Question

Suppose that X, which represents the fraction of a container that is filled, has a probability density function (pdf) given by f(x;θ) = θx^(θ-1) for 0 < x < 1, where θ > 0. Let X₁, ..., Xₙ be a random sample from this distribution.

(a) Demonstrate that the most powerful test for the null hypothesis H₀: θ = 1 against the alternative hypothesis Hₐ: θ = 2 rejects the null hypothesis if ∑ln(xi) ≥ c, where c is a constant.

(b) Is the test described in (a) a uniformly most powerful (UMP) test for testing H₀: θ = 1 against Hₐ: θ > 1? Provide an explanation for your reasoning.

Answer 1

The most powerful test rejects the null hypothesis if the sum of the natural logarithms of the sample values is greater than or equal to a constant. The test described is not a uniformly most powerful test for the given hypotheses.

Step-by-step explanation:

To demonstrate that the most powerful test for the null hypothesis H₀: θ = 1 against the alternative hypothesis Hₐ: θ = 2 rejects the null hypothesis if ∑ln(xi) ≥ c, we start by finding the likelihood ratio test statistic. The likelihood ratio is given by λ = (θ₁/θ₂)ⁿ, where n is the sample size and θ₁ and θ₂ are the parameter values under the null and alternative hypotheses respectively. Taking the natural logarithm of both sides, we get ln(λ) = n(ln(θ₁) - ln(θ₂)). We can then rewrite this as ∑ln(xᵢ) ≥ c, where xᵢ represents each individual value in the sample.

Regarding part (b), the test described in part (a) is not a uniformly most powerful (UMP) test. A UMP test is one that is the most powerful for every value of the parameter θ under the alternative hypothesis. However, in this case, the UMP test depends on the specific value θ = 2. Therefore, the test described in part (a) is not UMP for testing H₀: θ = 1 against Hₐ: θ > 1.