Final-prac - Stat/Math 415 Final Prac 1 Independent test Hypothesis H P ij = P i P.j H 1 P ij 6 = P i P.j i = 1 2 3 j = 1 2 Test Statistic Q 2 =(68

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Unformatted text preview: Stat/Math 415 Final Prac 1. Independent test Hypothesis H : P ij = P i. P .j H 1 : P ij 6 = P i. P .j i = 1 , 2 , 3 j = 1 , 2 Test Statistic Q 2 = (68- 124 220 375 ) 2 124 220 375 + ··· = 10 . 086 ∼ χ 2 2 Conclusion C = { Q > χ 2 . 05 , 2 = 5 . 99 } 10 . 086 > 5 . 99 , reject H Therefore Major choice is different by gender. 2. Independent two sample t-test a H : μ P = μ C H 1 : μ P 6 = μ C Assumption Two independent distributions have an equal variance. Test Statistic T = 193- 174 q 24 × 68 2 +20 × 44 2 44 ( 1 25 + 1 21 ) ∼ t 44 Conclusion C = {| T | > t . 025 , 44 = 2 . 015368 } 1 . 1 2 . 015368 . Fail to reject H Or, we can use Z-test instead of t by Large sample theory ( n 1 + n 2 > 30 ) Then we don’t need to assume about an equal variance Z = 193- 174 q 68 2 25 + 44 2 21 = 1 . 141289 , C = {| Z | > 1 . 96 } Our conclusion is the same as t-test. b A 95% CI for mean difference is {| (193- 174)- ( μ P- μ C ) q 24 × 68 2 +20 × 44 2 44 ( 1 25 + 1 21 ) | ≤ 2 . 015368...
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This note was uploaded on 05/17/2008 for the course MATH 415 taught by Professor Yuzhang during the Spring '08 term at Pennsylvania State University, University Park.

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Final-prac - Stat/Math 415 Final Prac 1 Independent test Hypothesis H P ij = P i P.j H 1 P ij 6 = P i P.j i = 1 2 3 j = 1 2 Test Statistic Q 2 =(68

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