Hw3anskey

Hw3anskey - Peter Zeitz HW3 Answer Key Part I: Multiple...

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Unformatted text preview: Peter Zeitz HW3 Answer Key Part I: Multiple Choice 1) d 2) c n = 500 ^ p = 0 : 75 z 1 & : 05 2 = 1 : 96 & & x = p ^ p (1 & ^ p ) p n h ^ p & z 1 & : 05 2 & & x ; ^ p + z 1 & : 05 2 & & x i = [0 : 713 ; : 787] 3) c 4) b n = z 2 & 2 x E 2 = (2 : 58) 2 (18 : 65) 2 4 : 8 2 = 100 : 49 round up - > 101 5) c since E & & X = E & 1 n P X i = 1 n P E [ X i ] = 1 n n = 6) d 7) b 8) c 7.94 = 24 , & = 7 , n = 100 = ) & x = = 24 : & & x = & p n = 7 p 100 = 0 : 7 The sampling distribution of & x is approximately normal because n > 30 : 7.96a . For & x = 22 : z = (& x & ) & & x = 22 & 24 : 7 = & 2 : 86 : P (& x < 22) = P ( z < & 2 : 86) = 0 : 0021 7.96b. For & x = 23 : z = (& x & ) & & x = 23 & 24 : 7 = & 1 : 43 : For & x = 26 : z = (& x & ) & & x = 26 & 24 : 7 = 2 : 86 P ( z < 2 : 86) & P ( z < & 1 : 43) = 0 : 9215 7.96c. P (& x within 1 minute of ) = P (23 & x 25) For & x = 23 : z = (& x & ) & & x = 23 & 24 : 7 = & 1 : 43 : For & x = 25 : z = (& x & ) & & x = 25 & 24 : 7 = 1 : 43 P ( z < 1 : 43) & P ( z < & 1 : 43) = 0 : 8472 7.96d. P (& x greater than by 2 minutes or more ) = P (& x 26) For & x = 26 : z = (& x & ) & & x = 26 & 24 : 7 = 2 : 86 : P ( z 2 : 86) = 1 & P ( z < 2 : 86) = : 0021 7.99 p = 0 : 88 , q = 0 : 12 ; n = 80 : & p = p = 0 : 88 , & & p = p pq n = q (0 : 88)(0 : 12) 80 = : 036 np = (80) (0 : 88) = 70 : 4 > 5 , nq = (80) (0 : 12) = 9 : 6 > 5 Since np and nq are both greater than 5, the sampling distribution of ^ p is approximately normal....
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This note was uploaded on 04/01/2008 for the course ECON 41 taught by Professor Guggenberger during the Winter '07 term at UCLA.

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Hw3anskey - Peter Zeitz HW3 Answer Key Part I: Multiple...

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