40 2 10 1 10 2 10 1 1 4 d a consultant has suggested

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̅ ) = 40, ∑(? 𝑖 − ? ̅ ) 2 10 𝑖=1 = 10 ∑ (? 𝑖 − ? ̅ ) 2 𝑛=10 𝑖=1 = 177.6, ??? = 2.2 ? 0 = 10.2, ? 1 = 4 d) A consultant has suggested, on the basis of previous experience, that the mean number of broken ampules should not exceed 9.0 when no transfers are made . Conduct an appropriate test, using ? = 0.025 . State the alternatives, decision rule, and conclusion. What is the P-value of the test? ? = 0.025
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24 1. Hypothesis ? 0 : ? 0 ≤ 9 ? 1 : ? 0 > 9 2. Test statistic ? 0 = ? 0 − ? 00 ?(? 0 ) = 10.2 − 9 0.6633 = 1.809 ? 2 (? 0 ̂ ) = ??? ( 1 ? + ? ̅ 2 (? 𝑖 − ? ̅ ) 2 𝑛 𝑖=1 ) = 2.2 ( 1 10 + 1 2 10 ) = 0.44 ?(? 0 ) = 0.6633 3. Decision: Reject ? 0 if ? 0 > ? (1−𝛼,𝑛−2) , 1.809 ≯ ? (0.975,8) = 2.306 Then not reject ? 0 p-value= ?(? (𝑛−2) > ? 0 ) = (1 − ?(? (𝑛−2) < 1.809)) = (1 − 0.945) = 0.055 ≮ 0.025 , then we not reject ? 0 . at ? = 0.05 ? 0 − ? (1− 𝛼 2 ,𝑛−2) ?(? 0 ) ≤ ? 0 ≤ ? 0 + ? (1−𝛼/2,𝑛−2) ?(? 0 ) ? (1− 𝛼 2 ,𝑛−2) = ? (0.975,8) = 2.306 10.2 − 2.306 ∗ 0.6633 ≤ ? 0 ≤ 10.2 + 2.306 ∗ 0.6633 8.76 ≤ ? 0 ≤ 11.728
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25 Analysis of Variance Source DF Seq SS Contribution Adj SS Adj MS F-Value P-Value Regression 1 160.000 90.09% 160.000 160.000 72.73 0.000 Xi 1 160.000 90.09% 160.000 160.000 Error 8 17.600 9.91% 17.600 2.200 Total 9 177.600 100.00% Coefficients Term Coef SE Coef 95% CI T-Value P-Value VIF Constant 10.200 0.663 (8.670, 11.730) 15.38 0.000 Xi 4.000 0.469 (2.918, 5.082) 8.53 0.000 1.00 Regression Equation Yi = 10.200 + 4.000 Xi
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26 Q2.25. Refer to Airfreight breakage Problem 1.21. a. Set up the ANOVA table. Which elements are additive? b. Conduct an F test to decide whether or not there is a linear association between the number of times a carton is transferred and the number of broken ampules; control the ? risk at 0.05. State the alternatives, decision rule, and conclusion. c. Obtain the t* statistic for the test in part (b) and demonstrate numerically its equivalence to the F* statistic obtained in part (b). ? ̅ = ?, ? ̅ = ??. ? , (? 𝒊 − ? ̅ ) ?=?? 𝒊=? (? 𝒊 − ? ̅ ) = ??, ∑ (? 𝒊 − ? ̅ ) ? ?? 𝒊=? = ?? ∑ (? 𝒊 − ? ̅ ) ? ?=?? 𝒊=? = ???. ?, 𝑴?𝑬 = ?. ?, ? ? = ??. ?, ? ? = ? ? 𝑖 ? 𝑖 (? 𝑖 − ? ̅ ) (? 𝑖 − ? ̅ ) (? 𝑖 − ? ̅ ) 2 (? 𝑖 − ? ̅ ) ∗ (? 𝑖 − ? ̅ ) (? 𝑖 − ? ̅ ) 2 ? 𝑖 ̂ (? 𝑖 − ? 𝑖 ̂ ) 2 1 16 0 1.8 0 0 3.24 14.2 3.24 0 9 -1 -5.2 5.2 1 27.04 10.2 1.44 2 17 1 2.8 2.8 1 7.84 18.2 1.44 0 12 -1 -2.2 2.2 1 4.84 10.2 3.24 3 22 2 7.8 15.6 4 60.84 22.2 0.04 1 13 0 -1.2 0 0 1.44 14.2 1.44 0 8 -1 -6.2 6.2 1 38.44 10.2 4.84 1 15 0 0.8 0 0 0.64 14.2 0.64 2 19 1 4.8 4.8 1 23.04 18.2 0.64 0 11 -1 -3.2 3.2 1 10.24 10.2 0.64 10 142 0 0 40 10 177.6 142 17.6
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27 ∑(? 𝑖 − ? 𝑖 ̂ ) 2 = 17.6 ANOVA TABLE Source of Variation d.f SS MS F p-value Regression 1 SSR= 177.6 − 17.6 = 160 ??? = 160 160 2.2 = 72.72 0.00 Error 8 SSE= 17.6 ??? = 17.6 8 = 2.2 Total 9 SSTo= 177.6 ? = 0.05 1. Hypothesis ? 0 : ? 1 = 0 ? 1 : ? 1 ≠ 0 2. Test statistic ? = 72.72 3. Decision: Reject ? 0 if ? > ? (1−𝛼,1,𝑛−2) , 72.72 > ? (0.95,1,8) = 5.31 Then reject ? 0 p-value= ?(? (1,𝑛−2) > ? ) = (1 − ?(? (1,8) < 72.72)) = (1 − 0.9999) = 0.0001 < 0.05 , then we reject ? 0 .
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28 Analysis of Variance Source DF Adj SS Adj MS F-Value P-Value Regression 1 160.000 160.000 72.73 0.000 Xi 1 160.000 160.000 72.73 0.000 Error 8 17.600 2.200 Lack-of-Fit 2 0.933 0.467 0.17 0.849 Pure Error 6 16.667 2.778 Total 9 177.600 ? = 8.528 , (? ) 2 = (8.528) 2 = 72.72 = ? Q2.26. Refer to Plastic hardness Problem 1.22. a. Set up the ANOVA table. b. Test by means of an F test whether or not there is a linear association between the hardness of the plastic and the elapsed time. Use a = .01. State the alternatives, decision rule, and conclusion. Analysis of Variance Source DF Adj SS Adj MS F-Value P-Value Regression 1 5297.51 5297.51 506.51 0.000 Xi 1 5297.51 5297.51 506.51 0.000 Error 14 146.43 10.46 Lack-of-Fit 2 17.67 8.84 0.82 0.462 Pure Error 12 128.75 10.73 Total 15 5443.94
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29 ? = 0.01 1. Hypothesis ? 0 : ?
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