https://file.aiquickdraw.com/mathbot/file/17300143838969h43syhb.jpg

Let's go through each problem in the set and outline the key points for solving them.

### Problem 1
1. **Distribution**: \(X_1, X_2, \dots, X_n\) is a sample from a Bernoulli distribution with pmf:
   \[
   f(x; \theta) = \theta^x (1 - \theta)^{1 - x}, \quad x = 0, 1
   \]
   where \(\theta\) is either \(\frac{1}{2}\) (null hypothesis \(H_0\)) or \(\frac{1}{3}\) (alternative hypothesis \(H_1\)).

2. **Goal**: Show that the set
   \[
   C = \left\{ (x_1, \dots, x_n) : \sum_{i=1}^n x_i \geq k \right\}
   \]
   forms the critical region for the most powerful test of size \(\alpha\) for testing \(H_0\) versus \(H_1\).

3. **Approach**:
   - Use the **Neyman-Pearson Lemma**, which gives the most powerful test by comparing the likelihood ratio to a threshold.
   - Calculate the likelihood ratio for the Bernoulli distribution with hypotheses values.
   - Determine the value of \(k\) such that the test has size \(\alpha\).

### Problem 2
1. **Distribution**: \(X_1, X_2, \dots, X_{10}\) is a sample from a Poisson distribution with mean \(\theta\).

2. **Goal**: Show that
   \[
   C = \left\{ (x_1, \dots, x_{10}) : \sum_{i=1}^{10} x_i \geq 3 \right\}
   \]
   is the critical region of a most powerful test of size \(\alpha = 0.0803\) for testing \(H_0: \theta = 0.1\) versus \(H_0: \theta = 0.5\).

3. **Approach**:
   - Apply the **Neyman-Pearson Lemma** for Poisson distributions.
   - Compute the probability of the test statistic under \(H_0\) to verify that the size is \(0.0803\).
   - Determine whether \(\sum_{i=1}^{10} X_i\) satisfies the necessary conditions for the test.

### Problem 3
1. **Distribution**: \(X_1, X_2, \dots, X_n\) is a sample from a normal distribution \(N(\theta, 2)\) with unknown mean \(\theta\).

2. **Goal**: Derive the most powerful test of size \(\alpha\) for testing \(H_0: \theta = -1\) versus \(H_1: \theta = 1\).

3. **Approach**:
   - Use the **Neyman-Pearson Lemma** again to derive the likelihood ratio test.
   - Simplify the test to its implementable form, focusing on the statistic that depends linearly on \(\theta\).
   - Determine the critical region by finding the appropriate test statistic and comparing it to a threshold for size \(\alpha\).

### Problem 4
1. **Distribution**: A single observation \(X\) from the density function:
   \[
   f(x; \theta) = (2 \theta x + 1 - \theta), \quad 0 < x < 1
   \]
   with \(-1 \leq \theta \leq 1\).

2. **Goal**: Find the most powerful size \(\alpha\) test of \(H_0: \theta = 0\) versus \(H_1: \theta = 1\).

3. **Approach**:
   - Compute the likelihood ratio for the density under \(H_0\) and \(H_1\).
   - Apply the **Neyman-Pearson Lemma** to determine the critical region in terms of \(X\) for a test of size \(\alpha\).
   - Verify that the resulting test satisfies the Neyman-Pearson criterion.

---

Would you like me to go into detailed solutions for each problem?

---

### Follow-up Questions

1. How does the Neyman-Pearson Lemma determine the form of a most powerful test?
2. What are the steps to calculate the likelihood ratio for hypothesis testing?
3. In Problem 2, why is the sum of Poisson-distributed random variables used as a test statistic?
4. How do you determine the test size \(\alpha\) in the context of these hypothesis tests?
5. What role does the critical region play in the context of most powerful tests?

### Tip

For hypothesis tests involving composite hypotheses, consider how the likelihood ratio varies across the parameter space to identify uniformly most powerful (UMP) tests if possible.

Let X1, X2, ..., Xn be a random sample from a Bernoulli distribution...

Explore hypothesis testing using Neyman-Pearson Lemma to derive the most powerful tests for Bernoulli, Poisson, and normal distributions with varying hypotheses. Detailed solutions for critical region and likelihood ratio calculation are provided.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation