Dr. Hackney STA Solutions pg 171

Dr. Hackney STA Solutions pg 171 - 10-14 Solutions Manual...

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10-14 Solutions Manual for Statistical Inference b. The mean is μ = r (1 - p ) /p , and a little algebra will verify that the variance, r (1 - p ) /p 2 can be written r (1 - p ) /p 2 = μ + μ 2 /r . Thus r n r ± (1 - p ) r - p ¯ x 1 - p ² = n μ - ¯ x p μ + μ 2 /r . The confidence interval is found by setting this equal to z α/ 2 , squaring both sides, and solving the quadratic for μ . The endpoints of the interval are r (8¯ x + z 2 α/ 2 ) ± q rz 2 α/ 2 q 16 r ¯ x + 16¯ x 2 + rz 2 α/ 2 8 r - 2 z 2 α/ 2 . For the continuity correction, replace ¯ x with ¯ x +1 / (2 n ) when solving for the upper endpoint, and with ¯ x - 1 / (2 n ) when solving for the lower endpoint. c. We table the endpoints for α = . 1 and a range of values of r . Note that r = is the Poisson, and smaller values of r give a wider tail to the negative binomial distribution. r lower bound upper bound 1 22 . 1796 364 . 42 5 36 . 2315 107 . 99 10 38 . 4565 95 . 28 50 40 . 6807 85 . 71 100 41 . 0015 84 . 53 1000 41 . 3008 83 . 46 41 . 3348 83 . 34 10 . 43 a. Since P ³ X i X i = 0 ! = (1 - p ) n = α/ 2 p = 1 - α 1 /n and P ³ X i X i = n ! = p n = α/ 2 p = α 1 /n , these endpoints are exact, and are the shortest possible. b. Since p [0 , 1], any value outside has zero probability, so truncating the interval shortens
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