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 distribution of the test statistic.
Given
\[ X_i\sim \chi_m^2 \] Since \(X_1,X_2,X_3,X_4\) are independent, their sum follows a chi-square distribution with degrees of freedom added.
So,
\[ \sum_{i=1}^{4}X_i\sim \chi_{4m}^2 \]

Step 2: Find Type-I error probability \(\alpha\).
Under \(H_0:m=1\),
\[ \sum_{i=1}^{4}X_i\sim \chi_4^2 \] The critical region is
\[ \sum_{i=1}^{4}X_i>6 \] Therefore,
\[ \alpha=P\left(\chi_4^2>6\right) \] Using the given value,
\[ \chi^2_{4,0.1991}=6 \] we get
\[ \alpha=0.1991 \]

Step 3: Find Type-II error probability \(\beta\).
Under \(H_1:m=2\),
\[ \sum_{i=1}^{4}X_i\sim \chi_8^2 \] Type-II error means failing to reject \(H_0\) when \(H_1\) is true.
So,
\[ \beta=P\left(\sum_{i=1}^{4}X_i\leq 6\mid m=2\right) \] \[ \beta=P\left(\chi_8^2\leq 6\right) \] Using the given value,
\[ \chi^2_{8,0.6472}=6 \] This gives
\[ P(\chi_8^2>6)=0.6472 \] Hence,
\[ \beta=1-0.6472 \] \[ \beta=0.3528 \]

Step 4: Check option (A).
Now,
\[ \frac{3}{4}\alpha+\frac{1}{4}\beta = \frac{3}{4}(0.1991)+\frac{1}{4}(0.3528) \] \[ =0.149325+0.0882 \] \[ =0.237525 \] Thus,
\[ 0.20<0.237525<0.25 \] Therefore, option (A) is true.

Step 5: Check remaining options.
Option (B) says
\[ \alpha>0.20 \] But
\[ \alpha=0.1991<0.20 \] So, option (B) is false.
Option (C) says
\[ \beta<0.30 \] But
\[ \beta=0.3528>0.30 \] So, option (C) is false.
The power of the test is
\[ 1-\beta=1-0.3528 \] \[ =0.6472 \] Since
\[ 0.6472>\frac{1}{2} \] option (D) is false.

Step 6: Final conclusion.
Hence, the correct statement is
\[ \boxed{(A)} \]