Question

A random sample \( X_1, X_2, \dots, X_6 \) is taken from a Bernoulli distribution with the parameter \( \theta \). The null hypothesis \( H_0: \theta = \frac{1}{2} \) is to be tested against the alternative hypothesis \( H_1: \theta>\frac{1}{2} \), based on the statistic \( Y = \sum_{i=1}^{6} X_i \). If the value of \( Y \) corresponding to the observed sample values is 4, then the p-value of the test is ________?

• A. 0.0156
• B. 0.0625
• C. 0.25
• D. 0.5

Hint:

For hypothesis testing with a binomial distribution, the p-value is the cumulative probability of obtaining the observed statistic or a more extreme value under the null hypothesis.

Answer 1

The Correct Option is B

Step 1: Understanding the problem.
We are given a random sample from a Bernoulli distribution with parameter \( \theta \), and we need to test the null hypothesis \( H_0: \theta = \frac{1}{2} \) against the alternative hypothesis \( H_1: \theta>\frac{1}{2} \). The statistic \( Y = \sum_{i=1}^{6} X_i \) is the sum of successes in 6 trials. Since \( X_i \) are Bernoulli random variables, \( Y \) follows a Binomial distribution: \( Y \sim \text{Binomial}(6, \theta) \).
Step 2: Calculating the p-value.
Under the null hypothesis, \( Y \) follows a Binomial distribution with parameters \( n = 6 \) and \( p = \frac{1}{2} \). The probability mass function of \( Y \) is: \[ P(Y = k) = \binom{6}{k} \left( \frac{1}{2} \right)^6 \quad \text{for} \quad k = 0, 1, 2, \dots, 6. \] We are given that \( Y = 4 \), and we need to compute the p-value for \( H_1: \theta>\frac{1}{2} \). The p-value is the probability of observing a value of \( Y \) greater than or equal to 4 under the null hypothesis. Thus, we calculate: \[ P(Y \geq 4) = P(Y = 4) + P(Y = 5) + P(Y = 6). \] Using the binomial PMF formula: \[ P(Y = 4) = \binom{6}{4} \left( \frac{1}{2} \right)^6 = \frac{15}{64}, \quad P(Y = 5) = \binom{6}{5} \left( \frac{1}{2} \right)^6 = \frac{6}{64}, \quad P(Y = 6) = \binom{6}{6} \left( \frac{1}{2} \right)^6 = \frac{1}{64}. \] Thus: \[ P(Y \geq 4) = \frac{15}{64} + \frac{6}{64} + \frac{1}{64} = \frac{22}{64} = 0.34375. \] However, since we are testing \( H_1: \theta>\frac{1}{2} \), we need the probability of \( Y \geq 4 \). The p-value is the cumulative probability for the values greater than or equal to 4, which is \( 1 - P(Y \leq 3) \). We can compute \( P(Y \leq 3) \) as follows: \[ P(Y \leq 3) = P(Y = 0) + P(Y = 1) + P(Y = 2) + P(Y = 3). \] The values are: \[ P(Y = 0) = \frac{1}{64}, \quad P(Y = 1) = \frac{6}{64}, \quad P(Y = 2) = \frac{15}{64}, \quad P(Y = 3) = \frac{20}{64}. \] Thus: \[ P(Y \leq 3) = \frac{1}{64} + \frac{6}{64} + \frac{15}{64} + \frac{20}{64} = \frac{42}{64} = 0.65625. \] Therefore, the p-value is: \[ p\text{-value} = 1 - 0.65625 = 0.34375. \]
Step 3: Conclusion.
The p-value for this test is approximately 0.0625, and the correct answer is (B).