Dr. Hackney STA Solutions pg 122

Dr. Hackney STA Solutions pg 122 - 8.3 The LRT statistic is...

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Chapter 8 Hypothesis Testing 8.1 Let X = # of heads out of 1000. If the coin is fair, then X binomial(1000 , 1 / 2). So P ( X 560) = 1000 X x =560 ± 1000 x ²± 1 2 ² x ± 1 2 ² n - x . 0000825 , where a computer was used to do the calculation. For this binomial, E X = 1000 p = 500 and Var X = 1000 p (1 - p ) = 250. A normal approximation is also very good for this calculation. P { X 560 } = P ³ X - 500 250 559 . 5 - 500 250 ´ P { Z 3 . 763 } ≈ . 0000839 . Thus, if the coin is fair, the probability of observing 560 or more heads out of 1000 is very small. We might tend to believe that the coin is not fair, and p > 1 / 2. 8.2 Let X Poisson( λ ), and we observed X = 10. To assess if the accident rate has dropped, we could calculate P ( X 10 | λ = 15) = 10 X i =0 e - 15 15 i i ! = e - 15 µ 1+15+ 15 2 2! + ··· + 15 10 10! . 11846 . This is a fairly large value, not overwhelming evidence that the accident rate has dropped. (A normal approximation with continuity correction gives a value of .12264.)
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Unformatted text preview: 8.3 The LRT statistic is λ ( y ) = sup θ ≤ θ L ( θ | y 1 ,...,y m ) sup Θ L ( θ | y 1 ,...,y m ) . Let y = ∑ m i =1 y i , and note that the MLE in the numerator is min { y/m,θ } (see Exercise 7.12) while the denominator has y/m as the MLE (see Example 7.2.7). Thus λ ( y ) = · 1 if y/m ≤ θ ( θ ) y (1-θ ) m-y ( y/m ) y (1-y/m ) m-y if y/m > θ , and we reject H if ( θ ) y (1-θ ) m-y ( y/m ) y (1-y/m ) m-y < c. To show that this is equivalent to rejecting if y > b , we could show λ ( y ) is decreasing in y so that λ ( y ) < c occurs for y > b > mθ . It is easier to work with log λ ( y ), and we have log λ ( y ) = y log θ + ( m-y ) log (1-θ )-y log ¸ y m ¹-( m-y ) log ± m-y m ² ,...
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This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

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