MATH2009 Engineering ModellingInformation Technology and Mathematical SciencesEXAMINATIONMATH2009 Engineering ModellingThis paper is for Mawson Lakes students.Examination Duration: 180 minutesReading Time: 10 minutesExam Conditions:Extra 10 mins per hour and use of English language or bilingual print dictionary (noannotations) for ENTEXT studentsThis exam is part open bookMaterials Permitted In The Exam Venue:(No electronic aids are permitted e.g. laptops, phones)2 x A4 double sided handwritten notes permittedNon programmable calculator permittedMaterials To Be Supplied To Students:2 x Number of 16pg answer booklets per studentInstructions To Students:Answer all 9 questions.Write answers to sections A and B in different booklets.Mark each booklet clearly A or B, as applicable.Student Number |__|__|__|__|__|__|__|__|__|Last Name _______________________First Name _______________________Calculator Details Make _________________Model _________________For Examiner Use OnlyQuestion MarkTotal ________Study Period 2 Exams, 2019This exam paper must not be removed from the venueSection A, statistics. Mark your answer booklet with an ‘A’.50 marks this section.1. In a semi conductor manufacturing plant the finished semiconductor is wired toa frame. The variables in this study are the pull strength (the amount of forcerequired to break the bond), the wire length and the height of the die. We wantto find a model relating the pull strength to the wire length and die height. Thereis no physical mechanism that can be applied here so we use the measured data ofa sample of 25 to construct an empirical relationship. The output from a Minitabrun for this multiple regression is given below.(a) Test the hypothesis for determining if there is a connection between the response variable and the collection of possible predictor variables, indicatingwhere you obtain the p-value for the test.(b) Given that there is a connection, test the individual possible predictors andthe constant for inclusion in the model.(c) From that analysis, write down the model.(d) Give the value of the coefficient of determination and what it means.[2+6+4+3=15 marks]Regression Analysis: Pull Strength versus Wire Length, DieHeightAnalysis of VarianceSource DF Adj SS Adj MS F-Value P-ValueRegression 2 5990.8 2995.39 572.17 0.000Wire Length 1 4507.5 4507.53 861.01 0.000Die Height 1 104.9 104.92 20.04 0.000Error 22 115.2 5.24Total 24 6105.9Model SummaryS R-sq R-sq(adj) R-sq(pred)2.28805 98.11% 97.94% 97.44%CoefficientsTerm Coef SE Coef T-Value P-Value VIFConstant 2.26 1.06 2.14 0.044Wire Length 2.7443 0.0935 29.34 0.000 1.17Die Height 0.01253 0.00280 4.48 0.000 1.1722. Breakdowns on a production line are random events in continuous time, and thenumber of breakdowns in a given time is described by the Poisson distribution. Theaverage breakdown rate is one every 250 hours.(a) What is the probability that there will be precisely three breakdowns in oneweek of 168 hours?(b) What is the chance of there being at least one breakdown in one week of 168hours?(c) What is the chance of there being no breakdown for a week, but at least onein the following week?[2+2+2=6 marks]3. The times between breakdown on a production line are Exponential, with the average time between breakdowns being 250 hours.(a) What is the probability that the next breakdown will occur within the nextweek of 168 hours?(b) What is the probability that there will be no breakdown for a week, but atleast one in the following week?(c) If there has been no breakdown for a week, what is the probability that therewill be at least one breakdown in the following week?[2+2+2=6 marks]34. Find below the descriptive statistics and histograms for the winter daily total solarradiation and summer daily total radiation (in MJ=m:2) for Adelaide airport, aswell as the graph of the empirical distribution of the summer daily totals.(a) What statistical measures would you use to describe the centre and spread ofthe data in each case? Give reasons.(b) If X denotes the daily winter solar total, determine the value x such thatP(X < x) = 0:05. Hint: From the answers in the previous question, youshould be able to make an assumption about the distribution of this variable.(c) If Y denotes the summer daily total solar radiation, given the graph of theempirical cumulative distribution function (CDF) of the summer daily totals,determine y such that P(Y < y) = 0:05.[4+3+3=10 marks]Descriptive Statistics: Winter Daily SolarVariable N Mean StDev Minimum Q1 Median Q3 MaximumDaily Solar 920 9.62 3.46 0.071 7.13 9.53 11.72 18.74Descriptive Statistics: Summer Daily SolarVariable N Mean StDev Minimum Q1 Median Q3 MaximumDaily Solar 900 26.93 7.05 0.32 24.11 28.92 32.15 35.410.0 2.5 5.0 7.5 10.0 12.5 15.0 17.5706050403020100Mean 9.628StDev 3.465N 920Winter Daily SolarFrequencyHistogram of Winter Daily SolarNormal40 6 12 18 24 30 36 429080706050403020100Mean 26.94StDev 7.058N 900Summer Daily SolarFrequencyHistogram of Summer Daily SolarNormal0.00%5.00%10.00%15.00%20.00%25.00%30.00%35.00%40.00%45.00%50.00%55.00%60.00%65.00%70.00%75.00%80.00%85.00%90.00%95.00%100.00%105.00%0 5 10 15 20 25 30 35 40Solar Energy (MJ/sq.m.)Figure 1: Summer Solar CDF55. Given below are descriptive statistics, as well as boxplots, of the moisture content(by percent) for random samples of different fruits and vegetables, probability plotsfor the two variables, and output from the test of equality of population means.Descriptive Statistics: Fruits, VegetablesStatisticsVariable N N* Mean SE Mean StDev Minimum Q1 Median Q3 MaximumFruits 8 0 82.88 2.44 6.90 72.00 76.00 85.00 87.75 92.00Vegetables 9 0 91.67 1.25 3.74 85.00 88.50 92.00 95.00 96.00959085807570Fruits VegetablesBoxplot of Fruits, Vegetables60 72 84 96 10899959080706050403020105 180 90 10099959080706050403020105 1Mean 82.88StDev 6.896N 8AD 0.274P-Value 0.559FruitsMean 91.67StDev 3.742N 9AD 0.287P-Value 0.536VegetablesFruitsPercentVegetablesProbability Plot of Fruits, VegetablesNormal – 95% CI6TestNull hypothesis H₀: μ₁ – µ₂ = 0Alternative hypothesis H₁: μ₁ – µ₂ ≠ 0T-Value DF P-Value-3.21 10 0.009(a) Comment, from examining the boxplots, whether you think we will reject H0that the population mean moisture contents are equal. Why?(b) Comment on what test we should conduct, giving reasons with reference to theprobability plots.(c) Perform the hypothesis test for equality of population means.(d) Construct the 95% confidence interval for the difference in population means.Is this consistent with the result of the hypothesis test? Why?[3+3+4+3=13 marks]7Section B, calculus (mathematics). Start a new answer booklet and mark it with a ‘B’.50 marks this section.6. Suppose n houses are located at the distinct points (x1; y1); (x2; y2); : : : ; (xn; yn). Apower substation must be located at a point such that the sum of squares of thedistances between the houses and the substation is minimised. Let (x; y) denotethe coordinates of the substation. If s is the sum of squares of the distances of thesubstation from the houses, then s = (x – x1)2 + (x – x2)2 + : : : + (x – xn)2 + (y –y1)2 + (y – y2)2 + : : : + (y – yn)2.(a) Find the location of the substation in the case that n = 3 and the houses arelocated at positions (0; 0); (2; 0); (1; 1):(b) Find the location of the substation in the case that n = 3 and the houses arelocated at positions (x1; y1); (x2; y2); (x3; y3):(c) Without performing any more calculations, predict what the optimaldistance would be for n houses.[4+4+4=12 marks]7. Suppose the temperature in Celsius at a point (x; y) in a region R is given byT (x; y) = x + 3y, where R is bounded by y = x2 and y = 8 – x2.(a) Find where the upper and lower boundaries meet each other, and sketch regionR.(b) Calculate the area of R.(c) Calculate the average temperature in R.Note. The average value of a function f on a two dimensional region R of area A,is given by A1 RRR f(x; y)dydx:[4 + 4 + 4 = 12 marks]8. Consider an object moving along a parametric curve whose position vector at time tis given byr(t) = 3t2 i + (t2 – 2) j + (1 – t2) k:(a) Compute the velocity v(t) and the acceleration a(t) of the object.(b) Compute the object’s speed v(t) and the arclength s of the part of the curvebetween t = 0 and t = 2.(c) Find the unit tangent vector T(t).(d) Show that the curvature κ of the curve is zero, and explain what this meansabout the shape of the curve.[3 + 3 + 2 + 3 = 11 marks]89. Consider the hemi-ellipsoid described by z = g(x; y) = p45 – 2×2 – 4y2.(a) Show that g(2; -3) = 1. (b) Write down expressions for the partial derivativesand .@g@g@x(c) Evaluate the first partial derivatives at x = 2; y = -3.@y (d) Find an equation of the plane tangent to the surface z = g(x; y) at (2; -3; 1).(e) Find an expression for the gradient vector rg(x; y) and evaluate rg(0; 3):(f) Compute the directional derivative Dug(0; 3) of g at the point P (0; 3) in thedirection of the vector v from P (0; 3) to Q(2; 1).(g) In what direction(s) is the directional derivative of g at (0; 3) equal to zero?[1 + 2 + 2 + 3 + 2 + 3 + 2 = 15 marks]END OF EXAMINATION QUESTIONSStatistical formulas and tables follow on succeeding pages : : :9Areas under the Gaussian curveProbabilities that a Unit Gaussianvariable lies between 0 and z. pppp p p p p p p p p p p p p pp p p p p p p p p p p p p p p pp p p p p p p p p p p p p p p pp p p p p p p p p p p p p p pp p p p p p p p p p p p pp p p p p p p p p p p pp p p p p p p p p p pp p p p p p p p pp p p p p p p p–6 0z Second decimal place of z0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09z0.00.10.20.30.40.50.60.70.80.91.01.11.21.31.41.51.61.71.81.92.02.12.22.32.42.52.62.72.82.93.03.13.23.33.43.53.63.73.83.90.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.03590.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.07530.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.11410.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.15170.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.18790.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.22240.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.25490.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.28520.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.31330.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.33890.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.36210.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.38300.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.40150.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.41770.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.43190.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.44410.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.45450.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.46330.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.47060.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.47670.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.48170.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.48570.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.48900.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.49160.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.49360.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.49520.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.49640.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.49740.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.49810.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.49860.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.49900.4990 0.4991 0.4991 0.4991 0.4992 0.4992 0.4992 0.4992 0.4993 0.49930.4993 0.4993 0.4994 0.4994 0.4994 0.4994 0.4994 0.4995 0.4995 0.49950.4995 0.4995 0.4995 0.4996 0.4996 0.4996 0.4996 0.4996 0.4996 0.49970.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.49980.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.4998 0.49980.4998 0.4998 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.49990.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.49990.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.4999 0.49990.5000 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 10Student’s t-DistributionCritical values of t –6p p p pp p pp p pp pp pp 0t Right-tail probabilities0.100 0.050 0.025 0.010 0.0051200νν12345678910111213141516171819202122232425262728293040608010012013.078 6.314 12.706 31.821 63.6571.886 2.920 4.303 6.965 9.9251.638 2.353 3.182 4.541 5.8411.533 2.132 2.776 3.747 4.6041.476 2.015 2.571 3.365 4.0321.440 1.943 2.447 3.143 3.7071.415 1.895 2.365 2.998 3.4991.397 1.860 2.306 2.896 3.3551.383 1.833 2.262 2.821 3.2501.372 1.812 2.228 2.764 3.1691.363 1.796 2.201 2.718 3.1061.356 1.782 2.179 2.681 3.0551.350 1.771 2.160 2.650 3.0121.345 1.761 2.145 2.624 2.9771.341 1.753 2.131 2.602 2.9471.337 1.746 2.120 2.583 2.9211.333 1.740 2.110 2.567 2.8981.330 1.734 2.101 2.552 2.8781.328 1.729 2.093 2.539 2.8611.325 1.725 2.086 2.528 2.8451.323 1.721 2.080 2.518 2.8311.321 1.717 2.074 2.508 2.8191.319 1.714 2.069 2.500 2.8071.318 1.711 2.064 2.492 2.7971.316 1.708 2.060 2.485 2.7871.315 1.706 2.056 2.479 2.7791.314 1.703 2.052 2.473 2.7711.313 1.701 2.048 2.467 2.7631.311 1.699 2.045 2.462 2.7561.310 1.697 2.042 2.457 2.7501.303 1.684 2.021 2.423 2.7041.296 1.671 2.000 2.390 2.6601.292 1.664 1.990 2.374 2.6391.290 1.660 1.984 2.364 2.6261.289 1.658 1.980 2.358 2.6171.282 1.645 1.960 2.327 2.5764030201512100 To obtain a critical valuefor a number of degrees offreedom ν not provided inthe table, interpolate usingthe values of 1200=ν shownin the right hand column.Thus for instance to obtainthe critical value for aone-sided test at asignificance level of 0.025with 45 degrees offreedom, calculate1200=45 = 26:67;and the required criticalvalue2:000+26:67 – 2030 – 20(2:021-2:000)= 2:014:Student’s” t-distributionis asymptotic to theNormal (Gaussian)distribution, so criticalvalues shown in the lastrow are for a Normallydistributed variable.11Useful FormulasSome notation µσpnmxX¯population meanpopulation standard deviationpopulation proportionsample sizemargin of errorobserved value of random variable Xmean of random variable Xx¯observed sample meanSssample standard deviationobserved sample standard deviation p^ sample proportionTest statisticsInference for µ when σ known: Z =X¯ – µσ=pn ∼ N(0; 1)Inference for µ when σ unknown: T =X¯ – µS=pn ∼ t(n – 1)Inference for p: Z = pp(1 p^–pp)=n ∼ N(0; 1) approx.Inference for µ1 – µ2, assume σ1 6= σ2T =X¯1 – X¯2 – (µ1 – µ2)pS12=n1 + S22=n2 ∼ t(ν) approx.where ν is computed by the Welch-Satterthwaite formula.Inference for p1 – p2: Z = pp¯p^(1 1 –p^p¯2)(1 -=n (p11-+ 1 p2=n ) 2) ∼ N(0; 1) approx.p¯ =x1 + x2n1 + n2under H0 : p1 = p2( x1 and x2 are the numbers of successes in samples of size n1 and n2.)Confidence intervalsConfidence interval for µ when σ known: ¯ x ± z∗pσnConfidence interval for µ when σ unknown: ¯ x ± t∗psnConfidence intervals for p: ^ p ± z∗rp(1n- p)Use 0:5 (conservative) or ^ p in place of the unknown p in this formula.12 End.
