Step 1: Concept Introduction:
To determine the Best Critical Region (BCR) for a simple null hypothesis test against a simple alternative, the Neyman-Pearson Lemma is applied. This lemma indicates the BCR is defined by the likelihood ratio.
Step 2: Core Formula:
According to the Neyman-Pearson Lemma, the BCR with a size of \(\alpha\) is a region W, defined by:
\[ W = \left\{ \mathbf{x} : \frac{L(\theta_1 | \mathbf{x})}{L(\theta_0 | \mathbf{x})} \ge k \right\} \]
where \(L\) represents the likelihood function. The value of \(k\) is selected such that \(P(\mathbf{X} \in W | \theta_0) = \alpha\).
Step 3: Detailed Solution:
In this case, \(\theta_0 = 1\) and \(\theta_1 = 2\). The probability density function (PDF) is given by \(f(x;\theta) = \frac{1}{\sqrt{2\pi}}e^{-(x-\theta)^2/2}\).The likelihood function for a sample size of 2 is \(L(\theta) = f(x_1;\theta)f(x_2;\theta)\).The likelihood ratio is expressed as:
\[ \frac{L(2)}{L(1)} = \frac{e^{-(x_1-2)^2/2}e^{-(x_2-2)^2/2}}{e^{-(x_1-1)^2/2}e^{-(x_2-1)^2/2}} = \exp\left[-\frac{1}{2}\left(\sum(x_i-2)^2 - \sum(x_i-1)^2\right)\right] \]
Simplifying the exponent:
\[ \sum(x_i-2)^2 - \sum(x_i-1)^2 = \sum(x_i^2-4x_i+4) - \sum(x_i^2-2x_i+1) = \sum(-2x_i+3) = -2\sum x_i + 2(3) = -2(x_1+x_2)+6 \]
The likelihood ratio test \(\frac{L(2)}{L(1)} \ge k\) is then:
\[ \exp\left[-\frac{1}{2}(-2(x_1+x_2)+6)\right] \ge k \]
\[ \exp(x_1+x_2-3) \ge k \]
Applying the natural logarithm to both sides:
\[ x_1+x_2-3 \ge \ln(k) \implies x_1+x_2 \ge 3+\ln(k) \]
Therefore, the BCR takes the form \(x_1+x_2 \ge k'\), where \(k'\) is a constant.To determine \(k'\), we use the test size, \(\alpha\). Given that \(P(Z>1.64) = \alpha\), then \(\alpha \approx 0.05\).\[ \alpha = P(\text{Reject } H_0 | H_0 \text{ is true}) = P(X_1+X_2 \ge k' | \theta=1) \]
Under \(H_0: \theta=1\), \(X_1, X_2\) are independent and identically distributed (i.i.d.) with a \(N(1, 1)\) distribution.Let \(S = X_1+X_2\). S follows a Normal distribution with:- Mean: \(E(S) = E(X_1) + E(X_2) = 1+1=2\)- Variance: \(\text{Var}(S) = \text{Var}(X_1) + \text{Var}(X_2) = 1+1=2\)Thus, under \(H_0\), \(S \sim N(2, 2)\).Standardizing the probability:
\[ P\left( \frac{S - E(S)}{\sqrt{\text{Var}(S)}} \ge \frac{k' - 2}{\sqrt{2}} \right) = \alpha \]
\[ P\left( Z \ge \frac{k' - 2}{\sqrt{2}} \right) = \alpha \]
Given \(P(Z \ge 1.64) = \alpha\), we have:
\[ \frac{k' - 2}{\sqrt{2}} = 1.64 \]
\[ k' - 2 = 1.64\sqrt{2} \approx 1.64 \times 1.4142 = 2.3193 \]
\[ k' = 2 + 2.3193 = 4.3193 \]
Rounding to two decimal places, \(k' = 4.32\).
Step 4: Solution:
The best critical region is \( W = \{(x_1, x_2): x_1 + x_2 \ge 4.32\} \).