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Unformatted text preview: Stat231 William Marshall Stat231 William Marshall June 20, 2010 Stat231 William Marshall Week 8 Goals: More Confidence intervals CI Summary Intro to hypothesis testing Stat231 William Marshall Relative Likelihood The likelihood function L ( θ ) = f ( y 1 , y 2 ,..., y n ) The Maximum likelihood estimate, ˆ θ is the value of θ which maximizes L ( θ ) The relative likelihood function, a normalized version of the likelihood R ( θ ) = L ( θ ) L ( ˆ θ ) The relative likelihood gives context to the value of the likelihood and allows us to determine plausible values of θ Stat231 William Marshall Example 14 In an incoming inspection, a sample of 50 parts is randomly selected from a large batch and tested to see if all specifications are met. A total of 7 of the selected parts fail to meet the specifications. Find a 95% confidence interval for the proportion of parts in the batch that fail to meet specifications. Model: Y ∼ Bino (50 ,π ) Data: y = 7 What is the relative likelihood function for π ? Stat231 William Marshall Distribution Can replace estimate with estimator in relative likelihood R ( ˜ θ ) = L ( θ ) L ( ˜ θ ) What is the distribution of R ( ˜ θ )? Unproven result: For large values of n, 2 log R ( ˜ θ ) ∼ χ 2 1 Stat231 William Marshall Likelihood based CI Find an approximate 95% CI for π Using the tables P ( χ 2 1 < 3 . 84) = 0 . 95 Substitute 2...
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 Spring '10
 Marsh

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