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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) p n + c 2 2 n c r c 2 4 n 2 + p n (1 p n ) n 1 + c 2 n equation (2.20) p n c r p n (1 p n ) n and equation (2.22) g 1 g ( p n ) c n where g ( p ) = 2 sin 1 ( p ) and g 1 ( ) = sin 2 ( / 2) Any one of these three intervals answers this part of the question. (b) Using 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 = 1 that X c . That is Z c e x dx = e c and set this equal to and solve for c . The solution is c = log( ) = log(0 . 05) = 2 . 995732. (b) The Pvalue is P ( X x ), where = 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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 Spring '02
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