Assignment 12.docx - 5.1 a = x 1 x 2=2.35 z 0=2.01 0=0 x1...

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5.1 a) ´ x 1 − ´ x 2 = 2.35 z 0 = 2.01 θ 0 = 0 . σ Θ = ´ x 1 − ´ x 2 0 z 0 . σ Θ = 2.35 0 2.01 = 1.169154 b) This is a one-sided test because the P-value is found as the probability above z 0 = 2.01 in the standard normal distribution. P = 1 – Φ ( 2.01 ) = 0.0222 c) If α = 0.05, P = 0.0222 < 0.05. Therefore, we reject the null the hypothesis in favor of the alternative hypothesis. d) LCL = ´ x 1 − ´ x 2 z α / 2 σ 1 2 n 1 + σ 2 2 n 2 = 2.35 – (1.64485)(1.169154) = 0.4269 UCL = = ´ x 1 − ´ x 2 + z α / 2 σ 1 2 n 1 + σ 2 2 n 2 = 2.35 + (1.64485)(1.169154) = 4.2731
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Therefore, the 90% two-sided CI on the difference in means 0.4269 ≤μ 1 μ 2 4.2731 5.3 a) Parameter of interest: The mean net volume of filling plastic bottles Null Hypothesis: H 0 : μ 1 μ 2 = 0 or H 0 : μ 1 = μ 2 Alternative Hypothesis: H 1 : μ 1 ≠μ 2 Test Statistics: The test statistic is z 0 = μ 1 μ 2 0 σ 1 2 n 1 + σ 2 2 n 2 . α = 0.05 Reject H 0 if : P < 0.05 Computation: Z 0 = 16.015 16.005 0.0004 10 + 0.0006257 10 = 0.987 Conclusion: The P-value is P = 2 [1 - Φ ( 0.987 ) ] = 0.3271 Since P = 0.3271 > 0.05, fail to reject the null hypothesis H 0 : μ 1 = μ 2 The practical engineering conclusion is so both machines fill to the same mean net volume.
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b) β = Ф ( Z α / 2 0 σ 1 2 n 1 + σ 2 2 n 2 ) Ф ( Z α / 2 0 σ 1 2 n 1 + σ 2 2 n 2 ) . β = Ф ( 1.96 0.04 0.020 2 10 + 0.025 2 10 ) Ф ( 1.96 0.04 0.020 2 10 + 0.025 2 10 ) β = Ф ( 1.9909 ) Ф ( 5.9109 ) β = 0.023295 0.000 β = 1 0.023295 = 0.9767 c) ´ x 1 2 −´ x 2 2 Z α / 2 σ 1 2 n 1 + σ 2 2 n 2 ≤μ 1 μ 2 ´ x 1 2
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  • Fall '17
  • Dr Bheem

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