ams572_notes_12

ams572_notes_12 - AMS 572 Lecture Notes #12 Nov. 4th, 2010...

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AMS 572 Lecture Notes #12 Nov. 4 th , 2010 Ch. 9. Categorical Data Analysis Exact test on one population proportion: Data: sample of n, X are # of “Successes”, n-X are # of “Failures” ~ ( , ) X B n p ( ) (1 ) , 0,1,2, , x n x n P X x p p x n x -   = = - =     K (1)  0 0 0 : : a H p p H p p = 0 0 0 0 ( | : ) (1 ) n i n i i x n p value P X x H p p p p i - =   - = = = -     (2)  0 0 0 : : a H p p H p p = < 0 0 0 0 0 ( | : ) (1 ) x i n i i n p value P X x H p p p p i - =   - = = = -     (3)  0 0 0 : : a H p p H p p = 0 0 2*min{ ( | ), ( | )} p value P X x H P X x H - = . 1
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Inference on Two Population Proportions: 1 p , 2 p Independent samples, both are large e.g. Suppose we wish to compare the proportions of smokers among male and female students in SBU. Two large independent samples: Population 1: 1 p Population 2: 2 p Sample 1: 1 1 1 1 , , n x n x - Sample 2: 2 2 2 2 , , n x n x - ( 1 1 1 5, 5 x n x - , 2 2 2 5, 5 x n x - ) 1 Point estimator: · µ 1 2 1 2 1 2 1 2 x x p p p p n n - = - = - 2 By CLT, µ 1 1 1 1 1 (1 ) ( , ) p p p N p n - : 2 2 2 2 2 (1 ) ( , ) p p p N p n - : Q Two samples are independent µ 1 1 2 2 1 2 1 2 1 2 (1 ) (1 ) ( , ) p p p p p p N p p n n - - - - + : 3 P.Q. for 1 2 p p - : µ 1 2 1 2 1 1 2 2 1 2 ( ) (0,1) (1 ) (1 ) p p p p Z N p p p p n n - - - = - - + : not P.Q. µ µ µ * 1 2 1 2 1 1 2 2 1 2 ( ) (0,1) (1 ) (1 ) p p p p Z N p p p p n n - - - = - - + : P.Q. 4 100(1- α )% large samples CI for 1 2 p p - : 2
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µ * 1 2 1 2 1 2 2 2 2 2 1 ( ) ( ) P Z Z Z P p p Z S p p p p Z S α - = - = - - - - + Here, µ µ 1 1 2 2 1 2 (1 ) (1 ) p p p p S n n - - = + 5 Test General case: 0 1 2 1 2 : : a H p p H p p - = ∆ - µ µ µ 0 1 2 0 0 1 1 2 2 1 2 : (0,1) (1 ) (1 ) H p p T Z N p p p p n n - - ∆ = - - + : p-value=P(Z 0 Z ) At the significance level , we reject 0 H if 0 Z Z and p-value< α . When =0, one often uses the following test statistic µ µ µ 0 * 1 2 1 2 0 (0,1) 1 1 (1 )( ) H p p Z N p p n n - - = - + : here µ µ 1 1 2 2 1 2 1 2 1 2 n p n p X X p n n n n + + = = + + Example. A random sample of Democrats and a random sample of Republicans  were polled on an issue. Of 200 Republicans, 90 would vote yes on the issue; of  100 democrats, 58 would vote yes. Let p 1  and p 2  denote respectively the  proportions of all Democrats or all Republicans who would vote yes on this issue.  (a) Construct a 95% confidence interval for (p 1  - p 2 ) (b) Can we say that more Democrats than Republicans favor the issue at the  1% level of significance? Please report the p-value.  (c) Please write up the entire SAS program necessary to answer question  raised in (b). Please include the data step. 3
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ams572_notes_12 - AMS 572 Lecture Notes #12 Nov. 4th, 2010...

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