Question

If, \(x \ge 1\) is the critical region for testing \(H_0: \theta = 2\) against the alternate \(H_1: \theta = 1\). On the basis of a single observation from the population \(f(x;\theta) = \theta e^{-x\theta}; x>0, \theta>0\), then the size of Type II error is:

• A. \( \frac{1}{e} \)
• B. \( \frac{1}{e^2} \)
• C. \( \frac{e-1}{e} \)
• D. \( 1 - \frac{1}{e^2} \)

Hint:

Remember the relationship between the critical region and the acceptance region. The acceptance region is always the complement of the critical region. \(\beta\) is the probability of the outcome falling in the acceptance region, calculated under \(H_1\).

Answer 1

The Correct Option is C

Step 1: Definition of Type II Error:
A Type II error occurs when the null hypothesis (\(H_0\)) is not rejected, even though the alternative hypothesis (\(H_1\)) is true. The probability of committing a Type II error is represented by \(\beta\).

Step 2: Formula and Acceptance Region:
The formula for \(\beta\) is: \(\beta = P(\text{Fail to Reject } H_0 | H_1 \text{ is true})\). Given a critical region of \(x \ge 1\), the acceptance region is \(x<1\). We need to compute \(P(X<1)\) assuming \(H_1\) is true, where \(\theta = 1\).

Step 3: Calculation of \(\beta\):
The population follows an exponential distribution with rate parameter \(\theta\). Under \(H_1: \theta = 1\), the PDF is: \[ f(x; 1) = 1 . e^{-x . 1} = e^{-x}, \quad x>0 \] \(\beta\) is the probability of observing a value in the acceptance region \(x<1\), given \(\theta=1\): \[ \beta = P(X<1 | \theta=1) \] This is calculated by integrating the PDF under \(H_1\) from 0 to 1: \[ \beta = \int_{0}^{1} e^{-x} \,dx \] \[ = [-e^{-x}]_{0}^{1} \] \[ = (-e^{-1}) - (-e^{-0}) = -e^{-1} - (-1) = 1 - e^{-1} \] Which simplifies to: \[ \beta = 1 - \frac{1}{e} = \frac{e-1}{e} \]
Step 4: Conclusion:
Therefore, the probability of a Type II error is \( \frac{e-1}{e} \).