2008hw5_solutions

# 2008hw5_solutions - ST 314 HW#5 Solutions 3.76 n = 2000 p =...

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ST 314 HW #5 Solutions 3.76 n = 2000 p = probability of underfilling = .005 q = probability of not underfilling = .995 Y = number bottles underfilled μ = np = (2000)(.005) = 10 σ 2 = npq = 2000(.005)(.995) = 9.95 σ = 3.15 a Pr(Y 10) = Pr(Y* 10.5) = Pr[(y* – μ ) / σ (9.5 – 10) / 3.15] = Pr(Z 1.16) = .5636 b Pr(Y 15) = Pr(Y* 14.5) Pr(Y 15) = Pr[(y* – μ ) / σ (14.5 – 10) / 3.15] = Pr(Z 1.43) = 1-.9236 = .0764 c Pr(Y 20) = Pr(Y* 19.5) Pr(Y 20) = Pr[(y* – μ ) / σ (19.5 – 10) / 3.15] = Pr(Z 3.02) = 1-.9987 = .0013 d The approximate probability of finding 20 or more underfilled is only .13%. If the consumer group finds 20 or more underfilled, a rare event, then we might conclude that the probability of a underfilled bottle is greater than 0.5%. 4.33 a (1) H 0 : p = .10 H a : p > .10 p = true proportion of bores outside the specifications (2) Z = is the test statistic p 0 = .10 q 0 = 1-.10 = .90 n = 165 np 0 = 16.5 nq 0 = 148.5

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2008hw5_solutions - ST 314 HW#5 Solutions 3.76 n = 2000 p =...

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