1 2 2 8528 09758 2308 then reject p value 2 2 2 1 8 9

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(1− 𝛼 2 ,𝑛−2) , 8.528 > ? (0.975,8) = 2.308 Then reject ? 0 p-value= 2?(? (𝑛−2) > |? 0 |) = 2 (1 − ?(? (8) < 8.528)) = 2(1 − 0.9999) 0.0002 < 0.05 , then we reject ? 0 . 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 72.73 0.000 Error 8 17.600 9.91% 17.600 2.200 Lack-of-Fit 2 0.933 0.53% 0.933 0.467 0.17 0.849 Pure Error 6 16.667 9.38% 16.667 2.778 Total 9 177.600 100.00% Model Summary S R-sq R-sq(adj) PRESS R-sq(pred) 1.48324 90.09% 88.85% 25.8529 85.44% Coefficients
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18 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 H.W: Q2.7 Refer to Plastic hardness Problem 1.22. a. Estimate the change in the mean hardness when the elapsed time increases by one hour. Use a 99 percent confidence interval. Interpret your interval estimate. b. The plastic manufacturer has stated that the mean hardness should increase by 2 Brinell units per hour. Conduct a two-sided test to decide whether this standard is being satisfied; use ? = ?. ?? . State the alternatives, decision rule, and conclusion. What is the P-value of the test?
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19 Chapter 2 We assume that the normal error regression model is applicable. This model is: ? 𝑖 = ? 0 + ? 1 ? 𝑖 + 𝜀 𝑖 where: ? 0 and ? 1 , are parameters ? 𝑖 are known constants 𝜀 𝑖 are independent ? (0, 𝜎 2 ) ?(? 𝑖 ) = ? 0 + ? 1 ? 𝑖 Sampling Distribution of ? ? ̂ ? 1 ̂ = ? 1 = ∑ (? 𝑖 − ? ̅ )(? 𝑖 − ? ̅ ) (? 𝑖 − ? ̅ ) 2 𝑛 𝑖=1 𝑛 𝑖=1 ?(? 1 ̂ ) = ? 1 𝜎 2 (? 1 ̂ ) = 𝜎 2 (? 𝑖 − ? ̅ ) 2 𝑛 𝑖=1 ? 2 (? 1 ̂ ) = ??? (? 𝑖 − ? ̅ ) 2 𝑛 𝑖=1
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20 ? 1 − ? 1 ?(? 1 ) ~? (𝑛−2) Confidence Interval for ? ? ? [? 1 − ? (1− 𝛼 2 ,𝑛−2) ?(? 1 ) ≤ ? 1 ≤ ? 1 + ? (1−𝛼/2,𝑛−2) ?(? 1 )] = 1 − ? C.I (1 − ?)% for ? 1 ? 1 − ? (1− 𝛼 2 ,𝑛−2) ?(? 1 ) ≤ ? 1 ≤ ? 1 + ? (1−𝛼/2,𝑛−2) ?(? 1 ) Tests Concerning ? ? 1. Hypothesis ? 0 : ? 1 = ? 10 ? 1 : ? 1 ≠ ? 10 ? 0 : ? 1 = ? 10 ? 1 : ? 1 > ? 10 ? 0 : ? 1 = ? 10 ? 1 : ? 1 < ? 10 2. Test statistic ? 0 = ? 1 − ? 10 ?(? 1 ) 3. Decision: Reject ? 0 if |? 0 | > ? (1− 𝛼 2 ,𝑛−2) ? 0 > ? (1−𝛼,𝑛−2) ? 0 < ? (𝛼,𝑛−2) P-value: Reject ? 0 if ? − ????? < ? p-value= 2?(? (𝑛−2) > |? 0 |) p-value = ?(? (𝑛−2) > ? 0 ) ? − ????? = ?(? (𝑛−2) < ? 0 ) Sampling Distribution of ? ? ̂ ? 0 ̂ = ? 0 = ? ̅ − ? 1 ? ̅
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21 ?(? 0 ̂ ) = ? 0 𝜎 2 (? 0 ̂ ) = 𝜎 2 ( 1 ? + ? ̅ 2 (? 𝑖 − ? ̅ ) 2 𝑛 𝑖=1 ) ? 2 (? 0 ̂ ) = ??? ( 1 ? + ? ̅ 2 (? 𝑖 − ? ̅ ) 2 𝑛 𝑖=1 ) ? 0 − ? 0 ?(? 0 ) ~? (𝑛−2) Confidence Interval for ? ? ? [? 0 − ? (1− 𝛼 2 ,𝑛−2) ?(? 0 ) ≤ ? 0 ≤ ? 0 + ? (1−𝛼/2,𝑛−2) ?(? 0 )] = 1 − ? C.I (1 − ?)% for ? 0 ? 0 − ? (1− 𝛼 2 ,𝑛−2) ?(? 0 ) ≤ ? 0 ≤ ? 0 + ? (1−𝛼/2,𝑛−2) ?(? 0 )
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22 Tests Concerning ? ? 1. Hypothesis ? 0 : ? 0 = ? 00 ? 1 : ? 0 ≠ ? 00 ? 0 : ? 0 = ? 00 ? 1 : ? 0 > ? 00 ? 0 : ? 1 = ? 00 ? 1 : ? 1 < ? 00 2. Test statistic ? 0 = ? 0 − ? 00 ?(? 0 ) 3. Decision: Reject ? 0 if |? 0 | > ? (1− 𝛼 2 ,𝑛−2) ? 0 > ? (1−𝛼,𝑛−2) ? 0 < ? (𝛼,𝑛−2) P-value: Reject ? 0 if ? − ????? < ? p-value= 2?(? (𝑛−2) > |? 0 |) p-value = ?(? (𝑛−2) > ? 0 ) ? − ????? = ?(? (𝑛−2) < ? 0 ) ? = ? 0 + ? 1 ? ANOVA TABLE Source of Variation d.f SS MS F p-value Regression 1 SSR= ∑(? 𝑖 ̂ − ? ̅ ) 2 ??? = ??? 1 ??? ??? Error n-2 SSE= ∑(? 𝑖 − ? 𝑖 ̂ ) 2 ??? = ??? ? − 2 Total n-1 SSTo= ∑(? 𝑖 − ? ̅ ) 2
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23 1. Hypothesis ? 0 : ? 1 = 0 (Non liner) ? 1 : ? 1 ≠ 0 2. Test statistic ? = ??? ??? 3. Decision: Reject ? 0 if ? > ? (1−𝛼,1,𝑛−2) P-value: Reject ? 0 if ? − ????? < ? ? − ????? = ?(? (1,𝑛−2) > ? ) Q2.6 . Refer to Airfreight breakage Problem 1.21. ? ̅ = 1, ? ̅ = 14.2 , ∑ (? 𝑖 − ? ̅ ) 𝑛=10 𝑖=1 (? 𝑖 − ?
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