Quiz Practice Questions 1 (Attendance 2) for Statistics 512Applied Regression AnalysisMaterial Covered: Chapters 1,2 Neter et al. and KuhnBy: Friday, 12th September, Fall 2003These are practice questions for the quiz. All questions for the quiz are based on thequestions given below. These practice questions are not to be handed in.Applied Linear Statistical Models (Neter et al.) Questions.Chapter Problem(s) hints1, pages 36–43 1.61.19,1.23 Grade point average data1.22,1.26 Plastic hardness data2, pages 86–94 2.7(a),(b); 2.16, 2.26 Plastic hardness dataApplied Statistics and the SAS Programming Language (Cody and Smith)Chapter Problem(s) hints5, pages 135–137 (5–1), (5–3), (5–5) XY Z data(1.6) normal regression model(a) Use your calculators to sketch E{Y } = 200 + 5X (use “Y=”, WINDOWand GRAPH), where, at X = 10, 20, 30, normal–shaped curves are drawn,with standard deviation σ = 4.(b) Parameter β0is the mean response at X = 0,parameter β1is the change in the mean response for a unit change in theindependent variable, X(1.19) Grade point average data, qz1-1-19,23Cut and paste the following program into SAS1.*PRACTICE QUIZ 1, 1-19, PAGES 36-43;DATA GPAVSCORE; INPUT GPA SCORE;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 GLM DATA=GPAVSCORE; TITLE 'REGRESSION LINE FOR GPA VS SCORE DATA'; MODEL GPA=SCORE; OUTPUT OUT=GPAVSCORE2 PREDICTED=PRED RESIDUALS=RESID;RUN;*1.19(B) REGRESSION PLOT;SYMBOL1 V=CIRCLE C=BLACK I=R;SYMBOL2 V=DOT C=BLACK I=R;PROC GPLOT DATA=GPAVSCORE2; TITLE '1.19(B) PLOT GPA VS SCORE'; PLOT GPA*SCORE;RUN;*1.23(A) RESIDUALS;PROC PRINT DATA=GPAVSCORE2; TITLE '1.23(A) RESIDUALS PP 36-43'; VAR SCORE GPA PRED RESID;RUN;QUIT;(a) From SAS output,b0= −1.69956, b1= 0.83991,ˆY = −1.69956 + 0.83991X(b) From the SAS output, it looks like the regression does fit the data well.(c) Let X = 5.0 and soˆYh= −1.69956 + 0.83991X = −1.69956 + 0.83991(5.0) = 2.5(d) b1= 0.839911Cutting and pasting will save you the time of typing the program in SAS.(1.22) Plastic hardness data, qz1-1-22,26*PRACTICE QUIZ 1, 1-22,26; *PLASTIC HARDNESS DATA, PAGES 36-43;DATA HARDVTIME; INPUT HARD TIME;DATALINES;199 16205 16196 16200 16218 24220 24215 24223 24237 32234 32235 32230 32250 40248 40253 40246 40;PROC GLM DATA=HARDVTIME; TITLE 'REGRESSION LINE FOR HARDNESS VS TIME DATA'; MODEL HARD=TIME; OUTPUT OUT=HARDVTIME2 PREDICTED=PRED RESIDUALS=RESID;RUN;*1.22(A) REGRESSION PLOT;SYMBOL1 V=CIRCLE C=BLACK I=R;SYMBOL2 V=DOT C=BLACK I=R;PROC GPLOT DATA=HARDVTIME2; TITLE '1.19(B) PLOT HARDNESS VS TIME'; PLOT HARD*TIME;RUN;*1.23(A) RESIDUALS;PROC PRINT DATA=HARDVTIME2; TITLE '1.26(A) RESIDUALS PP 36-43'; VAR TIME HARD PRED RESID;RUN;QUIT;(a) From SAS output,ˆY = 168.6 + 2.034375X(b) At X = 40,ˆYh= 168.6 + 2.034375(40) = 249.975(c) b1= 2.0344(1.23) more Grade point average data; see SAS program in 1.19 above(a) The residuals are given on the SAS output. They do sum to zero.(b) The error mean square or residual mean square, MSE , is is given on theSAS output.E{MSE } = σ2= 0.189, σ =√0.189 = 0.435(1.26) more Plastic hardness data; see SAS program in 1.22 above(a) The residuals are given on the SAS output. They do sum to zero.(b) The error mean square or residual mean square, MSE , is is given on theSAS output.E{MSE } = σ2= 10.459, σ =√0.189 = 3.234The last estimate is expressed in Brinell units.(2.7) more Plastic hardness data; inference on β1, qz1-2-7,16,26Both SAS and the TI–83 calculator can be used here.*PRACTICE QUIZ 1, 2-7,16,26, PAGES 86-94;DATA HARDVTIME; INPUT HARD TIME; TCRIT = TINV(0.995,14); FCRIT = FINV(0.99,1,14);DATALINES;199 16205 16196 16200 16218 24220 24215 24223 24237 32234 32235 32230 32250 40248 40253 40246 40;*2.7(A) CI SLOPE, 2.26(B);PROC PRINT DATA=HARDVTIME; VAR HARD TIME TCRIT FCRIT AVE;RUN;PROC REG DATA=HARDVTIME OUTEST=EST TABLEOUT ALPHA=0.01; TITLE '2.7(A) CI SLOPE PP 86-94'; MODEL HARD=TIME; TEST TIME = 2;RUN;PROC PRINT DATA=EST; TITLE '2.7(A) CI SLOPE PP 86-94';RUN;QUIT;(a) Use PRGM REGINF, where {B0,B1,MN,PI} is {0,1,0,0};NULL B1 can be whatever you want (let’s make it 1) since we want aconfidence interval and not do a test;ALPHA is 0.01 (since we want a 99 percent confidence interval) to deter-mineb1± t(1 − α/2; n − 2)s{b1} = 2.034 ± (2.977)(0.09) = (1.77, 2.3)(b) Use PRGM REGINF again, only let NULL B1 = 2H0: β1= 2 versus Ha: β16= 2t∗=b1−β10s{b1}=2.034−20.09= 0.38, t = 2.977since t∗= 0.38 < t = 2.977, then data supports β1= 2p–value is 0.71(2.16) more Plastic hardness data; mean response E{Yi}The TI–83 calculator is better suited to deal with these questions than SAS.(a) Use PRGM REGINF, where {B0,B1,MN,PI} is {0,0,1,0};NULL MN can be whatever you want (let’s make it 1) since we want aconfidence interval and not do a test;X is 30, the elapsed time, XhM can be whatever you want (let’s make it 1) since we are not calculatinga prediction interval;ALPHA is 0.02 (since we want a 98 percent confidence interval) to deter-mineˆYh± t(1 − α/2; n − 2)s{ˆYh} = 229.631 ± (2.624)(0.83) = (227.46, 231.81)(b) Use PRGM REGINF, where {B0,B1,MN,PI} is {0,0,1,1};NULL MN can be whatever you want (let’s make it 1) since we want aconfidence interval and not do a test;X is 30, the elapsed time, XhM is one (1) since we want the prediction interval for one new observation;ALPHA is 0.02 (since we want a 98 percent prediction interval) to deter-mineˆYh±t(1−α/2; n−2)s{predmean} = 229.631±(2.624)(3.34) = (220.87, 238.39)(c) Use PRGM REGINF, where {B0,B1,MN,PI} is {0,0,1,1};NULL MN can be whatever you want (let’s make it 1) since we want aconfidence interval and not do a test;X is 30, the elapsed time, XhM is ten (10) since we want the prediction interval for the mean of ten newobservations;ALPHA is 0.02 (since we want a 98 percent prediction interval) to deter-mineˆYh±t(1−α/2; n−2)s{predmean} = 229.631±(2.624)(1.32) = (226.18, 233.09)(d) Yes, the prediction interval in part (c) is narrower than the predictioninterval in part (d). Yes, it should be narrower since we are determiningthe mean of ten observations, which are less variable than one observation.(e) Use PRGM REGSCIPI,X is 30, the elapsed time, XhM can be whatever you want (let’s make it 1) since we are not calculatinga prediction interval;K can be whatever you want (let’s make it 1)