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hw111soluNew

# hw111soluNew - Assignment 1 Answer Keys 1 H0 B = A vs H1 B...

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Assignment 1 Answer Keys 1. H 0 : μ B = μ A vs H 1 : μ B 6 = μ A a) There are 6 3 = 20 ways to allocate 3 A ’s and 3 B ’s to the six plants. The 20 possible average differences between B and A are - 2 . 67 , - 2 , - 1 . 33 , - 1 . 33 , - 1 . 33 , - . 67 , - . 67 , - . 67 , 0 , 0 , 0 , 0 , . 67 , . 67 , . 67 , 1 . 33 , 1 . 33 , 1 . 33 , 2 , 2 . 67 , with each being equally likely. The distribution can be represented by graphics such as histogram or simply by the probability mass function. Let δ denote the average difference. Then, p ( δ ) = - 2 . 67 1 / 20 - 2 1 / 20 - 1 . 33 3 / 20 - 0 . 67 3 / 20 0 4 / 20 0 . 67 3 / 20 1 . 33 3 / 20 2 1 / 20 2 . 67 1 / 20 b) δ obs =1.33. Because it is a two-sided test, P - value = P ( δ 1 . 33 or δ ≤ - 1 . 33) = 10 20 = . 50 c) s 2 A = 3, s 2 B = 2 . 33, s 2 pool = 2 . 67, and ¯ y B - ¯ y A = 1 . 33. The observed test statistic T obs = ¯ y B - ¯ y A s pool q 1 / 3 + 1 / 3 = 1 P - value = P ( T ≤ - 1 or T 1 | t (4)) = . 374 . d) Using the usual significance level α , we fail to reject H 0 in both b) and c). The t tests

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