1Quiz Practice Questions 7 (Attendance 14) for Statistics 512Applied Regression AnalysisMaterial Covered: Chapter 15 Neter et al and KuhnBy: Friday, 5th December, 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) hints15, pages 654–659 15.1 why is a regression model appropriate?15.13 Property assessments15.22 Microcomputer Assembly15.24 Grade Point AveragePractice Quiz Questions 7 (Attendance 14), 15.1 2(15.1) correlation or regression model?A regression model is probably appropriate here where machining time wouldbe the response variable in the regression model. The other variable, weight ofrough casting block, is fixed ahead of time and so most certainly would be thepredictor variable.Practice Quiz Questions 7 (Attendance 14), 15.13 3(15.13) statistical inference*PRACTICE QUIZ 7 15.13 PROPERTY ASSESSMENTS, PAGE 655;DATA PROPERTY; INPUT ASSESS PRICE;DATALINES;13.9 28.616 34.710.3 2111.8 25.516.7 36.812.5 2410 19.111.4 22.513.9 28.312.2 2515.4 31.114.8 29.614.9 35.112.9 3015.8 36.2;SYMBOL1 V=STAR C=BLACK;PROC GPLOT DATA=PROPERTY; *scatterplot of assessment values and price; TITLE '15.13 PROPERTY ASSESSMENTS, PAGE 655'; PLOT PRICE*ASSESS;RUN;PROC CORR DATA=PROPERTY; *estimate r12; VAR ASSESS PRICE;RUN;QUIT;(a) Bivariate normal?Yes, the data does seem to indicate a bivariate normal because the datais “football–shaped”. However, it would be better to graph a three–dimensional data set to see if the data follows a bivariate normal shape.(b) Correlation coefficientlooking at the SAS output, r12= 0.95285r12is a biased estimator of ρ12r212is an estimator of ρ212that measures either the relative reduction in thevariability of Y1associated with the use of variable Y2or the other wayaround(c) Test if ρ126= 01. Statement.H0: ρ12= 0 versus Ha: ρ126= 02. Test.t∗=r12√n − 2q1 − r212=0.95285√15 − 2√1 − 0.952852= 11.32The critical value at α = 0.05 is |t(0.95, n − 2)| = |t(0.95, 15 − 2)| =|3.012| = 3.0123. Conclusion.Since the test statistic, 11.32, is larger than the critical value, 3.012,we reject the null hypothesis that H0: ρ12= 0.Practice Quiz Questions 7 (Attendance 14), 15.13 4(d) Test if ρ126= 0.6It is not appropriate to uset∗=r12√n − 2q1 − r212to test H0: ρ12= 0.6 versus Ha: ρ126= 0.6because this test statistic is used to testH0: ρ12= 0 versus Ha: ρ126= 0 instead.Practice Quiz Questions 7 (Attendance 14), 15.22 5(15.22) Microcomputer assembly.Notice, in this question, a lot of information is already given; in other words,SAS is not required.regression Y1on Y2Y1on Y3Y1on Y2, Y3SSR 7436 66 7439SSE 392 7762 389Total 7828 7828 7828(a) Coefficient of multiple correlation and determinationr21.23=SSR(Y2, Y3)SSTO(Y1)=7, 4397, 828= 0.9503The r21.23= 0.9503 measures the proportion of the sums of squares “ex-plained” by the Y2, Y3variables, as compared to the total sums of squaresas “explained” by all of the variables, Y1, Y2, Y3variables. In this case, alarge proportion, 95%, of the total variability is explained by Y2, Y3.(b) Test at 1% H0: ρ1.23= 0 versus Ha: ρ1.236= 0.1. Statement.H0: ρ1.23= 0 versus Ha: ρ1.236= 02. Test.F∗= r21.231 − r21.23! n − q − 1q!=0.95031 − 0.950330 − 2 − 12= 258.13The critical value at α = 0.01 is F (0.99, q, n−q −1) = F (0.99, 2, 27) =5.493. Conclusion.Since the test statistic, 258.13, is larger than the critical value, 5.49,we reject the null hypothesis that H0: ρ1.23= 0. In other words, dataindicates the variables Y2, Y3explain more than zero (no) proportionof the total variability in the model.(c) Test at 1% H0: ρ12.3= 0 versus Ha: ρ12.36= 01. Statement.H0: ρ12.3= 0 versus Ha: ρ12.36= 02. Test.Sincer212.3=SSE (Y3) − SSE (Y2, Y3)SSE (Y3)=7762 − 3897762= 0.9499Practice Quiz Questions 7 (Attendance 14), 15.22 6thent∗=r12.3√n − q − 1q1 − r212.3=√0.9499√30 − 2 − 1√1 − 0.9499= 22.63The critical value at α = 0.01 is t(1 −α/2, n −q −1) = t(0.995, 27) =2.7713. Conclusion.Since the test statistic, 22.63, is larger than the critical value, 2.771,we reject the null hypothesis that H0: ρ12.3= 0. That is,(d) 99% interval estimation of ρ12.3Sincez0=12loge1 + r12.31 − r12.3=12loge 1 +√0.94991 −√0.9499!= 2.1772andσ{z0} =1√n − q − 3=1√30 − 1 − 3= 0.1961an approximate 99% confidence interval of ρ12.3is given byz0± z(1 − α/2)σ{z0} = 2.1772 ± 2.576(0.1961) = (1.6715, 2.6819)and so sincer12.3=e2z0− 1e2z0+ 1then e2z0− 1e2z0+ 1,e2z0− 1e2z0+ 1!= e2(1.6715)− 1e2(1.6715)+ 1,e2(2.6819)− 1e2(2.6819)+ 1!= (0.932, 0.991)Practice Quiz Questions 7 (Attendance 14), 15.24 7(15.24) Grade Point Average*PRACTICE QUIZ 7, 15.24 GRADE POINT AVERAGE, PAGE 658;DATA GPA; INPUT GPA_FRESH TEST;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;PROC REG DATA=GPA; TITLE '15.24 GRADE POINT AVERAGE, PAGE 658'; MODEL GPA_FRESH = TEST; OUTPUT OUT=GPA2 R=RESIDUALS;RUN;DATA GPA3; SET GPA2; ABS_RESIDUALS=ABS(RESIDUALS);RUN;PROC CORR DATA=GPA3 SPEARMAN; *spearman rank of abs residuals and X; VAR ABS_RESIDUALS TEST;RUN;QUIT;(a) test of constant variance or not1. Statement.H0: no association between residuals and Xversus Ha: association between residuals and Xor: H0: constant variance wrt to X versusHa: variance varies with X2. Test.Since, from the SAS output, rS= −0.3708, the test statistic ist∗=rS√n − 2q1 − r2S==−0.3708√20 − 2q1 − (−0.3708)2= −1.6939Also, the critical value at α = 0.05 is t(1 −α/2, n −2) = t(0.975, 18) =2.1013. Conclusion.Since the test statistic, |t∗| = |−1.6939| = 1.6939, is smaller than thecritical value, 2.101, we accept the null hypothesis of no associationbetween residuals