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Solution_HW3

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

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Solution to Homework Assignment 3 1 Question 1 a. Since g 0 ( p ) = { p (1 - p ) } - 1 / 3 , using delta method, we have that V ar { g ( p ) } = { g 0 ( π ) } 2 V ar ( p ) = { pi (1 - π ) } - 2 / 3 { π (1 - π ) } n = { p (1 - p ) } 1 / 3 n . b. From part a., we have that g ( p 1 ) a N g ( π 1 ) , { π 1 (1 - π 1 ) } 1 / 3 n 1 , g ( p 2 ) a N g ( π 2 ) , { π 2 (1 - π 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 0 : π 1 = π 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 π 1 and π 2 , so a test statistic for testing the null hypothesis H 0 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 0 at 2-sided significance level α if | T | ≥ z α/ 2 . Note : Another possible test statistic for testing H 0 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 0 : π 1 = π 2 , T 1 also has the asymptotic standard normal distribution and will give more evidence against (rejecting) H 0 . 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 : π 1 = π 1 A , π 2 = π 2 A is T a N ˆ g ( π 1 A ) - g ( π 2 A ) p 4 { ¯ π (1 - ¯ π ) } 1 / 3 /n , { π 1 A (1 - π 1 A ) } 1 / 3 + { π 2 A (1 - π 2 A ) } 1 / 3 2 { ¯ π (1 - ¯ π ) } 1 / 3 ! , 1

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where ¯ π = ( π 1 A + π 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 { π 1 (1 - π 1 ) } 1 / 3 n + 2 { π 2 (1 - π 2 ) } 1 / 3 n , 1 . d. The sample size calculated based on the test statistic T has to satisfy that φ = z α/ 2 + z β σ * , where φ = g ( π 1 A ) - g ( π 2 A ) p 4 { ¯ π (1 - ¯ π ) } 1 / 3 /n σ 2 * = { π 1 A (1 - π 1 A ) } 1 / 3 + { π 2 A (1 - π 2 A ) } 1 / 3 2 { ¯ π (1 - ¯ π ) } 1 / 3 . That is, n { g ( π 1 A ) - g ( π 2 A ) } 2 4 { ¯ π (1 - ¯ π ) } 1 / 3 = ( z α/ 2 + z β σ * ) 2 .
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