hw11sol_fall03 - AMS 310.01 FALL 2003 Homework#11 Solutions...

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Unformatted text preview: AMS 310.01 FALL 2003 Homework #11 Solutions 7.56 n=8 In this problem, the alternative is two sided and = 0.05 so we use Table 8(c) with (a) d (b) d (c) d (d) d (e) d = 0.50 , = 0.75 , = 1.00 , = 1.25 , = 1.50 , P (Type II error) = 0.68 P (Type II error) = 0.40 P (Type II error) = 0.18 P (Type II error) = 0.05 P (Type II error) = 0.01 << Using the formula >> P (Type II error) = P (acceptance H 0 | H1 is true) * Acceptance region of Ho : - z / 2 < x - 0 < z / 2 / n 0 - z / 2 < x < 0 + z / 2 n n P (acceptance H 0 | H1 is true) = P( 0 - z / 2 < x < 0 + z / 2 | = 1 ) n n - - 0 - z / 2 1 x - 0 + z / 2 1 n n 1 = P( < < ) / n / n / n - 1 x - 1 0 - 1 = P( 0 - z / 2 < < + z / 2 ) / n / n / n - 1 - 1 = ( 0 + z / 2 ) - ( 0 - z / 2 ) / n / n ...
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