ORF 245 Fundamentals of Engineering Statistics. Final Exam

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1 Priceto Uiversity Departmet of Operatios Research ad Fiacial Egieerig ORF 45 Fudametals of Egieerig Statistics Fial Exam May, 008 7:30pm-0:30pm PLEASE DO NOT TURN THIS PAGE AND START THE EXAM UNTIL YOU ARE TOLD TO DO SO. Istructios: This exam is ope book ad ope otes. Calculators are allowed, but ot computers or the use of statistical software packages. Write all your work i the space provided after each questio. There are questios o both sides of each page. Explai as thoroughly ad as clearly as possible all your steps i aswerig each questio. Full or partial credit ca oly be grated if itermediate steps are clearly idicated. Name: Pledge: I pledge my hoor that I have ot violated the hoor code durig this examiatio. Sigature: : () 6: (5) : () : (06) 7: (0) : (0) 3: (0) 8: (0) 3: (0) 4: (05) 9: (0) 4: () 5: (05) 0: (08) 5: (0) Total: (75)

2 Descriptive Statistics: ) Let x ad s deote the sample mea ad variace for the sample x,..., x ad let x + ad s + deote these quatities whe a additioal observatio x + is added to the sample. a) (4 pts.) Show how x + ca be computed from x ad x +. + x+ = xi = xi + x+ = x + + i= + i= + x + ( ) b) (8 pts.) Show that s ( ) + = ( ) s + x+ x + so that s + ca be computed from x +, x, ad s. Item dropped do ot grade.

3 ) Cosider the followig histogram that shows the time i moths that articles submitted to a certai scietific joural i 00 took to be reviewed for publicatio. a) (3 pts.) Which class iterval cotais the media review time? Readig the approximate areas uder the histogram (prob/cum): 0-: 0.7/0.7; -: 0./0.37; -3: 0.05/0.475; 3-4: 0./0.585; 4-5: 0.07/0.655; 5-6: 0.08/0.735; 6-7: 0.075/0.80; 7-8: 0.5/0.935; 8-9: 0.065/.0 The media review time falls i the 3 to 4 moth category. b) (3 pts.) Which class iterval cotais the third quartile of the review times? The third quartile of the review times falls i the 6 to 7 moth category. (A aswer of 5 to 6 moths is also acceptable, give the ucertaity i the readig of areas i the histogram; but 7 to 8 moths is ot acceptable.) 3

4 Probability: 3) Items are ispected for flaws by two quality ispectors. If a flaw is preset, it will be detected by the first ispector with probability 0.9, ad by the secod ispector with probability 0.7. Assume that the ispectors fuctio idepedetly. a) (4 pts.) If a item has a flaw, what is the probability that it will be foud by at least oe of the ispectors? { } Let I : evet that ispector i fids a flaw, i=, i c c = = = Pr( I flaw) 0.9, Pr( I flaw) 0., Pr( I flaw) 0.7, Pr( I flaw) = 0.3 Pr(flaw foud by at least oe ispector) = Pr( I I flaw) = Pr( I flaw) + Pr( I flaw) Pr( I I flaw) = = 0.97 b) (6 pts.) Assume that both ispectors ispect every item ad that if a item has o flaw, the either ispector will detect a flaw. Assume also that the probability that a item has a flaw is 0.0. If a item is passed by both ispectors, what is the probability that it actually has a flaw? c c Pr( I o flaw) = 0, Pr( I o flaw) =, Pr( I o flaw) = 0, Pr( I o flaw) = Pr(a item passed by both ispectors is actually flawed) = c c c c c c I c c I I c c Pr( I flaw) Pr (flaw I I) = Pr( ) Pr( I I ) = Pr( I I flaw) Pr(flaw) + Pr( I I o flaw) Pr(o flaw) c c I c c c c = Pr( I flaw) Pr( I flaw) Pr(flaw) + Pr( I o flaw) Pr( I o flaw) Pr(o flaw) = = c c Pr( I flaw) = Pr( I flaw) Pr( I flaw) Pr(flaw) = = c c Fially: Pr (flaw I I) = = c c Alterative solutio: diagram tree combied with coditioal probability. 4

5 4) (5 pts.) A ur cotais 3 red balls ad 7 black balls. Players A ad B withdraw balls from the ur cosecutively util a red ball is selected. Namely, A draws the first ball, the B draws the secod oe, the A agai, ad so o, util the first oe of them draws a red ball. If there is o replacemet of the draw balls, fid the probability that A selects the red ball. Pr(A selects red ball) = Pr(red o st draw) + Pr(first red o 3rd draw) + Pr(first red o 5th draw) + Pr(first red o 7th draw) = = = = or 58.33% Alterative solutio: = = Radom Variables: 5) (5 pts.) Two types of cois are produced at a factory: a fair coi ad a biased oe that comes up heads 55 percet of the time. We have a coi from this factory but do ot kow whether it is a fair coi or a biased oe. I order to ascertai which type of coi we have, we will perform the followig statistical test: we will toss the coi 000 times. If the coi lads o heads 55 or more times, the we will coclude that it is a biased coi, whereas, if it lads heads less tha 55 times, the we will coclude that it is the fair coi. If the coi is actually fair, what is the probability that we will reach a false coclusio? [Hit: use the Normal approximatio with cotiuity correctio.] Let X be the # of heads i 000 tosses of a fair coi The X Bi(000,0.5) X N(500, 50) Pr(test yields false coclusio) = Pr( X 55) = Pr Z = Φ (.5495) = = or 6.06% 50 5

6 6) (5 pts.) A bus travels betwee two cities A ad B, which are 00 miles apart. If the bus has a breakdow, the distace from the breakdow to city A has a uiform distributio over (0, 00). There is a bus service statio i city A, i B, ad i the ceter of the route betwee A ad B. It is suggested that it would be more efficiet to have the three statios located 5, 50, ad 75 miles, respectively, from A. Do you agree? Why? [Hit: compare the expected distace that the bus would have to be towed, from the breakdow poit to the earest service statio.] Let X be the distace from A to where the bus breaks dow: X Uif(0,00) Let Y be the distace from the breakdow poit to the earest service statio i case X if 0 X 5 50 X if 5 < X 50 The Y = is uiformly distributed i each of these itervals X 50 if 50 < X X if 75 < X 00 EY = E[ X 0 X 5] Pr(0 X 5) + E[50 X 5 < X 50] Pr(5 < X 50) + EX [ < X 75] Pr(50 < X 75) + E[00 X 75 < X 00] Pr(75 < X 00) = ( ) (6.5 50) ( ) 0.5 EY =.5 Now let Z be the distace from the breakdow poit to the earest service statio i case 5 X if 0 X 5 X 5 if 5 X 37.5 < 50 X if 37.5 < X 50 The Z = is uiformly distributed i each of these itervals X 50 if 50 < X X if 6.5 < X 75 X 75 if 75 < X 00 EZ = E[5 X 0 X 5] Pr(0 X 5) + E[ X 5 5 < X 37.5] Pr(5 < X 37.5) + E[50 X 37.5 < X 50] Pr(37.5 < X 50) + E[ X < X 6.5] Pr(50 < X 6.5) + E[75 X 6.5 < X 75] Pr(6.5 < X 75) + E[ X < X 00] Pr(75 < X 00) = (5.5) (3.5 5) ( ) ( ) ( ) ( ) 0.5 EZ = As EZ < EY, the havig service statios at 5, 50 ad 75 miles IS more efficiet. Alterate solutios: computig the expected values as itegrals rather tha coditioal expectatios; or graphig the distaces ad computig the areas uder the graphs (but, i this case, the areas have to be proportioal to the values above). 6

7 Joit Probability Distributios: 7) Choose a umber X at radom from the set of umbers {,,3,4,5 }. Now choose a umber at radom from the subset o larger tha X, that is, from {,..., X }. Call this secod umber Y. a) (0 pts.) Fid the joit probability mass fuctio of X ad Y. X py ( y) Y p ( x) X b) (7 pts.) Fid the expected value ad the variace of Y EY = EY = Var( Y ) = ( ) + ( ) + (3 ) (4 ) + (5 ) = Var( Y ) = c) (3 pts.) Are X ad Y idepedet? Explai. Note that pxy, (5,5) = px(5) py(5) = = Sice there is at least oe pair of values ( xy, ) for which p ( x, y) p ( x) p ( y), the X ad Y are NOT idepedet. XY, X Y 7

8 Statistical Estimatio: 8) (0 pts.) Maximum likelihood estimates possess the property of fuctioal ivariace, which meas that if ˆ θ is the MLE of θ, ad h( θ ) is ay fuctio of θ, the h( ˆ θ ) is the MLE of h( θ ). Give a radom sample X,..., X from a geometric distributio with parameter p, fid the MLE of the odds ratio p ( p). Let X, X,..., X be a radom sample of variable distributed as a Geom( p) The: p ( x) = ( p) p, for x 0 X x The joit p.m.f. of X,..., X is give by: x x x X,..., X (,..., ; ) ( ) ( )... ( ) = p x x p p p p p p p = ( p) The likelihood fuctio is thus: l[ p ( x,..., x ; p)] = ( x )l( p) + l p X,..., X i i x i i p The MLE for the parameter p is obtaied by derivatio of the likelyhood fuctio with respect to p: ( ) d l[ px,..., X ( x,..., x ; )] p x i i pˆ pˆ = 0 + = 0 = or = dp pˆ pˆ pˆ x pˆ x i i 8

9 Cofidece Itervals: 9) Let X represet the umber of evets that are observed to occur i uits of time or space, ad assume that X Poisso λ, where λ is the mea umber of evets that ( ) occur i oe uit of time or space. Assume that is large, so that X N λ, λ. A suitable estimator of λ is give by ˆ λ = X, with stadard error SE( ˆ λ) = λ. a) (4 pts.) Assumig that X is large, what is the distributio of ˆλ? (Name the distributio ad tell the values of its parameters.) ˆ E( λ) = EX = λ ˆ λ N, ˆ λ Var( λ) = Var( X ) = X ( ) ( λ λ ) b) (4 pts.) Use the distributio foud i the previous item ad the fact that SE ( ˆ λ) ˆ λ to derive a expressio for the 00( α ) % cofidece iterval for λ. ( ˆ Give that λ λ) N(0,), the the 00(- α)% CI for λ is give by: ˆ λ ˆ ( λ z ˆ, ˆ z ˆ α λ λ + α λ ) c) (4 pts.) A 5 ml sample of a certai suspesio is foud to cotai 300 particles. The mea umber of particles per ml i the suspesio is 60, give or take ˆ λ = 300 = 60 ad SE( ˆ λ) 60 = = d) (4 pts.) After 4 miutes, a geologist couted 56 particles emitted from a certai radioactive rock. Fid a 95% cofidece iterval for the rate of emissios i uits of particles per miute. ˆ λ = 56 = 64 ad SE( ˆ λ) 64 = 4 ad z0.05 = Thus the 95% CI for λ is: , = (56.6,7.84) ( ) 9

10 e) (4 pts.) For how may miutes should particles be couted so that the 95% cofidece iterval specifies the rate to withi ± particle per miute? ˆ We wat z λ ˆ 0.05 = = λz0.05 = = 45.9 For 46 miutes. 0) A sample of seve cocrete blocks had their compressive stregth measured i MPa. The results were 367.6, 4.5, 38.7, 93.6, 406., 45.7, ad Te thousad bootstrap samples were geerated from these data, ad the bootstrap sample meas were arraged i order. Refer to the smallest mea as Y, the secod smallest as Y, ad so o, with the largest beig Y0000. Assume that Y 50 = 83.4, Y 5 = 83.4, Y 00 = 9.5, Y 0 = 9.5, Y 50 = 305.5, Y 5 = 305.5, Y 500 = 38.5, Y 50 = 38.5, Y 9500 = 449.7, Y 950 = 449.7, Y 9750 = 46., Y 975 = 46., Y 9900 = 476., Y 990 = 476., Y 9950 = 483.8, ad Y 995 = a) (4 pts.) Compute the 95% bootstrap cofidece iterval for the mea compressive stregth. Y + Y Y + Y = % CI for the mea =, (305.5,46.) b) (4 pts.) Was this a parametric or a oparametric bootstrap procedure? Explai. Noparametric: the te thousad samples were geerated through radom samplig, with replacemet, from the give sample, without ay iformatio o the distributio of the populatio uderlyig the sample. 0

11 Tests of Hypothesis: ) A article by Abdel-Aty et al. i the Joural of Trasportatio Egieerig presets a tabulatio of types of car crashes by the age of the driver over a three-year period i Florida. Here is the table: Age of drivers 5-4 years 5-64 years Total # of accidets 8,486 9,70 # of accidets i driveways 4,43 0,70 a) (4 pts.) The differece betwee the proportios of driveway accidets for drivers aged 5-4 ad drivers aged 5-64 is 0.6 %, give or take %. pˆ5 = 443 = pˆ5 = 070 = pˆ5 pˆ5 = or 0.6% pˆ5( pˆ5) pˆ5( pˆ5) SE( pˆ5 pˆ5) = + = = or % b) (4 pts.) Ca you coclude that driveway accidets amog 5-4 year-olds i FL are ideed likely to be proportioately higher tha driveway accidets amog 5-64 year-old Floridias? State the hypotheses clearly ad aswer this questio usig the P-value. H : p p 0 H : p p > pˆ5 pˆ 5 ( p5 p5) z = where pˆ pool = = SE( pˆ ) pool SE ( pˆ ) ˆ ( ˆ ) pool = ppool p pool = z = =.94, thus P-value = Pr( Z.94) = or 0.6% reject H at sigificace level % 0 Thus: youger Floridias do have a higher rate of driveway accidets tha older oes c) (4 pts.) Assumig that youg drivers i Florida do preset a higher proportio of driveway accidets tha older drivers, does this mea that youger Floridia drivers should be required to take a special course o how to drive o driveways, but ot older drivers? Explai. Though statistically speakig youger Floridias do have a higher rate of driveway accidets tha older oes, practically speakig the differece is too small (0.6%) to justify differetiated drivig traiig for the two groups. This is thus a typical case i which statistical sigificace does ot traslate ito practical sigificace.

12 ) A egieer claims that a ew type of hard disk for laptops lasts loger tha the old type. Idepedet radom samples of 75 of each of the two types are chose, ad the sample meas ad stadard deviatios of their lifetimes are computed: New: X = 4387 h s = 5 h Old: X = 460 h s = 3 h a) (4 pts.) Ca you coclude that the mea lifetime of ew hard disks is greater tha that of the old hard disks? State the hypotheses clearly ad aswer this questio at the % sigificace level. Item dropped do ot grade. b) (4 pts.) If the ew hard disks have ideed a mea lifetime 40 h loger tha the old oes, what is the probability ( β ) that the test performed i the previous item will icur ito error of type II (that is, failig to reject )? H 0 Item dropped do ot grade. c) ( pts.) Recompute the probability of error type II for the case of the ew hard disks havig a mea lifetime 80 h loger tha the old oes. Item dropped do ot grade.

13 Correlatio ad Liear Regressio: 3) A chemical egieer is studyig the effect of temperature ad stirrig rate o the yield of a certai product. The process is ru 6 times, at the settigs idicated i the followig table. The uits for yield are percet of a theoretical maximum. The matrix of sample correlatio coefficiets amog the variables i questio is as follows: a) (5 pts.) Based o the aalysis of sample correlatio above, would you try ad fit a multiple liear regressio model i which the yield is the respose variable ad temperature ad stirrig rates are the covariates? Explai. No, it is ot advisable to fit a model where both covariates are used, because there is a high level of liear correlatio betwee temperature ad stirrig rate (0.9064). This is kow as multicolliearity, ad it will cofoud the least squares estimatio of the liear regressio coefficiets. 3

14 b) (5 pts.) Fid the 95% cofidece iterval for the coefficiet of correlatio betwee the stirrig rate ad the yield. What assumptios did you make i order to compute this cofidece iterval? Assumig that stirrig rate ad yield come from a bivariate ormal distributio, the, + r + ρ by the Fisher trasformatio: V = l N l, r ρ 3 c e c e ad a 95% CI for ρ will be give by,, where c 0.05 / 3 ad 0.05 / 3 c c = v z c = v+ z e + e Thus: for v= l = , z0.05 =.96 c = / 3 = 0.43 ad c = / 3 =.59 Ad fially, the 95% CI for ρ is: (0.407, 0.909). 4) The chemical egieer from the previous questio has decided to calibrate a simple liear regressio model with the yield as the respose variable ( Y ) ad stirrig rate as the covariate ( X ). The results of the calibratio obtaied through Excel are: a) ( pts.) What proportio of the observed variatio i yield ca be attributed to the simple liear regressio relatioship betwee yield ad stirrig rate? r = = % b) (5 pts.) Ca you say that a icrease of 0 rpm i the stirrig rate will produce a icrease i yield of at least %? State the hypotheses clearly ad aswer this questio at the 5% sigificace level. H t : β 0. H : β > = 0. 0 ˆ β β = 0 = =.58 SE ˆ β P-value = Pr( T.58) = or 7.4% caot reject H at 5% 4 0 Thus, we caot say that 0 rpm will icrease yield by at least %. 4

15 c) (5 pts.) Costruct the 95% cofidece iterval for the predictio of the yield percetage that correspods to a stirrig rate of 55 rpm. I order to compute this iterval, you may eed the followig additioal iformatio: * * Give x = 55 ad yˆ = x, the yˆ = = 78.7 * * * * The 95% CI for y is: yˆ ± t SE ( yˆ x ), where 0.05,4 * * * ( x x ) SE pred ( yˆ x ) = ˆ σ + + S xx pred Computig: Syy ( r ) ˆ σ = = 34.5 ( 0.564) =.70 4 * * (55 45) SE pred ( yˆ x = 55) = = Fially: * yˆ t SE * * ( yˆ x ) = = ,4 * * * yˆ + t SE ( yˆ x ) = = ,4 pred pred Thus, the 95% CI * * for ( y x = 55) is: (7.5,84.9) 5

16 Multiple Liear Regressio: 5) A study was made i which data was obtaied to relate y = specific surface area 3 ( cm /g ) to x = % NaOH used as a pretreatmet chemical ad x = treatmet time (mi) for a batch of pulp. The followig R output resulted from a request to fit the Y = β + β x + β x +ε. model 0 a) (6 pts.) Fill i the blaks i the tables above by computig the followig values: the coefficiets of determiatio regular ad adjusted, the regressio sum of squares, the mea sums of squares regressio ad residuals, ad the value of the F statistics. Show your computatios. Item dropped do ot grade. b) ( pts.) What proportio of observed variatio i specific surface area ca be explaied by the model relatioship? Item dropped do ot grade. 6

17 c) (4 pts.) Does the chose model appear to specify a useful relatioship betwee the respose ad the covariates? Explai. Item dropped do ot grade. d) (4 pts.) Provided that % NaOH remais i the model, would you suggest that the covariate treatmet time be elimiated? Explai. Item dropped do ot grade. e) (4 pts.) Calculate a 95% cofidece iterval for the expected chage i specific surface area associated with a icrease of % i NaOH whe treatmet time is held fixed. Item dropped do ot grade. 7