Final Exam, with Answer Key - Statistics for Engineering | STAT 4706, Exams of Statistics

Download Final Exam, with Answer Key - Statistics for Engineering | STAT 4706 and more Exams Statistics in PDF only on Docsity! << f Stat 4706 Name Exam 1 Spring 2000 Show all work on these pages for full credit. You are allowed to use a calculator, the tables provided, and a formula sheet (one 8 1/2 x 11 sheet, both sides, formulas only.) Otherwise, this is a closed book exam. 1. An industrial engineer wishes to establish the relationship between y = cost per production of a batch of laminated wafers, in dollars x = size of the batch, measured in number of wafers For n= 20 batches, the following calculations resulted: Sux = 75,710.0 Syy = 5,946,702.0 Sxy = 659,553.0 Ya = 3,340.0 >» = 37,130.0 (a) Give the Least Squares regression equation. 6595532, _ ga Ai = WIE A= Ue @ 0.91 (22GB) = 856.5 - B.21(6Y oe 2 56S - HSE = 01.67 A Ya ¥Ol b7 thi E (b) Give the regression variance, s? 2 (454553) xi 4553) t Jeg, A, = F956 18 - —aeaee 7570 L [stal7e> -$7¢8, 740-4] gle ‘ 7 , £009 EF b 48 = MME 3. Continued. . (a) Based on the experimental data, give the regression function that explains the cost of this software development. A Ye GH t G63 Ey tO18 Ky +0, 20k (b) Do the 3 regressor variables collectively have a significant effect of explaining the cost of this software development? Justify your answer. fF xltbi3on wt a P- walee of 0.0607 wheck (c) Give the regression variance, 5? : 2 as = 346.3 (d) If xz is the last variable to be entered in the regression model, would it have a significant effect in explaining the cost of developing this software? Justify your answer. to, Me fiat of hipre tao 2,707 wih 2 Fe wate Ff 8,4837 wndeiatecy 1. y mt enrhenee. eT Va far wn feed am syplasing Te cont of dewclaping tin. afftone phan 7, wok ty tae, sally x fa mets  4, The engineering consulting firm (Problem 3) felt that an improvement of the regression model could be made by the inclusion of other regressors. Two additional (secret) variables were included and a stepwise selection procedure was run to see if a "better" model could be obtained. The SAS output of this procedure is: Cost of Software Development Stepwise Procedure for Dependent variable ¥ Step 1 Varisble x3 Entered R-aquare = 0.82478630 C(p) = 18.20077678 OF Sum of Squares Mean Square F Prob r Regreasion a 15705,57436542 15705.57436542 51.78 06,0003 Error ia 3336.42794227 303.31072202 Total 12 19041. 99230769 Standard Type II Varieble Error Sum of Squares F ProboF IWTERCEP 12.40428457 6. 96912667 960.69180139 3.17 0.1027 x3 2.01281099 0,44074902 © 15705.57436542 Step 2 Variable x4 Entered R-mquare = 0.90598356 51.78 0.0001 c(p) = 7.59543473 oF Sum of Squares Mean Square F ProboF Regreasion 2 17251.73202617 —-8625..26601309 48.18 9.0002 Error 10 1790,26028152 179.02602815 total az 19042.99230769 Parmmster Standard ‘Type II Variable Eatimate Error Sum of Squares F ProbF INTERCEP -17.99417635 11.64743361 427 29695) 2.39 0.1534 x3 2.75265331 0.60182066 = -3745.20637482 20.92 0.0010 xa -0.73367784 0,24365289 = -1566.15766075 a.64 0.0148 Step 2 Variable X5 Entered R-square = 0,92541601 Clip) = 6.57867465 or Sum of Squares Mean Square F Proper Regression 3 17621.76436345 9 5873,92152115 37.22 0.0002 Error 2 1420.22774424 157.80308269 Total 12 19041.99230765 Parameter Standard ‘ype IT Variable Estimate Error sun of Squares YF Prob>F INTERCEP -25.56821922 12.00188288 716.17816153 4.54 0.0620 x 3.42038073 0.71371711 = -3624.20008550 22.97 09,0010 m +0.93410669 0.26845760 © 1910.54398281 12.11 0.0069 Comt of Software Development 2 5 =0.01774399 0,01150745 270.03253728 2.34 0.1602 R-square = 0,90598356 Cp) = 7.59543473 or Sum of Squares Mean Square F Prob>r Regression a 47251.73202617 862586601309 48.28 0.0002 Error 20 1790.26028152 179.02602815 Fotal 12 19041.99230769 4. Continued i Pacamater Standard Variable Estimate Error ¥ Probr INTERCEP -27,.99417635 11.64743361 2.39) 0.1534 x 2.75265331 0, 69182066 20.92 0.9020 Fey -0,73367784 0.24965289 «= -1546.15766075 9.64 0,0148 | All variables left in the model are significant at the 0.1500 level. The stepwise mathod terminated because the next variable to be entered was just removed. Sumary of Stepwiee Procedure for Dependent Verieble ¥ variable member Partial = Model Step Entered Removed In RD Rea tp) F OProb>F 1 x3 1 «0.8248 «0.9248 «18.2008 «= 51.7805 9.0001 | 2 x 2 0,0812 0.9060 = 7.5954 8.6365 0.0148 3 25 30,0194 «0.9254 = 6.8787 2.3449 0.1601 ‘ aS 2 0.0194 0.9060 7, 5954 2.3449 0.1602 (a) Give the final model obtained by the stepwise procedure. He LIDGI PE BAF Beg — OTF Eye What is the R? value for this model? 2 i Rae, Yee What is the regression variance for this final model? ! AL 247%03 (b) If possible, give the simplest model that has an R? value of at least 0.90. : A -ITGG + A.IS Vg ~ F073 Key