11_01_2ans - STAT 400 Examples for(2 Fall 2011 The sample...

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Unformatted text preview: STAT 400 Examples for 11/01/2011 (2) Fall 2011 The sample proportion : n x p = ˆ where x is the number of elements in the sample found to belong to the category of interest (the number of "successes"), and n is the sample size. E( P ˆ ) = p , Var( P ˆ ) = ( ) n p p- ⋅ 1 , SD( P ˆ ) = ( ) 1 n p p- ⋅ . A large-sample confidence interval for the population proportion p is n p p z p - ± ⋅ ⋅ ˆ ˆ ˆ 1 α 2 . 1. Just prior to an important election, in a random sample of 749 voters, 397 preferred Candidate Y over Candidate Z. Construct a 90% confidence interval for the overall proportion of voters who prefer Candidate Y over Candidate Z. X = 397. n = 749. 749 397 X p ˆ = = n = 0.53. The confidence interval : n ) p ˆ 1 ( p ˆ z p ˆ 2- ⋅ ⋅ ± α . 90% confidence level α = 0.10 α 2 = 0.05. 2 z α = 1.645. 749 ) 47 )( 53 ( 645 1 53 . . . . ⋅ ± 0.53 ± 0.03 ( 0.50 , 0.56 ) Upper-tail probability 0.25 0.20 0.15 0.10 0.05 0.025 0.02 0.01 0.005 0.0025 0.001 0.0005 z 0.674 0.841 1.036 1.282 1.645 1.960 2.054 2.326 2.576 2.807 3.091 3.291 50% 60% 70% 80% 90% 95% 96% 98% 99% 99.5% 99.8% 99.9% Confidence level 1 ½ . Let X have a Binomial distribution with parameters n and p . Recall that ( ) 1 X p p n p n-- has an approximate Standard Normal N ( 0, 1 ) distribution, provided that n is large enough, and ( ) α α α- <-- <- ≈ 1 1 X P 2 2 z p p n p n z . Show that an approximate 100 ( 1 – α ) % confidence interval for p is ( ) n z n z n p p z n z p 2 2 2 2 2 2 2 2 1 4 1 2 ˆ ˆ ˆ α α α α + +- ± + , where n p X ˆ = ....
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11_01_2ans - STAT 400 Examples for(2 Fall 2011 The sample...

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