Chapter 1 Q1.19) Grade poit average. The director of admissios of a small college selected 120 studets at radom from the ew freshma class i a study to determie whether a studet's grade poit average (OPA) at the ed of the freshma year (Y) ca be predicted from the ACT test score (X). The results of the study follow. Assume that first-order regressio model (1.1) is appropriate. a. Obtai the least squares estimates of β 0 ad β 1, ad state the estimated regressio fuctio. b. Plot the estimated regressio fuctio ad the data."does the estimated regressio fuctio appear to fit the data well? c. Obtai a poit estimate of the mea freshma OPA for studets with ACT test score X = 30. d. What is the poit estimate of the chage i the mea respose whe the etrace test score icreases by oe poit? Solutio: X = 24.725, Y = 3.07405 =120 (X i X ) (Y i Y ) = 92.40565 =120 (X i X ) 2 = 2379.925 =120 (Y i Y ) 2 = 49.40545 b 1 = β 1 = =120 (X i X )(Y i Y ) =120 (X i X ) 2 = 92.40565 2379.925 = 0.038827 b 0 = β 0 = Y b 1 X = 3.07405 0.038827 24.725 = 2.114049 1
Y = 2.114 + 0.0388 X At X=30 Y ĥ = 2.114 + 0.0388 (30) = 3.278863 whe the etrace test score icreases by oe poit, the mea respose icrease by 0.038827. Q1.20) Copier maiteace. The Tri-City Office Equipmet Corporatio sells a imported copier o a frachise basis ad performs prevetive maiteace ad repair service o this copier. The data below have bee collected from 45 recet calls o users to perform routie prevetive maiteace service; for each call, X is the umber of copiers serviced ad Y is the total umber of miutes spet by the service perso. Assume that first-order regressio model (1.1) is appropriate. )مصنع يعمل على الصنعة الوقائية( هو عدد الناسخات الخدمات X.هو العدد اإلجمالي للدقائق التي يقضيها الشخص الخدمة Y a. Obtai the estimated regressio fuctio. b. Plot the estimated regressio fuctio ad the data. How well does the estimated regressio fuctio fit the data? c. Iterpret b 0 i your estimated regressio fuctio. Does b 0 provide ay relevat iformatio here? Explai. d. Obtiu a poim estimate of the mea service time whe X = 5 copiers are serviced. Solutio: X = 5.11111, Y = 76.26667 2
=120 (X i X ) (Y i Y ) = 5118.667 =120 (X i X ) 2 = 340.4444 =120 (Y i Y ) 2 = 80376.8 b 1 = β 1 = =120 (X i X )(Y i Y ) =120 (X i X ) 2 = 15.03525 b 0 = β 0 = Y b 1 X = 0.58016 Y = 0.58016 + 15.03525 X At X=5 Y ĥ = 0.58016 + 15.03525 (5) = 74.59608 Q1.21) (H.W) Airfreight breakage. A substace used i biological ad medical research is shipped by airfreight to users i cartos of 1,000 ampules. The data below, ivolvig 10 shipmets, were collected o the umber of times the carto was trasferred from oe aircraft to aother over the shipmet route (X) ad the umber of ampules foud to be broke upo arrival (Y). Assume that first-order regressio model (1.1) is appropriate. 3
a. Obtai the estimated regressio fuctio. Plot the estimated regressio fuctio ad the data. Does a liear regressio fuctio appear to give a good fit here? b. Obtai a poit estimate of the expected umber of broke ampules whe X = 1 trasfer is made. c. Estimate the icrease i the expected umber of ampules broke whe there are 2 trasfers as compared to 1 trasfer. d. Verify that your fitted regressio lie goes through the poit (X, Y ). Q1.22) Plastic hardess. Refer to Problems 1.3 ad 1.14. Sixtee batches of the plastic were made, ad from each batch oe test item was molded. Each test item was radomly assiged to oe of the four predetermied time levels, ad the hardess was measured after the assiged elapsed time. The results are show below; X is the elapsed time i hours? ad Y is hardess i Briell uits. Assume that first-order regressio model (1.1) is appropria'te. a. Obtai the estimated regressio fuctio. Plot the estimated regressio fuctio ad the data. Does a liear regressio fuctio appear to give a good fit here? b. Obtai a poit estimate of the mea hardess whe X = 40 hours. c. Obtai a poit estimate of the chage i mea hardess whe X icreases by 1 hour. Solutio: X = 28, Y = 225.5625 =120 (X i X ) (Y i Y ) = 2604 =120 (X i X ) 2 = 1280 4
=120 (Y i Y ) 2 = 5443.938 b 1 = β 1 = =120 (X i X )(Y i Y ) =120 (X i X ) 2 = 2.034375 b 0 = β 0 = Y b 1 X = 168.6 Y = 168.6 + 2.034375 X At X=40 Y ĥ = 168.6 + 2.034375 (40) = 249.975 Q1.24) Refer to Copier maiteace Problem 1.20. a Obtai the residuals e i ad the sum of the squared residuals e i 2. What is the relatio betwee the sum of the squared residuals here ad the quatity Q i (1.8)? b. Obtai poit estimates of σ 2 ad. I what uits is σ expressed? e i 2 = 3416.377 e i 2 = Q σ 2 = e i 2 2 = 3416.377 = 79.45063 = MSE 43 σ = MSE = 79.45063 5
Q1.25) (H.W) Refer to Airfreight breakage Problem 1.21. a. Obtai the residual for the first case. What is its relatio to e 1? b. Compute e i 2 ad MSE. What is estimated by MSE? Q1.26) (H.W) Refer to Plastic hardess Problem 1.22. a. Obtai the residuals ej. Do they sum to zero i accord with (1.17)? b. Estimate σ 2 ad. I what uits is σ expressed? Q1.21) Solutio (H.W) Airfreight breakage. A substace used i biological ad medical research is shipped by airfreight to users i cartos of 1,000 ampules. The data below, ivolvig 10 shipmets, were collected o the umber of times the carto was trasferred from oe aircraft to aother over the shipmet route (X) ad the umber of ampules foud to be broke upo arrival (Y). Assume that first-order regressio model (1.1) is appropriate. a. Obtai the estimated regressio fuctio. Plot the estimated regressio fuctio ad the data. Does a liear regressio fuctio appear to give a good fit here? Y = 10.2 + 4.0X b. Obtai a poit estimate of the expected umber of broke ampules whe X = 1 trasfer is made. If X=1 The Y ĥ = 10.2 + 4.0(1) = 14.20 c. Estimate the icrease i the expected umber of ampules broke whe there are 2 trasfers as compared to 1 trasfer. Y h2 = 10.2 + 4.0(2) = 18.20 Y h1 = 10.2 + 4.0(1) = 14.20 Y h2 Y h1 = b 1 = 4.0 d. Verify that your fitted regressio lie goes through the poit (X, Y ). X = 1, Y = 14.2 (X, Y ) = (1, 14.2) 6
If X=1 The Y ĥ = 10.2 + 4.0(1) = 14.20 The we ca say the regressio lie goes through the poit (X, Y ) = (1, 14.2) 7
Chapter 2 We assume that the ormal error regressio model is applicable. This model is: Y i = β 0 + β 1 X i + ε i where: β 0 ad β 1, are parameters X i are kow costats ε i are idepedet N (0, σ 2 ) Samplig Distributio of β 1 β 1 = b 1 = (X i X )(Y i Y ) (X i X ) 2 E(Y i ) = β 0 + β 1 X i E(β ) 1 = β 1 σ 2 (β ) 1 = s 2 (β ) 1 = σ 2 (X i X ) 2 MSE (X i X ) 2 b 1 β 1 s(b 1 ) ~t ( 2) Cofidece Iterval for β 1 P [b 1 t (1 α 2, 2)s(b 1) β 1 b 1 + t (1 α/2, 2) s(b 1 )] = 1 α 8
C.I (1 α)% for β 1 b 1 t (1 α 2, 2)s(b 1) β 1 b 1 + t (1 α/2, 2) s(b 1 ) Tests Cocerig β 1 1. Hypothesis H 0 : β 1 = β 10 H 1 : β 1 β 10 2. Test statistic H 0 : β 1 = β 10 H 1 : β 1 > β 10 H 0 : β 1 = β 10 H 1 : β 1 < β 10 T 0 = b 1 β 10 s(b 1 ) 3. Decisio: Reject H 0 if T 0 > t α (1 2, 2) T 0 > t (1 α, 2) T 0 < t (α, 2) P-value: Reject H 0 if p value < α p-value=2p(t ( 2) > T 0 ) p value = P(t ( 2) > T 0 ) p value = P(t ( 2) < T 0 ) Page 114 Q2.4. Refer to Grade poit average Problem 1.19. a. Obtai a 99 percet cofidece iterval for β 1. Iterpret your cofidece iterval. Does it iclude zero? Why might the director of admissios be iterested i whether the cofidece iterval icludes zero? Solutio: By usig Miitab: Stat Regressio Regressio Fit Regressio Mode 9
10
Regressio Aalysis: Yi versus Xi Aalysis of Variace Source DF Seq SS Cotributio Adj SS Adj MS F-Value P-Value Regressio 1 3.588 7.26% 3.588 3.5878 9.24 0.003 Xi 1 3.588 7.26% 3.588 3.5878 9.24 0.003 Error -2=118 45.818 92.74% SSE=45.818 MSE=0.3883 Lack-of-Fit 19 6.486 13.13% 6.486 0.3414 0.86 0.632 11
Pure Error 99 39.332 79.61% 39.332 0.3973 Total 119 49.405 100.00% Model Summary S R-sq R-sq(adj) PRESS R-sq(pred) 0.623125 7.26% 6.48% 47.6103 3.63% Coefficiets Term Coef SE Coef 99% CI T-Value P-Value VIF Costat 2.114 0.321 ( 1.274, 2.954) 6.59 0.000 Xi 0.0388 0.0128 (0.0054, 0.0723) 3.04 0.003 1.00 Regressio Equatio Yi = 2.114 + 0.0388 Xi 99% C.I for β 1 : b 1 t (1 α 2, 2)s(b 1) β 1 b 1 + t (1 α/2, 2) s(b 1 ) Iterpret your cofidece iterval. Does it iclude zero? No 0.0054 β 1 0.0723 Why might the director of admissios be iterested i whether the cofidece iterval icludes zero? If the C.I of β 1 iclude zero, the β 1 ca tack zero ad β 1 = 0 b. Test, usig the test statistic t*, whether or ot a liear associatio exists betwee studet's ACT score (X) ad GPA at the ed of the freshma year (Y). Use a level of sigificace of 0.01 State the alteratives, decisio rule, ad coclusio. 12
α = 0.01 1. Hypothesis H 0 : β 1 = 0 H 1 : β 1 0 2. Test statistic T 0 = b 1 β 10 = b 1 s(b 1 ) s(b 1 ) = 0.0388 0.0128 = 3.04 3. Decisio: Reject H 0 if T 0 > t α (1, 2), 3.04 > t (0.995,118) = 1.70943 2 The reject H 0 c. What is the P-value of your test i part (b)? How does it support the coclusio reached i part (b)? p-value= 0.003<0.01, the we reject H 0. 13
Q2.5. Refer to Copier maiteace Problem 1.20. =45 = 45, X i 45 45 45 2 = 230, Y i = 3432, X i = 1516, X i Y i SSE = 3416.377 = 22660 a. Estimate the chage i the mea service time whe the umber of copiers serviced icreases by oe. Use a 90 percet cofidece iterval. Iterpret your cofidece iterval. 90% C.I for β 1 : b 1 t (1 α 2, 2)s(b 1) β 1 b 1 + t (1 α/2, 2) s(b 1 ) α = 1 0.9 = 0.1 b 1 = (X i X )(Y i Y ) (X i X ) 2 = (X iy i X Y i X i Y + X Y ) (X 2 = X iy i X Y i 2X X i + X 2 ) 2 X i X 2 s 2 MSE 3416.377/(45 2) (b 1 ) = (X i X ) 2 = 1516 45 5.1111 2 = 0.23337 s(b 1 ) = 0.2337 = 0.48308 t (1 α 2, 2) = t (0.95,43) = 1.68107 b 1 t (1 α 2, 2)s(b 1) = 15.035 1.68107 0.48308 = 14.222 b 1 + t (1 α 2, 2)s(b 1) = 15.035 + 1.68107 0.48308 = 15.84709 14.222 β 1 15.847 = 22660 45 5.1111 76.2667 1516 45 5.1111 2 = 15.035 b. Coduct a t test to determie whether or ot there is a liear associatio betwee X ad Y here; cotrol the α a risk at 0.01. State the alteratives, decisio rule, ad coclusio. What is the P-value of your test? α = 0.01 1. Hypothesis 14
H 0 : β 1 = 0 H 1 : β 1 0 2. Test statistic T 0 = b 1 β 10 = b 1 s(b 1 ) s(b 1 ) = 15.035 0.48308 = 31.123 3. Decisio: Reject H 0 if T 0 > t α (1, 2), 31.123 > t (0.995,43) = 2.695 2 The reject H 0 p-value=2p(t ( 2) > T 0 ) = 2 (1 P(t ( 2) < 31.123)) = 2(1 1) 0.00 < 0.01, the we reject H 0. c. Are your results i parts (a) ad (b) cosistet? Explai. Yes, the C.I of β 1 does ot iclude zero, ad we reject H 0. d. The maufacturer has suggested that the mea required time should ot icrease by more tha 14 miutes for each additioal copier that is serviced o a service call. Coduct a test to decide whether this stadard is beig satisfied by Tri-City. Cotrol the risk of a Type I error at 0.05. State the alteratives, decisio rule, ad coclusio. What is the P-value of the test? α = 0.05 1. Hypothesis H 0 : β 1 14 H 1 : β 1 > 14 2. Test statistic T 0 = b 1 β 10 = b 1 14 15.035 14 = = 2.143 s(b 1 ) s(b 1 ) 0.48308 3. Decisio: Reject H 0 if T 0 > t (1 α, 2), 2.143 > t (0.95,43) = 1.861 The reject H 0 p-value=p(t ( 2) > T 0 ) = (1 P(t ( 2) < 2.143)) = (1 0.981) = 0.019 < 0.05, the we reject H 0. 15
Q2.6. Refer to Airfreight breakage Problem 1.21. =10 X = 1, Y = 14.2, (X i X ) (Y i Y ) = 40 10 (X i X ) 2 = 10, MSE = 2.2 a. Estimate β 1 with a 95 percet cofidece iterval. Iterpret your iterval estimate. 95% C.I for β 1 : b 1 t (1 α 2, 2)s(b 1) β 1 b 1 + t (1 α/2, 2) s(b 1 ) α = 1 0.95 = 0.05 b 1 = β 1 = s 2 (b 1 ) = =120 (X i X )(Y i Y ) =120 (X i X ) 2 = 4 MSE (X i X ) 2 s(b 1 ) = 0.22 = 0.469 = 2.2 10 = 0.22 t (1 α 2, 2) = t (0.975,8) = 2.306 b 1 t (1 α 2, 2)s(b 1) = 4 2.306 0.469 = 2.918 b 1 + t (1 α 2, 2)s(b 1) = 4 + 2.306 0.469 = 5.081 2.918 β 1 5.081 b. Coduct a t test to decide whether or ot there is a liear associatio betwee umber of times a carto is trasferred (X) ad umber of broke ampules (Y). Use a level of sigificace of 0.05. State the alteratives, decisio rule, ad coclusio. What is the P-value of the test? α = 0.05 16
1. Hypothesis H 0 : β 1 = 0 H 1 : β 1 0 2. Test statistic T 0 = b 1 β 10 = b 1 s(b 1 ) s(b 1 ) = 4 0.469 = 8.528 3. Decisio: Reject H 0 if T 0 > t α (1, 2), 8.528 > t (0.975,8) = 2.308 2 The reject H 0 p-value=2p(t ( 2) > T 0 ) = 2 (1 P(t (8) < 8.528)) = 2(1 0.9999) 0.0002 < 0.05, the we reject H 0. Aalysis of Variace Source DF Seq SS Cotributio Adj SS Adj MS F-Value P-Value Regressio 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% Coefficiets 17
Term Coef SE Coef 95% CI T-Value P-Value VIF Costat 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 Regressio Equatio Yi = 10.200 + 4.000 Xi H.W: Q2.7 Refer to Plastic hardess Problem 1.22. a. Estimate the chage i the mea hardess whe the elapsed time icreases by oe hour. Use a 99 percet cofidece iterval. Iterpret your iterval estimate. b. The plastic maufacturer has stated that the mea hardess should icrease by 2 Briell uits per hour. Coduct a two-sided test to decide whether this stadard is beig satisfied; use α = 0. 01. State the alteratives, decisio rule, ad coclusio. What is the P-value of the test? 18
Chapter 2 We assume that the ormal error regressio model is applicable. This model is: Y i = β 0 + β 1 X i + ε i where: β 0 ad β 1, are parameters X i are kow costats ε i are idepedet N (0, σ 2 ) E(Y i ) = β 0 + β 1 X i Samplig Distributio of β 1 β 1 = b 1 = (X i X )(Y i Y ) (X i X ) 2 E(β ) 1 = β 1 σ 2 (β ) 1 = s 2 (β ) 1 = σ 2 (X i X ) 2 MSE (X i X ) 2 19
b 1 β 1 s(b 1 ) ~t ( 2) Cofidece Iterval for β 1 P [b 1 t (1 α 2, 2)s(b 1) β 1 b 1 + t (1 α/2, 2) s(b 1 )] = 1 α C.I (1 α)% for β 1 b 1 t (1 α 2, 2)s(b 1) β 1 b 1 + t (1 α/2, 2) s(b 1 ) Tests Cocerig β 1 1. Hypothesis H 0 : β 1 = β 10 H 1 : β 1 β 10 2. Test statistic H 0 : β 1 = β 10 H 1 : β 1 > β 10 H 0 : β 1 = β 10 H 1 : β 1 < β 10 T 0 = b 1 β 10 s(b 1 ) 3. Decisio: Reject H 0 if T 0 > t α (1 2, 2) T 0 > t (1 α, 2) T 0 < t (α, 2) P-value: Reject H 0 if p value < α p-value=2p(t ( 2) > T 0 ) p-value= P(t ( 2) > T 0 ) p value = P(t ( 2) < T 0 ) Samplig Distributio of β 0 β 0 = b 0 = Y b 1 X 20
E(β ) 0 = β 0 σ 2 (β ) 0 = σ 2 ( 1 X 2 + (X i X ) 2 ) s 2 (β ) 0 = MSE ( 1 X 2 + (X i X ) 2 ) b 0 β 0 s(b 0 ) ~t ( 2) Cofidece Iterval for β 1 P [b 0 t (1 α 2, 2)s(b 0) β 0 b 0 + t (1 α/2, 2) s(b 0 )] = 1 α C.I (1 α)% for β 0 b 0 t (1 α 2, 2)s(b 0) β 0 b 0 + t (1 α/2, 2) s(b 0 ) 21
Tests Cocerig β 1 1. Hypothesis H 0 : β 0 = β 00 H 1 : β 0 β 00 2. Test statistic H 0 : β 0 = β 00 H 1 : β 0 > β 00 H 0 : β 1 = β 00 H 1 : β 1 < β 00 T 0 = b 0 β 00 s(b 0 ) 3. Decisio: Reject H 0 if T 0 > t α (1 2, 2) T 0 > t (1 α, 2) T 0 < t (α, 2) P-value: Reject H 0 if p value < α p-value=2p(t ( 2) > T 0 ) p-value= P(t ( 2) > T 0 ) p value = P(t ( 2) < T 0 ) Y h = b 0 + b 1 X h ANOVA TABLE Source of Variatio d.f SS MS F p-value Regressio 1 SSR= (Y i Y ) 2 MSR = SSR MSR 1 MSE Error -2 SSE= (Y i Y ) 2 i MSE = SSE 2 Total -1 SSTo= (Y i Y ) 2 22
1. Hypothesis H 0 : β 1 = 0 (No lier) H 1 : β 1 0 2. Test statistic F = MSR MSE 3. Decisio: Reject H 0 if F > F (1 α,1, 2) P-value: Reject H 0 if p value < α p value = P(F (1, 2) > F ) Q2.6. Refer to Airfreight breakage Problem 1.21. X = 1, Y = 14.2, =10 (X i X ) (Y i Y ) = 40, (X i X ) 2 = 10 =10 10 (Y i Y ) 2 = 177.6, MSE = 2.2 b 0 = 10.2, b 1 = 4 d) A cosultat has suggested, o the basis of previous experiece, that the mea umber of broke ampules should ot exceed 9.0 whe o trasfers are made. Coduct a appropriate test, usig α = 0.025. State the alteratives, decisio rule, ad coclusio. What is the P-value of the test? α = 0.025 23
1. Hypothesis H 0 : β 0 9 H 1 : β 0 > 9 2. Test statistic T 0 = b 0 β 00 = 10.2 9 s(b 0 ) 0.6633 = 1.809 s 2 (β ) 0 = MSE ( 1 X 2 + (X i X ) 2 ) = 2.2 ( 1 10 + 12 10 ) = 0.44 s(b 0 ) = 0.6633 3. Decisio: Reject H 0 if T 0 > t (1 α, 2), 1.809 t (0.975,8) = 2.306 The ot reject H 0 p-value=p(t ( 2) > T 0 ) = (1 P(t ( 2) < 1.809)) = (1 0.945) = 0.055 0.025, the we ot reject H 0. at α = 0.05 b 0 t (1 α 2, 2)s(b 0) β 0 b 0 + t (1 α/2, 2) s(b 0 ) t (1 α 2, 2) = t (0.975,8) = 2.306 10.2 2.306 0.6633 β 0 10.2 + 2.306 0.6633 8.76 β 0 11.728 24
Aalysis of Variace Source DF Seq SS Cotributio Adj SS Adj MS F-Value P-Value Regressio 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% Coefficiets Term Coef SE Coef 95% CI T-Value P-Value VIF Costat 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 Regressio Equatio Yi = 10.200 + 4.000 Xi 25
Q2.25. Refer to Airfreight breakage Problem 1.21. a. Set up the ANOVA table. Which elemets are additive? b. Coduct a F test to decide whether or ot there is a liear associatio betwee the umber of times a carto is trasferred ad the umber of broke ampules; cotrol the α risk at 0.05. State the alteratives, decisio rule, ad coclusio. c. Obtai the t* statistic for the test i part (b) ad demostrate umerically its equivalece to the F* statistic obtaied i part (b). =10 X = 1, Y = 14. 2, (X i X ) (Y i Y ) = 40, 10 (X i X ) 2 = 10 =10 (Y i Y ) 2 = 177. 6, MSE = 2. 2, b 0 = 10. 2, b 1 = 4 X i Y i (X i X ) (Y i Y ) (X i X ) 2 (X i X ) (Y i Y ) (Y i Y ) 2 Y i (Y i Y ) 2 i 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 26
(Y i Y ) 2 i = 17.6 ANOVA TABLE Source of Variatio d.f SS MS F p-value Regressio 1 SSR=177.6 17.6 = 160 MSR = 160 160 2.2 = 72.72 0.00 Error 8 SSE=17.6 MSE = 17.6 8 = 2.2 Total 9 SSTo= 177.6 α = 0.05 1. Hypothesis H 0 : β 1 = 0 H 1 : β 1 0 2. Test statistic F = 72.72 3. Decisio: Reject H 0 if F > F (1 α,1, 2), 72.72 > F (0.95,1,8) = 5.31 The reject H 0 p-value=p(f (1, 2) > F ) = (1 P(F (1,8) < 72.72)) = (1 0.9999) = 0.0001 < 0.05, the we reject H 0. 27
Aalysis of Variace Source DF Adj SS Adj MS F-Value P-Value Regressio 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 t = 8.528, (t ) 2 = (8.528) 2 = 72.72 = F Q2.26. Refer to Plastic hardess Problem 1.22. a. Set up the ANOVA table. b. Test by meas of a F test whether or ot there is a liear associatio betwee the hardess of the plastic ad the elapsed time. Use a =.01. State the alteratives, decisio rule, ad coclusio. Aalysis of Variace Source DF Adj SS Adj MS F-Value P-Value Regressio 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 28
α = 0.01 1. Hypothesis H 0 : β 1 = 0 H 1 : β 1 0 2. Test statistic F = 506.51 3. Decisio: p-value=0.000<0.01, the we reject H 0. 29
Prove that 2 Q = ε i = (Y i β 0 β 1 X i ) 2 Q = 0, Q = 0 β 0 β 1 Q = 2 (Y β i b 0 b 1 X i ) ( 1) = 0 0 (Y i b 0 b 1 X i ) = 0 (Y i ) (b 0 ) (b 1 X i ) = 0 (Y i ) b 0 b 1 (X i ) = 0 (Y i ) = b 0 + b 1 (X i ) (1) Q = 2 [(Y β i b 0 b 1 X i )( X i )] = 0 0 (Y i X i b 0 X i b 1 X 2 i ) = 0 30
(Y i X i ) b 0 (X i ) b 1 (X 2 i ) = 0 (Y i X i ) = b 0 (X i ) + b 1 (X 2 i ) (2) By solvig 1 ad 2 together (Y i ) = b 0 + b 1 (X i ) (Y i X i ) = b 0 (X i ) + b 1 (X 2 i ) From 1 Y = b 0 + b 1 X b 0 = Y b 1 X (Y i X i ) = (Y β 1 X ) (X i ) + b 1 (X 2 i ) (Y i X i ) = Y (X i ) + b 1 [ (X 2 i ) X (X i )] (Y i X i ) Y (X i ) = b 1 [ (X 2 i ) X (X i )] 31
(Y i X i ) Y X = b 1 [ (X 2 i ) X 2] b 1 = (Y ix i ) [ (X 2 i ) ( X i ) 2 (X i 2 ) (X i X )(Y i Y ) (X i X ) 2 ( (X i X ) 2 ) 2 Y X X 2 ] = (X i X )(Y i Y ) (X i X ) 2 = (X i X )(Y i Y ) (X i X ) 2 (X i X ) 4 e i = 0 e i = (Y i Y ) i = (Y i b 0 b 1 X i ) = (Y i Y + b 1 X b 1 X i ) = (Y i Y ) b 1 (X i X ) = 0 0 = 0 32
b 0 = Y b 1 X e i X i = 0, e i X i = [(Y i b 0 b 1 X i )X i ] = (Y i X i b 0 X i b 1 X 2 i ) = (Y i X i ) b 0 (X i ) b 1 (X 2 i ) = (Y i X i ) (Y b 1 X ) (X i ) b 1 (X 2 i ) = (Y i X i ) Y (X i ) + b 1 X (X i ) b 1 (X 2 i ) = (Y i X i ) Y X + b 1 X 2 b 1 (X 2 i ) = [ (Y i X i ) Y X ] b 1 [ (X 2 i ) X 2] = (X i X )(Y i Y ) b 1 (X i X ) 2 = (X i X )(Y i Y ) Y i = (b 0 + b 1 X i ) = (Y b 1 X + b 1 X i ) = Y + b 1 (X i X ) = Y i (X i X )(Y i Y ) (X i X ) 2 (X i X ) 2 = 0 Var ( (X i X )(Y i Y ) (X i X ) 2 ) (X i X )(Y i Y ) (X i X ) 2 = (X i X )Y i Y (X i X ) (X i X ) 2 = (X i X )Y i (X i X ) 2 33
Var ( (X i X )(Y i Y ) (X i X ) 2 ) = Var ( (X i X )Y i (X i X ) 2 ) = Var ( (X i X ) (X i X ) 2 = (X i X ) 2 ( (X i X ) 2 ) 2 σ2 (X = σ 2 i X ) 2 ( (X i X ) 2 ) 2 = σ 2 (X i X ) 2 Y i (X i X ) ) = ( (X i X ) 2 2 ) Var(Y i ) Prove that SSTo=SSR+SSE. L.H.S= SSTo = (Y i Y ) 2 = (Y i Y i + Y i Y ) 2 = ((Y i Y ) i + (Y i Y )) 2 = [(Y i Y ) 2 i + (Y i Y ) 2 + 2(Y i )(Y i i Y )] (Y i Y ) i = e i (Y i )(Y i i Y ) e i Y i The = e i = 0 = e i (Y i Y ) (Y i )(Y i i Y ) = 0 The = (Y i Y ) 2 i = e i Y i Y e i + (Y i Y ) 2 34 + 2 (Y i )(Y i i Y )
SSTo = (Y i Y ) 2 i (Y i Y ) 2 i + (Y i Y ) 2 = SSE & (Y i Y ) 2 SSTo = SSR + SSE = L. H. S = SSR 35
Chapter 2 2.13 Refer to Grade poit average. Calculate R 2. What proportio of the variatio i Y is accouted for by itroducig X ito the regressio model? From page 98 X = 24. 725, =120 (X i X ) 2 = 2379. 925 Aalysis of Variace Source DF Adj SS Adj MS F-Value P-Value Regressio 1 SSR=3.588 3.5878 9.24 0.003 Xi 1 3.588 3.5878 9.24 0.003 Error 118 SSE=45.818 MSE=0.3883 Lack-of-Fit 19 6.486 0.3414 0.86 0.632 Pure Error 99 39.332 0.3973 Total 119 SSTo=49.405 Model Summary S R-sq R-sq(adj) R-sq(pred) 0.623125 7.26% 6.48% 3.63% R 2 = SSR SSTo = 3.588 49.405 = 0.0726 R 2 = 1 SSE SSTo = 1 45.818 = 1 0.9274 = 0.0726 49.405 This meas that 7.26% of chage i the mea freshma OPA for studets is by ACT test score 36
a. Obtai a 95 percet iterval estimate of the mea freshma OPA for studets whose ACT test score is 28. Iterpret your cofidece iterval. From page 76- to 79 Y ĥ = b 0 + b 1 X h s 2 (Y ĥ ) = MSE ( 1 + (X h X ) 2 (X i X ) 2 ) Y ĥ ± t (1 α 2 ; 2) s(y ĥ) α = 0.05, α 2 = 0.025 At X h = 28 Y ĥ = 2.114 + 0.0388 (28) = 3.2012 s 2 (Y ĥ ) = MSE ( 1 + (X h X ) 2 (X i X ) 2 ) s 2 (Y ĥ ) = MSE ( 1 + (X h X ) 2 (X i X ) 2 ) = 0. 3883 ( 1 120 s(y ĥ ) = 0.007776 = 0.0706 t (1 α ; 2) = t(0.975; 118) = 1.9807 2 3.22012 ± 1.9807(0.0706) (28 24.725)2 + ) = 0.004986 2379.925 3.0614 < E(Y h ) < 3.3410 37
b. Mary Joes obtaied a score of 28 o the etrace test. Predict her freshma OPA-usig a 95 percet predictio iterval. Iterpret your predictio iterval. s 2 (Y ) ew = MSE (1 + 1 + (X h X ) 2 (X i X ) 2 ) = 0. 3883 (1 + 1 120 s(y ) ew = 0.6271 3.22012 ± 1.9807(0.6271) s 2 (Y ) ew = MSE (1 + 1 + (X h X ) 2 (X i X ) 2 ) Y ĥ ± t (1 α 2 ; 2) s(y ) ew (28 24.725)2 + ) = 0.39328 2379.925 1.9594 < Y h(ew) < 4.4430 c. Is the predictio iterval i part (b) wider tha the cofidece iterval i part (a)? Should it be? هل فترة الثقة للتنبؤ في الجزء )ب( أوسع من فترة الثقة في الجزء )أ( هل يجب أن تكون Yes, Yes 38
2.15. Refer to Airfreight breakage Problem 1.21. 10 X = 1, (X i X ) 2 = 10 ANOVA TABLE Source of Variatio d.f SS MS F p-value Regressio 1 SSR=160 MSR = 160 72.72 0.00 Error 8 SSE=17.6 MSE = 2.2 Total 9 SSTo= 177.6 a. Because of chages i airlie routes, shipmets may have to be trasferred more frequetly tha i the past. Estimate the mea breakage for the followig umbers of trasfers: X = 2, 4. Use separate 99 percet cofidece itervals. Iterpret your results. At X h = 2 Y ĥ = 10.2 + 4 (2) = 18.2 s 2 (Y ĥ ) = MSE ( 1 + (X h X ) 2 (X i X ) 2 ) = 2. 2 ( 1 10 s(y ĥ ) = 0.44 = 0.6633 t (1 α ; 2) = t(0.995; 8) = 3.355 2 18.2 ± 3.355(0.6633) At X h = 4 (2 1)2 + ) = 0.44 10 15.976 < E(Y h ) < 20.424 39
Y ĥ = 10.2 + 4 (4) = 26.2 s 2 (Y ĥ ) = MSE ( 1 + (X h X ) 2 (X i X ) 2 ) = 2. 2 ( 1 10 s(y ĥ ) = 2.2 = 1.483 t (1 α ; 2) = t(0.995; 8) = 3.355 2 26.2 ± 3.355(1.483) (4 1)2 + ) = 2.2 10 12.748 < E(Y h ) < 23.652 We coclude that the mea umber of ampules foud to be broke upo arrival whe 2 trasfers from oe aircraft to aother over the shipmet route of 2 are produced is somewhere betwee 15.976 ad 20.424 ampules أن متوسط عدد أمبوالت وجدت منكسره عند وصولهم عندما تم نقله عبر 2 مرات من طائرة واحدة إلى آخر عبر مسار الشحنة, بين 15.976 و 20.424 أمبوله. We coclude that the mea umber of ampules foud to be broke upo arrival whe 4 trasfers from oe aircraft to aother over the shipmet route are produced is somewhere betwee 12.748 ad 23.652 ampules. أن متوسط عدد أمبوالت وجدت منكسره عند وصولهم عندما تم نقله عبر 4 مرات من طائرة واحدة إلى آخر عبر مسار الشحنة, بين 12.748 و 23.652 أمبوله. b. The ext shipmet will etail two trasfers. Obtai a 99 percet predictio iterval for the umber of broke ampules for this shipmet. Iterpret your predictio iterval. s 2 (Y ) ew = MSE (1 + 1 + (X h X ) 2 (X i X ) 2 ) = 2. 2 (1 + 1 (2 1)2 + ) = 2.64 10 10 s(y ĥ ) = 2.64 = 1.6248 18.2 ± 3.355(1.6248) 40
12.748 < Y h(ew) < 23.652 With cofidece coefficiet 0.99, we predict that the mea umber of ampules foud to be broke upo arrival whe 2 trasfers from oe aircraft to aother over the shipmet route of 2 are produced is somewhere betwee 12.748 ad 23.652 ampules. Refer to Plastic hardess. Graph Boxplot simple X ok Boxplot of Xi 40 35 30 Xi 25 20 15 41
Graph Dotplot simple X ok Dotplot of Xi 16 20 24 28 32 36 40 Xi 42
Graph time series simple X ok Time Series Plot of Xi 40 35 30 Xi 25 20 15 2 4 6 8 Idex 10 12 14 16 43
For test ormality of residuals Graph probability plot sigle (distributio Normal) X ok If p-value >0.05, the it is ormal 44
Refer to Grade poit average. Graph Boxplot simple X ok Boxplot of Xi 35 30 Xi 25 20 15 45
Graph Dotplot simple X ok Dotplot of Xi 15 18 21 24 27 30 33 Xi 46
Graph time series simple X ok Time Series Plot of Xi 35 30 Xi 25 20 15 1 12 24 36 48 60 Idex 72 84 96 108 120 47
For test ormality of residuals Graph probability plot sigle (distributio Normal) X ok At 0.01 it is ormal but at 0.05, it is ot ormal 48
Chapter 5 Q5.1. For the matrices below, obtai (1) A +B, (2) A - B, (3) AC, (4) AB', (5) B'A. 1 4 1 3 A = [ 2 6] B = [ 1 4] C = [ 3 8 1 5 4 0 ] 3 8 2 5 Solutio: 1 4 1 3 1 + 1 4 + 3 2 7 A + B = [ 2 6] + [ 1 4] = [ 2 + 1 6 + 4] = [ 3 10] 3 8 2 5 3 + 2 8 + 5 5 13 1 4 1 3 1 1 4 3 0 1 A B = [ 2 6] [ 1 4] = [ 2 1 6 4] = [ 1 2] 3 8 2 5 3 2 8 5 1 3 1 4 AC = [ 2 6] [ 3 8 1 1 3 + 4 5 1 8 + 4 4 1 1 + 4 0 23 24 1 5 4 0 ] = [ 2 3 + 6 5 2 8 + 6 4 2 1 + 6 0] = [ 36 40 2] 3 8 3 3 + 8 5 3 8 + 8 4 3 1 + 8 0 49 56 3 B = [ 1 1 2 3 4 5 ] 1 4 (A 3 2 B 2 3 ) 3 3 = [ 2 6] [ 1 1 2 1 1 + 4 3 1 1 + 4 4 1 2 + 4 5 13 17 22 3 4 5 ] = [ 2 1 + 6 3 2 1 + 6 4 2 2 + 6 5] = [ 20 26 34] 3 8 3 1 + 8 3 3 1 + 8 4 3 2 + 8 5 27 35 46 (B 2 3 A 3 2 ) 2 2 = [ 1 1 2 1 4 3 4 5 ] [ 2 6] = [ 1 1 + 1 2 + 2 3 1 4 + 1 6 + 2 8 3 1 + 4 2 + 5 3 3 4 + 4 6 + 5 8 ] = [ 9 26 26 76 ] 3 8 49
Q5.4. Flavor deterioratio. The results show below were obtaied i a small-scale experimet to study the relatio betwee storage temperature (X) ad umber of weeks before flavour deterioratio of a food product begis to occur (Y). i 1 2 3 4 5 X i 8 4 0-4 -8 Y i 7.8 9.0 10.2 11.0 11.7 ο F of Assume that first-order regressio model (2.1) is applicable. Usig matrix methods, fid (1) Y Y, (2) X X, (3) X Y. Y 1 = X 2 Β 2 1 + ε 1 E(Y 1 ) = X 2 Β 2 1 Β 2 1 = (X X) 1 X Y V(Β) = MSE(X X) 1 MSE = e e 2 C1 C2 C3 1 8 7.8 1 2 4 9.0 1 3 0 10.2 1 4-4 11.0 1 5-8 11.7 1 1 8 7.8 1 4 9.0 X = 1 0, Y = 10.2, X = [ 1 1 1 1 1 1 4 11.0 8 4 0 4 8 ] [ 1 8] [ 11.7] 50
X i (X X) 2 2 = [ ] = [1 1 1 1 1 2 X i X 8 4 0 4 8 ] i [ = [ 5 0 0 160 ] (X X) 1 = 1 2 X i X i [ ] X i = X i 2 X i X i = 5 160 0 = 800 (X X) 1 = 1 800 [160 0 0 5 ] = [0.2 0 0 0.00625 ] 1 8 1 4 1 0 1 4 1 8] 7.8 9.0 10.2 Y i (X 2 Y 1 ) 2 1 = [ ] = [ 1 1 1 1 1 X i Y 8 4 0 4 8 ] i 11.0 [ 11.7] Β 2 1 = (X X) 1 X Y = [ 0.2 0 0 0.00625 ] [ 49.7 39.2 ] = [ 0.2 49.7 + 0 39.2 0 49.7 + 0.00625 39.2 ] = [ 9.94 0.245 ] = [ 1 + 1 + 1 + 1 + 1 8 + 4 + 0 4 8 8 + 4 + 0 4 8 8 8 + 4 4 + 0 0 + 4 4 + 8 8 ] 7.8 + 9 + 10.2 + 11 + 11.7 = [ 8 7.8 + 4 9 + 0 10.2 4 11 8 11.7 ] = [ 49.7 39.2 ] Y = 9.940 0.245X 1 8 9.94 0.245 8 7.98 1 4 Y 1 = X 2 Β 2 1 = 1 0 [ 9.94 9.94 0.245 4 8.96 1 4 0.245 ] = 9.94 0.245 0 = 9.94 9.94 + 0.245 4 10.92 [ 1 8] [ 9.94 + 0.245 8] [ 11.9 ] 7.8 7.98 7.8 7.98 9.0 8.96 9 8.96 e 1 = Y 1 Y 1 = 10.2 9.94 = 10.2 9.94 = 11.0 10.92 11 10.92 [ 11.7] [ 11.9 ] [ 11.7 11.9] [ 0.18 0.04 0.26 0.08 0.2 ] 51
e 1 e 1 = [ e 2 i ] = [ 0.18 0.04 0.26 0.08 0.2] 0.18 0.04 0.26 0.08 = [0.148] [ 0.2 ] MSE = 0.148 = 0.049333 3 V(Β) = MSE(X X) 1 = 0.049333 [ 0.2 0 0 0.00625 ] = [0.009867 0 0 0.000308 ] V [ β 0 β ] ) 0 cov(β, 0 β ) 1 1 cov(β, 0 β ) 1 Var(β ) 1 7.8 9.0 Y 1 Y 1 = [ y 2 i ] = [7.8 9.0 10.2 11.0 11.7] 10.2 = [503.77] 11.0 [ 11.7] MTB > Copy C3 C1 m1 MTB > Prit m1 X 2 Data Display Matrix M1 1 8 1 4 1 0 X 2 1-4 1-8 MTB > tra m1 m2 MTB > prit m2 X 2 52
Data Display Matrix M2 1 1 1 1 1 8 4 0-4 -8 X 2 MTB > mult m2 m1 m3 MTB > prit m3 (X X) 2 2 Data Display Matrix M3 (X X) 2 2 = [ X i X i X i 2 ] 5 0 0 160 MTB > iver m3 m4 MTB > prit m4 1 (X X) 2 2 Data Display Matrix M4 0.2 0.00000 0.0 0.00625 MTB > copy c2 m5 MTB > Prit m5 Y 1 53
Data Display Matrix M5 7.8 9.0 Y 1 10.2 11.0 11.7 MTB > mult m2 m5 m6 MTB > prit m6 X 2 Y 1 = X Y 2 1 Data Display Matrix M6 49.7-39.2 MTB > mult m4 m6 m7 MTB > prit m7 (X X) 1 2 2 (X Y) 2 1 = Β 2 1 Data Display Matrix M7 9.940 Β 2 1-0.245 Y = 9.940 0.245X 54
MTB > tra m5 m13 MTB > prit m13 Data Display Matrix M13 7.8 9 10.2 11 11.7 MTB > mult m13 m5 m14 Aswer = 503.7700 MTB > mult m1 m7 m8 MTB > prit m8 Y 1 = X 2 Β 2 1 Data Display Matrix M8 7.98 8.96 9.94 10.92 11.90 MTB > copy m8 c4 55
MTB > Let c5 = 'y'-c4 MTB > copy c5 m9 e MTB > tra m9 m10 MTB > prit m10 e Data Display Matrix M10-0.18 0.04 0.26 0.08-0.2 MTB > mult m10 m9 m11 e e Aswer = 0.1480 MSE=0.1480/3=0.049333 MSE = e e 2 MTB > mult 0.049333 m4 m12 MTB > prit m12 V(Β) = MSE(X X) 1 Data Display Matrix M12 0.0098666 0.0000000 0.0000000 0.0003083 56
V [ β 0 β ] = [ Var(β ) 0 cov(β, 0 β ) 1 1 cov(β, 0 β ) 1 Var(β ) ] 1 Regressio Aalysis: y versus x Aalysis of Variace Source DF Adj SS Adj MS F-Value P-Value Regressio 1 9.6040 9.60400 194.68 0.001 x 1 9.6040 9.60400 194.68 0.001 Error 3 0.1480 0.04933 Total 4 9.7520 Model Summary S R-sq R-sq(adj) R-sq(pred) 0.222111 98.48% 97.98% 94.11% Coefficiets Term Coef SE Coef T-Value P-Value VIF Costat 9.9400 0.0993 100.07 0.000 x -0.2450 0.0176-13.95 0.001 1.00 Regressio Equatio y = 9.9400-0.2450 x 57
H.W Q5.2 For the matrices below, obtai (1) A + C, (2) A C, (3) B A, (4) AC, (5) C A. 2 1 6 3 8 A = [ 3 5 ] B = [ 9 ] C = [ 8 6 ] 5 7 3 5 1 4 8 1 2 4 Q5.5 Cosumer fiace. The data below show, for a cosumer fiace compay operatig i six cities, the umber of competig loa compaies operatig i the city (X) ad the umber per thousad of the compay's loas made i that city that are curretly deliquet (Y); i 1 2 3 4 5 6 X i 4 1 2 3 3 4 Y i 16 5 10 15 13 22 Assume that first-order regressio model (2.1) is applicable. Usig matrix methods, fid (1) Y Y, (2) X X, (3) X Y. 58