sol4 - HOMEWORK #4 SOLUTIONS IE121 8-7 a) = P( X > 185...

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HOMEWORK #4 SOLUTIONS IE121 8-7 a) α = P( X > 185 when μ = 175) = - - 10 / 20 175 185 10 / 20 175 X P = P(Z > 1.58) = 1 - P(Z 1.58) = 1 - 0.94295 = 0.05705 b) β = P( X 185 when μ = 195) = - - 10 / 20 195 185 10 / 20 195 X P = P(Z - 1.58) = 1 - P(Z 1.58) = 0.05705. 8.8 a) Since x >185, we reject the null hypothesis and conclude that the mean foam height is greater than 175 mm. b) P( X > 190 when μ = 175) = P X - - 175 20 10 190 175 20 10 / / = P(Z > 2.37) = 1 - P(Z 2.37) = 1 - 0.99111 = 0.00889. The probability that a value of at least 190 mm would be observed (if the true mean height is 175 mm) is only 0.00889. Thus, the sample value of x = 190 mm would be an unusual result. 8-10 a) ( 29 0571 . 0 175 | = = μ c X P - 16 / 20 175 c Z P = P(Z 1.58) Thus, 1.58 = 16 / 20 175 - c , and c = 182.9 b) If the true mean foam height is 195 mm, then β = P( X 182.9 when μ
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sol4 - HOMEWORK #4 SOLUTIONS IE121 8-7 a) = P( X > 185...

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