/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 64 Suppose that \(X\), the fraction... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that \(X\), the fraction of a container that is filled, has pdf \(f(x ; \theta)=\theta x^{\theta-1}\) for \(00\) ), and let \(X_{1}, \ldots, X_{n}\) be a random sample from this distribution. a. Show that the most powerful test for \(H_{0}: \theta=1\) versus \(H_{\mathrm{a}}: \theta=2\) rejects the null hypothesis if \(\Sigma \ln \left(x_{i}\right) \geq c .\) b. Is the test of (a) UMP for testing \(H_{0}: \theta=1\) versus \(H_{\mathrm{a}}: \theta>1\) ? Explain your reasoning. c. If \(n=50\), what is the (approximate) value of \(c\) for which the test has significance level \(.05 ?\)

Short Answer

Expert verified
a. Reject if \(\Sigma \ln(x_i) \geq c\). b. No, it's not UMP for \(\theta > 1\). c. Use CLT to find \(c\) for \(\alpha = 0.05\).

Step by step solution

01

Understand the Likelihood Ratio Test (LRT)

The Likelihood Ratio Test is used to compare two hypotheses, the null hypothesis and an alternative, by considering the likelihoods of the observed data under each hypothesis. We calculate the likelihood function for both the null and alternative hypotheses and form their ratio.
02

Write the Likelihood Functions

For the given pdf, the likelihood function under null hypothesis \(H_0: \theta = 1\) is \( L(\theta = 1) = \prod_{i=1}^{n} x_i^{0} = 1 \). For the alternative hypothesis \(H_a: \theta = 2\), the likelihood function is \( L(\theta = 2) = 2^n \prod_{i=1}^{n} x_i \).
03

Form the Likelihood Ratio

The likelihood ratio \( \lambda \) is given by the ratio of the likelihoods under the null and alternative hypotheses: \( \lambda = \frac{L(\theta = 1)}{L(\theta = 2)} = \frac{1}{2^n \prod_{i=1}^{n} x_i} \).
04

Simplify the Rejection Criterion

For the most powerful test, the rejection region is based on \( \ln(\lambda) \). Simplifying \( \ln(\lambda) \), we have \( - \ln(2^n) - \sum_{i=1}^{n} \ln(x_i) \). We reject \(H_0\) if \( \sum_{i=1}^{n} \ln(x_i) \geq c \), where \( c \) is a suitable constant.
05

Check Uniformly Most Powerful (UMP) Criterion

To check if the test is UMP for \(H_0: \theta = 1\) versus \(H_a: \theta > 1\), we note that the UMP test should be valid for all \(\theta > 1\). However, the derived test is specific to the alternative \(\theta = 2\), therefore the test is not UMP for all \(\theta > 1\).
06

Determine Critical Value \(c\) for Significance Level

For \(n = 50\) and significance level \(\alpha = 0.05\), calculate \(c\) by ensuring that \(P\left(\sum \ln(x_i) \geq c \mid \theta = 1\right) = 0.05\). Using the Central Limit Theorem, the distribution of \( \sum \ln(x_i) \) is approximately normal with mean \(-50\) and variance \(50\). Find \(c\) using a normal distribution table or calculator to satisfy this condition.

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Key Concepts

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

Probability Density Function
A Probability Density Function (PDF) is a fundamental concept in statistics used to describe the likelihood of a continuous random variable. It characterizes the distribution of a random variable by assigning probabilities to intervals of values rather than distinct outcomes.
For any continuous random variable, the PDF is a function that can be integrated over an interval to find the probability that the variable falls within that range. The area under the entire curve of the PDF is equal to 1, indicating total certainty that the random variable takes on a value within the entire range.
In this exercise, the PDF is given by \( f(x; \theta) = \theta x^{\theta-1} \) for \(0 < x < 1\), where \(\theta > 0\). This shows that the shape of the PDF, and thus the probabilities, change according to different values of \(\theta\). This particular PDF assumes all values are equally probable within the range when \(\theta=1\), but this changes as \(\theta\) increases, making the likelihood ratio test necessary for comparing hypotheses.
Significance Level
The significance level, often denoted as \(\alpha\), is a critical threshold in hypothesis testing. It helps determine whether a null hypothesis should be rejected. In statistical tests, the significance level represents the probability of rejecting a true null hypothesis, known as Type I error.
Choosing a significance level is crucial because it dictates the risk of a false positive. Common practice sets \(\alpha\) at 0.05, meaning there is a 5% risk of concluding that an effect exists when it does not.
In our task, the significance level of 0.05 is used to find the critical value \(c\) for which the test will reject the null hypothesis. This involves adjusting \(c\) such that the probability of the test statistic exceeding \(c\), given the null is true, is equal to \(\alpha\). This ensures the findings are statistically significant without being too liberal with false positives.
Central Limit Theorem
The Central Limit Theorem (CLT) is a pivotal concept in statistics that explains how the mean of a large number of independent and identically distributed (i.i.d.) random variables will be approximately normally distributed, regardless of the original distribution of the variables.
This theorem holds immense value in simplifying complex statistical problems. By relying on the normal approximation, statisticians can make inferences about populations and calculate probabilities that would otherwise be impractical.
In this problem, the CLT allows us to approximate the distribution of \( \Sigma \ln(x_i) \) as normal. Given \(n=50\), the sum's distribution is approximately normal with mean \(-50\) and variance \(50\). This approximation is crucial for determining the critical value \(c\) to ensure the hypothesis test maintains the specified significance level.
Uniformly Most Powerful Test
The concept of a Uniformly Most Powerful (UMP) Test is important when testing statistical hypotheses. A test is UMP if it is the most powerful for a given significance level across all possible values of a parameter in an alternative hypothesis.
In hypothesis testing, constructing a test that remains most powerful across entire parameter ranges is challenging, thus having immense practical significance when achievable.
In our exercise, the test derived is not UMP for \(H_0: \theta = 1\) versus \(H_a: \theta > 1\) because it is tailored specifically for \(\theta = 2\). A UMP test would need to be most powerful for all values \(\theta > 1\), which requires different conditions and cannot be limited to a single point alternative.

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Most popular questions from this chapter

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