Question

Let 𝑥₁, 𝑥₂, 𝑥₃ and 𝑥₄ be observed values of a random sample from an 𝑁(𝜃, 𝜎 ² ) distribution, where 𝜃∈ℝ and 𝜎>0 are unknown parameters. Suppose that 𝑥̅=\(\frac{1}{4} ∑^4_{i=1} 𝑥_𝑖 = 3.6 \) and \(\frac{1}{3} ∑^4_{i=1} (𝑥_𝑖-\overline{x} )^2= 20.25\) . For testing the null hypothesis 𝐻₀ ∶ 𝜃=0 against 𝐻₁ ∶𝜃≠0, the 𝑝-value of the likelihood ratio test equals

• A. 0.712
• B. 0.208
• C. 0.104
• D. 0.052

Answer 1

The Correct Option is B

To solve this problem, we will conduct a likelihood ratio test to test the null hypothesis \( H_0: \theta = 0 \) against the alternative hypothesis \( H_1: \theta \neq 0 \). We are provided with a sample mean \( \overline{x} = 3.6 \) and the sample variance \( \frac{1}{3} \sum_{i=1}^4 ( x_i - \overline{x} )^2 = 20.25 \).

The sample variance can be calculated using \( S^2 = \frac{1}{n-1} \sum_{i=1}^{n} ( x_i - \overline{x} )^2 \), where \( n \) is the sample size. Given \( \frac{1}{3} \sum_{i=1}^{4} ( x_i - \overline{x} )^2 = 20.25 \), solving for \( S^2 \) yields \( S^2 = 20.25 \).

We have the following data:

• Sample Size \( n = 4 \)
• Sample Mean \( \overline{x} = 3.6 \)
• Sample Variance \( S^2 = 20.25 \)

Since \( x_i \) are from the \( N(\theta, \sigma^2) \) distribution, we assume the normal distribution under the null hypothesis \( H_0: \theta = 0 \). The test statistic under the null hypothesis is:

\[T = \frac{\overline{x} - 0}{S / \sqrt{n}}\]

Substituting the known values:

\[T = \frac{3.6 - 0}{\sqrt{20.25/4}}\]\[T = \frac{3.6}{\sqrt{5.0625}}\]\[T = \frac{3.6}{2.25} \approx 1.6\]

The test statistic \( T \) follows a \( t \)-distribution with \( n-1 = 3 \) degrees of freedom. To find the \( p \)-value, we look at the \( t \)-distribution table or use statistical software to find the probability:

\[P(T > 1.6) + P(T < -1.6)\]

From statistical tables or software, the two-tailed \( p \)-value for \( T = 1.6 \) with 3 degrees of freedom is approximately \( 0.208 \).

Thus, the correct \( p \)-value of the likelihood ratio test is 0.208, which matches the given correct option.