Kurskod: TAMS28 Matematisk statistik/provkod: TEN 208-08-20 08:00-2:00 Examiator: Zhexia Liu Tel: 0700895208 You are permitted to brig: a calculator, ad formel -och tabellsamlig i matematisk statistik Scores ratig: 8- poits givig rate 3; 5-45 poits givig rate 4; 5-8 poits givig rate 5 3 poits Eglish Versio A populatio X is discrete ad has a probability mass fuctio as follows X 2 3 px 04 02 04 A radom sample {X,, X 00 is take from this populatio Fid the probability P X + + X 00 20 Solutio µ = 04 + 2 02 + 3 04 = 2, σ 2 = 2 04 + 2 2 02 + 3 2 04 µ 2 = 08 σ = 08944 P X + + X 00 20 = P X 20/00 = P X µ σ/ 2 µ σ/ = P N0, 2 = Φ2 = 08686 = 034 2 3 poits Assume that X has a probability desity fuctio fx = cx 2 + 2x, for 0 x 2 2 p Fid the costat c 22 2p Fid the coditioal probability P X < 5 X Solutio 2 = 2 0 cx2 + 2xdx = c[x 3 /3 + x 2 ] 2 0 = 20c/3 c = 3/20 = 05 22 P X < 5 X = P X < 5 P X = c[x3 /3 + x 2 ] 5 c[x 3 /3 + x 2 ] 2 = 5 cx 2 + 2xdx 2 cx2 + 2xdx = c 2042 c 5334 = 038 3 3 poits A populatio X has a probability desity fuctio fx = θ + x θ for x 0 It is kow that the parameter θ is either 2, 3 or 4 A sample {02, 08 is take from this populatio Based o this sample, fid a poit estimate of θ usig Maximum-Likelihood method Solutio The likelihood fuctio is Lθ = fx fx 2 = θ + x θ θ + x 2 θ = θ 2 26 θ Sice θ ca oly take 2, 3 or 4, we just put these three values i Lθ ad fid which is the maximum L2 = 03969, L3 = 0435, L4 = 03403 ˆθ = 3 /3
4 3 poits Oe has measured the levels of a certai cotamiatio i soil samples take partly i the viciity of a idustry, ad partly i a purely area with the correspodig soil The results are: i x i s i Near idustry 8 80 049 Clea area 9 0 046 Oe ca assume that the values for ear idustry are observatios from X Nµ, σ, ad the values i clea area are observatios from X 2 Nµ 2, σ which is idepedet of X 4 5p Costruct a oe-sided 95% cofidece iterval of µ µ 2 i the form a, + 42 5p Costruct a oe-sided 99% cofidece iterval of σ 2 i the form 0, b Solutio 4 I µ µ 2 = x x 2 t α + 2 2 s +, + 2 = 80 0 75 04742 8 + 9, + = 07 0403, + = 0297, +, where s 2 = s2 + s2 2 + 2 2 = 02249 s = 04742 42 I σ 2 = 0, + 2 2s 2 χ 2 α + 2 2 8 + 9 202249 = 0, χ 2 099 8 + 9 2 = 0, 33735 = 0, 0645 523 5 3 poits Assume that Let Y = Y = X + Y 2 Solutio It is easy to see that X 0 X = N, X 2 5 2 Fid the probability P Y 35 + Y 2 32 Y = Y N Y 2 2, 85 4 2 2 9 4 2 2 9 Now let Z = Y + Y 2 = AY where the matrix A = The the mea of Z is µ Z = Aµ Y = 2 = 297, 85 ad the variace of Z is So Z N297, 7 Thus σz 2 = AC Y A = 4 2 = 7 2 9 P Y + Y 2 32 = P N0, 32 297/ 7 = Φ056 = 0723 2/3
6 3 poits A dice has bee throw 20 times with the followig result: Outcome 2 3 4 5 6 Frequecy 0 32 2 8 23 25 Does the result idicate that the dice is ubalaced? Perform a χ 2 test with a 5% sigificace level Solutio It is the clear that the test statistic is ad the rejectio regio is H 0 : p = p 2 = p 3 = p 4 = p 5 = p 6 = /6 agaist H : some p i /6 T S = 6 y i p i 2 = 73, p i i= C = χ 2 α6, + = 070, + Sice T S C, we reject H 0, amely, the dice is ideed ubalaced 3/3
Svesk versio 3 poäg E populatio X är diskret och har e saolikhetsfuktio eligt följade X 2 3 px 04 02 04 Ett stickprov {X,, X 00 tas frå dea populatio Beräka saolikhete P X + + X 00 20 2 3 poäg Atag att X har e täthetsfuktio fx = cx 2 + 2x, för 0 x 2 2 p Beräka kostate c 22 2p Beräka de betigade saolikhete P X < 5 X 3 3 poäg E populatio X har e täthetsfuktio fx = θ + x θ för x 0 Det är kät att parameter θ är atige 2, 3 eller 4 Ett stickprov {02, 08 tas frå dea populatio Baserat på detta stickprov, beräka e puktskattig av θ geom att aväda Maximum Likelihood-metode 4 3 poäg Ma har mätt haltera av e viss föroreig i jordprover taga dels i ärhete av e idustri dels i ett retömråde med motsvarade jordmå Resultate är: i x i s i Nära idustri 8 80 049 Rea området 9 0 046 Ma ka ata att värdea ära idustri är observatioer frå X Nµ, σ, och värdea i rea området är observatioer frå X 2 Nµ 2, σ som är oberoede av X 4 5p Kostruera ett esidigt 95% kofidesitervall av µ µ 2 i forme a, + 42 5p Kostruera ett esidigt 99% kofidesitervall av σ 2 i forme 0, b 5 3 poäg Atag att Låt Y = Y = X + Y 2 6 3 poäg X 0 X = N, X 2 5 2 Beräka saolikhete P Y 35 + Y 2 32 E tärig har kastats 20 gåger med edaståede resultat: 4 2 2 9 Utfall 2 3 4 5 6 Frekves 0 32 2 8 23 25 Tyder resultatet på att tärig är obalaserad? Utför ett χ 2 test vid 5% sigifikasivå /
TAMS28: Notatios ad Formulas by Xiagfeg Yag Basic probability ad radom variables rv Coditioal probability P A B = P A B P B 2 Discrete rv X has a pmf fx = P X = x satisfyig fx 0 ad fxi =, X x x2 x fx fx fx2 fx Mea or Expected value or Expectatio µ = EX = xifxi; Variace σ 2 = V X = EX 2 µ 2 = x 2 i fxi µ2 3 Cotiuous rv X has a pdf fx satisfyig fx 0 ad fxdx =, b P a < X < b = fxdx a Mea or Expected value or Expectatio µ = EX = xfxdx; Variace σ 2 = V X = EX 2 µ 2 = x2 fxdx µ 2 4 Cumulative distributio fuctio cdf of a rv X is F x = P X x 5 If X ad Y are rv, a, b ad c are scalars, the EaX + by + c = aex + bey + c If X ad Y are idepedet rv, a, b ad c are scalars, the V ax + by + c = a 2 V X + b 2 V Y 2 Special rv X N0, : stadard Normal, EX = 0 ad V X = X Nµ, σ : geeral Normal, EX = µ ad V X = σ 2, X µ σ N0, X Bi, p : Biomial has a pmf fk = P X = k = p k k p k, k = 0,, 2,, EX = p, V X = p p Normal approximatio to Biomial Bi, p N p, p p, if p p 0 X P oµ : Poisso has a pmf fk = P X = k = e µ µ k k!, k = 0,, 2, EX = µ, V X = µ Normal approximatio to Poisso P oµ Nµ, µ, if µ 5 Poisso approximatio to Biomial Bi, p P o p, if 0 ad p 0 X Expλ : Expoetial has a pdf fx = { λe λx, x 0, 0, otherwise EX = 2 λ, V X = λ X Ua, b or X Rea, b: Uiform has a pdf { fx = b a, a < x < b, 0, otherwise EX = a + b b a2, V X = 2 2 3 Cetral Limit Theorem CLT Suppose that a populatio has mea= µ ad variace= σ 2 A radom sample {X, X2,, X from this populatio is give The for large 30, X µ σ/ N0, If the populatio is Normal Nµ, σ, the holds for ay Uderstad that µ = E X ad σ/ 2 = V X /4 4 Several otatios i statistics 4 Sample mea X = X+X2+X = Xi x+x2+x, ad x = = xi 42 Sample variace S 2 = Xi 2 X, ad s 2 = xi x 2 43 A poit estimate of a ukow parameter θ obtaied by Methods of Momets is deoted as ˆθMM 44 A poit estimate of a ukow parameter θ obtaied by Maximum Likelihood method is deoted as ˆθML Capital letters such as X 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 5 Cofidece Itervals CI CI-: α cofidece iterval for a populatio mea µ ay If populatio X Nµ, σ ad σ is kow, the X µ σ/ N0, ad Iµ = x z α/2 σ, x + z α/2 σ X µ 2 30 For arbitrary populatio X σ is kow or ukow, it holds that σ/ N0, ad σ σ ˆσ ˆσ Iµ = x z α/2, x + z α/2 or Iµ = x z α/2, x + z α/2 3 ay If populatio X Nµ, σ ad σ is ukow, the X µ S/ T ad Iµ = x t α/2 s, x + t α/2 s CI- : α cofidece iterval for two idepedet populatio meas µx µy ay, If populatios X NµX, σx ad Y NµY, σy are idepedet, σx ad σy are kow, IµX = σ X x 2 ȳ z µy α/2 X Ȳ µx µy N0,, ad σ 2 X, x ȳ + z α/2 σ 2 X 2, 30 For arbitrary idepedet populatios X ad Y σx, σy are kow or ukow, it holds that X Ȳ µx µy N0,, ad σ 2 X IµX = x ȳ z µy α/2 σ 2 X or IµX = ˆσ X x 2 ȳ z µy α/2, x ȳ + z α/2 + ˆσ2 Y, x ȳ + z α/2 σ 2 X ˆσ X 2 + ˆσ2 Y 3 ay, If populatios X NµX, σx ad Y NµY, σy are idepedet, σx = σy are ukow, X Ȳ µx µy S + T + 2, where S 2 = S2 X + S2 Y, + 2 IµX µy = x ȳ t α/2 + 2 s +, x ȳ + t α/2 + 2 s + Remark: if S 2 is obtaied from k samples, the the degrees of freedom is + + k k 2/4
CI-2: α cofidece iterval for a populatio variace σ 2 or σ A populatio X Nµ, σ ad a sample {x,, x, the S2 σ 2 χ2 ad s 2 s 2 s 2 s 2 I σ 2 = χ 2 α/2, or Iσ = χ 2 α/2 χ 2 α/2, χ 2 α/2 CI-2 : α cofidece iterval for 2 same populatio variace σ 2 or σ -st populatio X Nµ, σ ad a sample {x,, x, 2-d populatio Y Nµ2, σ ad a sample {y,, y, k-th populatio Z Nµk, σ ad a sample {z,, zk, the ˆσ 2 = s 2 = s2 ++k s2 k ++k k ad ++k ks2 σ 2 χ2 + + k k, so + + k ks 2 I σ 2 = χ 2 α/2 + + k k, + + k ks 2 χ 2 α/2 + + k k CI-3: α cofidece iterval with ormal approximatios 3 A ormal to a Biomial X Bi, p N p, p p, if p p 0 Therefore, X p ˆP p = N0,, where ˆP = X/ p p p p/ Thus Ip = ˆp z α/2 ˆp ˆp/, ˆp + z α/2 ˆp ˆp/ 3 Two ormals to two Biomials X Bi, p N p, p p, if p p 0 Y Bi, p2 N p2, assume that X ad Y are idepedet p2 p2, if p2 p2 0 The, ˆP ˆP2 p p2 p p + p2 p2 N0,, where ˆP = X/ ad ˆP2 = Y/ Thus Ip p2 = ˆp ˆp2 z α/2 ˆp ˆp + ˆp2 ˆp2, ˆp ˆp2 + z α/2 ˆp ˆp + ˆp2 ˆp2 3/4 32 A ormal to a Poisso If X P oµ with a radom sample {X,, X, the X + + X P oµ Nµ, µ, if µ 5 X + + X µ µ = X µ µ/ N0,, ad Iµ = x z α/2 x/, x + z α/2 x/ 6 Hypothesis Testig HT There are three differet pairs: H0 : θ = θ0 or H : θ < θ0 H0 : θ = θ0 or H : θ > θ0 H0 : θ = θ0 H : θ θ0 H0 is true H0 is false ad θ = θ reject H0 type I error or sigificace level α power hθ do t reject H0 α type II error βθ = hθ TS = test statistic reject H0 if TS C C = rejectio regio HT-: HT for a populatio mea µ ay If populatio X Nµ, σ ad σ is kow, the X µ σ/ N0, ad H : µ < µ0 H : µ > µ0 H : µ µ0 σ/, C =, zα; σ/, C = zα, + ; σ/, C =, z α/2 z α/2, + X µ 2 30 For arbitrary populatio X σ is kow or ukow, it holds that σ/ N0, ad σ/ H : µ < x µ0 or ˆσ/, C =, zα; µ0 σ/ H : µ > x µ0 or ˆσ/, C = zα, + ; µ0 σ/ H : µ x µ0 or ˆσ/, C =, z α/2 z α/2, + µ0 3 ay If populatio X Nµ, σ ad σ is ukow, the X µ S/ T ad H : µ < µ0 H : µ > µ0 H : µ µ0 s/, C =, tα ; s/, C = tα, + ; s/, C =, t α/2 t α/2, + Similarly: HT- from CI-, HT-2 from CI-2, HT-2 from CI-2 ad HT-3 from CI-3 4/4