1Quiz Practice Questions 6 (Attendance 12) for Statistics 512Applied Regression AnalysisMaterial Covered: Chapter 13By: Friday, 21st November, 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.All hints are taken from Kuhn’s Workbook, unless otherwise specified.Chapter Problem(s) hints13, pages 559–566 13.1, 13.2 intrinsically linear?13.5,13.6,13.7,13.8,13.9 home computersPractice Quiz 6 (Attendance 12), 13.1 2(13.1) intrinsically linear?(a) f = exp(γ0+ γ1X)intrinsically linear becauseln Y = ln exp(γ0+ γ1X)Y0= γ0+ γ1Xwhere Y0= ln f(b) f = γ0+ γ1(γ2)X1γ3X2nonlinear because, there does not appear to be a transformation to convertf into the form,Y = γ0+ γ1X01+ γ2X02(c) f = γ0+γ1γ0Xnonlinear because, althoughY = γ0+ γ01Xwhere Y = f and γ01=γ1γ0, this is linear in the variables, not the parameters,where two parameters have been condensed into one.Practice Quiz 6 (Attendance 12), 13.2 3(13.2) intrinsically linear?(a) f = exp(γ0+ γ1logeX)intrinsically linear becauseln Y = ln exp(γ0+ γ1logeX)Y0= γ0+ γ1logeX= γ0+ γ1X0where Y0= ln f and X0= logeX. The resulting transformed functionis linear in the parameters, although not linear in either of the variables,(X, Y ).(b) f = γ0(X1)γ1(X2)γ2intrinsically linear becauseln Y = ln γ0(X1)γ1(X2)γ2Y0= ln γ0+ γ1ln X1+ γ2ln X2Y0= γ00+ γ1X01+ γ2X02where Y0= ln f, γ00= ln γ0, X01= ln X1and X02= ln X2(c) f = γ0− γ1(γ2)Xnonlinear there does not appear to be any transformation to make it linearin the variables.Practice Quiz 6 (Attendance 12), 13.5 4(13.5) Home Computers*PRACTICE QUIZ 6, 13.5 NONLINEAR REGRESSION;DATA COMPUTERS; INPUT CITYIND PERCBUY PRICE; L_PERCBUY = LOG(PERCBUY);DATALINES;0 0.9 10 0.8 2.50 0.65 50 0.46 100 0.34 200 0.26 300 0.17 400 0.15 500 0.06 750 0.04 1001 0.93 11 0.77 2.51 0.63 51 0.5 101 0.3 201 0.24 301 0.19 401 0.12 501 0.08 751 0.05 100;PROC REG DATA=COMPUTERS; TITLE '13.5(A) INITIAL ESTIMATES: LINEAR REGRESSION'; MODEL L_PERCBUY = PRICE;RUN;PROC NLIN DATA=COMPUTERS; TITLE '13.5(B) NONLINEAR REGRESSION, HOME COMPUTERS'; PARMS A=0 B=0.03020 C=0.70682; MODEL PERCBUY = A + C*EXP(-B*PRICE); OUTPUT OUT=PLOT P=PPERCBUY R=RESIDUALS;RUN;QUIT;(a) Initial Estimates.Initial estimates for the parameters, γ0, γ1and γ2, of the exponentialmodel,Yi= γ0+ γ2exp(−γ1Xi) + εiare obtained by using the linear regression,Y0i= β0+ β2Xiwhere Y0i= ln Yi, β0= ln γ2and β1= −γ1and so, from the SAS output, the initial estimates forγ0is: g(0)0= 0γ1is: g(0)1= −b1= −(−0.03020) = 0.03020γ2is: g(0)2= exp(b0) = exp(−0.34698) = 0.70682(b) Nonlinear Least Squares Analysis.From SAS, using the starting valuesg(0)0= 0, g(0)1= 0.03020 and g(0)2= 0.70682, givesg0= 0.08360, g1= 0.06412, g2= 0.83053In other words, the estimated nonlinear function isˆYi= 0.08360 + 0.83053 exp(−0.06412Xi)Practice Quiz 6 (Attendance 12), 13.6 5(13.6) Home Computers*PRACTICE QUIZ 6, 13.6 PLOTS OF NONLINEAR REGRESSION;DATA COMPUTERS; INPUT CITYIND PERCBUY PRICE; L_PERCBUY = LOG(PERCBUY);DATALINES;0 0.9 10 0.8 2.50 0.65 50 0.46 100 0.34 200 0.26 300 0.17 400 0.15 500 0.06 750 0.04 1001 0.93 11 0.77 2.51 0.63 51 0.5 101 0.3 201 0.24 301 0.19 401 0.12 501 0.08 751 0.05 100;PROC NLIN DATA=COMPUTERS OUTEST=EST; TITLE '13.6(A) NONLINEAR REGRESSION, HOME COMPUTERS'; PARMS A=0 B=0.03020 C=0.70682; MODEL PERCBUY = A + C*EXP(-B*PRICE); OUTPUT OUT=PLOT P=PPERCBUY R=RESIDUALS;RUN;*sort data, in descending order by price, so gplot looks OK;PROC SORT DATA=PLOT OUT=SPLOT; TITLE '13.6(A) NONLINEAR REGRESSION PLOT, HOME COMPUTERS'; BY PRICE;RUN;SYMBOL1 V=STAR C=BLACK;SYMBOL2 V=POINT C=BLACK I=SPLINE L=1;PROC GPLOT DATA=SPLOT; TITLE '13.6(A) NONLINEAR REGRESSION PLOT, HOME COMPUTERS'; PLOT PERCBUY*PRICE=1 PPERCBUY*PRICE=2 / OVERLAY;RUN;PROC GPLOT DATA=SPLOT; TITLE '13.6(B) RESIDUALS VERSUS PREDICTED'; PLOT RESIDUALS*PPERCBUY / VREF=0;RUN;PROC GPLOT DATA=SPLOT; TITLE '13.6(B) RESIDUALS VERSUS PRICE'; PLOT RESIDUALS*PRICE / VREF=0;RUN;PROC UNIVARIATE DATA=PLOT NORMAL PLOT; TITLE '13.6(B) RESIDUAL NORMAL PROBABILITY PLOT'; VAR RESIDUALS;RUN;QUIT;(a) Nonlinear Regression PlotFrom the SAS output, the following nonlinear regressionˆYi= 0.08360 + 0.83053 exp(−0.06412Xi)does, in fact, appear to fit the data well.(b) Various Residual Plots.(i) Both residual plots (versus predicted,ˆY and versus price, X) seemto indicate a sinusodal pattern and so seem to indicate non–constantvariance.(ii) The normal probability plot, though, seems to be near to a straightline and so indicates normality in the residual data.Practice Quiz 6 (Attendance 12), 13.7 6(13.7) Home Computers*PRACTICE QUIZ 6, 13.7 LACK OF FIT TEST;DATA COMPUTERS; INPUT CITYIND PERCBUY PRICE; L_PERCBUY = LOG(PERCBUY);DATALINES;0 0.9 10 0.8 2.50 0.65 50 0.46 100 0.34 200 0.26 300 0.17 400 0.15 500 0.06 750 0.04 1001 0.93 11 0.77 2.51 0.63 51 0.5 101 0.3 201 0.24 301 0.19 401 0.12 501 0.08 751 0.05 100;PROC NLIN DATA=COMPUTERS; TITLE '13.7 NONLINEAR REGRESSION, HOME COMPUTERS'; PARMS A=0 B=0.03020 C=0.70682; MODEL PERCBUY = A + C*EXP(-B*PRICE); OUTPUT OUT=PLOT P=PPERCBUY R=RESIDUALS;RUN;PROC ANOVA DATA=PLOT; TITLE '13.7 LACK OF FIT TEST OF NONLINEAR REGRESSION'; CLASS PRICE; MODEL PERCBUY = PRICE; RUN;QUIT;1. Statement.The statement of the test isH0: E{Y } = γ0+ γ2exp(−γ1X) versus H0: E{Y } 6= γ0+ γ2exp(−γ1X)2. Test.The test statistic1isF∗=SSE (R) − SSE (F )dfR− dfF÷SSE (F )dfF=SSE − SSPE(n − 2) − (n − c)÷SSPEn − c=SSLFc − 2÷SSPEn − c=0.02794 − 0.00387÷0.003810=1From SAS,SSE = 0.0279 from the ”reduced” nonlinear model, Yi= γ0+ γ2exp(−γ1Xi) + εi,and SSPE = 0.00380 from the ’full” ANOVA model, Yij= µj+ εij.Practice Quiz 6 (Attendance 12), 13.7 7(circle one) 9.075 /