Kurskod: TAMS Provkod: TENB 2 March 205, 4:00-8:00 Examier: Xiagfeg Yag (Tel: 070 2234765). Please aswer i ENGLISH if you ca. a. You are allowed to use: a calculator; formel -och tabellsamlig i matematisk statistik (from MAI); TAMS : Notatios ad Formulae (by Xiagfeg Yag), a dictioary. b. Scores ratig: 8- poits givig rate 3;.5-4.5 poits givig rate 4; 5-8 poits givig rate 5. (3 poits) Eglish Versio (o Swedish Versio) At a certai gas statio, 40% of the customers use regular gas, 35% use plus gas, ad 25% use premium. Of those customers usig regular gas, oly 30% fill their taks. Of those customers usig plus gas, 60% fill their taks, whereas of those usig premium, 50% fill their taks. (.). (p) What is the probability that the ext customer will request plus gas ad fill the tak? (.2). (p) What is the probability that the ext customer fills the tak? (.3). (p) If oe kows that the ext customer fills the tak, what is the probability that regular gas is requested? Solutio. What we ca get from the problem is the followig: (.). (.2). (.3). P (regular) 40%, P (plus) 35%, P (premium) 25%, P (fill regular) 30%, P (fill plus) 60%, P (fill premium) 50%. P (plus ad fill) P (plus) P (fill plus) 35% 60% 2%. P (fill) P (fill regular) P (regular) + P (fill plus) P (plus) + P (fill premium) P (premium) 30% 40% + 60% 35% + 50% 25% 45.5%. P (regular ad fill) P (regular fill) P (fill) P (fill regular) P (regular) P (fill) 30% 40% 45.5% 2/45.5 26.37%. 2 (3 poits) Aie ad Alvie have agreed to meet betwee 5:00 pm ad 6:00 pm for dier at a local health-food restaurat. Let X Aie s arrival time ad Y Alvie s arrival time. Suppose that the joit probability desity fuctio (joit pdf) for the two-dimesio radom variable (X, Y ) is f(x, y), if 5 x 6 ad 5 y 6. (2.). (p) Fid the margial pdf f X (x) of X ad the margial pdf f Y (y) of Y. (2.2). (p) What is the probability that they both arrive betwee 5:5 pm ad 5:45 pm? (2.3). (p) If the first oe to arrive will wait oly 0 miutes before leavig to eat elsewhere, what is the probability that they have dier at the local health-food restaurat? (Hit: the evet of iterest is A {(x, y) : x y 6 }.) Page /5
Solutio. (2.) f X (x) f(x, y)dy 6 5 dy, 5 x 6. Similarly, f Y (y) f(x, y)dx 6 5 dx, 5 y 6. (2.2) P (both 5 : 5 5 : 45) P (5.25 X 5.75 ad 5.25 Y 5.75) 5.75 5.75 5.25 5.25 f(x, y)dxdy 0.25. (2.3) The probability is P (A) f(x, y)dxdy /36 0.3056. A (draw the graph, the you will see clearly). 3 (3 poits) A biary commuicatio chael trasmits a sequece of bits (0s ad s). Suppose that for ay particular bit trasmitted, there is a 0% chace of a trasmissio error (a 0 becomig a, or a becomig a 0). Assume that bit errors occur idepedetly of oe aother. Now we cosider trasmittig 000 bits, What is the probability that at most 25 trasmissio errors occur? Solutio. We have two differet methods to solve the problem. The first method is the usual oe: CLT. Let X i record the possible error of the i-th bit, so X i 0 p(x) 0.9 0. So we ca easily get µ E(X i ) 0. ad σ 2 V (X i ) 0.09. Thus the probability is P (at most 25 trasmissio errors occur) P (X + X 2 +... + X 000 25) P ( X + X 2 +... + X 000 000 P ( X µ σ/ 25 000 ) P ( X 0.25) 0.25 0. 0.09/ 000 ) P (N(0, ) < 2.64) 0.9959. The secod method is Normal approximatio to a Biomial radom variable. If we use X total umber of errors, the X Bi(000, 0.). Thus P (at most 25 trasmissio errors occur) P (X 25) P (N(000 0., 000 0. 0.9) 25) P (N(00, 90) 25) 25 00 P (N(0, ) ) P (N(0, ) < 2.64) 0.9959. 90 Page 2/5
4 (3 poits) Let X deote the proportio of allotted time that a radomly selected studet speds workig o a certai test. Suppose that the pdf of X is f(x) (θ + ) x θ, 0 x, where θ > is a ukow parameter. A radom sample of 3 studets yields data: {0.7, 0.92, 0.83}. (4.). (p) Fid a poit estimate ˆθ MM of θ usig Method of Momets. (4.2). (2p) Fid a poit estimate ˆθ ML of θ usig Maximum-Likelihood method. Solutio. (4.). For Method of Momets, the first equatio is E(X) x. The mea E(X) ca be calculated as E(X) 0 x(θ + ) x θ dx (θ + )/(θ + 2). By solvig E(X) x, we have θ (2 x )/( x) which yields ˆθ MM (2 x )/( x). From the data, x 0.7+0.92+0.83 3 0.82, thus ˆθ MM (2 0.82 )/( 0.82) 0.64/0.8 3.56. (4.2). For the Maximum-Likelihood method, we write the likelihood fuctio as L(θ) f(x ) f(x 2 )... f(x ) (θ + ) (x... x ) θ. Maximizig L(θ) is equivalet to maximize l L(θ) where l L(θ) l(θ + ) + θ l(x... x ). By d l L(θ) dθ 0, we have θ+ + l(x... x ) 0, therefore (The secod derivative d2 l L(θ) dθ 2 ˆθ ML l(x... x ) 3 l(0.54256) 3 0.622 3.9. < 0 which yields that ˆθ ML is ideed a maximal poit) 5 (3 poits) Oe has measured the same physical quatity five idepedet times with Method A ad four idepedet times with Method B, ad the results are: Method A: x.0635 s 2 x 3.7 0 Method B: ȳ.055 s 2 y 5.6 0 We assume that the sample for Method A is from N(µ x, σ 2 ), ad the sample for Method B is from from N(µ y, σ 2 ). (5.). (p) Costruct a (two-sided) 95% cofidece iterval of µ x. (5.2). (p) Do you thik µ x µ y? Aswer this usig a (two-sided) 95% cofidece iterval of µ x µ y. (5.3). (p) Costruct a upper 95% cofidece iterval (oe-sided) for σ i the form (0, b). Solutio. (5.) Sice σ is ukow, a 95% cofidece iterval of µ x would be s x s I µx ( x t α/2 ( ), x + t α/2 ( ) x ) 3.7 0 3.7 0 (.0635 t 0.025 (5 ),.0635 + t 0.025 (5 ) ) 5 5 (.0635 2.78 0.86 0 3,.0635 + 2.78 0.86 0 3 ) (.0635 0.00239,.0635 + 0.00239) (.06,.06589). Page 3/5
(5.2) A 95% cofidece iterval of µ x µ y is where I µx µ y (( x ȳ) t α/2 ( + 2 2) s +, ( x ȳ) + t α/2 ( + 2 2) s + ) 2 2 ( x ȳ).0635.055 0.02; t α/2 ( + 2 2) t 0.025 (5 + 4 2) 2.36; s ( s 2 )s 2 x + ( 2 )s 2 y (5 ) 3.7 0 + (4 ) 5.6 0 3.6 0 + 2 2 5 + 4 2 /7 2.25 0 3 + 2 5 + 4 0.67. Thus I µx µ y (0.02 0.003365, 0.02 + 0.003365) (0.008635, 0.05365). Sice 0 / I µx µ y, we thik µ x µ y. (5.3) Sice σ ca be from Method A or Method B, we may solve this problem i three differet ways ad all are correct! st way: σ from Method A. Oe-sided cofidece iterval for σ 2 is I σ 2 (0, ( ) s 2 x χ 2 α ( ) ) (0, (5 ) 3.7 0 χ 2 0.05 (5 ) ) (0, 4.8 0 ) (0, 20.845 0 ). 0.7 Thus oe-sided cofidece iterval for σ is I σ (0, 20.845 0 ) (0, 0.0045656). 2d way: σ from Method B. Oe-sided cofidece iterval for σ 2 is I σ 2 (0, ( 2 ) s 2 y χ 2 α ( 2 ) ) (0, (4 ) 5.6 0 χ 2 0.05 (4 ) ) (0, 6.8 0 ) (0, 48 0 ). 0.35 Thus oe-sided cofidece iterval for σ is I σ (0, 48 0 ) (0, 0.0069282). 3rd way: σ from both Method A ad Method B (did t talk about this i the lectures, but iclude here just for your referece). Oe-sided cofidece iterval for σ 2 is I σ 2 (0, ( + 2 2) s 2 χ 2 α ( + 2 2) ) (0, (5 + 4 2) 4.54 0 χ 2 0.05 (5 + 4 2) ) (0, 3.6 0 ) (0, 4.56 0 ). 2.7 Thus oe-sided cofidece iterval for σ is I σ (0, 4.56 0 ) (0, 0.00386). 6 (3 poits) The PCB-cocetratio is measured for 0 fish i a lake, ad the results are:.5, 0.8,.6, 9.4, 2.4,.4, 2.2,.0, 0.6, 0.8 We assume that these observatios are from a ormal populatio N(µ, 0.8). Previous measuremets show that the average PCB-cocetratio for the fish is 0.8, but it is suspected that the cocetratio ow becomes higher i the lake. (6.). (p) Test the followig hypotheses with a sigificace level α 0.05 : H 0 : µ 0.8 versus H a : µ > 0.8. (6.2). (2p) For the test i (6.), what is the power whe the actual µ.0? Page 4/5
Solutio. (6.) Sice the populatio variace is kow σ 2 0.8, accordig to H a the rejectio regio C (z α, + ) (z 0.05, + ) (.65, + ). The test statistic is Sice T S / C, we do t reject H 0. (6.2) The power is T S x µ 0 σ/.7 0.8 0.8/ 0.3. h() P (reject H 0 whe H 0 is wrog ad µ ) P ( X µ 0 σ/ >.65 whe µ ) (eed to chage X µ 0 σ/ to X µ σ/ sice X µ σ/ N(0, )) P ( X µ σ/ + µ µ 0 σ/ >.65 whe µ ) 0.8 P (N(0, ) + >.65) 0.8/ 0 P (N(0, ) >.65 0.7027) P (N(0, ) > 0.95) 0.8289 0.7. Page 5/5