Quiz Practice Questions 4 (Attendance 8) for Statistics 512Applied Regression AnalysisMaterial Covered: Chapter 9 Neter et al. and KuhnBy: Friday, 24th October, 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.Applied Linear Statistical Models(Neter et al.) Questions.Chapter Problem(s) hints9, pages 392–399 9.6, 9.10, 9.16 Chemical Shipment dataApplied Statistics and the SAS Programming Language(Cody and Smith) Questions.No questions.(9.6) Chemical Shipment data, qz4-9-6-chem-partial*PRACTICE QUIZ 4, PARTIAL RESID/REGR PLOTS, PP 392-399;DATA CHEMICAL; INPUT NUMBER WEIGHT TIME;DATALINES;7 5.11 5818 16.72 1525 3.2 4114 7.03 9311 10.98 1015 4.04 3823 22.07 2039 7.03 7816 10.62 1175 4.76 4417 11.02 12112 9.51 1126 3.79 5012 6.45 828 4.6 4815 13.86 12717 13.03 14021 15.21 1556 3.64 3911 9.57 90;PROC REG DATA = CHEMICAL; TITLE '9.6(A,B) PARTIAL RESIDUAL PLOTS'; MODEL TIME = NUMBER WEIGHT / PARTIAL;RUN;PROC REG DATA = CHEMICAL; TITLE '9.6(C) REGRESSION OF PARTIAL RESIDUALS'; MODEL TIME WEIGHT = NUMBER; OUTPUT OUT = OUTPLOT R = RTIME RWEIGHT;RUN;PROC REG DATA = OUTPLOT; TITLE '9.6(C) REGRESSION OF PARTIAL RESIDUALS'; MODEL RTIME = RWEIGHT;RUN;SYMBOL1 V=STAR C=BLACK;PROC GPLOT DATA=OUTPLOT; TITLE '9.6(C) PARTIAL RES Y|X1 VS PARTIAL RES X2|X1'; PLOT RTIME*RWEIGHT; RUN;QUIT;(a) Partial Residual Plots.The partial residual plots are given in the SAS output.One plot is e(Y |X1) vs e(X2|X1) (TIME vs NUMBER)The other plot is e(Y |X2) vs e(X1|X2) (TIME vs WEIGHT)(b) Interpretation of Partial Residual Plots.The attached partial residual plots for the two regression coefficients (ig-nore the one against the intercept) both indicate a linear term in the sec-ond variable (either X1or X2) may be a useful addition to the regressionmodel already containing the first variable (either X2or X1). Both plotsalso seem to reveal one outlying point.(c) Regression of Partial Residuals.The regression of Y on X1isˆY (X1) = −2.0558 + 8.10973X1and so the partial residuals ei(Y |X1) = Yi−ˆYi(X1) (which is plotted above)The regression of X2on X1isˆX2(X1) = −1.05916 + 0.85472X1and so ei(X2|X1) = Yi−ˆYi(X1)The regression of the residuals are e(Y |X1) = 5.07969e(X2|X1)This is the same partial residual plot as one of the partial residual plotsgiven in part (a); it appears as the “TIME versus WEIGHT” plot.(9.10) Chemical Shipment data, qz4-9-10-chem-leverage*PRACTICE QUIZ 4, LEVERAGE, PP 392-399;DATA CHEMICAL; INPUT NUMBER WEIGHT TIME;DATALINES;7 5.11 5818 16.72 1525 3.2 4114 7.03 9311 10.98 1015 4.04 3823 22.07 2039 7.03 7816 10.62 1175 4.76 4417 11.02 12112 9.51 1126 3.79 5012 6.45 828 4.6 4815 13.86 12717 13.03 14021 15.21 1556 3.64 3911 9.57 90;PROC REG DATA = CHEMICAL; TITLE '9.10(A) STUDENTIZED DELETED RESIDUALS'; TITLE2 '9.10(B) OUTLYING X USING LEVERAGES'; MODEL TIME = NUMBER WEIGHT / R INFLUENCE;RUN;PROC GPLOT DATA = CHEMICAL; TITLE '9.10(C) SCATTERPLOT, NEW OBS OUTLIER?'; PLOT NUMBER*WEIGHT / VREF = 0;RUN;*9.10(C) NEW X OUTLIER USING LEVERAGE?;PROC IML; USE CHEMICAL; READ ALL VAR {'NUMBER'} INTO X1; READ ALL VAR {'WEIGHT'} INTO X2; READ ALL VAR {'TIME'} INTO Y; N = NROW(X1); X = J(N,1,1)||X1||X2; C = INV(X`*X); XNEW = {1, 15, 8}; HNEW = XNEW`*C*XNEW; PRINT HNEW;RUN;*CHEMICAL DATA WITHOUT OBS 7;DATA CHEMICAL7; INPUT NUMBER WEIGHT TIME;DATALINES;7 5.11 5818 16.72 1525 3.2 4114 7.03 9311 10.98 1015 4.04 389 7.03 7816 10.62 1175 4.76 4417 11.02 12112 9.51 1126 3.79 5012 6.45 828 4.6 4815 13.86 12717 13.03 14021 15.21 1556 3.64 3911 9.57 90;*CHEMICAL DATA WITHOUT OBS 12;DATA CHEMICAL12; INPUT NUMBER WEIGHT TIME;DATALINES;7 5.11 5818 16.72 1525 3.2 4114 7.03 9311 10.98 1015 4.04 3823 22.07 2039 7.03 7816 10.62 1175 4.76 4417 11.02 1216 3.79 5012 6.45 828 4.6 4815 13.86 12717 13.03 14021 15.21 1556 3.64 3911 9.57 90;PROC IML; USE CHEMICAL; READ ALL VAR {'NUMBER'} INTO X1; READ ALL VAR {'WEIGHT'} INTO X2; READ ALL VAR {'TIME'} INTO Y; N = NROW(X1); X = J(N,1,1)||X1||X2; B = INV(X`*X)*X`*Y; YHAT = X*B; *REMOVE OBS 7, RECALC REGRESS, USE ORIG X!; USE CHEMICAL7; READ ALL VAR {'NUMBER'} INTO X71; READ ALL VAR {'WEIGHT'} INTO X72; READ ALL VAR {'TIME'} INTO Y7; N7 = NROW(X71); X7 = J(N7,1,1)||X71||X72; B7 = INV(X7`*X7)*X7`*Y7; YHAT7 = X*B7; *ABSOLUTE PERCENT DIFFERENCE FOR OBS 7; ABSPERDIFF7 = SUM(ABS((YHAT7 - YHAT)/YHAT)*100)/N; PRINT ABSPERDIFF7; *REMOVE OBS 12, RECALC REGRESS, USE ORIG X!; USE CHEMICAL12; READ ALL VAR {'NUMBER'} INTO X121; READ ALL VAR {'WEIGHT'} INTO X122; READ ALL VAR {'TIME'} INTO Y12; N12 = NROW(X121); X12 = J(N7,1,1)||X121||X122; B12 = INV(X12`*X12)*X12`*Y12; YHAT12 = X*B12; *ABSOLUTE PERCENT DIFFERENCE FOR OBS 12; ABSPERDIFF12 = SUM(ABS((YHAT12 - YHAT)/YHAT)*100)/N; PRINT ABSPERDIFF12;RUN;QUIT;(a) Studentized Deleted Residuals For Y Outliers.The studentized deleted residuals (RStudent in SAS output) are0.4268, −0.8005, 0.4815, . . . , −1.0258, −0.6141.The studentized deleted residual associated with observation 12, t12≈3.626, appears to be a Y outlier because it is over three (studentizeddeleted residual) standard deviations (3.626 to be exact) in size. Use theBonferroni outlier test procedure to investigate this.1. Statement.H0: (largest) observation Y12is