Discussion
Let \(X_{1}, X_{2}, X_{3}, X_{4}\) be a random sample of size \(4\) from a \(\chi_{m}^{2}\) distribution, where \(m \in \mathbb{N}\) is an unknown parameter. To test \(H_{0}: m = 1\) against \(H_{1}: m = 2\), the critical region \(\sum_{i=1}^{4} X_{i} \gt 6\) is being used. If \(\alpha\) and \(\beta\) denote the probabilities of Type-I error and Type-II error, respectively, then which one of the following statements is true?
[You may use \(\chi_{1,0.0143}^{2} = \chi_{2,0.0498}^{2}\) \(= \chi_{4,0.1991}^{2} = \chi_{8,0.6472}^{2} = 6\)]
Answer: A
The true statement is \(0.20 \lt \frac{3}{4}\alpha + \frac{1}{4}\beta \lt 0.25\).
The sum of independent chi-square variables is chi-square with the degrees of freedom added. Under \(H_{0}\), \(\sum X_{i} \sim \chi_{4}^{2}\); under \(H_{1}\), \(\sum X_{i} \sim \chi_{8}^{2}\). From the values given:
\[ \begin{aligned} \alpha &= P(\chi_{4}^{2} \gt 6) = 0.1991 \\ 1 - \beta &= P(\chi_{8}^{2} \gt 6) = 0.6472 \\ \beta &= 0.3528 \end{aligned} \]
\[ \begin{aligned} &\tfrac{3}{4}(0.1991) + \tfrac{1}{4}(0.3528) \\ &\approx 0.1493 + 0.0882 = 0.2375 \end{aligned} \]
which lies between 0.20 and 0.25. The other statements fail: \(\alpha = 0.1991\) is just below 0.20, \(\beta = 0.3528\) is above 0.30, and the power, 0.6472, exceeds \(\frac{1}{2}\). The values for 1 and 2 degrees of freedom are distractors, since the test uses the sum of four observations.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development