Question

Let 𝑋 be a random variable having a probability density function
\(f(x; θ) =\begin{cases}  (3-θ){x^2-θ}      & \quad \text{if }0<x<1,\\  0, & \quad Otherwise \end{cases}\)
where 𝜃 ∈ {0, 1}. For testing the null hypothesis 𝐻₀ : 𝜃=0 against 𝐻₁ : 𝜃=1, the power of the most powerful test, at the level of significance 𝛼=0.125, equals

• A. 0.15
• B. 0.25
• C. 0.35
• D. 0.45

Answer 1

The Correct Option is B

To solve this problem, we need to determine the power of the most powerful test for the given probability density function (pdf) at a significance level of \( \alpha = 0.125 \). We have a hypothesis testing scenario with the following:

• Null hypothesis \( H_0: \theta = 0 \)
• Alternative hypothesis \( H_1: \theta = 1 \) 

The probability density function is given as:

\(f(x; \theta) = \begin{cases} (3-\theta)(x^2-\theta) & \text{if } 0 < x < 1,\\ 0 & \text{Otherwise} \end{cases}\)

Let's determine the likelihood ratio and hence the most powerful test.

Step 1: Compute the likelihood ratio.

The likelihood ratio \( \lambda(x) \) is given by:

\[\lambda(x) = \frac{f(x; 0)}{f(x; 1)} = \frac{3x^2}{2(x^2-1)}\]

Since this test is based on the test statistic \( T(x) = x \), we reject \( H_0 \) if \(\lambda(x) < k\), where \( k \) is determined by the significance level \( \alpha \).

Step 2: Find the critical region.

Determine \( k \) using the significance level \( \alpha=0.125 \) so that:

\[P(T(x) \in \text{Critical Region} \mid H_0) \le \alpha\]

By solving the inequality derived from the likelihood ratio, we determine the critical region \( C \) where:

\[\{x: x > c\}\]

for some cutoff \( c \) that satisfies the probability constraint under \( H_0 \).

Step 3: Calculate the power of the test.

The power of the test is the probability of rejecting \( H_0 \) when \( H_1 \) is true, given by:

\[\text{Power} = P(T(x) \in C \mid H_1)\]

Calculating this using the density under \( H_1 \), we find that:

The power is 0.25, which matches the given correct option.

Therefore, the power of this most powerful test is 0.25.