Question

Let X₁, X₂ ,…, X₂₅ be a random sample from a 𝑁(μ, 1) distribution, where μ ∈ \(\R\) is unknown. Consider testing of the hypothesis H₀ : μ = 5.2 against H₁ : μ = 5.6. The null hypothesis is rejected if and only if \(\frac{1}{25}\sum^{25}_{i=1}X_i > k\) , for some constant k. If the size of the test is 0.05, then the probability of type-II error equals __________ (round off to 2 decimal places)

Answer 1

Correct Answer: 0.34

To solve this problem, we need to determine the probability of a Type-II error given the hypotheses: \(H_0: \mu = 5.2\) and \(H_1: \mu = 5.6\). The sample size is 25 from \(N(\mu, 1)\), and the test statistic is \(\bar{X} = \frac{1}{25}\sum_{i=1}^{25}X_i\). The hypothesis \(H_0\) is rejected if \(\bar{X} > k\), with the given test size (Type-I error rate) being 0.05.

1. **Determine k using Type-I error probability**: Under \(H_0\), \(\bar{X} \sim N(5.2, \frac{1}{25})\). We need \(P(\bar{X} > k \mid H_0)=0.05\). This requires finding the critical value \(k\) such that:

1.1 Standardize: \(Z = \frac{\bar{X} - 5.2}{\frac{1}{\sqrt{25}}}\), thus \(Z \sim N(0, 1)\).

1.2 Solve \(P(Z > \frac{k-5.2}{0.2}) = 0.05\). Using the standard normal table, \(P(Z > 1.645) = 0.05\).

1.3 So, \(\frac{k-5.2}{0.2} = 1.645\) leads to \(k = 5.2 + 1.645 \times 0.2 = 5.529\).

2. **Probability of Type-II error under \(H_1\)**: Assume \(H_1: \mu = 5.6\), then \(\bar{X} \sim N(5.6, \frac{1}{25})\).

2.1 Standardize for this new distribution: \(Z = \frac{\bar{X} - 5.6}{0.2}\).

2.2 Probability of Type-II error: \(P(\bar{X} \leq k \mid H_1)\), where \(k = 5.529\).

2.3 Calculate: \(P\left(\frac{\bar{X} - 5.6}{0.2} \leq \frac{5.529 - 5.6}{0.2}\right) = P(Z \leq -0.355)\).

2.4 Using standard normal tables, \(P(Z \leq -0.355) \approx 0.361\).

3. **Verify the result falls in the specified range**: 0.361 is approximately 0.36.

Thus, the probability of Type-II error is approximately 0.36.