Assignment 10.docx

# Assignment 10.docx - z 0.05 2 4.44 =20 hours alpha=0.05 n=...

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4.44 σ = 20 hoursalpha = 0.05 n = ( z 0.05 2 σ E ) 2 n = ( 1.96 ( 20 ) 5 ) 2 n = 61.46 4.46 σ = 1000 = 31.6228 psi alpha = 0.01 n = ( z 0.005 2 σ E ) 2 n = ( 2.576 ( 31.6228 ) 15 ) 2 n = 29.49 4.48 a) P value = 2 P ( 2.48 ) = between 0.02 0.05 Using the table corresponding t 0 = 2.48 liesbetweent 0.025,9 = 2.262 ( ¿ α = 2 × 0.025 = 0.05 ) and corresponding ¿ t 0.01,9 = 2.821 ¿ 02) b) Pvalue = 2 P ( | 3.95 | ) = between 0.002 0.005 Using the table corresponding | t 0 | = 3.95 liesbetweent 0.0025,9 = 3.690 ( ¿ α = 2 × 0.0025 = 0.005 ) and corresponding ¿ t 0.001,9 = 4.297 ¿ 002) c) Pvalue = 2 P ( 2.69 ) = between 0.02 0.05

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Using the table corresponding t 0 = 2.69 liesbetweent 0.025,9 = 2.262 ( ¿ α = 2 × 0.025 = 0.05 ) and corresponding ¿ t 0.01,9 = 2.821 ¿ 02) d) Pvalue = 2 P ( 1.88 ) = between 0.05 0.1 Using the table corresponding t 0 = 1.88 liesbetweent 0.05,9 = 1.833 ( ¿ α = 2 × 0.05 = 0.10 ) and corresponding ¿ t 0.025 , 9 = 2.262 ¿ 05) e) Pvalue = 2 P ( | 1.25 | ) = between 0.2 0.5 Using the table corresponding | t 0 | = 1.25 liesbetweent 0.25,9 = 0.703 ( ¿ α = 2 × 0.25 = 0.5 ) and corresponding ¿ t 0.10 , 9 = 1.383 ¿ 2) 4.50 a) s = ^ σ ´ x ( n ) = 0.4673 ( 25 ) = 2.3365 CI = ´ X t α 2 ,n 1 ( S n ) t α 2 , n 1 = ( CI ´ X ) n s = ¿ t α 2 , n 1 = ( 91.6160 92.5805 ) 25 2.34 = 2.0609 CI = ´ X + t α 2 , n 1 ( S n ) = 92.5805 + 2.0609 ( 2.34 25 ) CI = 93.5450
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• Fall '17
• Dr Bheem

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