Quiz Practice Questions 3 (Attendance 6) for Statistics 512Applied Regression AnalysisMaterial Covered: Chapters 7 Neter et al. and KuhnBy: Friday, 10th October, Fall 2003These are practice questions for the quiz. One or more questions 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.1. Applied Linear Statistical Models(Neter et al.) Questions.Chapter Problem(s) hints7, pages 317–324 7.4, 7.13, 7.17, 7.25, 7.41 Chemical Shipment data2. Applied Statistics and the SAS Programming Language(Cody and Smith) Questions.No questions.(7.4) Chemical Shipment, qz3-7-4-chem-extSS*PRACTICE QUIZ 3, EXTRA SS ETC, PP 317-323;DATA CHEMICAL; INPUT NUMBER WEIGHT TIME; NMWGT = NUMBER * WEIGHT;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;*7.4 EXTRA SS;PROC REG DATA=CHEMICAL; TITLE '7.4 EXTRA SS, X2 AND X1|X2'; MODEL TIME = WEIGHT NUMBER / SS1;RUN;PROC REG DATA=CHEMICAL; TITLE '7.4 EXTRA SS, X1 AND X2|X1'; MODEL TIME = NUMBER WEIGHT / SS1;RUN;QUIT;(a) ANOVA table, X2, X1|X2Source Sum Of Squares Degrees of Freedom Mean SquaresRegression 40496 2 20248X239309 1 39309X1|X21187.61 1 1187.61Error 536.47 17 31.56Total 41033 19(b) Test that X1can be dropped from the model1. Statement.H0: β1= 0 versus H1: β16= 02. Test.The partial F∗test statistic isF∗=SSE (R) − SSE(F )dfR− dfF÷SSE (F )dfF=SSE (X1) − SSE (X1, X2)(n − 2) − (n − 3)÷SSE (X1, X2)n − 3=SSR(X1|X2)1÷SSE (X1, X2)n − 3=1187.611÷536.46817= 37.63The upper critical value at α = 0.05, with (1, 17) df is 6.043. Conclusion.Since F∗= 37.63 > 6.04, we reject the null that β1= 0.(c) Comparing Extra Sums of SquaresSource Sum Of Squares Degrees of Freedom Mean SquaresRegression 40496 2 20248X138658 1 38658X2|X11838.20 1 1838.20Error 536.47 17 31.56Total 41033 19Using this ANOVA table and the one above, notice thatSSR(X1) + SSR(X2|X1) = 38658 + 1838.20= 40496 = SSR(X1, X2)= 39309 + 1187.61= SSR(X2) + SSR(X1|X2)Notice that X2“explains more” of the regression than X1becauseSSR(X2) = 39309 > SSR(X1) = 38658.(7.13) Chemical Shipment continued, qz3-7-13-chem-corr*PRACTICE QUIZ 3, 7.13, CORRELATION, PP 317-323;DATA CHEMICAL; INPUT NUMBER WEIGHT TIME; NMWGT = NUMBER * WEIGHT;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;*7.13 CORRELATION;PROC CORR DATA=CHEMICAL; TITLE '7.13 CORRELATION MATRIX, Y1, Y2, 12'; VAR TIME NUMBER WEIGHT NMWGT;RUN;PROC REG DATA=CHEMICAL; TITLE '7.13 PARTIAL CORRELATION, Y2.1'; MODEL TIME = NUMBER;RUN;PROC REG DATA=CHEMICAL; TITLE '7.13 PARTIAL CORRELATION, Y1.2'; MODEL TIME = WEIGHT;RUN;PROC REG DATA=CHEMICAL; TITLE '7.13 OVERALL CORRELATION'; MODEL TIME = NUMBER WEIGHT;RUN;QUIT;(a) Full Correlations.From the correlation matrix given by the SAS program,r2Y 1= 0.972= 0.94r2Y 2= 0.982= 0.96r212= 0.932= 0.86It is “good” that the response, Y , is strongly correlated with (“describedby”) both the predictor variables. It is “bad” that the two predictorsvariables are also strongly correlated with one another; this indicates theymight be describing the same thing.(b) Partial Correlation, r2Y 1.2From the ANOVA table given above for X2, X1|X2,Source Sum Of Squares Degrees of Freedom Mean SquaresRegression 40496 2 20248X239309 1 39309X1|X21187.61 1 1187.61Error 536.47 17 31.56Total 41033 19and from the following ANOVA table given in the current SAS program,Source Sum Of Squares Degrees of Freedom Mean SquaresRegression, X239309 1 39309Error 1724.08 18 95.78Total 41033 19we findr2Y 1.2=SSR(X1|X2)SSE (X2)=1187.611724.08= 0.69This says, roughly, that the “extra” correlation between the response andthe first predictor, given that the second predictor is already in the model,is fairly strong.(c) Partial Correlation, r2Y 2.1Also, from the ANOVA table given above for X1, X2|X1,Source Sum Of Squares Degrees of Freedom Mean SquaresRegression 40496 2 20248X138658 1 38658X2|X11838.20 1 1838.20Error 536.47 17 31.56Total 41033 19and from the following ANOVA table given in the current SAS program,Source Sum Of Squares Degrees of Freedom Mean SquaresRegression, X138658 1 38658Error 2374.67 18 131.93Total 41033 19we findr2Y 2.1=SSR(X2|X1)SSE (X1)=1838.202374.67= 0.77This says, roughly, that the “extra” correlation between the response andthe second predictor, given that the first predictor is already in the model,is fairly strong (in fact, stronger than the other way around).(d) Overall CorrelationFinally, from the “overall” regression,Source Sum Of Squares Degrees of Freedom Mean SquaresRegression 40496 2 20248Error 536.47 17 31.56Total 41033 19we findR2= 0.99This says, roughly, that “overall”, with both predictors “equally” ac-counted for, the response is very strongly correlated to the predictors.(7.17) Chemical Shipment continued, qz3-7-17-chem-trans*PRACTICE QUIZ 3, 7.17, STAND REG, PP 317-323;DATA CHEMICAL; INPUT NUMBER WEIGHT TIME; NMWGT = NUMBER * WEIGHT;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;*7.17(A) STANDARDIZED REGRESSION;PROC UNIVARIATE DATA=CHEMICAL NOPRINT; TITLE '7.17 STANDARDIZED REGRESSION'; VAR TIME; OUTPUT OUT=OUTTIME MEAN=MEANTIME STD=STDTIME NOBS=NTIME;RUN;PROC UNIVARIATE DATA=CHEMICAL NOPRINT; TITLE '7.17 STANDARDIZED REGRESSION'; VAR WEIGHT; OUTPUT OUT=OUTWEIGHT MEAN=MEANWEIGHT STD=STDWEIGHT NOBS=NWEIGHT;RUN;PROC