Final Exam - Engineering Statistics | STAT 305, Exams of Statics

Download Final Exam - Engineering Statistics | STAT 305 and more Exams Statics in PDF only on Docsity! Statistics 305 Final Exam December 12, 2007 Instructor: Dr. Wu Name Note: Points for each question are indicated in the left margin in squate brackets. J. The data used in creating the attached one page of regression analysis printout come from a study of a nitride etch process on a single wafer plasma etcher. The process variables studied were 2 = power applied to the cathode (W) £9 = pressure in the reaction chamber (mTorr) <3 = gap between the anode and the cathode (cm) 4 = flow of the reactant gas Cos and the response variable was y = selectivity of the process (SiN /polysilicon). Note that the “Stdirr Pred y” column on the output is for the second regression analysis. Use the first regression analysis output in answering the questions (a}—(d) below. Note that Zy = 0.9636 and S\(x3i — Fs)? = 0.3055. (a) What fraction of observed raw variation in y is explained by a linear equation in £3? 2] 2 , R= 0777 (b) What is the sample correlation between y and 23? PI ~ JR = -a99F (Same Sign aA that of 4) (c) Give a 95% upper contidence bound for the increase in mean value of selectivity of the process for a 0.2 cm increase in gap between the anode and the cathode. (No nced to simplify.) 7 s 5 PEP 8 g2(& +f Ee = 0.2 (1 086 tiBs Fe) QOD of ty % A833 {d) Give a 95% two-sided prediction interval for the next selectivity of the process when gap between the anode and the cathode equals 1.1 em. (No need to simplify.) ia $b BE a No WEE a0) 7 Yt Su Sts * Se; -3F 2.262 SE 2,529 —hOkE OD 4 2,202: 4.0967. fiber Ged POSS The following questions (e)—(j) are based on the multiple linear regression (MLR) model vi = Bo + Bro + Bora + Baza; + Bara: + ei. Use the attached printout in answering these questions. (e) What is the sample correlation between y and 7? 2 Ss . FSDS . NR? = fa9e¢ =O ~F2 (f) Give the fitted value and residual corresponding to the third ebservation (with 2 = 275,29 = 550,23 = 12,24 = 200 and y = 1.10). (No‘need to simplify.) Ml Dabo tb XH hare +b:%3 Aho XE ) = 2 2b 96H (0.00327) (228) (~a 00/24) (E50) + (AOR): (62. Hf (0, 000532.) C200) Cay = 10S (g) Give a 95% two-sided prediction interval for the next selectivity of the process corresponding to the conditions of the third observation. (No need to simplify.) Dt t/ BFF ir AP QI) of te Ds 2967 ee 3 #2, G47 [aasoly* 4 (0,.0923)% (Sa given in purl Ff) ahve) (h) Give a 95% lower confidence bound for 3. (No need to simplify.) 4 da 7 € Cop lh) QI of te es APES 4 08¢1— Ch 942) (0. OPE2) {i} Find the value of an F statistic and its degrees of freedom for testing whether all the predictors £1,Z2,23, and z4 can be dropped from this MLR. model. What is your conclusion? R] Observed F = _ 44 27 ate 4% Conclusion (circle only one): (a) all the predictors should be dropped not all the predictors should be drapped 4. The diameters of five ball bearings were measured using two different, kinds of calipers. The data are given below. Ball Bearing 1 2 3 4 5 Caliper 1 2700 275.275.290.285 BE, = .279 8; = .0082 Caliper 2 260.270) 270) 280.280 y= 272 sz = 0084 (a) Give a 95% upper prediction bound for the diameter measurement using Caliper 1 on a single additional ball bearing of this type. (No need to simplify.) (5) Rte Je = 279 #2, 122 (20082 f 14- ee QUIS) of ty rs 2,732 (b) Give a 95% two-sided confidence interval for the mean difference in the diameter measurements using the two calipers on ball bearings of this type. (No need to simplify.) [6] a) dz Ae se '5> f= Sof = 0,007 ~ % 1010 (005 C08” 010» 005” 2 jos 2 oe Sf pk GM) 7 5x0 dt t Stile Sy 50,002 FE. Of. 000278 0007 277 an (c) What assumptions have to be made in order to obtain the confidence interval in part (b)? Pl 7b lifeenes ar srmably Ah bake! {d) Use the five-step significance-testing format to assess the strength of evidence to support that there is a mean difference in the diameter measurements using the two calipers on bal] bearings of this type. 8 gh _“e=0 O44: pro @ jit stttiche, TH sor Aiforena debibuhen, tp Lege uber of [7 a Evoke jer fra, @ Ai sangh gies To Siz = 071 © Povebit = POT) xF7NF2FCT PRIO <S2:(/- 995) =4.0f Weiant plow 72 ee Wane er A 5. An experiment was made in order to determine spring constants for three different types of steel springs. Experimental values of spring constants were obtained for n1 = 7 springs of type 1, 2 = 6 springs of type 2, and ny = 6 springs of type 3, using an 8.8 Ib actual load. The data, some summary statistics, and the analysis of variance table are given below. Type | Springs Type 2 Springs ‘Type 3 Springs 1,99, 2.06, 1.99 2.85, 2.74, 2.74 2.10, 2.01, 1.93 1.94, 2.05, 1.88 2.63, 2.74, 2.80 2.02, 2.10, 2.05 2.30 F, = 2.030 Fg = 2.750 Ty = 2.035 By = 0.134 Bq = 0.074 33 = 0.064 Analysis of Variance Source DF Sum of Squares Mean Square F Ratio Treatments 2 2.41463 1.05731 108.62 Error 16 0.15575 0.00973 C.Total 18 2.27038 Let j1, #2, and yg be the true spring constants for spring types 1, 2, and 3, respectively. {a} Give a 90% two-sided confidence interval for the quantity fa, — zee + 23). (No need to simplify.) ® F-LGD tts [Fe Ge FS Map ty ra. PS ATES 78 2303 LCR LNG fa Jl pp da LE (b) Give the value of an F statistic and its degrees of freedom for testing Hy : 1 = 2 = fag versus Ha: not Ho. iD Fak 62 dbf +6) (c) Give a 95% two-sided confidence interval for jg — yg based on the last two samples only. (No need to simplify.) , , oe ete 2 " 2 DS taps _— SOPD*S COR |, gon zi SP apes HS Sp JG = 4 eeP2. AOI) Ff Le FS 2,222 (2.75- nana) bo 228) (2, 062) Jett y x1 x2 x3 x4 StdErr Pred y 1.63 275 450 0.8 125 0.0275693 1.37 275 500 1 160 0,02317067 11 275 550 1.2 200 0.04325283 1.58 300 450 1 200 0.03328395 1.26 300 500 1.2 125 0,03277322 1.72 275 450 0.8 125 0.0275693 1,65 300 550 0.8 160 © 0.03297551 1.42 325 450 1.2 160 0.03909274 1.69 325 500 0.8 200 0.03929552 1.54 325 550 1 125 = 0.03879703 1.72 275 450 0.8 125 0.0275693 Response y Whole Model x3 Summary. of Fit RSquare 0.798787 RSquare Adj 0.77643 > f s Root Mean Square Error 0.0967 Sta 305 Mean of Response 1.516364 tT an Observations (or Sum Wgts) 11 Firal Ban Analysis of Varlance Source DF Sum of Squares Mean Square F Ratio (Dec 12, 2007) Model 1 0,33409621 0.334096 35.7287 ‘ Error 9 0.08415833 0.009351 Prob > F ¢, Total 10 0.41825455 0.0002" Parameter Estimates Term Estimate Std Error t Ratio Probaltl Intercept 25241667 0.171106 14.75 <,0001* x3 1.045833 0.174966 -5.98 0.0002* Response y Whole Modet xt x2. x3. x4 Summary of Fit RSquare 0.964037 RSquare Adj 0.940062 Root Mean Square Error 0.050069 Mean of Response 1.516364 Observations (or Sum Wgts) a Analysis of Variance Source DF Sum of Squares Mean Square F Ratio Modet 4 0.40321283 0.100803 40.2095 Error 6 0.01504172 0.002507 Prob > F C. Total 10 0.41825455 0.0002* Parameter Estimates Term Estimate Std Error t Ratio Probaltl Intercept 2.2895549 0.254409 9.00 0.0001* al 0.0032704 0,000762 4.29 0.0051* x2 -0,001331 0,000381 -3.49 0.0129° x3 -1,041201 0,095263 -10.93 <,0001* x4 -0,000532 0.000509 1.04 0.3365