Question

Let \( X \sim \text{Bin}(3, \theta) \), where \( \theta \in (0,1) \) is an unknown parameter. For testing
\[ H_0: \frac{1}{4} \leq \theta \leq \frac{3}{4} \quad \text{against} \quad H_1: \theta < \frac{1}{4} \quad \text{or} \quad \theta > \frac{3}{4}, \]
consider the test \[ \phi(x) = \begin{cases} 1 & \text{if } x \in \{0, 3\}, \\ 0 & \text{if } x \in \{1, 2\}. \end{cases} \]
The size of the test \( \phi \) is ___________ (rounded off to two decimal places).

Hint:

For binomial tests, the size of the test is computed by integrating the probabilities over the relevant range of parameter values under the null hypothesis.

Answer 1

Step 1: Define the size of the test.
The size of the test is the probability of rejecting \( H_0 \) when \( H_0 \) is true. This is the probability that \( X \in \{0, 3\} \) when \( \frac{1}{4} \leq \theta \leq \frac{3}{4} \). In other words, we need to compute: \[ \text{Size of the test} = \Pr(X = 0 \mid \frac{1}{4} \leq \theta \leq \frac{3}{4}) + \Pr(X = 3 \mid \frac{1}{4} \leq \theta \leq \frac{3}{4}). \] Step 2: Calculate the probability mass function for \( X \).
For a binomial distribution \( X \sim \text{Bin}(3, \theta) \), the probability mass function is: \[ \Pr(X = x) = \binom{3}{x} \theta^x (1 - \theta)^{3 - x}. \] Thus, for \( X = 0 \) and \( X = 3 \): \[ \Pr(X = 0) = \binom{3}{0} \theta^0 (1 - \theta)^3 = (1 - \theta)^3, \] \[ \Pr(X = 3) = \binom{3}{3} \theta^3 (1 - \theta)^0 = \theta^3. \] Step 3: Integrate over the range \( \frac{1}{4} \leq \theta \leq \frac{3}{4} \).
The size of the test is: \[ \int_{1/4}^{3/4} \Pr(X = 0 \mid \theta) d\theta + \int_{1/4}^{3/4} \Pr(X = 3 \mid \theta) d\theta. \] Substitute the expressions for \( \Pr(X = 0) \) and \( \Pr(X = 3) \): \[ \int_{1/4}^{3/4} (1 - \theta)^3 d\theta + \int_{1/4}^{3/4} \theta^3 d\theta. \] Using standard integration, we find: \[ \int_{1/4}^{3/4} (1 - \theta)^3 d\theta \approx 0.140625, \quad \int_{1/4}^{3/4} \theta^3 d\theta \approx 0.140625. \] Thus, the size of the test is approximately: \[ 0.140625 + 0.140625 = 0.28125. \] Thus, the size of the test is \( \boxed{0.28} \).