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Unformatted text preview: Testing Equality of Two Proportions p 1 = p 2 Testing H : p 1 = p 2 . θ = p 1 p 2 Independent samples Y 1 ,Y 2 . Y 1 ∼ binomial( n 1 ,p 1 ) ,Y 2 ∼ binomial( n 2 ,p 2 ) ˆ Θ = ˆ P 1 ˆ P 2 = Y 1 n 1 Y 2 n 2 Large sample sizes n 1 ,n 2 . Use normal approximation. ˆ P i ∼ N p i , p i (1 p i n i , i = 1 , 2 ˆ Θ ∼ N p 1 p 2 , p 1 (1 p 1 ) n 1 + p 2 (1 p 2 ) n 2 Standard deviation σ ˆ Θ : r p 1 (1 p 1 ) n 1 + p 2 (1 p 2 ) n 2 1 Test statistic Z = ˆ Θ θ σ ˆ Θ = ( ˆ P 1 ˆ P 2 ) ( p 1 p 2 ) r p 1 (1 p 1 ) n 1 + p 2 (1 p 2 ) n 2 has approximately standard normal dis tribution. Problem: What use for p 1 , p 2 in de nominator? Practical solution: Use estimates ˆ p 1 , ˆ p 2 Better estimate: Under H , p 1 and p 2 are equal to same quantity p = p 1 = p 2 . Pool data from both samples. ˆ P = X 1 + X 2 n 1 + n 2 2 Z = ( ˆ P 1 ˆ P 2 ) ( p 1 p 2 ) r p (1 p ) n 1 + p (1 p ) n 2 = ( ˆ P 1 ˆ P 2 ) ( p 1 p 2 ) q ˆ P (1 ˆ P ) r 1 n 1 + 1 n 2 Example 4.64 p 269Example 4....
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 Fall '10
 drera
 Normal Distribution, Variance, Chisquare distribution, p1

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