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# Quiz5 - Quiz 5 We have two independent samples X 1,K X n ~...

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Quiz 5 We have two independent samples 1 2 1 1 1 , , ~ ( , ) iid n X X N μ σ K and 2 2 1 2 2 , , ~ ( , ) iid n Y Y N μ σ K , where 2 2 2 1 2 σ σ σ = = and 1 1 n n n = = . For the hypothesis of 0 1 2 1 2 : 0 : 0 a H H μ μ μ μ - = - = ∆ < a) Please derive the general formula for power calculation for the pooled variance t-test based on an effect size of EFF = σ at the significance level of α and with a sample size of n per group. Recall - Definition: Effect size = EFF = σ b) With a sample size of 15 per group, α = 0.05, and an estimated effect size of -1.5, please calculate the power of your pooled variance t-test. Solution: a) T.S : 0 T = 0 2 2 1 2 ( ) 0 ( ) ~ 1 1 2 H n p p X Y X Y t S S n n n - - - - = + At α=0.05, reject 0 H in favor of a H iff 0 2 2, n T t α - ≤ - Power = 1-β = P(reject 0 H | a H ) = 0 2 2, 1 2 ( | : 0) n a P T t H α μ μ - ≤ - - = ∆ < = 2 2, 1 2 ( ) ( | : ) 2 n a p X Y P t H S n α μ μ - - ≤ - - = ∆ = 2 2, 1 2 ( ) ( | : ) 2 2 n a p p X Y P t H S S n n α μ μ - - - ∆ ≤ - - - = ∆ 2 2, 1 2 ( * | : ) 2 n a n P T t Eff H α μ μ - ≤ - - - = ∆ (Effect size= σ p S ) Note: the T statistic above follows a t-distribution with (2n-2) degrees of freedom. b) With n = 15, α = 0.05, Eff = -1.5, the power is calculated as follows: Power (Eff = 1.5) = 28,0.05 1 2 15 ( 1.5* | : ) 2 a P T t

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