2 1 1 ℎ 2 2 1 ℎ 1 2 2 2 1 1 ℎ 2 2 1 1 1 5 2

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? 2 (? 𝑛𝑒𝑤 ̂ ) = ??? (1 + 1 ? + (? − ? ̅ ) 2 (? 𝑖 − ? ̅ ) 2 𝑛 𝑖=1 ) ? ̂ ± ? (1 − ? 2 ; ? − 2) ?(? 𝑛𝑒𝑤 ̂ ) ? 2 (? 𝑛𝑒𝑤 ̂ ) = ??? (1 + 1 ? + (? − ? ̅ ) 2 (? 𝑖 − ? ̅ ) 2 𝑛 𝑖=1 ) = ?. ???? (1 + 1 120 + (28 − 24.725) 2 2379.925 ) = 0.39328 ?(? 𝑛𝑒𝑤 ̂ ) = 0.6271 3.22012 ± 1.9807(0.6271) 1.9594 < ? ℎ(𝑛𝑒𝑤) < 4.4430 c. Is the prediction interval in part (b) wider than the confidence interval in part (a)? Should it be? ؤبنتلل ةقثلا ةرتف له يف ءزجلا ( ب ) عسوأ نم ةرتف ةقثلا يف ءزجلا ( أ ) ؟ له بجي نأ ؟نوكت Yes, Yes
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39 2.15. Refer to Airfreight breakage Problem 1.21. ? ̅ = 1, ∑(? 𝑖 − ? ̅ ) 2 10 𝑖=1 = 10 ANOVA TABLE Source of Variation d.f SS MS F p-value Regression 1 SSR= 160 ??? = 160 72.72 0.00 Error 8 SSE= 17.6 ??? = 2.2 Total 9 SSTo= 177.6 a. Because of changes in airline routes, shipments may have to be transferred more frequently than in the past. Estimate the mean breakage for the following numbers of transfers: X = 2, 4. Use separate 99 percent confidence intervals. Interpret your results. At ? = 2 ? ̂ = 10.2 + 4 (2) = 18.2 ? 2 (? ̂ ) = ??? ( 1 ? + (? − ? ̅ ) 2 (? 𝑖 − ? ̅ ) 2 𝑛 𝑖=1 ) = ?. ? ( 1 10 + (2 − 1) 2 10 ) = 0.44 ?(? ̂ ) = √0.44 = 0.6633 ? (1 − ? 2 ; ? − 2) = ?(0.995; 8) = 3.355 18.2 ± 3.355(0.6633) 15.976 < ?(? ) < 20.424 At ? = 4
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40 ? ̂ = 10.2 + 4 (4) = 26.2 ? 2 (? ̂ ) = ??? ( 1 ? + (? − ? ̅ ) 2 (? 𝑖 − ? ̅ ) 2 𝑛 𝑖=1 ) = ?. ? ( 1 10 + (4 − 1) 2 10 ) = 2.2 ?(? ̂ ) = √2.2 = 1.483 ? (1 − ? 2 ; ? − 2) = ?(0.995; 8) = 3.355 26.2 ± 3.355(1.483) 12.748 < ?(? ) < 23.652 We conclude that the mean number of ampules found to be broken upon arrival when 2 transfers from one aircraft to another over the shipment route of 2 are produced is somewhere between 15.976 and 20.424 ampules نأ طسوتم ددع تلاوبمأ تدجو هرسكنم دنع مهلوصو امدنع مت ربع هلقن 2 تارم نم ةرئاط ةدحاو ىلإ رخآ ربع راسم ةنحشلا نيب , 15.976 و 20.424 .هلوبمأ We conclude that the mean number of ampules found to be broken upon arrival when 4 transfers from one aircraft to another over the shipment route are produced is somewhere between 12.748 and 23.652 ampules. نأ طسوتم ددع تلاوبمأ تدجو هرسكنم دنع مهلوصو امدنع مت ربع هلقن 4 تارم نم ةرئاط ةدحاو ىلإ رخآ ربع راسم ةنحشلا نيب , 12.748 و 23.652 .هلوبمأ b. The next shipment will entail two transfers. Obtain a 99 percent prediction interval for the number of broken ampules for this shipment. Interpret your prediction interval. ? 2 (? 𝑛𝑒𝑤 ̂ ) = ??? (1 + 1 ? + (? − ? ̅ ) 2 (? 𝑖 − ? ̅ ) 2 𝑛 𝑖=1 ) = ?. ? (1 + 1 10 + (2 − 1) 2 10 ) = 2.64 ?(? ̂ ) = √2.64 = 1.6248 18.2 ± 3.355(1.6248)
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41 12.748 < ? ℎ(𝑛𝑒𝑤) < 23.652 With confidence coefficient 0.99, we predict that the mean number of ampules found to be broken upon arrival when 2 transfers from one aircraft to another over the shipment route of 2 are produced is somewhere between 12.748 and 23.652 ampules. 40 35 30 25 20 15 Xi Boxplot of Xi
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42 40 36 32 28 24 20 16 Xi Dotplot of Xi
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43 16 14 12 10 8 6 4 2 40 35 30 25 20 15 Index Xi Time Series Plot of Xi
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44 For test normality of residuals ????ℎ → ??????𝑖?𝑖?? ???? → ?𝑖?𝑔?? → (?𝑖???𝑖???𝑖?? ??????) → ? → ??
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