Question

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 \(\sum_{i=1}^{4}X_i>6\) 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?

• A. \(0.20<\dfrac{3}{4}\alpha+\dfrac{1}{4}\beta<0.25\)
• B. \(\alpha>0.20\)
• C. \(\beta<0.30\)
• D. The power of the test lies in the interval \(\left(0,\dfrac{1}{2}\right)\)

Hint:

For independent chi-square random variables, the sum is again chi-square with degrees of freedom equal to the sum of the individual degrees of freedom.

Answer 1

The Correct Option is A

Step 1: Find the test statistic law.
Each $X_i\sim\chi^2_m$, so the sum of four is $\chi^2_{4m}$.

Step 2: Get the size $\alpha$.
Under $H_0$ ($m=1$) the sum is $\chi^2_4$, and $\alpha=P(\chi^2_4>6)=0.1991$.

Step 3: Get $\beta$.
Under $H_1$ ($m=2$) the sum is $\chi^2_8$, and $\beta=P(\chi^2_8\le6)=1-0.6472=0.3528$.

Step 4: Test the options.
$\frac34\alpha+\frac14\beta=0.1493+0.0882=0.2375$, which sits in $(0.20,0.25)$, so (A) holds. Also $\alpha<0.20$, $\beta>0.30$, and power $=0.6472>\frac12$, ruling out (B), (C), (D).

Step 5: Conclude.
\[ \boxed{(A)} \]