Question:
Let \( X_1, X_2, X_3, X_4 \) be a random sample from a discrete distribution with the probability mass function
\[
P(X = 0) = 1 - P(X = 1) = 1 - p, \quad 0<p<1.
\]
To test the hypothesis
\[
H_0 : p = \frac{3}{4} \quad \text{against} \quad H_1 : p = \frac{4}{5},
\]
consider the test:
Reject \( H_0 \) if \( X_1 + X_2 + X_3 + X_4>3 \).
Let the size and power of the test be denoted by \( \alpha \) and \( \gamma \), respectively. Then \( \alpha + \gamma \) (round off to 2 decimal places) equals ________
Let \( X_1, X_2, X_3, X_4 \) be a random sample from a discrete distribution with the probability mass function
\[
P(X = 0) = 1 - P(X = 1) = 1 - p, \quad 0<p<1.
\]
To test the hypothesis
\[
H_0 : p = \frac{3}{4} \quad \text{against} \quad H_1 : p = \frac{4}{5},
\]
consider the test:
Reject \( H_0 \) if \( X_1 + X_2 + X_3 + X_4>3 \).
Let the size and power of the test be denoted by \( \alpha \) and \( \gamma \), respectively. Then \( \alpha + \gamma \) (round off to 2 decimal places) equals ________
Show Hint
The size of the test \( \alpha \) is the probability of rejecting \( H_0 \) when \( H_0 \) is true, and the power \( \gamma \) is the probability of rejecting \( H_0 \) when \( H_1 \) is true. The sum \( \alpha + \gamma \) provides insight into the overall effectiveness of the test.
Updated On: Dec 12, 2025
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Correct Answer: 0.68
Solution and Explanation
Step 1: Understand the problem setup.
We are given a random sample \( X_1, X_2, X_3, X_4 \) from a discrete distribution with two possible values \( X = 0 \) and \( X = 1 \), with probabilities \( P(X = 0) = 1 - p \) and \( P(X = 1) = p \).
The test rejects the null hypothesis \( H_0 \) if the sum of the random sample exceeds 3, i.e., if \( X_1 + X_2 + X_3 + X_4>3 \). We are asked to find the size \( \alpha \) and the power \( \gamma \) of the test, and then compute \( \alpha + \gamma \).
Step 2: Calculate the size of the test (\( \alpha \)).
The size \( \alpha \) is the probability of rejecting \( H_0 \) when \( H_0 \) is true. Under \( H_0 : p = \frac{3}{4} \), the number of successes \( X_1 + X_2 + X_3 + X_4 \) follows a binomial distribution with parameters \( n = 4 \) and \( p = \frac{3}{4} \). We reject \( H_0 \) if \( X_1 + X_2 + X_3 + X_4>3 \).
Thus, we need to compute the probability of getting 4 successes, which is: \[ \alpha = P(X_1 + X_2 + X_3 + X_4>3 \mid p = \frac{3}{4}). \] The probability mass function of the binomial distribution is: \[ P(X = k) = \binom{4}{k} \left( \frac{3}{4} \right)^k \left( \frac{1}{4} \right)^{4-k}. \] For \( X_1 + X_2 + X_3 + X_4 = 4 \), we have: \[ P(X_1 + X_2 + X_3 + X_4 = 4) = \binom{4}{4} \left( \frac{3}{4} \right)^4 = 1 \times \left( \frac{3}{4} \right)^4 = \frac{81}{256}. \] Thus, \( \alpha = P(X_1 + X_2 + X_3 + X_4 = 4) = \frac{81}{256} \).
Step 3: Calculate the power of the test (\( \gamma \)).
The power \( \gamma \) is the probability of rejecting \( H_0 \) when \( H_1 \) is true. Under \( H_1 : p = \frac{4}{5} \), the number of successes \( X_1 + X_2 + X_3 + X_4 \) follows a binomial distribution with parameters \( n = 4 \) and \( p = \frac{4}{5} \). We reject \( H_0 \) if \( X_1 + X_2 + X_3 + X_4>3 \).
Thus, we need to compute the probability of getting 4 successes, which is: \[ \gamma = P(X_1 + X_2 + X_3 + X_4>3 \mid p = \frac{4}{5}). \] For \( X_1 + X_2 + X_3 + X_4 = 4 \), we have: \[ P(X_1 + X_2 + X_3 + X_4 = 4) = \binom{4}{4} \left( \frac{4}{5} \right)^4 = 1 \times \left( \frac{4}{5} \right)^4 = \frac{256}{625}. \] Thus, \( \gamma = P(X_1 + X_2 + X_3 + X_4 = 4) = \frac{256}{625} \).
Step 4: Calculate \( \alpha + \gamma \).
Now, we can compute: \[ \alpha + \gamma = \frac{81}{256} + \frac{256}{625}. \] Finding a common denominator: \[ \alpha + \gamma = \frac{81}{256} + \frac{256}{625} = \frac{50625 + 65536}{160000} = \frac{116161}{160000}. \] Thus, \( \alpha + \gamma \approx 0.726 \).
Final Answer: \[ \boxed{0.73}. \]
We are given a random sample \( X_1, X_2, X_3, X_4 \) from a discrete distribution with two possible values \( X = 0 \) and \( X = 1 \), with probabilities \( P(X = 0) = 1 - p \) and \( P(X = 1) = p \).
The test rejects the null hypothesis \( H_0 \) if the sum of the random sample exceeds 3, i.e., if \( X_1 + X_2 + X_3 + X_4>3 \). We are asked to find the size \( \alpha \) and the power \( \gamma \) of the test, and then compute \( \alpha + \gamma \).
Step 2: Calculate the size of the test (\( \alpha \)).
The size \( \alpha \) is the probability of rejecting \( H_0 \) when \( H_0 \) is true. Under \( H_0 : p = \frac{3}{4} \), the number of successes \( X_1 + X_2 + X_3 + X_4 \) follows a binomial distribution with parameters \( n = 4 \) and \( p = \frac{3}{4} \). We reject \( H_0 \) if \( X_1 + X_2 + X_3 + X_4>3 \).
Thus, we need to compute the probability of getting 4 successes, which is: \[ \alpha = P(X_1 + X_2 + X_3 + X_4>3 \mid p = \frac{3}{4}). \] The probability mass function of the binomial distribution is: \[ P(X = k) = \binom{4}{k} \left( \frac{3}{4} \right)^k \left( \frac{1}{4} \right)^{4-k}. \] For \( X_1 + X_2 + X_3 + X_4 = 4 \), we have: \[ P(X_1 + X_2 + X_3 + X_4 = 4) = \binom{4}{4} \left( \frac{3}{4} \right)^4 = 1 \times \left( \frac{3}{4} \right)^4 = \frac{81}{256}. \] Thus, \( \alpha = P(X_1 + X_2 + X_3 + X_4 = 4) = \frac{81}{256} \).
Step 3: Calculate the power of the test (\( \gamma \)).
The power \( \gamma \) is the probability of rejecting \( H_0 \) when \( H_1 \) is true. Under \( H_1 : p = \frac{4}{5} \), the number of successes \( X_1 + X_2 + X_3 + X_4 \) follows a binomial distribution with parameters \( n = 4 \) and \( p = \frac{4}{5} \). We reject \( H_0 \) if \( X_1 + X_2 + X_3 + X_4>3 \).
Thus, we need to compute the probability of getting 4 successes, which is: \[ \gamma = P(X_1 + X_2 + X_3 + X_4>3 \mid p = \frac{4}{5}). \] For \( X_1 + X_2 + X_3 + X_4 = 4 \), we have: \[ P(X_1 + X_2 + X_3 + X_4 = 4) = \binom{4}{4} \left( \frac{4}{5} \right)^4 = 1 \times \left( \frac{4}{5} \right)^4 = \frac{256}{625}. \] Thus, \( \gamma = P(X_1 + X_2 + X_3 + X_4 = 4) = \frac{256}{625} \).
Step 4: Calculate \( \alpha + \gamma \).
Now, we can compute: \[ \alpha + \gamma = \frac{81}{256} + \frac{256}{625}. \] Finding a common denominator: \[ \alpha + \gamma = \frac{81}{256} + \frac{256}{625} = \frac{50625 + 65536}{160000} = \frac{116161}{160000}. \] Thus, \( \alpha + \gamma \approx 0.726 \).
Final Answer: \[ \boxed{0.73}. \]
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