Question

Let X₁, X₂, X₃, X₄ be a random sample from a distribution with the probability mass function
\(f(x) = \begin{cases}    \theta^x(1-\theta)^{1-x}, & x=0,1 \\     0, & \text{otherwise}, \end{cases}\)
where θ ∈ (0, 1) is unknown. Let 0 < α ≤ 1. To test the hypothesis \(H_0:\theta=\frac{1}{2}\) against \(H_1:\theta>\frac{1}{2},\), consider the size α test that rejects H₀ if and only if \(\sum^4_{i=1} 𝑋𝑖 ≥ k_α\), for some 𝑘_α ∈ {0, 1, 2, 3, 4}. Then for which one of the following values of α, the size α test does NOT exist ?

• A. \(\frac{1}{16}\)
• B. \(\frac{1}{4}\)
• C. \(\frac{11}{16}\)
• D. \(\frac{5}{16}\)

Answer 1

The Correct Option is B

Given a random sample \(X_1, X_2, X_3, X_4\) from a distribution with the probability mass function:

\(f(x) = \begin{cases} \theta^x(1-\theta)^{1-x}, & x=0,1 \\ 0, & \text{otherwise} \end{cases}\)

We need to test the hypothesis:

• \(H_0: \theta = \frac{1}{2}\)
• \(H_1: \theta > \frac{1}{2}\)

The test rejects \(H_0\) if and only if \(\sum_{i=1}^{4} X_i \geq k_{\alpha}\), where \(k_{\alpha} \in \{0, 1, 2, 3, 4\}\).

The distribution of \(\sum_{i=1}^{4} X_i\) under \(H_0\) (\(\theta = \frac{1}{2}\)) is Binomial with parameters \(n=4\) and \(p=\frac{1}{2}\). The probability mass function of a Binomial random variable \(Y\) with parameters \(n\) and \(p\) is:

• \(P(Y=k) = \binom{n}{k} p^k (1-p)^{n-k}\)

Substituting \(n = 4\) and \(p = \frac{1}{2}\), we find:

• \(P(Y=k) = \binom{4}{k} \left(\frac{1}{2}\right)^4 = \frac{1}{16} \binom{4}{k}\)

We need \(\alpha\) equal to the probability of rejection, so calculate:

• \(P(\sum_{i=1}^{4} X_i \geq k_{\alpha})\)

Calculate cumulative probabilities:

• \(P(\sum=0) = \frac{1}{16}\)
• \(P(\sum \leq 1) = \frac{1}{16} + \frac{4}{16} = \frac{5}{16}\)
• \(P(\sum \leq 2) = \frac{1}{16} + \frac{4}{16} + \frac{6}{16} = \frac{11}{16}\)
• \(P(\sum \leq 3 ) + P(\text{all 4}) = 1\)

We can see which values are feasible for \(1-\alpha\). Calculating, \(P(\text{sum} \ge k)\) for \(\alpha = \frac{1}{4}, k=3\) is not a possible cutoff:

• The cutoff probabilities for correct \(\alpha = \frac{1}{4}, k=3\) yields no such test.

Thus, the value of \(\alpha\) for which the size \(\alpha\) test does not exist is \(\frac{1}{4}\).