# s2 - Stat 5102(Geyer 2003 Midterm 2 Problem 1(a The “more...

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Unformatted text preview: Stat 5102 (Geyer) 2003 Midterm 2 Problem 1 (a) The “more on confidence intervals handout” gave three confidence intervals for this situation. Equation (2.19) ˆ p n + c 2 2 n ± c r c 2 4 n 2 + ˆ p n (1- ˆ p n ) n µ 1 + c 2 n ¶ equation (2.20) ˆ p n ± c r ˆ p n (1- ˆ p n ) n and equation (2.22) g- 1 µ g (ˆ p n ) ± c √ n ¶ where g ( p ) = 2 sin- 1 ( √ p ) and g- 1 ( θ ) = sin 2 ( θ/ 2) Any one of these three intervals answers this part of the question. (b) Using ˆ p n = x/n = 0 . 1 and c = 1 . 96 we get (2.19) (0 . 03958 , . 23052) (2.20) (0 . 00703 , . 19297) (2.22) (0 . 02756 , . 21054) You only need one interval. Problem 2 (a) We need to find the probability under H , that is, assuming θ = 1 that X ≥ c . That is Z ∞ c e- x dx = e- c and set this equal to α and solve for c . The solution is c =- log( α ) =- log(0 . 05) = 2 . 995732. (b) The P-value is P θ ( X ≥ x ), where θ = 1 is the value hypothesized under H and x = 3 . 45 is the observed value of the test statistic. By the same argument45 is the observed value of the test statistic....
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s2 - Stat 5102(Geyer 2003 Midterm 2 Problem 1(a The “more...

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