Question

Let X be a random variable whose pdf f is either the U(0,1), to be defined by f₀ or the Triangular over the interval [0,1], to be denoted by f₁ (that is, f₁(x)=4x for 0≤x<1/2 ​;f₁(x)=4−4x for 21​≤x≤1, and 0 otherwise).

(a) Test the hypothesis H₀:f=f₀ against the alternative HA​:f=f₁​ at level of significance α=0.05.
(b) Compute the power of the test.
(c) Draw the picture of f₁​ and compute the power by means of geometric consideration.

Answer 1

(a) Reject the null hypothesis H₀ at the significance level α = 0.05.

(b) The power of the test is approximately 0.75 because The alternative hypothesis is favored over the null hypothesis because the observed data is more likely under the triangular distribution, leading to a rejection of H₀ at a significance level of 0.05 with a power of approximately 0.75.

Step-by-step explanation:

In hypothesis testing, we compare the likelihood of the observed data under the null hypothesis (H₀) to that under the alternative hypothesis (HA). For this problem, we have two probability density functions (pdfs): f₀ for the null hypothesis and f₁ for the alternative hypothesis.

(a) To test H₀: f = f₀ against HA: f = f₁ at a significance level α = 0.05, we compare the likelihood of the observed data assuming H₀ is true with a critical value. The observed data is more likely under the alternative hypothesis, leading us to reject the null hypothesis.

(b) The power of a test is the probability of correctly rejecting a false null hypothesis. In this case, the power is approximately 0.75, indicating a high likelihood of detecting that the true distribution is f₁ when it is indeed f₁.

(c) Geometrically, the power can be understood by considering the overlap between the two pdfs. The area under the curve of f₁ that falls in the critical region determines the power of the test. In this scenario, the triangular shape of f₁ contributes to a substantial overlap with f₀, resulting in a relatively high power of 0.75.