Question

Let X₁, X₂, X₃, X₄ be a random sample from a Poisson distribution with unknown mean λ > 0. For testing the hypothesis
H₀ : λ = 1 against H₁ : λ = 1.5,
let β denote the power of the test that rejects H₀ if and only if \(∑_{i=1}^4 𝑋_𝑖  ≥ 5\). Then which one of the following statements is true ?

• A. β > 0.80
• B. 0.75 < β ≤ 0.80
• C. 0.70 < β ≤ 0.75
• D. 0.65 < β ≤ 0.70

Answer 1

The Correct Option is C

The problem involves testing a hypothesis about the mean of a Poisson distribution. We have a random sample \(X_1, X_2, X_3, X_4\) from a Poisson distribution with an unknown mean λ. Let us test the hypothesis:

• \(H_0\ : \lambda = 1\)
• \(H_1\ : \lambda = 1.5\)

The decision rule is to reject \(H_0\) if and only if \(∑_{i=1}^4 𝑋_𝑖 \ge 5\).

Step-by-Step Solution

1. For a Poisson distribution with parameter \(\lambda\), the sum of \(n\) independent Poisson random variables, each with parameter \(\lambda\), follows a Poisson distribution with parameter \(n\lambda\). Here, the sum \(\sum_{i=1}^4 X_i\) follows \(\text{Poisson}(4 \lambda)\).
2. Under \(H_1: \lambda = 1.5\), the sum follows \(\text{Poisson}(4 \times 1.5 = 6)\).
3. We calculate the power of the test, denoted by \(\beta\), which is the probability of rejecting \(H_0\) when \(H_1\) is true. This is given by:

\(\beta = P(\sum_{i=1}^4 X_i \ge 5 \mid \lambda = 1.5)\)

1. We need to find:

\(\beta = 1 - [P(\sum_{i=1}^4 X_i = 0) + P(\sum_{i=1}^4 X_i = 1) + \ldots + P(\sum_{i=1}^4 X_i = 4)] \mid \lambda = 1.5\)

1. Using the Poisson probability mass function:

P\(\sum_{i=1}^4 X_i = k \mid \lambda = 1.5 = \frac{e^{-6} \cdot 6^k}{k!}\)

1. Calculate this for \(k = 0, 1, 2, 3, 4\):
   • \(k = 0\): \(P(k = 0) = \frac{e^{-6} \cdot 6^0}{0!} = e^{-6}\)
   • \(k = 1\): \(P(k = 1) = \frac{e^{-6} \cdot 6^1}{1!} = 6 e^{-6}\)
   • \(k = 2\): \(P(k = 2) = \frac{e^{-6} \cdot 6^2}{2!} = 18 e^{-6}\)
   • \(k = 3\): \(P(k = 3) = \frac{e^{-6} \cdot 6^3}{3!} = 36 e^{-6}\)
   • \(k = 4\): \(P(k = 4) = \frac{e^{-6} \cdot 6^4}{4!} = 54 e^{-6}\)
2. Summing these probabilities:

\(\text{Total} = e^{-6} (1 + 6 + 18 + 36 + 54)\)

This results in:

\(\text{Total} = e^{-6} \times 115\)

1. Then \(\beta = 1 - e^{-6} \times 115\).
2. Considering the exponential decay and summation, \(\beta\) lies between 0.70 and 0.75.

Thus, the correct answer is:

• \(0.70 < \beta \leq 0.75\)