Math Problem Statement
Solution
The problem involves hypothesis testing for a normal distribution. Here's the breakdown:
Problem Context:
- are random samples from a normal distribution with mean and variance .
- Sample size .
- Significance level .
- Null hypothesis: .
- Alternative hypothesis: .
- We are to find the most powerful test (Neyman-Pearson Lemma).
Steps to Solve:
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Statistical Distribution:
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Under the null hypothesis , the test statistic is:
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Under , .
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Critical Region:
- The Neyman-Pearson Lemma tells us that the most powerful test is based on the likelihood ratio: for some critical value .
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Finding the Critical Value:
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We reject if falls in the critical region defined such that the probability of Type I error () is:
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Since under :
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From the standard normal distribution, the critical value is:
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Test Procedure:
- Reject if .
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Power of the Test:
- To calculate the power, compute:
- Under , :
- From standard normal tables:
Summary:
- The critical region is .
- The power of the test is approximately 0.9996.
Would you like a step-by-step explanation of any specific part?
Related Questions:
- What is the significance of the Neyman-Pearson Lemma in hypothesis testing?
- How do you interpret the power of a hypothesis test?
- What happens if we change the sample size ? How does it affect the critical value and power?
- Can you explain the relationship between and in hypothesis testing?
- How would the results change if the variance were unknown?
Tip:
When conducting hypothesis tests, always sketch the sampling distributions under and to visualize the critical regions and probability of errors.
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Math Problem Analysis
Mathematical Concepts
Statistical Hypothesis Testing
Normal Distribution
Neyman-Pearson Lemma
Formulas
Z = (X̄ - μ) / (σ / √n)
P(Reject H0 | H0) = α
Critical Value = μ0 + zα × σ
Power = P(Reject H0 | HA)
Theorems
Neyman-Pearson Lemma
Suitable Grade Level
University Level (Statistics)
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