ORF 245 Fundamentals of Engineering Statistics. Midterm Exam 2

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1 Priceto Uiversit Departmet of Operatios Research ad Fiacial Egieerig ORF 45 Fudametals of Egieerig Statistics Midterm Eam April 17, 009 :00am-:50am PLEASE DO NOT TURN THIS PAGE AND START THE EXAM UNTIL YOU ARE TOLD TO DO SO. Istructios: This eam is ope book ad ope otes. Calculators are allowed, but ot computers or the use of statistical software packages. Write all our work i the space provided after each questio. Use the other side of the page, if ecessar. Eplai as thoroughl ad as clearl as possible all our steps i aswerig each questio. Full or partial credit ca ol be grated if itermediate steps are clearl idicated. Name: Pledge: I pledge m hoor that I have ot violated the hoor code durig this eamiatio. Pledge: Sigature: 1: (0) : (0) 3: (0) 4: (0) 5: (0) Total: (0)

2 1. (0 pts.) Automobiles arrive at a vehicle equipmet ispectio statio accordig to a Poisso process with a rate of α = per hour. Suppose that with probabilit 0.5 a arrivig vehicle will have o equipmet violatios. a) What is the probabilit that eactl te arrive durig the et hour ad all te have o violatios? P(X = ad o violatios) = P(o violatios X = ) P(X = ) =( 0.5) () e! = ( )(0.15) = b) For a fied, show that the probabilit that arrive durig the et hour, of which te have o violatios, is give b e (5).!( )! P( arrive ad eactl have o violatios) = P(eactl have o violatios arrive) P( arrive) () = P( successes i trials whe p =.5) e! () e (5) = (.5) (.5) e =.!!( )! (5) 5 c) Usig the result from part (b) ad the fact that = e show that u= 0 ( u)! the probabilit that te o-violatio cars arrive durig the et e 5 hour is give b 5 (ote that this meas that the # of oviolatio cars arrivig durig the et hour follows a Poisso! process with a rate of 5 per hour). e (5) P(eactl without a violatio) = =!( )! = e 5 (5) = u e 5 (5) e 5 5 = e! = ( )!! u= 0 ( u)!! e 5 5 =.! u

3 . (0 pts.) a) A statisticia reports that the life, i thousads of miles, of a certai tpe of electroic cotrol for locomotives has a approimate logormal distributio with parameters μ =.36 ad, where he has take commo (base-) logarithms. Compute the fractio of these cotrols that would fail o a 80,000-mile warrat. σ = 0.30 P( X 80) = P(log X log 80) = P(log X ) = P Z =Φ( 1.040) = or 14.9% 0.30 b) We usuall have about five power outages i a four-week period. Makig appropriate distributioal assumptios, what is the chace that we get through a week without a outage? Solutio A: assume that the time betwee outages is a epoetial radom variable with rate λ, give i # of outages per week ˆ λ = 5 4 = 1.5 outages/week; PX ( t) = 1 e PX ( 1) = 1 PX ( 1) = 1 (1 e ) = e = or 8.65% ˆ λt Solutio B: assume that outages o a da follow a Beroulli process, that is, there is a probabilit p of a outage happeig o a da ad a probabilit (1 p) of o outages happeig o a da; assume further that the occurrece of outages o successive das are idepedet (this solutio is less 'iterestig' tha A, but still plausible) ( ˆ) ( ) pˆ = 5/ 8 = P(o outages i 7 das) = 1 p = =.814 =.53 or 5.3% c) Still i the same case as part (b), with probabilit 75 percet how log ca we go without a outage? Solutio A: 1.5c PX ( c) = 0.75 e = c= l(0.75) c = 0.3 weeks or 1.61 das Solutio B: ( ˆ ) P(0 outages i das) =.75 1 p = = betwee 1 ad das

4 3. (0 pts.) I geetic ivestigatios oe frequetl samples from a m biomial distributio f( ) = p ( 1 p) m ecept that observatios of = 0 are impossible; so, i fact, the samplig is from the coditioal m m p (trucated) distributio ( 1 p), for { 1,,..., m m}. Fid the 1 ( 1 p) maimum likelihood estimator of p i the case m = for samples of size. 1 p (1 p) p (1 p) For m=, px ( ) = =, for 1 or = = 1 (1 p) p Thus the joit p.m.f. of the radom sample X, X,..., X is: (1 ) 1 p p p (1 p) p (1 p) px1,..., X ( 1,..., ; ) p =... 1 p p p + i = p (1 p) = 1 i ( p) Takig the logarithm of the likelihood fuctio, we get: 1 ( ) ( ) ( ) ( ) l f( p) = l + + l l 1 l p p p + Now take the derivative of the log fuctio with respect to p ad set it to 0: ( )( ˆ)( ˆ) ( i ) ( = ) dl f( p) 1 1 = = 0 dp pˆ 1 pˆ pˆ ( ) ( ) ( ) ( ) ( ) ( + ) + 1 p p pˆ pˆ + pˆ 1 pˆ = 0 pˆ = + pˆ = pˆ =. Alterate likelihood fuctio: p (1 p) p (,..., ; p) = ( 1 ( 1 p) ) X1,..., X 1 dl f( p) 1 = 0 dp pˆ... pˆ =. ( ) ( ) 1 (1 pˆ ) = 0 1 pˆ 1 1 ˆ ( p)

5 4. (0 pts.) Whe the populatio distributio is ormal ad is large, the sample stadard deviatio S has approimatel a ormal distributio with ES ( ) σ ad var( S) σ ( ). We alread kow that i this case, for a, X is ormal with E( X) = μ ad var( X) = σ. a) Assumig that the uderlig distributio is ormal, what is a approimatel ubiased estimator ˆ θ of the 99 th percetile θ = μ+.33σ? ( X +.33S ) = E( X ) +.33E( S) μ +. 33σ E, so ˆ = X +. 33S approimatel ubiased. θ is b) Whe the X i s are ormal, it ca be show that X ad S are idepedet rv s (oe measures locatio whereas the other measures spread). What is the estimated stadard error SE ( ˆ θ ) for the estimator ˆ θ of part (a)? var (.33 ) var( ).33 var( ) σ X S X S σ + = + +. The estimated stadard s error (stadard deviatio) is c) If soil ph is ormall distributed i a certai regio ad 64 soil samples ield = 6.33, s = 0.16, does this provide strog evidece for cocludig that at most 99% of all possible samples would have a ph of less tha 6.75? Write the appropriate hpothesis testig for the parameter θ, that is, the 99 th percetile, ad perform the test usig α = This problem has bee wrogl stated. The iteded questio was to ask for evidece that at least 99% of all possible samples have a ph of less tha I this case, the solutio is that more tha 99% of all soil samples have ph less tha 6.75 iff the 99 th percetile is less tha Thus we wish to test H0 : μ +.33σ 6.75 vs. Ha : μ +.33σ < H 0 will be rejected at level.0f z Sice ( +.33s) = = = s /.0385 z, is ot rejected at the.01 level. The 99 percetile is ot sigificatl less tha As it is stated, though, the actual test (which becomes perfuctor) is H 0 : μ +.33 σ vs. H a : μ +.33σ > H0 will be rejected at level.0f z.33. Sice z = 1., there is o evidece at all to reject H 0. Give the circumstaces, either aswer will be acceptable for this problem! H 0 th

6 5. (0 pts.) A egieer claims that a ew tpe of power suppl for home computers lasts loger tha the old tpe. Idepedet radom samples of 75 of each of the two tpes are chose, ad the sample meas ad the stadard deviatios of their lifetimes are computed: New: 1 = 4387h s 1 = 5h Old: = 460h s = 31h a) Ca ou coclude at the 5% sigificace level that the mea lifetime of the ew power supplies is more tha 50h loger tha that of the old power supplies? State the appropriate hpothesis ad show our computatios. H : μ μ 50 vs H : μ μ > Δ0 s1 s 1) zobs =, where SE = + SE m SE = + = zobs = = z = z > z reject H at the 5% sigificace level 0.05 obs ( z obs ) ( ) ) P-value = 1 Φ = 1 Φ = = 0.06 P-value=.6% reject H 0 b) What is the power of the test performed i part (a) if the actual mea lifetime of the ew power supplies were 0h loger tha that of the old power supplies? Show our computatios. Power at Δ= 0 is give b: Δ Δ f( Δ ) = 1 Φ z0.05 f(0) SE = Φ f (0) = 1 Φ = = ( ) The power of the test if the actual differece were 0h is 35.3%. c) Assumig that our coclusio i part (a) was affirmative, would it be justifiable for the compa to sped millios of dollars i a ew marketig strateg cetered o the icreased lifetime of the power supplies? Though statisticall sigificat, I do t thik that i practical terms a improvemet of 50h i a mea lifetime of about 4300h is a sufficiet reaso for spedig millios of $ i a ew marketig campaig that would claim solel that the mea lifetime of the power supplies has bee icreased b 1.%. However, if ou thik strogl about the statistical sigificace, I will accept a dissetig opiio

Common Large/Small Sample Tests 1/55

Common Large/Small Sample Tests 1/55 Commo Large/Small Sample Tests 1/55 Test of Hypothesis for the Mea (σ Kow) Covert sample result ( x) to a z value Hypothesis Tests for µ Cosider the test H :μ = μ H 1 :μ > μ σ Kow (Assume the populatio