hw05-sol - Homework 5 Solutions 9-10 a) = P( X 98.5) + P( X...

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Unformatted text preview: Homework 5 Solutions 9-10 a) = P( X 98.5) + P( X > 101.5) = 9 / 2 100 5 . 98 9 / 2 100 X P + 9 / 2 100 5 . 101 9 / 2 100 X P = P(Z 2.25) + P(Z > 2.25) = (P(Z - 2.25)) + (1 P(Z 2.25)) = 0.01222 + 1 0.98778 = 0.01222 + 0.01222 = 0.02444 b) = P(98.5 X 101.5 when = 103) = 9 / 2 103 5 . 101 9 / 2 103 9 / 2 103 5 . 98 X P = P( 6.75 Z 2.25) = P(Z 2.25) P(Z 6.75) = 0.01222 0 = 0.01222 c) = P(98.5 X 101.5 when = 105) = 9 / 2 105 5 . 101 9 / 2 105 9 / 2 105 5 . 98 X P = P( 9.75 Z 5.25) = P(Z 5.25) P(Z 9.75) = 0 0 = 0. The probability of accepting the null hypothesis when it is actually false is smaller in part c since the true mean, = 105, is further from the acceptance region. A larger difference exists. A larger difference exists....
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hw05-sol - Homework 5 Solutions 9-10 a) = P( X 98.5) + P( X...

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