Kapitel 1: 4. 2/50. Det här är klassisk sannolikhet eftersom vi tittar på ett parti där vi känner till exakt hur många som är fel. 6. b) (s,s,s) (s,s,f) (s,f,s) (s,f,f) (f, s, s) (f,s,f) (f,f,s) (f,f,f) c) Nej, de är inte disjunkta d) A1' är komplementet till A1, alltså händelsen att första skotten missade. (f, s, s) (f, s, f) (f, f, s) (f,f,f) e) Händelsen att första skottet träffar och de andra och tredje missar. (s, f, f) f) Klassisk sannolikhet kan användas eftersom alla händelser har lika stor sannolikhet att inträffa. 1/8 Kapitel 2 2. a) 10/35 b) 24/35 6. 5% and 10% 8. a) 0.15 b) 0.40 36. 0.0917 40. P(A|F) = 12/34 P(B|F) = 8/34 P(C|F) = 9/34 P(D|F) = 5/34 Kapitel 3 2) Diskret 4) Inte diskret 6) Diskret 8. a) 0.03 b) 0.02, 0.05, 0.1, 0.3, 0.7, 0.9, 0.97, 1 c) 0.65 d) 0.30, 0.10, nej de är inte samma e) 0, 1 16. E(X) = np = 2.7 var(X) = npq = 0.27 20. var(c) = 0 var(cX) = c^2var(X) 24. E(X)=13 E(X^2) = 325 Var(X) = 156 42. a) 0.1216 b) 0.8784 c) P(X>4) = 0.0432 44. a) 0.7639 b) P(X=0) = 0.0134 46. p=1/5, n=25 62. a) P( <= 4) = 0.947 b) 6 C) P( >= 12) = 2% => Not a big probability to see such a rare occurence 68. k = 1 Kapitel 4 4. a) b) 0.415 c) 34. a) a^-6 b) 1/2 42. a) 0.9544 b) 0.96 c) 100.6 mg/100 ml d) 44. a) P(X <= 1875) = 0.0228 -> Unusual event b) P(X >= 1878) = 0 (almost zero) -> Very unusual event 52. a) 0.1112 jämfört med 0.1071 b) 0.5512 jämfört med 0.5725 c) 0.8888 jämfört med 0.8929 d) 0.1215 jämfört med 0.1304 54. a) n(1-p) = 6 > 5 -> Normal approximation allowed b) sigma = 2.324, mu = 54 c) P(X >= 59) = 2.64% d) P(X <= 58) = 81% Kapitel 5 2. a) b) X has marginal density as hypergeometric distribution with N = 7, n = 4 and r = 3. Y has marginal density as hypergeometric distribution with N = 7, n = 4 and r = 4 c) No, they are not independent. Show that there are values where f(x,y) are not equal to f(x)f(y) 8. a) c = 1/6640 b) 37.35% c) f(x) = c*(8x+6) f(y) = c*40*(2y+81) d) P(Y >= 1) = 84/166 e) P(X >= 20) = 123/166 f) 20. a) negative, warmer temperature gives a faster ignition. b) E(X) = 6580/249 E(Y) = 502/498 E(XY) = 6620/249 C(X,Y) = - 3200/62001 42. a) (y-1)/ln(y) b) 1/(2*ln(2)) c) x/2 d) 3/8 Kapitel 6 18. b) s2 = 1 505 155.6 s = 1226.87 Kapitel 7 26. muhat = xbar sigmahat^2 = M2 - M1^2 muhat is unbiased but sigmahat^2 is biased 50. a) E(X) = 5/2, Var(X) = 1.25 b) c) Kapitel 8 2. a) 20.42855 b) [13.91, 33.47] c) [3.73 , 5.79] d) If x hat and S would be the true mean and standard deviation it would be unusual for the surround luminance to exceed 18. 14. [10.5199 , 11.4801] 18. mu <= 0.04357 with 95% confidence. This is far below 0.07 and therefore it seems like the water mets the criterion. Kapitel 10 18. [-6.59, -1.16] Since negative values in all of the confidence interval it is reasonable to believe that mu1 < mu2. In other words river-level terrain seem to have a greater change in temperature. Kapitel 11 20. 95% Konfidensinterval [-0.86, 1.30] 22. a) 5 b) 95% prediction interval [3.70 , 6.31] 30. b0 = -0.229 b1 = 0.9927 32. a) mu = 14.66 b) 90% konfidensintervall [14.59, 14.73] 46. a) No apparent problem. b) Variance seems to vary c) Model misspecification (the data seems to be nonlinear so a linear regression is not enough) d) Gaps in data. Since there is a huge gap in the data it is not reasonable to try to fit a linear model. Either data is missing in the gap so we do not know if the data will exhibit a linear trend or not. Kapitel 12 1. Degree 1. 3. (a) sum_i x_1i = 6, sum_i x_2i = 25, sum_i x_1i*x_2i = 50, sum_i x_1i*y_i = 44, sum_i x_1i^2 = 20, sum_i x_2i^2 = 209, sum_i y_i = 24, sum_i x_2i y_i = 200 5. b0 + b1*x1 + b2*x2 + b3*x3, b3 = b12, x3 = x1*x2 7. (a) | 8 464.4 | * | b0 | = | 46.2 | | 464.4 32089.96 | | b1 | | 3173.17 | (b) b0 = 0.2177, b1 = 0.0957 14. a) [1, 0, 8 ; 1,2,9; 1,4,8] b) [3,6,25 ; 6,20,50 ; 25,50,209] c) [24;44;200] d) Compare normal equations with exercise 3 e) Multiply with X'X and see that the result is the identity matrix f) Insert the given values and see that they add up 38. [1,1,1;1,2,4;1,3,9;1,4,16;1,5,25;1,6,36;1,7,49] 40. T = 3.6 p = 13.07% If your answer is not exactly the same it might be due to roundoff. 42. [26.280, 50.254] If your answer is not exactly the same it might be due to roundoff. Kapitel 13 4. a) H0: mu1 = mu2 = mu3 = mu4 H1: mu_i1 different from mu_i2, some 1 <= i1, i2 <= 4 b) T = 6.3343, p = 0.0017 c) För minst två av grupperna skiljer sig medelvärdena åt. Kapitel 14 4. H0: (alphabeta)ij = 0, i=1,2,3, j=1,2,3 H1: (alphabeta)ij är skilt från noll, för något par (i,j) 6. T = 1.0405 p = 0.4138 8. T = 130.24 p = 2*10^-11 26. 16, 32, 128 34. Jämför med tabell 14.8 (a) index i vektorerna motsvarar [A B AB C AC BC ABC D AD BD ABD CD ACD BCD ABCD]: SS = MS = [133.66 4.35 2.53 101.53 3.5115 9.9015 1.2015 1.3615 1.2015 1.2015 5.61 4.6516 1.2015 5.2815 4.6515] SSE = 4.15 F = MS/(SSE/2^4) = [515.32 16.78 9.76 391.45 13.54 38.17 4.63 5.25 4.63 4.63 21.63 17.99 4.63 20.36 17.93] (b) index i vektorerna motsvarar [A B AB C AC BC D AD BD CD]: SS = MS = [133.66 4.35 2.53 101.53 3.5115 9.9015 1.3615 1.2015 1.2015 4.6516] SSE = 22.0960 F = MS/(SSE/(2^4+5)) = [127.0302 4.1342 2.4045 96.4939 3.3373 9.4104 1.2940 1.1419 1.1419 4.4209] (c) index i vektorerna motsvarar [A B AB C AC BC ABC D AD BD ACD]: SS = MS = [133.66 4.35 2.53 101.53 3.5115 9.9015 1.2015 1.3615 1.2015 1.2015 1.2015] SSE = 24.345 F = MS/(SSE/(2^4+4)) = [109.8062 3.5737 2.0785 83.4104 2.8848 0.9871 1.1185 0.9871 0.9871 0.9871]