, where we decide to reject \(H_0\), relies on this likelihood ratio. If \(Y\) is calculated to be less than a critical value \(c\), identified through predetermined significance levels, the null hypothesis is rejected.
Gamma Distribution
The gamma distribution is a two-parameter family of continuous probability distributions defined by a shape parameter \(\alpha\) and a scale parameter \(\beta\).

In our problem, the gamma distribution is characterized by \(\alpha = 2n\) because we sum \(n\) gamma random variables each with \(\alpha = 2\), and the scale parameter \(\beta = \theta\). The probability density function for a gamma distribution is given by \[ f_Y(y) = \frac{\theta^{2n} y^{2n-1} e^{-\theta y}}{(2n-1)!}, \text{ for } y > 0. \]

The statistic \(Y\) defined as \(\Sigma X_{i}\), where each \(X_i\) follows a gamma distribution with specified parameters, inherits its gamma distribution nature with updated parameters. Understanding this concept is vital for analyzing the properties of \(Y\), such as calculating probabilities and setting the critical region in hypothesis testing.
Type I Error
In hypothesis testing, a Type I error occurs when we incorrectly reject a true null hypothesis. This is often denoted by the significance level, \(\alpha\), which represents the probability of making such an error.

In our example, a Type I error would happen if we reject \(H_0: \theta = 1\) even though it is true. We are working with a significance level of \(0.05\), meaning there is a 5% chance of incorrectly rejecting \(H_0\). This probability is used to determine the critical region boundary \(c\), such that \(P(Y < c | \theta = 1) = 0.05\).

Controlling the Type I error rate is fundamental in designing statistically sound experiments and analyses, ensuring that conclusions drawn have a defined tolerance for error.
Power Function
The power function of a statistical test is a vital tool that measures the test's ability to correctly reject a false null hypothesis (\(H_0\)) against an alternative hypothesis (\(H_1\)). Essentially, it gives us the probability that the test will identify an effect when there is one to be detected.

For the exercise at hand, the power function \(\gamma(\theta)\) is calculated as the probability that \(Y\) is less than the critical value \(c\) given \(\theta > 1\). This is expressed by the integral \[ \gamma(\theta) = \int_0^c \frac{\theta^{2n} y^{2n-1} e^{-\theta y}}{(2n-1)!} \, dy. \]

The concept of a power function helps us understand a test's effectiveness. A higher power means a better chance of detecting the true effect. By maximizing the power while controlling for Type I errors, we can establish a uniformly most powerful (UMP) test, which guarantees the best performance under the given conditions.

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Most popular questions from this chapter

Let \(X\) and \(Y\) have the joint pdf. $$f\left(x, y ; \theta_{1}, \theta_{2}\right)=\frac{1}{\theta_{1} \theta_{2}} \exp \left(-\frac{x}{\theta_{1}}-\frac{y}{\theta_{2}}\right), \quad 0

Let \(X_{1}, X_{2}, \ldots, X_{25}\) denote a random sample of size 25 from a normal distribution \(N(\theta, 100)\). Find a uniformly most powerful critical region of size \(\alpha=0.10\) for testing \(H_{0}: \theta=75\) against \(H_{1}: \theta>75\)

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample from a normal distribution \(N(\theta, 100) .\) Show that \(C=\left\\{\left(x_{1}, x_{2}, \ldots, x_{n}\right): c \leq \bar{x}=\sum_{1}^{n} x_{i} / n\right\\}\) is a best critical region for testing \(H_{0}: \theta=75\) against \(H_{1}: \theta=78\). Find \(n\) and \(c\) so that $$P_{H_{0}}\left[\left(X_{1}, X_{2}, \ldots, X_{n}\right) \in C\right]=P_{H_{0}}(\bar{X} \geq c)=0.05$$ and $$P_{H_{1}}\left[\left(X_{1}, X_{2}, \ldots, X_{n}\right) \in C\right]=P_{H_{1}}(\bar{X} \geq c)=0.90$$

. Let \(X\) and \(Y\) have a joint bivariate normal distribution. An observation \((x, y)\) arises from the joint distribution with parameters equal to either $$\mu_{1}^{\prime}=\mu_{2}^{\prime}=0,\quad\left(\sigma_{1}^{2}\right)^{\prime}=\left(\sigma_{2}^{2}\right)^{\prime}=1, \quad \rho^{\prime}=\frac{1}{2}$$ or $$\mu_{1}^{\prime \prime}=\mu_{2}^{\prime \prime}=1,\left(\sigma_{1}^{2}\right)^{\prime \prime}=4, \quad\left(\sigma_{2}^{2}\right)^{\prime \prime}=9, \rho^{\prime \prime}=\frac{1}{2}$$ Show that the classification rule involves a second-degree polynomial in \(x\) and \(y\).

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

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