Question

. Let \(X\) have a pdf of the form \(f(x ; \theta)=\theta x^{\theta-1}, 0

Short answer

The power function for the given hypothesis test set up is \(\beta(\theta) = \{\frac{9}{32}\) for \(\theta=1\) and \(\frac{27}{64}\) for \(\theta=2\)\}.

Step-by-step solution

Understand the Hypotheses and the Test Statistic

The null hypothesis \(H_{0}\) and the alternative hypothesis \(H_{1}\) are given by \(\theta=1\) and \(\theta=2\) respectively. The test statistic is \(x_{1}x_{2}\) and the critical region is defined such that if \(x_{1}x_{2}\) is less than or equal to \(\frac{3}{4}\), we reject the null hypothesis \(H_{0}\)

Calculating the Power Function

The power function of a hypothesis test is the probability of correctly rejecting \(H_{0}\) when \(H_{1}\) is true. It is denoted as \(\beta(\theta)\) and is computed as follows:First, consider the scenario where \(H_{0}\) is true, i.e., \(\theta=1\). Then, the pdf of \(x_{1}x_{2}\) is \(f(x_{1}x_{2}; \theta=1)= x_{1}x_{2}\) for \(0 < x_{1},x_{2} < 1\). The probability of falling in the critical region is given by the integral over the critical region:\(P_{\theta=1}(x_{1}x_{2} \leq \frac{3}{4}) = \int_{0}^{\sqrt[4]{3/4}} \int_{0}^{\sqrt[4]{3/4}} x_{1}x_{2} dx_{2} dx_{1} = \frac{9}{32}\). For \(H_{1}\), where \(\theta=2\), the pdf is \(f(x_{1}x_{2}; \theta=2)=2x_{1}x_{2}\). The probability of falling in the critical region is \(P_{\theta=2}(x_{1}x_{2} \leq \frac{3}{4}) = \int_{0}^{\sqrt[4]{3/4}} \int_{0}^{\sqrt[4]{3/4}} 2x_{1}x_{2} dx_{2} dx_{1} = \frac{27}{64}\). The power function is then given by: \[\beta(\theta) = \begin{cases}\frac{9}{32} & \text{if } \theta=1, \rac{27}{64} & \text{if } \theta=2.\end{cases}\]

Interpretation of the Power Function

The values of the power function for each possible value of \(\theta\) gives the probability of rejecting \(H_{0}\) when it should be rejected and that of rejecting it when it should not be. For \(\theta=1\), which details when \(H_{0}\) is true, there is lower power at \(\frac{9}{32}\). This means the likelihood of a Type I error (rejecting \(H_{0}\) when it should not be) exists. On the other hand, for \(\theta=2\) i.e., \(H_{1}\) is true, the power of the test is \(\frac{27}{64}\). This higher power signifies that the test is more likely to correctly reject \(H_{0}\) in favour of \(H_{1}\), which describes the purpose of a hypothesis test.

Key concepts

Power Function

In hypothesis testing, the power function is a crucial concept. It represents the probability that a test will correctly reject a null hypothesis when the alternative hypothesis is true. This is important because it tells us how well the test can detect an effect when there is one to be found.
The power function is denoted by \( \beta(\theta) \), where \( \theta \) is the parameter in question. In the context of our exercise, this involves testing the hypotheses \( H_{0}: \theta=1 \) and \( H_{1}: \theta=2 \).
To calculate the power function, we determine the probability that the test statistic falls within the critical region under the true alternative hypothesis. Specifically, when \( \theta=2 \), we integrate the Probability Density Function (PDF) over the critical region and find:

• For \( \theta=1 \), the power is \( \frac{9}{32} \).
• For \( \theta=2 \), the power is \( \frac{27}{64} \).

This indicates that the test is more sensitive to changes under \( \theta=2 \), as the power is higher at this value, indicating a better ability to detect when \( H_{0} \) should indeed be rejected.

Probability Density Function (PDF)

The Probability Density Function (PDF) is a mathematical function that describes the likelihood of a continuous random variable to take a particular value. It plays a pivotal role in statistical inferences, especially in hypothesis testing.
The PDF for this particular problem is given by \( f(x; \theta) = \theta x^{\theta - 1} \) for \( 0 < x < 1 \), and \( \theta \) takes values in the set \( \{1, 2\} \). This tells us how the values of \( X \) are distributed given \( \theta \).
The PDF helps determine the probability associated with the critical region, which is essential for evaluating the power function. When testing between these hypotheses, adjust the PDF accordingly:

• If \( \theta = 1 \), the PDF simplifies to \( f(x) = x^{0} = 1 \), a uniform distribution over the interval \((0,1)\).
• If \( \theta = 2 \), the PDF becomes \( f(x) = 2x \), giving more weight to values closer to 1.

By integrating the PDF over the critical region, we find the probabilities needed to calculate the power function.

Critical Region

In hypothesis testing, the critical region or rejection region is the set of all values of the test statistic that leads to rejecting the null hypothesis. Defining this region correctly is essential to ensure that the test has the correct balance of power and significance.
In the given exercise, the critical region is defined as \( C=\{(x_1, x_2) : x_1 x_2 \geq 3/4 \} \). This means that if the product of our sample values \( x_1 \) and \( x_2 \) falls within this region, we reject \( H_0: \theta=1 \) in favor of \( H_1: \theta=2 \).
This region is chosen based on the distribution and the desired significance level of the test. The power of a test is connected to how this critical region is defined. A larger critical region generally increases the power of a test, but it must be carefully balanced against the risk of Type I errors (false positives).
In conclusion, understanding how to define and use the critical region is key to effectively interpreting the results of a hypothesis test. The critical region distinguishes which outcomes point towards significant evidence against the null hypothesis, helping to make decisions based on statistical data.