Solution_HW3 - Solution to Homework Assignment 3 1 Question...

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Unformatted text preview: Solution to Homework Assignment 3 1 Question 1 a. Since g ( p ) = { p (1- p ) }- 1 / 3 , using delta method, we have that V ar { g ( p ) } = { g ( ) } 2 V ar ( p ) = { pi (1- ) }- 2 / 3 { (1- ) } n = { p (1- p ) } 1 / 3 n . b. From part a., we have that g ( p 1 ) a N g ( 1 ) , { 1 (1- 1 ) } 1 / 3 n 1 , g ( p 2 ) a N g ( 2 ) , { 2 (1- 2 ) } 1 / 3 n 2 , where p 1 and p 2 are sample proportions from treatment 1 and 2, respectively. Under the null hypothesis H : 1 = 2 , we know that the pooled sample proportion p = ( p 1 n 1 + p 2 n 2 ) / ( n 1 + n 2 ) is a better estimator for both 1 and 2 , so a test statistic for testing the null hypothesis H would be T = g ( p 1 )- g ( p 2 ) p { p (1- p ) } 1 / 3 /n 1 + { p (1- p ) } 1 / 3 /n 2 . With equal allocation of patients to the two treatments, i.e. n 1 = n 2 = n/ 2, we have the test statistic be written as T = g ( p 1 )- g ( p 2 ) p 4 { p (1- p ) } 1 / 3 /n , where p = ( p 1 + p 2 ) / 2. Clearly, the test statistic described above has asymptotic standard normal distribution N (0 , 1). Therefore, we reject H at 2-sided significance level if | T | z / 2 . Note : Another possible test statistic for testing H is T 1 = g ( p 1 )- g ( p 2 ) q { p 1 (1- p 1 ) } 1 / 3 n 1 + { p 2 (1- p 2 ) } 1 / 3 n 2 = g ( p 1 )- g ( p 2 ) q 2 { p 1 (1- p 1 ) } 1 / 3 n + 2 { p 2 (1- p 2 ) } 1 / 3 n . Under H : 1 = 2 , T 1 also has the asymptotic standard normal distribution and will give more evidence against (rejecting) H . However, the test statistic T has more stable denominator as it uses larger sample to estimate the unknown parameters and hence will have a better asymptotic normal distribution than T 1 . c. The asymptotic distribution of T under H A : 1 = 1 A , 2 = 2 A is T a N g ( 1 A )- g ( 2 A ) p 4 { (1- ) } 1 / 3 /n , { 1 A (1- 1 A ) } 1 / 3 + { 2 A (1- 2 A ) } 1 / 3 2 { (1- ) } 1 / 3 ! , 1 where = ( 1 A + 2 A ) / 2. Similarly, we may derive the asymptotic distribution for T 1 as T 1 a N g ( p 1 )- g ( p 2 ) q 2 { 1 (1- 1 ) } 1 / 3 n + 2 { 2 (1- 2 ) } 1 / 3 n , 1 . d. The sample size calculated based on the test statistic T has to satisfy that = z / 2 + z * , where = g ( 1 A )- g ( 2 A ) p 4 { (1- ) } 1 / 3 /n 2 * = { 1 A (1- 1 A ) } 1 / 3 + { 2 A (1- 2 A ) } 1 / 3 2...
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This note was uploaded on 01/15/2012 for the course PHC 6020 taught by Professor Staff during the Fall '11 term at University of Florida.

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Solution_HW3 - Solution to Homework Assignment 3 1 Question...

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