Question:medium

If, \(1 \le x \le 1.5\) is the critical region for testing the null hypothesis \(H_0: \theta=1\) against the alternative hypothesis \(H_1: \theta=2\) on the basis of a single observation from the population, \( f(x;\theta) = \begin{cases} \frac{1}{\theta} & ; 0 \le x \le \theta \\ 0 & ; \text{otherwise} \end{cases} \), then the power of the test, is

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To calculate the power of a test, always use the parameter value from the alternative hypothesis (\(H_1\)). To calculate the size of the test (Type I error, \(\alpha\)), use the parameter value from the null hypothesis (\(H_0\)).
Updated On: Feb 18, 2026
  • \( \frac{3}{4} \)
  • \( \frac{1}{2} \)
  • \( \frac{4}{5} \)
  • \( \frac{1}{4} \)
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The Correct Option is D

Solution and Explanation

Step 1: Concept Overview:
The power of a statistical test represents the likelihood of correctly rejecting the null hypothesis (\(H_0\)) when the alternative hypothesis (\(H_1\)) is actually true. This is determined by calculating the probability of an observation falling within the critical region, based on the parameter value specified by the alternative hypothesis.

Step 2: Core Formula:
Power is defined as \(P(\text{Reject } H_0 | H_1 \text{ is true})\). Given a critical region of \(1 \le x \le 1.5\), the test's power is \(P(1 \le X \le 1.5)\), calculated using the distribution under \(H_1\).

Step 3: Detailed Calculation:
Under the alternative hypothesis \(H_1: \theta = 2\), the population's probability density function (PDF) is: \[ f(x; 2) = \begin{cases} \frac{1}{2} & ; 0 \le x \le 2
0 & ; \text{otherwise} \end{cases} \] This represents a uniform distribution over the interval [0, 2]. The critical region for rejecting \(H_0\) is \(1 \le x \le 1.5\). The test's power is the probability of observing a value \(x\) within this region, given that \(\theta=2\). \[ \text{Power} = P(1 \le X \le 1.5 | \theta=2) \] This probability is found by integrating the PDF under \(H_1\) across the critical region: \[ \text{Power} = \int_{1}^{1.5} f(x; 2) \,dx = \int_{1}^{1.5} \frac{1}{2} \,dx \] \[ = \frac{1}{2} [x]_{1}^{1.5} \] \[ = \frac{1}{2} (1.5 - 1) = \frac{1}{2} (0.5) = \frac{1}{4} \]
Step 4: Solution:
The power of the test is \( \frac{1}{4} \).
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