Solution: X = , Y = = =

Transcription

1 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 = , Y = =120 (X i X ) (Y i Y ) = =120 (X i X ) 2 = =120 (Y i Y ) 2 = b 1 = β 1 = =120 (X i X )(Y i Y ) =120 (X i X ) 2 = = b 0 = β 0 = Y b 1 X = =

2 Y = X At X=30 Y ĥ = (30) = whe the etrace test score icreases by oe poit, the mea respose icrease by 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 = , Y =

3 =120 (X i X ) (Y i Y ) = =120 (X i X ) 2 = =120 (Y i Y ) 2 = b 1 = β 1 = =120 (X i X )(Y i Y ) =120 (X i X ) 2 = b 0 = β 0 = Y b 1 X = Y = X At X=5 Y ĥ = (5) = 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

4 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 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 = =120 (X i X ) (Y i Y ) = 2604 =120 (X i X ) 2 =

5 =120 (Y i Y ) 2 = b 1 = β 1 = =120 (X i X )(Y i Y ) =120 (X i X ) 2 = b 0 = β 0 = Y b 1 X = Y = X At X=40 Y ĥ = (40) = Q1.24) Refer to Copier maiteace Problem 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 = e i 2 = Q σ 2 = e i 2 2 = = = MSE 43 σ = MSE =

6 Q1.25) (H.W) Refer to Airfreight breakage Problem 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 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 = X b. Obtai a poit estimate of the expected umber of broke ampules whe X = 1 trasfer is made. If X=1 The Y ĥ = (1) = c. Estimate the icrease i the expected umber of ampules broke whe there are 2 trasfers as compared to 1 trasfer. Y h2 = (2) = Y h1 = (1) = 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

7 If X=1 The Y ĥ = (1) = The we ca say the regressio lie goes through the poit (X, Y ) = (1, 14.2) 7

8 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

9 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 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 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 10

11 Regressio Aalysis: Yi versus Xi Aalysis of Variace Source DF Seq SS Cotributio Adj SS Adj MS F-Value P-Value Regressio % Xi % Error -2= % SSE= MSE= Lack-of-Fit %

12 Pure Error % Total % Model Summary S R-sq R-sq(adj) PRESS R-sq(pred) % 6.48% % Coefficiets Term Coef SE Coef 99% CI T-Value P-Value VIF Costat ( 1.274, 2.954) Xi (0.0054, ) Regressio Equatio Yi = 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 β 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

13 α = Hypothesis H 0 : β 1 = 0 H 1 : β Test statistic T 0 = b 1 β 10 = b 1 s(b 1 ) s(b 1 ) = = Decisio: Reject H 0 if T 0 > t α (1, 2), 3.04 > t (0.995,118) = 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

14 Q2.5. Refer to Copier maiteace Problem =45 = 45, X i = 230, Y i = 3432, X i = 1516, X i Y i SSE = = 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 = = 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 /(45 2) (b 1 ) = (X i X ) 2 = = s(b 1 ) = = t (1 α 2, 2) = t (0.95,43) = b 1 t (1 α 2, 2)s(b 1) = = b 1 + t (1 α 2, 2)s(b 1) = = β = = 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 State the alteratives, decisio rule, ad coclusio. What is the P-value of your test? α = Hypothesis 14

15 H 0 : β 1 = 0 H 1 : β Test statistic T 0 = b 1 β 10 = b 1 s(b 1 ) s(b 1 ) = = Decisio: Reject H 0 if T 0 > t α (1, 2), > t (0.995,43) = The reject H 0 p-value=2p(t ( 2) > T 0 ) = 2 (1 P(t ( 2) < )) = 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 State the alteratives, decisio rule, ad coclusio. What is the P-value of the test? α = Hypothesis H 0 : β 1 14 H 1 : β 1 > Test statistic T 0 = b 1 β 10 = b = = s(b 1 ) s(b 1 ) Decisio: Reject H 0 if T 0 > t (1 α, 2), > t (0.95,43) = The reject H 0 p-value=p(t ( 2) > T 0 ) = (1 P(t ( 2) < 2.143)) = ( ) = < 0.05, the we reject H 0. 15

16 Q2.6. Refer to Airfreight breakage Problem =10 X = 1, Y = 14.2, (X i X ) (Y i Y ) = (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 = = 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 = = = 0.22 t (1 α 2, 2) = t (0.975,8) = b 1 t (1 α 2, 2)s(b 1) = = b 1 + t (1 α 2, 2)s(b 1) = = β 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 State the alteratives, decisio rule, ad coclusio. What is the P-value of the test? α =

17 1. Hypothesis H 0 : β 1 = 0 H 1 : β Test statistic T 0 = b 1 β 10 = b 1 s(b 1 ) s(b 1 ) = = Decisio: Reject H 0 if T 0 > t α (1, 2), > t (0.975,8) = The reject H 0 p-value=2p(t ( 2) > T 0 ) = 2 (1 P(t (8) < 8.528)) = 2( ) < 0.05, the we reject H 0. Aalysis of Variace Source DF Seq SS Cotributio Adj SS Adj MS F-Value P-Value Regressio % Xi % Error % Lack-of-Fit % Pure Error % Total % Model Summary S R-sq R-sq(adj) PRESS R-sq(pred) % 88.85% % Coefficiets 17

18 Term Coef SE Coef 95% CI T-Value P-Value VIF Costat (8.670, ) Xi (2.918, 5.082) Regressio Equatio Yi = Xi H.W: Q2.7 Refer to Plastic hardess Problem 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 = State the alteratives, decisio rule, ad coclusio. What is the P-value of the test? 18

19 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

20 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 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

21 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

22 Tests Cocerig β 1 1. Hypothesis H 0 : β 0 = β 00 H 1 : β 0 β 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

23 1. Hypothesis H 0 : β 1 = 0 (No lier) H 1 : β 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 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 = State the alteratives, decisio rule, ad coclusio. What is the P-value of the test? α =

24 1. Hypothesis H 0 : β 0 9 H 1 : β 0 > 9 2. Test statistic T 0 = b 0 β 00 = s(b 0 ) = s 2 (β ) 0 = MSE ( 1 X 2 + (X i X ) 2 ) = 2.2 ( ) = 0.44 s(b 0 ) = Decisio: Reject H 0 if T 0 > t (1 α, 2), t (0.975,8) = The ot reject H 0 p-value=p(t ( 2) > T 0 ) = (1 P(t ( 2) < 1.809)) = ( ) = , 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) = β β

25 Aalysis of Variace Source DF Seq SS Cotributio Adj SS Adj MS F-Value P-Value Regressio % Xi % Error % Total % Coefficiets Term Coef SE Coef 95% CI T-Value P-Value VIF Costat (8.670, ) Xi (2.918, 5.082) Regressio Equatio Yi = Xi 25

26 Q2.25. Refer to Airfreight breakage Problem 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 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 = , 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

27 (Y i Y ) 2 i = 17.6 ANOVA TABLE Source of Variatio d.f SS MS F p-value Regressio 1 SSR= = 160 MSR = = Error 8 SSE=17.6 MSE = = 2.2 Total 9 SSTo= α = Hypothesis H 0 : β 1 = 0 H 1 : β Test statistic F = Decisio: Reject H 0 if F > F (1 α,1, 2), > 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)) = ( ) = < 0.05, the we reject H 0. 27

28 Aalysis of Variace Source DF Adj SS Adj MS F-Value P-Value Regressio Xi Error Lack-of-Fit Pure Error Total t = 8.528, (t ) 2 = (8.528) 2 = = F Q2.26. Refer to Plastic hardess Problem 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 Xi Error Lack-of-Fit Pure Error Total

29 α = Hypothesis H 0 : β 1 = 0 H 1 : β Test statistic F = Decisio: p-value=0.000<0.01, the we reject H 0. 29

30 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

31 (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

32 (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

33 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

34 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 ) (Y i )(Y i i Y )

35 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

36 Chapter 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 = , =120 (X i X ) 2 = Aalysis of Variace Source DF Adj SS Adj MS F-Value P-Value Regressio 1 SSR= Xi Error 118 SSE= MSE= Lack-of-Fit Pure Error Total 119 SSTo= Model Summary S R-sq R-sq(adj) R-sq(pred) % 6.48% 3.63% R 2 = SSR SSTo = = R 2 = 1 SSE SSTo = = = This meas that 7.26% of chage i the mea freshma OPA for studets is by ACT test score 36

37 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 = At X h = 28 Y ĥ = (28) = 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 ) = ( s(y ĥ ) = = t (1 α ; 2) = t(0.975; 118) = ± (0.0706) ( )2 + ) = < E(Y h ) <

38 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 ( (X h X ) 2 (X i X ) 2 ) = ( s(y ) ew = ± (0.6271) s 2 (Y ) ew = MSE ( (X h X ) 2 (X i X ) 2 ) Y ĥ ± t (1 α 2 ; 2) s(y ) ew ( )2 + ) = < Y h(ew) < 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

39 2.15. Refer to Airfreight breakage Problem X = 1, (X i X ) 2 = 10 ANOVA TABLE Source of Variatio d.f SS MS F p-value Regressio 1 SSR=160 MSR = Error 8 SSE=17.6 MSE = 2.2 Total 9 SSTo= 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 = (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 = t (1 α ; 2) = t(0.995; 8) = ± 3.355(0.6633) At X h = 4 (2 1)2 + ) = < E(Y h ) <

40 Y ĥ = (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 = t (1 α ; 2) = t(0.995; 8) = ± 3.355(1.483) (4 1)2 + ) = < E(Y h ) < 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 ad ampules أن متوسط عدد أمبوالت وجدت منكسره عند وصولهم عندما تم نقله عبر 2 مرات من طائرة واحدة إلى آخر عبر مسار الشحنة, بين و 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 ad ampules. أن متوسط عدد أمبوالت وجدت منكسره عند وصولهم عندما تم نقله عبر 4 مرات من طائرة واحدة إلى آخر عبر مسار الشحنة, بين و 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 ( (X h X ) 2 (X i X ) 2 ) = 2. 2 (1 + 1 (2 1)2 + ) = s(y ĥ ) = 2.64 = ± 3.355(1.6248) 40

41 < Y h(ew) < 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 ad ampules. Refer to Plastic hardess. Graph Boxplot simple X ok Boxplot of Xi Xi

42 Graph Dotplot simple X ok Dotplot of Xi Xi 42

43 Graph time series simple X ok Time Series Plot of Xi Xi Idex

44 For test ormality of residuals Graph probability plot sigle (distributio Normal) X ok If p-value >0.05, the it is ormal 44

45 Refer to Grade poit average. Graph Boxplot simple X ok Boxplot of Xi Xi

46 Graph Dotplot simple X ok Dotplot of Xi Xi 46

47 Graph time series simple X ok Time Series Plot of Xi Xi Idex

48 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

49 Chapter 5 Q5.1. For the matrices below, obtai (1) A +B, (2) A - B, (3) AC, (4) AB', (5) B'A A = [ 2 6] B = [ 1 4] C = [ ] Solutio: A + B = [ 2 6] + [ 1 4] = [ ] = [ 3 10] A B = [ 2 6] [ 1 4] = [ ] = [ 1 2] AC = [ 2 6] [ ] = [ ] = [ ] B = [ ] 1 4 (A 3 2 B 2 3 ) 3 3 = [ 2 6] [ ] = [ ] = [ ] (B 2 3 A 3 2 ) 2 2 = [ ] [ 2 6] = [ ] = [ ]

50 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 X i Y i 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 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 C X = 1 0, Y = 10.2, X = [ ] [ 1 8] [ 11.7] 50

51 X i (X X) 2 2 = [ ] = [ X i X ] i [ = [ ] (X X) 1 = 1 2 X i X i [ ] X i = X i 2 X i X i = = 800 (X X) 1 = [ ] = [ ] ] Y i (X 2 Y 1 ) 2 1 = [ ] = [ X i Y ] i 11.0 [ 11.7] Β 2 1 = (X X) 1 X Y = [ ] [ ] = [ ] = [ ] = [ ] = [ ] = [ ] Y = X Y 1 = X 2 Β 2 1 = 1 0 [ ] = = [ 1 8] [ ] [ 11.9 ] e 1 = Y 1 Y 1 = = = [ 11.7] [ 11.9 ] [ ] [ ] 51

52 e 1 e 1 = [ e 2 i ] = [ ] = [0.148] [ 0.2 ] MSE = = V(Β) = MSE(X X) 1 = [ ] = [ ] V [ β 0 β ] ) 0 cov(β, 0 β ) 1 1 cov(β, 0 β ) 1 Var(β ) Y 1 Y 1 = [ y 2 i ] = [ ] 10.2 = [503.77] 11.0 [ 11.7] MTB > Copy C3 C1 m1 MTB > Prit m1 X 2 Data Display Matrix M X MTB > tra m1 m2 MTB > prit m2 X 2 52

53 Data Display Matrix M 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 ] MTB > iver m3 m4 MTB > prit m4 1 (X X) 2 2 Data Display Matrix M MTB > copy c2 m5 MTB > Prit m5 Y 1 53

54 Data Display Matrix M Y MTB > mult m2 m5 m6 MTB > prit m6 X 2 Y 1 = X Y 2 1 Data Display Matrix M MTB > mult m4 m6 m7 MTB > prit m7 (X X) (X Y) 2 1 = Β 2 1 Data Display Matrix M Β Y = X 54

55 MTB > tra m5 m13 MTB > prit m13 Data Display Matrix M MTB > mult m13 m5 m14 Aswer = MTB > mult m1 m7 m8 MTB > prit m8 Y 1 = X 2 Β 2 1 Data Display Matrix M MTB > copy m8 c4 55

56 MTB > Let c5 = 'y'-c4 MTB > copy c5 m9 e MTB > tra m9 m10 MTB > prit m10 e Data Display Matrix M MTB > mult m10 m9 m11 e e Aswer = MSE=0.1480/3= MSE = e e 2 MTB > mult m4 m12 MTB > prit m12 V(Β) = MSE(X X) 1 Data Display Matrix M

57 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 x Error Total Model Summary S R-sq R-sq(adj) R-sq(pred) % 97.98% 94.11% Coefficiets Term Coef SE Coef T-Value P-Value VIF Costat x Regressio Equatio y = x 57

58 H.W Q5.2 For the matrices below, obtai (1) A + C, (2) A C, (3) B A, (4) AC, (5) C A A = [ 3 5 ] B = [ 9 ] C = [ 8 6 ] 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 X i Y i Assume that first-order regressio model (2.1) is applicable. Usig matrix methods, fid (1) Y Y, (2) X X, (3) X Y. 58