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final-prac

# final-prac - Stat/Math 415 Final Prac 1 Independent test...

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Stat/Math 415 Final Prac 1. Independent test Hypothesis H 0 : 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 0 . 05 , 2 = 5 . 99 } 10 . 086 > 5 . 99 , reject H 0 Therefore Major choice is different by gender. 2. Independent two sample t-test a H 0 : 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 0 . 025 , 44 = 2 . 015368 } 1 . 1 2 . 015368 . Fail to reject H 0 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 } = [ - 15 . 79625 , 53 . 79625] If we use Z table, then replace 2.015368 with 1.96. Since the interval includes 0, it is comparable to (a). 3. Estimation X iid N ( μ, σ 2 ) a Prove f ( x 1 , · · · , x n ) = Q f ( x i ) = (2 πθ ) - 2 n exp - ( x i - μ ) 2 2 θ l ( θ ) = log L ( θ | X ) = - n 2 log(2 πσ 2 ) - (

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