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HW9%20Solutions

# HW9%20Solutions - 9-20(a The critical values are given by...

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IE300/GE331 HW9 9-20 (a) The critical values are given by 𝜇𝜇 0 ± 𝑧𝑧 𝛼𝛼 /2 𝜎𝜎 / √𝑛𝑛 . Therefore, 𝑧𝑧 𝛼𝛼 /2 𝜎𝜎 / √𝑛𝑛 = 0.15. 𝑧𝑧 𝛼𝛼 /2 = 0.15 × 8 /0.25 = 1.6971, 𝛼𝛼 = 2 Φ ( 1.6971) = 0.0897 . Alternatively, 𝛼𝛼 = 𝑃𝑃 ( 𝑋𝑋 < 4.85) + 𝑃𝑃 ( 𝑋𝑋 > 5.15) = 𝑃𝑃 �𝑍𝑍 0 < 4.85 5 0.25 8 + 𝑃𝑃 �𝑍𝑍 0 > 5.15 5 0.25 8 = 0.0897 9-40 (a) The parameter of interest is the true mean tensile strength µ 𝐻𝐻 0 µ = 3500, 𝐻𝐻 1 µ 3500 α = 0.01, 𝑧𝑧 0 = 𝑥𝑥̅−𝜇𝜇 0 𝜎𝜎 √𝑛𝑛 . Reject 𝐻𝐻 0 if | 𝑧𝑧 0 | > 𝑧𝑧 α /2 where 𝑧𝑧 0.005 = 2.58 3450 = x , σ = 60, 𝑛𝑛 = 12, 𝑧𝑧 0 = 𝑥𝑥̅ − 𝜇𝜇 0 𝜎𝜎 √𝑛𝑛 = 2.8868 Reject the null hypothesis and conclude the true mean tensile strength is significantly different from 3500 at α = 0.01 . (b) Solve the equation | 𝑧𝑧 0 | = 𝑧𝑧 𝛼𝛼 /2 = 2.8868 , we obtain 𝛼𝛼 = 2 Φ ( 2.8868) = 0.004. When 𝛼𝛼 ≤ 0.004 , we fail to reject the null hypothesis. Otherwise, we reject the null hypothesis.
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