TAMS11: Probability ad Statistics Provkod: TENB 23 March 2016, 14:00-18:00 Examier: iagfeg Yag Tel: 070 0896661 Please aswer i ENGLISH if you ca a You are allowed to use: a calculator; formel -och tabellsamlig i matematisk statistik from MAI; TAMS 11: Notatios ad Formulae by iagfeg Yag, a dictioary b Scores ratig: 8-11 poits givig rate 3; 115-145 poits givig rate 4; 15-18 poits givig rate 5 1 3 poits Eglish Versio Let A ad B be idepedet evets ad 11 1p Fid P A B 12 2p Fid P A A B P A = P B = 05 2 3 poits A ur cotais 100 balls, of which 20 are white ad the rest black Oe radomly draw with replacemet 60 balls Fid the probability that there are at least 20 white balls 3 3 poits Assume that oe has obtaied the followig sample from a radom variable which is Nµ, σ 2 : x 1 = 20, x 2 = 18, x 3 = 24, x 4 = 17, x 5 = 21 31 1p Costruct a 99% two-sided cofidece iterval for µ if oe kows that σ = 2 32 1p Costruct a 99% two-sided cofidece iterval for µ if σ is ukow 33 1p Costruct a 99% oe-sided cofidece iterval for σ i the form 0, b 4 3 poits The followig give values: 385, 575, 484, are a sample from a distributio with a probability desity fuctio fx = 1 θ e x/θ, x > 0, θ > 0, where θ is ukow 41 15p Fid a poit estimate ˆθ ML of θ usig Maximum Likelihood-method 42 15p Fid a poit estimate ˆθ MM of θ usig Method of Momets Page 1/2
5 3 poits Assume that oe has obtaied the followig sample from a radom variable which is Nµ, 09 2 : 115, 94, 124 We wat to test the followig hypothesis with a sigificace level 005: H 0 : µ = 108, vs H a : µ > 108 51 1p Reject H 0? Use T S test statistic ad C rejectio regio 52 1p Reject H 0? Use p-value 53 1p Fid the power for µ = 110 6 3 poits Assume that, Y have the joit probability desity fuctio { e y, for 0 < x < y, fx, y = 0, otherwise 61 2p Fid the margial probability desity fuctios f x for ad f Y y for Y 62 1p Fid P > 2 Y < 4 Page 2/2
TAMS11: Probability ad Statistics Provkod: TENB 23 mars 2015, kl 14-18 Examiator: iagfeg Yag Tel: 070 0896661 Välige svara på ENGELSKA om du ka a Tillåta hjälpmedel är: e räkare; formel -och tabellsamlig i matematisk statistik frå MAI; TAMS 11: Notatios ad Formulae by iagfeg Yag; e ordbok b Betygsgräser: 8-11 poäg ger betyg 3; 115-145 poäg ger betyg 4; 15-18 poäg ger betyg 5 1 3 poäg Svesk Versio Låt A och B vara oberoede hädelser och 11 1p Beräka P A B 12 2p Beräka P A A B P A = P B = 05 2 3 poäg E ura iehåller 100 kulor, varav 20 är vita och reste svarta Ma drar på måfå och med återläggig 60 kulor Beräka saolikhete att få mist 20 vita kulor 3 3 poäg Ata att ma har fått följade stickprov på e stokastisk variabel som är Nµ, σ 2 : x 1 = 20, x 2 = 18, x 3 = 24, x 4 = 17, x 5 = 21 31 1p Beräka ett 99% tvåsidigt kofidesitervall för µ om ma vet att σ = 2 32 1p Beräka ett 99% tvåsidigt kofidesitervall för µ om σ är okät 33 1p Beräka ett 99% esidigt kofidesitervall för σ i form 0, b 4 3 poäg Följade giva värde: är ett stickprov frå e fördelig med täthetsfuktioe 385, 575, 484, fx = 1 θ e x/θ, x > 0, θ > 0, där θ är okät 41 15p Hitta e puktskattig ˆθ ML av θ geom att aväda Maximum Likelihood-metode 42 15p Hitta e puktskattig ˆθ MM av θ geom att aväda mometmetode Page 1/2
5 3 poäg Ata att ma har fått följade stickprov på e stokastisk variabel som är Nµ, 09 2 : Vi vill pröva på ivå 005: 115, 94, 124 H 0 : µ = 108, mot H a : µ > 108 51 1p H 0 förkastas? Aväd T S test statistic och C rejectio regio 52 1p H 0 förkastas? Aväd p-value 53 1p Beräka styrka för µ = 110 6 3 poäg Ata att, Y har täthetsfuktioe fx, y = { e y, för 0 < x < y, 0, aars 61 2p Bestäm de margiella täthetsfuktioera f x för och f Y y för Y 62 1p Beräka P > 2 Y < 4 Page 2/2
TAMS11: Notatios ad Formulas by iagfeg Yag 1 Basic probability 11 Coditioal probability P A B = P A B P B 12 Total probability P B = k i=1 P B AiP Ai where {Ai} are disjoit ad k i=1 Ai = S the sample space P B AjP Aj 13 Bayes Theorem P Aj B = k where {Ai} are i 12 i=1 P B AiP Ai 2 Radom variables rv 21 Discrete rv has a pmf px = P = x satisfyig px 0 ad pxi = 1, x1 x2 x px px1 px2 px Expectatio or Expected value or mea µ = E = xipxi; Variace σ 2 = V = E µ2 = E 2 µ 2 = x 2 i pxi xipxi 2 22 Cotiuous rv has a pdf fx satisfyig fx 0 ad fxdx = 1, b P a < < b = fxdx a Expectatio or Expected value or mea µ = E = xfxdx; Variace σ 2 = V = E µ2 = E 2 µ 2 = 2 x2 fxdx xfxdx 23 Cumulative distributio fuctio cdf of a rv is F x = P x 24 If ad Y are rv, a, b ad c are scalars, the Ea + by + c = ae + bey + c If ad Y are idepedet rv, a, b ad c are scalars, the V a + by + c = a 2 V + b 2 V Y 25 Two discrete rv ad Y form a two-dimesioal discrete rv, Y with a joit pmf px, y satisfyig px, y 0 ad pxi, yi = 1 xi yi The margial pmf of is px = y px, y; The margial pmf of Y is py y = x px, y; ad Y are idepedet if px, y = px py y Two cotiuous rv ad Y form a two-dimesioal cotiuous rv, Y with a joit pdf px, y satisfyig fx, y 0 ad fx, ydxdy = 1 The margial pdf of is fx = fx, ydy; The margial pdf of Y is fy y = fx, ydx; ad Y are idepedet if fx, y = fx fy y 3 Special rv 31 Several discrete rv Bi, p has a pmf px = P = x = p x 1 p x, x = 0, 1, 2,, x E = p, V = p 1 p There are idepedet ad idetical trials Each trial has success or failure, ad the probability of success is p Now = # of successes P oλ has a pmf px = P = x = e λ λ x, x = 0, 1, 2, x! 1/4 E = λ, V = λ = # of happeigs i a fixed period of time or a fixed legth, volume ad so o Hypergeometric has a pmf M N M x x px = P = x = N A box cotais N balls where M are white ad N M are black Now choose balls without replacemet The px is the probability that there are exactly x white balls i these selected balls 32 Several cotiuous rv Expλ has a pdf { λe λx, x 0, fx = 0, otherwise E = 1 λ, V = 1 2 λ = waitig time betwee two happeigs Nµ, σ 2 has a pdf fx = 1 σ 2π e x µ 2 2σ 2, < x < E = µ, V = σ 2 Whe we model large umber of observatios N0, 1 has a pdf fx = 1 2π e x2 2, < x < E = 0, V = 1 Ua, b has a pdf fx = { 1 b a, a < x < b, 0, otherwise E = a+b b a2 2, V = 12 4 Cetral Limit Theorem CLT Suppose that a populatio has mea= µ ad variace= σ 2 A radom sample {1, 2,, } from this populatio is give The for large 30, µ σ/ N0, 1 1 If the populatio is ormal, the 1 holds for ay Uderstad that µ = E ad σ/ 2 = V 5 Several otatios i statistics 51 Sample mea: = 1+2+ = i 52 Sample variace: S 2 = i 2 1 ; s 2 = xi x 2 1 ; x = x1+x2+x = xi 53 A poit estimator of a ukow parameter θ obtaied by Methods of Momets is deoted as ˆθMM 54 A poit estimator of a ukow parameter θ obtaied by Maximum Likelihood method is deoted as ˆθML Capital letters such as ad S 2 refer to the objects before measure/observe, ad they are i geeral rv Small letters such as x ad s 2 refer to the objects after measure/observe, ad they are i geeral scalars 2/4
6 Cofidece Itervals CI Part 1: 1 α CI of a populatio mea µ case 11 ay If populatio Nµ, σ 2 ad σ 2 is kow, the µ σ/ N0, 1 ad Iµ = x z α/2 σ, x + z α/2 σ case 12 30 For arbitrary populatio σ 2 is kow or ukow, it holds that Iµ = x z α/2 σ, x + z α/2 σ or Iµ = x z α/2 s, x + z α/2 s µ σ/ N0, 1 ad case 13 ay If populatio Nµ, σ 2 ad σ 2 is ukow, the µ S/ T 1 ad Iµ = x t α/2 1 s, x + t α/2 1 s Part 1 : 1 α CI of differeces of two idepedet populatio meas µ µy case 11 ay 1, 2 If populatios Nµ, σ 2, Y NµY, σ2 Y, ad σ2, σ2 Y are kow, the Ȳ µ µy N0, 1, ad σ 2 + σ2 Y 1 2 x Iµ = σ 2 ȳ z µy α/2 + σ2 Y, x ȳ + z σ 2 α/2 + σ2 Y 1 2 1 2 case 12 1, 2 30 For arbitrary populatios ad Y σ 2, σ2 Y are kow or ukow, it holds that Ȳ µ µy N0, 1, ad σ 2 + σ2 Y 1 2 Iµ µy = x ȳ z α/2 σ 2 1 + σ2 Y 2, x ȳ + z α/2 σ 2 1 + σ2 Y 2 or Iµ µy = x ȳ z α/2 s 2 1 + s2 Y 2, x ȳ + z α/2 s 2 1 + s2 Y 2 case 13 ay 1, 2 If populatios Nµ, σ 2, Y NµY, σ2 Y, ad σ2, σ2 Y are ukow, the case 131 if σ 2 = σ2 Y, the Ȳ µ µy S 1 1 + 1 2 T 1 + 2 2, where S 2 = 1 1S2 + 2 1S2 Y, ad 1 + 2 2 Iµ µy = x ȳ t α/2 1 + 2 2 s 1 1 + 1 2, x ȳ + t α/21 + 2 2 s 1 1 + 1 2 case 132 if σ 2 σ2 Y, the Ȳ µ µy T v, where v = S 2 + S2 Y 1 2 s 2 /1 + s 2 Y /2 2, ad s 2 /12 1 1 + s2 Y /22 2 1 Iµ µy = x ȳ t α/2 v s 2 1 + s2 Y 2, x ȳ + t α/2v s 2 1 + s2 Y 2 3/4 Part 2: 1 α CI of a populatio variace σ 2 for ay If populatio Nµ, σ 2 ad σ 2 is ukow, the 1S2 σ 2 χ2 1, ad 1s 2 1s 2 I σ 2 = χ 2 α/2 1, χ 2 1 α/2 1 Part 3: 1 α CI of a populatio proportio p whe ˆp 10 ad 1 ˆp 10 ˆP If the populatio is ukow with a populatio proportio a true rate p, the p N0, 1 ad p1 p/ ˆp1 ˆp ˆp1 ˆp Ip = ˆp z α/2, ˆp + z α/2 where ˆp is the sample proportio defied as ˆp = x/ Everythig writte above is for TWO-SIDED cofidece itervals We also have ONE-SIDED cofidece bouds 1 α upper cofidece boud is i the form, b : replace every α/2 by α; upper cofidece boud of populatio variace σ 2 is i the form 0, b sice σ 2 is always oegative upper cofidece boud of populatio proportio p is i the form 0, b sice p is always oegative 1 α lower cofidece boud is i the form a, + : also replace every α/2 by α lower cofidece boud of populatio proportio p is i the form a, 1 sice p is always 1 4/4