Assignment 9.docx - 4.34 a x =105.2 =5,n=8 =0.05 z...

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4.34 a) ´ x = 105.2, σ = 5, n = 8 α = 0.05 z 0.05 = 1.64485 95% lower one-sided confidence L = ´ x z α σ n = ( 105.2 )−( 1.64485 )( 5 ) 8 = 102.2923 (102.2923, ¿ z 0 = ´ X μ 0 σ n z 0 = 105.20 100 5 8 = 2.942 Pvalue = 1 ɸ ( z 0 ) P = 1 ɸ ( 2.94 ) P = 1 0.998359 = 0.001641 Since P = 0.001641 < 0.05, we reject the null hypothesis b) Pvalue = 2 [ 1 ɸ ( | z 0 | ) ] = ¿ P = 2 [ 1 ɸ ( 2.94 ) ] = ± 0.003282 Since P = ¿ ± 0.003282 < ¿ 0.05, we reject the null hypothesis c) CI = ´ X z α / 2 ( σ n ) ≤μ CI = 105.20 z 0.05 / 2 ( 5 8 ) ≤ μ CI = 105.20 ( 1.96 ) ( 5 8 ) ≤μ = 101.735 ≤μ 101.735 102.292: confidenceinterval islower for two sided
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4.36 a) we can get two equations ´ X z α 2 ( σ n ) = 38.02 ´ X = 38.02 + z α 2 ( σ n ) ´ X + z α 2 ( σ n ) = 61.98 ´ X = 61.98 z α 2 ( σ n ) 2 ´ X = 100 Solve for ´ X ´ X = 50 b) (3805, 61.98) is the 95% CI because the range is greater. A greater CI equates to a greater range. 4.38 a) Parameter : μ,the meanbreaking strengthof yarn H 0 : μ = 100 psi H 1 : μ > 100 psi . α = 0.05 Test Statistic : z 0 = ´ X μ 0 σ n Reject H 0 if P < 0.05 Computation : z 0 = 100.6 100 2 9 = 0.9 Pvalue = 1 ɸ ( z 0 ) = 1 ɸ ( 0.9 ) = 0.18406
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Conclusion: Since, P value 0.18406 ¿ 0.05: we fail to reject null hypothesis and conclude that
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  • Fall '17
  • Dr Bheem

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