Kurskod: TAMS28 MATEMATISK STATISTIK Provkod: TEN1 18 August 2017, 08:00-12:00. English Version
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1 Kurskod: TAMS28 MATEMATISK STATISTIK Provkod: TEN1 18 August 2017, 08:00-12:00 Examiner: Zhenxia Liu (Tel: ). Please answer in ENGLISH if you can. a. You are allowed to use a calculator, the formula and table collection edited by MAI. b. Scores rating: 8-11 points giving rate 3; points giving rate 4; points giving rate 5. 1 (4 points) English Version Suppose that a random variable X has the probability density function (1.1). (1p) Find the value of c. (1.2). (1p) Let A = {X > 1 2 }, calculate P (A). (1.3). (1p) Let B = {X < 1}, calculate P (B). (1.4). (1p) Calculate P (A B). f(x) = cx for 0 x 2. Solution. (1.1). c = 1/2. (1.2). P (A) = (1.3). P (B) = 1/4. (1.4). P (A B) = 3/4. 2 (3 points) Suppose that the distribution of a population X has the probability mass function (pmf) as follows X 0 1 p(x) 1/3 2/3 Let {X 1, X 2,..., X 180 } be a random sample from this population. (2.1). (1p) Find the mean µ = E(X) and the variance σ 2 = V (X). (2.2). (2p) Find the probability P (X 1 + X X ). Solution. (2.1). µ = E(X) = 2/3 and σ 2 = V (X) = 2/9. (2.2). P (X 1 + X X ) = P ( X ) = P ( X 127 µ σ/ n 180 µ σ ) P (N(0, 1) 1.11) = n 3 (3 points) Let {x 1, x 2,..., x n } be an observed sample from the population P o(2µ). (3.1). (1.5p) Find a point estimate ˆµ MM for µ using Method of Moments. (3.2). (1.5p) Find a point estimate ˆµ ML for µ using Maximum-Likelihood method. Solution. (3.1). For Method of Moments, we have E(X) = x. Then 2µ = x which yields ˆµ MM = x/2. (3.2). For the Maximum-Likelihood method, we write the likelihood function as L(µ) = f(x 1 ) f(x 2 )... f(x n ) = e 2nµ (2µ) n i=1 xi n i=1 x i! Page 1/4
2 Maximizing L(µ) is equivalent to maximize ln L(µ) where n n ln L(µ) = 2nµ + x i ln(2µ) ln( x i!). i=1 i=1 By d ln L(µ) dµ = 0, we have ˆµ ML = x 2. 4 (3 points) ( ) X Z = is a normal vector. Assume that X N(0, 1), Y N(1, 2), and cov(x, Y ) = 1. Y ( ) X Y (4.1). (2p) Determine the mean vector and the covariance matrix for. X + Y (4.2). (1p) Find P (3X > 2Y + 1). Solution. (4.1). The mean vector and covariance matrix for Z is ( ) 0 µ Z = and C 1 Z = ( ) Then we have ( ) ( ) ( ) X Y 1 1 X = X + Y 1 1 Y ( ) X Y Then the mean vector and covariance matrix for is X + Y µ = ( ) 1 1 µ 1 1 Z = ( ) 1 1 ( ) ( ) and C = C 1 1 Z = 1 1 ( ) (4.2). Notice that So we get 3X 2Y = ( 3 2 ) ( ) X N( 2, 13). Y P (3X > 2Y + 1) = P (N(0, 1) > 1 ( 2) 13 ) = 1 P (N(0, 1 < 0.83) = (3 points) Suppose that the distribution of heights of all adult Swedes is a normal distribution X N(µ 1, σ), and the distribution of heights of all adult Germans is also a normal distribution Y N(µ 2, σ). Assume that X and Y are independent. Now we choose two independent random samples from X and Y respectively, which yield the following data (in cm). Swedes: ; Germans: (5.1). (1p) Calculate sample standard deviations s X for Swedes and s Y for Germans. (5.2). (1p) Construct a 95% confidence interval for σ. (5.3). (1p) Test the hypothesis with a significance level α = 0.05 : H 0 : µ 1 = µ 2, H 1 : µ 1 > µ 2. Page 2/4
3 Solution. (5.1). x = 182, s x 4.06, ȳ = 180, s y = (5.2.). We can get the combined sample variance s 2 = (n1 1)s2 X +(n2 1)s2 Y n 1 1+n 2 1 = A (two-sided) 95% confidence interval for σ 2 is I σ 2 = ( (n 1 + n 2 2)s 2 χ 2 α/2 (n 1 + n 2 2), (n 1 + n 2 2)s 2 s2 χ 2 1 α/2 (n ) = (7 1 + n 2 2) 16, 7 s ), therefore a (two-sided) 95% confidence interval for σ is I σ (2.12, 6.53). (5.3.). Since σ is unknown, the rejection region is C = (t α (7), ) = (1.89, ). The test statistic is T S = s x ȳ (0) 1 = n Because the test statistic is NOT in the rejection region, we 1 n 2 7 (1/5+1/4) do NOT reject H 0. 6 (2 points) Suppose that we are studying mileage performance (in kilometers) on every tank of fuel (gasoline). The response variable Y = mileage. There are two possible regressors x 1 = horsepower and x 2 = temperature. David thinks that the mileage only depends on x 1, but Lisa thinks that the mileage depends on both x 1 and x 2. In this setting, we have two different models. (Lisa) Model 1: Y = β 0 + β 1x 1 + β 2x 2 + ε, where ε N(0, σ ) with unknown σ ; (David) Model 2: Y = β 0 + β 1 x 1 + ε, where ε N(0, σ) with unknown σ. Below is the output from Minitab based on some data. Model 1: Regression Analysis: Y versus horsepower (x1), temperature (x2) Y = horsepower (x1) temperature (x2) Constant horsepower (x1) temperature (x2) S = R-Sq = 99.3% R-Sq(adj) = 99.0% Source DF SS MS F P Regression Residual Error Model 2: Regression Analysis: Y versus horsepower (x1) Y = horsepower (x1) Constant horsepower (x1) ??? S = R-Sq = 99.2% R-Sq(adj) = 99.1% Source DF SS Regression Residual Error Page 3/4
4 (6.1). (1p) With a significance level 0.05, test H 0 : β 1 = β 2 = 0 against H 1 : at least one of β 1, β 2 0 (6.2). (1p) Would it be possible to remove x 2 with α = 5%? Motivate your answer briefly. SS Solution. (6.1). We can get T S = R /k SS E /(n k 1) and the rejection region C = (F 0.05 (k, n k 1), ) = (F 0.05 (2, 6), ) = (5.14, ), we reject H 0 since T S C. (6.2). Yes, it is since p value = > α. Page 4/4
5 Kurskod: TAMS28 MATEMATISK STATISTIK Provkod: TEN1 18 augusti 2017, kl Examinator: Zhenxia Liu (Tel: ). Vänligen svara på ENGELSKA om du kan. a. Tillåtna hjälpmedel är en räknare, formel -och tabellsamling utgiven av MAI. b. Betygsgränser: 8-11 poäng ger betyg 3; poäng ger betyg 4; poäng ger betyg 5. 1 (4 poäng) Den stokastiska variabeln X har täthetsfunktion (1.1). (1p) Bestäm värdet på c. (1.2). (1p) Låt A = {X > 1 2 }, beräkna P (A). (1.3). (1p) Låt B = {X < 1}, beräkna P (B). (1.4). (1p) Beräkna P (A B). 2 (3 poäng) Svensk version f(x) = cx för 0 x 2. Antag att fördelningen för en population X har massfunktionen (pmf) enligt följande X 0 1 p(x) 1/3 2/3 Låt {X 1, X 2,..., X 180 } vara ett slumpmässigt stickprov från populationen. (2.1). (1p) Beräkna väntevärdet µ = E(X) and och variansen σ 2 = V (X). (2.2). (2p) Beräkna sannolikheten P (X 1 + X X ). 3 (3 poäng) Låt {x 1, x 2,..., x n } vara ett observerat stickprov från populationen P o(2µ). (3.1). (1.5p) Hitta en punktskattning ˆµ MM för µ genom att använda momentmetoden. (3.2). (1.5p) Hitta en punktskattning ˆµ ML för µ genom att använda Maximum-Likelihood metoden. 4 (3 poäng) ( ) X Z = är en normal vektor. Antag att X N(0, 1), Y N(1, 2), och cov(x, Y ) = 1. Y ( ) X Y (4.1). (2p) Bestäm väntevärdesmatris och kovariansmatris för. X + Y (4.2). (1p) Beräkna P (3X > 2Y + 1). 5 (3 poäng) Antag att längderna på alla vuxna svenskar är normalfördelade X N(µ 1, σ), och att längderna på alla vuxna tyskar följer en normalfördelning Y N(µ 2, σ). Antag att X och Y är oberoende. Vi tar nu två oberoende stickprov från X och Y, respektive, och får följdande data. (i cm). Svenskar: ; Tyskar: Page 1/2
6 (5.1). (1p) Beräkna stickprovs standardavvikelse s X för Svenskar och s Y för Tyskar. (5.2). (1p) Konstruera ett 95% konfidensintervall för σ. (5.3). (1p) Pröva hypotesen på nivån α = 0.05 : 6 (3 poäng) H 0 : µ 1 = µ 2, H 1 : µ 1 > µ 2. Antag att vi studerar hur långt man kan åka (i kilometer) på varje tankning (av bensin). Responsvariabeln Y = sträcka. Det finns två möjliga förklaringsvariabler x 1 = antal hästkrafter och x 2 = temperatur. David tror att sträckan man kan åka bara beror på x 1, men Lisa tror att sträckan beror på både x 1 och x 2. I den här situationen har vi två olika modeller. (Lisa) Modell 1: Y = β 0 + β 1x 1 + β 2x 2 + ε, där ε N(0, σ ) och σ okänd, (David) Modell 2: Y = β 0 + β 1 x 1 + ε, där ε N(0, σ) och σ okänd. Nedan finns en utskrift från Minitab baserad på data. Model 1: Regression Analysis: Y versus horsepower (x1), temperature (x2) Y = horsepower (x1) temperature (x2) Constant horsepower (x1) temperature (x2) S = R-Sq = 99.3% R-Sq(adj) = 99.0% Source DF SS MS F P Regression Residual Error Model 2: Regression Analysis: Y versus horsepower (x1) Y = horsepower (x1) Constant horsepower (x1) ??? S = R-Sq = 99.2% R-Sq(adj) = 99.1% Source DF SS Regression Residual Error (6.1). (1p) Pröva på nivån 0.05 H 0 : β 1 = β 2 = 0 mot H 1 : minst en av β 1, β 2 0. (6.2). (1p) Vore det tänkbart att at bort x 2 med α = 5%? Motivera dit svar kortfattat. Page 2/2
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