Quiz Practice Questions 2 (Attendance 4) for Statistics 512Applied Regression AnalysisMaterial Covered: Chapters 4,5 Neter et al. and KuhnBy: Friday, 26th September, Fall 2003These are practice questions for the quiz. The quiz (not the practice questions) isworth 5% and marked out of 5 points. One or more questions is closely, but notnecessarily exactly, related to one or more of these questions will appear on the quiz.These practice questions are not to be handed in. Quizzes are to be done using Vistaon the Internet before 4am of the date of the quiz. Vista will not allow any quizto be done late. It is highly recommended that you complete this practice quiz, byhand, before logging onto Vista. The quiz is an individual one which means thateach student does this quiz by themselves without help from others.1. Applied Linear Statistical Models (Neter et al.) Questions.Chapter Problem(s) hints4, pages 171–174 4.5, 4.9, 4.18 Plastic hardness4.14, 4.15 Grade point average5, pages 211–214 5.1, 5.14, 5.17 various5.7, 5.26 Plastic hardness2. Applied Statistics and the SAS Programming Language (Cody and Smith) Ques-tions.No questions.(4.5) Plastic Hardness, family CIs for β0, β1, qz2-4-5-plastic-k-family-par-ciBoth SAS and the TI–83 calculator need to be used together here; that is, SASdoes not give you the complete answer.*PRACTICE QUIZ 2, 4-5, FAMILY CIs;*PLASTIC HARDNESS DATA, PAGES 171-174;DATA HARDVTIME; INPUT HARD TIME;DATALINES;199 16205 16196 16200 16218 24220 24215 24223 24237 32234 32235 32230 32250 40248 40253 40246 40;PROC REG DATA=HARDVTIME OUTEST=EST TABLEOUT ALPHA=0.1; TITLE '4.5(A) BONFERRONI JOINT CIs FOR B0 AND B1'; MODEL HARD = TIME; OUTPUT OUT=OUTPLOT PREDICTED=PRED RESIDUAL=RESID;RUN;PROC UNIVARIATE DATA=HARDVTIME; TITLE '4.5(B) AVERAGE OF X'; VAR TIME;RUN;QUIT;(a) Family Bonferroni Regression Parameters 90% CIFrom TI–83 (INVT 14 ENTER 0.975 ENTER)B = t(1 − α/2g; n − 2) = t(1 − 0.10/2(2); 16 − 2) = t(0.975; 14) = 2.145From SAS,1. Bonferroni CI for β0:b0= 168.6 and s{b0} = 0.6633,b0± Bs{b0} = 168.6 ± 2.145(2.657) = (162.901, 174.299)2. Bonferroni CI for β1:b1= 4 and s{b1} = 0.4690b1± Bs{b1} = 2.0344 ± 2.145(0.0904) = (1.840, 2.228)(b) Since the covariance1between intercept and slope isσ{b0, b1} = −Xσ2{b1}and, in this case, X = 28, b0and b1are negatively correlated.Also, the fact that the two parameters are negatively correlated is notreflected in the 2–family Bonferroni confidence interval calculated above.1(4.5), page 155, Neter et al.(c) In 90 percent of the samples, the two CIs capture (include) both parameters,whereas in 10 percent of the samples, either one or both of the CIs do notcapture (or include) their associated parameters.(4.9) Plastic Hardness, family CIs for expected responses, qz2-4-9-plastic-k-family-exp-ci*PRACTICE QUIZ 2, 4-9, K FAMILY RESPONSE CIs;*PLASTIC HARDNESS DATA, PAGES 171-174;DATA HARDVTIMEX; INPUT HARD TIME;DATALINES;199 16205 16196 16200 16218 24220 24215 24223 24237 32234 32235 32230 32250 40248 40253 40246 40. 20. 30. 40;DATA HARDVTIME X; SET HARDVTIMEX; IF HARD NE . THEN OUTPUT HARDVTIME; ELSE OUTPUT X;RUN;PROC REG DATA=HARDVTIME ALPHA=0.1 NOPRINT; TITLE '4.9(A) BONFERRONI JOINT CIs FOR MEAN'; MODEL HARD = TIME; OUTPUT OUT=OUTPLOT PREDICTED=PRED RESIDUAL=RESID;RUN;PROC REG DATA=HARDVTIMEX; MODEL HARD = TIME; OUTPUT OUT=PRED_DS(WHERE=(HARD=.)) P=PHAT STDP=STDP STDI=STDI;RUN;PROC PRINT DATA=PRED_DS;RUN;QUIT;(a) Family Bonferroni Responses 90% CIFrom TI–83B = t(1 − α/2g; n − 2) = t(1 − 0.10/2(3); 16 − 2) = t(0.9833; 14) = 2.360(INVT 14 ENTER 0.9833 ENTER)From SAS,1. Xh= 20:ˆYh= 209.2875 and s{ˆYh} = 1.0847ˆYh± Bs{ˆYh} = 209.2875 ± 2.360(1.0847) = (206.727, 211.847)2. Xh= 30:ˆYh= 229.6312 and s{ˆYh} = 0.8285ˆYh± Bs{ˆYh} = 229.6312 ± 2.360(0.8285) = (227.676, 231.586)3. Xh= 40:ˆYh= 249.9750 and s{ˆYh} = 1.3529ˆYh± Bs{ˆYh} = 249.9750 ± 2.360(1.3529) = (246.782, 253.166)In 90 percent of the samples, the three CIs capture (include) all expectedresponse parameters, whereas in 10 percent of the samples, either one ormore of the CIs do not capture (or include) their associated parameters.(b) Family Bonferroni vs Working–Hotelling Responses 90% CIFor Bonferroni, B = 2.360For Working–Hotelling,W =q2F (1 − α; 2, n − 2) =q2F (1 − 0.10; 2, 16 − 2) = 2.340(INVF 2 ENTER 14 ENTER 0.90 ENTER, then multiply by 2 and findthe square root)So, since B = 3.360 > W = 2.340, the family Bonferroni gives wider (lessefficient) CIs than the family Working–Hotelling 90% CIs.(c) Family Bonferroni vs Working–Hotelling Responses 90% CIFrom TI–83W = 2.340(INVF 2 ENTER 14 ENTER 0.90 ENTER)B = t(1 − α/2g; n − 2) = t(1 − 0.10/2(2); 16 − 2) = t(0.975; 14) = 2.145(INVT 14 ENTER 0.975 ENTER)Since W = 2.340 > B = 2.145, use B; notice, in this case, that two Xh,rather than three Xhare used and this is why B changes.From SAS,1. Xh= 30:ˆYh= 229.6312 and s{ˆYh} = 3.3385ˆYh± Bs{ˆYh} = 229.6312 ± 2.145(3.3385) = (222.470, 236.792)2. Xh= 40:ˆYh= 249.9750 and s{ˆYh} = 3.5056ˆYh± Bs{ˆYh} = 249.9750 ± 2.145(3.5056) = (242.455, 257.495)(4.14) Grade point average; origin model, qz2-4-14,15-gpa-origin*PRACTICE QUIZ 2, 4-14,15, PAGES 171-175;DATA GPAVSCOREX; INPUT GPA SCORE; LACKFIT = GPA;DATALINES;3.1 5.52.3 4.83 4.71.9 3.92.5 4.53.7 6.23.4 62.6 5.22.8 4.71.6 4.32 4.92.9 5.42.3 53.2 6.31.8 4.61.4 4.32 53.8 5.92.2 4.11.5 4.7. 4.7;*4.14, REGRESSION THROUGH ORIGIN;DATA GPAVSCORE X; SET GPAVSCOREX; IF GPA NE . THEN OUTPUT GPAVSCORE; ELSE OUTPUT X;RUN;PROC REG DATA=GPAVSCORE OUTEST=EST TABLEOUT ALPHA=0.05 ALL; TITLE1 '4.14(A),(B) REGRESSION THROUGH ORIGIN, INTERVAL EST B1'; MODEL GPA = SCORE / NOINT; OUTPUT OUT=OUTPLOT PREDICTED=PRED RESIDUAL=RESID;RUN;PROC REG DATA=GPAVSCOREX; TITLE '4.14(C) CONFIDENCE INTERVAL MEAN RESPONSE'; MODEL GPA = SCORE / NOINT; OUTPUT OUT=PRED_DS(WHERE=(GPA=.)) P=PHAT STDP=STDP STDI=STDI;RUN;PROC PRINT DATA=PRED_DS;RUN;*4.15, DIAGNOSTICS OF REGRESSION THROUGH ORIGIN;SYMBOL1 V=STAR C=BLACK;SYMBOL2 V=DOT C=BLACK I=R;PROC GPLOT DATA=OUTPLOT; TITLE '4.15(A) GPA VS SCORE, REGRESSION'; PLOT GPA*SCORE PRED*SCORE / OVERLAY;RUN;PROC GPLOT DATA=OUTPLOT; TITLE '4.15(B) RESIDUALS VS PREDICTED PLOT'; PLOT RESID*PRED;RUN;PROC GLM DATA=GPAVSCORE; TITLE '4.15(C) LACK OF FIT TEST'; CLASS SCORE; MODEL GPA = SCORE / SS1 NOINT;RUN;QUIT;(a) The regression through origin isˆY = 0.5061X(b) From the