exam2-sol

exam2-sol - 1.a: [12pt] Perform paired t-test. H0 : d = 0...

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1.a: [12pt] Perform paired t -test. H 0 : μ d = 0 vs. H a : μ d > 0. Test statistic is t = 3 . 3 - 0 4 . 7 / 10 = 2 . 22 with d.f. 9. The p -value is between 0.025 and 0.05. 1.b: [5pt] Difference (from ten couples) is a random sample from a normal distribution. 1.c: [8pt] 3 . 3 ± 1 . 833 × 4 . 7 10 = 3 . 3 ± 2 . 7 = [0 . 6 , 6 . 0], where 1 . 833 is t 9 , 0 . 05 . 2.a:[10pt] ˆ p = 1441 / 2000 = 0 . 7205 and z 0 . 01 = 2 . 33. 0 . 721 ± 2 . 33 × r 0 . 72 × 0 . 28 2000 = 0 . 721 ± 0 . 023 = [0 . 698 , 0 . 744] 2.b:[10pt] You can do it in two ways. The conservative way: 2 . 33 × r 0 . 5 × (1 - 0 . 5) n 0 . 01 = n 13573 The non-conservative way: 2 . 33 × r ˆ p × (1 - ˆ p ) n 0 . 01 = n 10933 3.a:[10pt] Perform the CI for two-sample pooled t -procedure. The estimated pooled variance is: s 2 p = (36 - 1) × 60 2 + (30 - 1) × 62 2 35 + 29 = 3710 . 56 Then 95% CI for ( μ 1 - μ 2 ) is (110 - 150) ± 2 . 009 q 3710 . 56 × ( 1 36 + 1 30 ) = - 40 ± 30 . 25 = [ - 70 . 25 , - 9 . 75], where 2 . 009 = t 50 , 0 . 025 t 64 , 0 . 025 . 3.b:[15pt] H 0 : μ 1 - μ 2 = 0 vs. H a : μ 1 - μ 2 < 0. The test-statistics is: t = 110 - 150 q 3710 . 56 × ( 1 36 + 1 30 ) = - 2 . 656 with d.f. = 35+29=64. The p -value is between 0.005 and 0.01. So we reject the null (at
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exam2-sol - 1.a: [12pt] Perform paired t-test. H0 : d = 0...

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