Övningar enligt 2a upplagan. Le1 2.1: 1/15 2.3: 0.45 2.4: (a) 0.4 (b) 0.2 (c) 0.2 (d) 0.1 (e) 0.8333 (f) 0.7143 2.5: 0.72 2.6: 0.6 2.7: 0.66 2.8: 0.9508 2.11: (a) 0.8488 (b) 0.4774 (c) 0.0012 2.14: (a) 0.006 (b) 0.000068 (c) 0.000051 (d) 0.000018 2.16: (a) 0.049 (b) 25/49 2.17: 1/3 2.32: Le2 3.1: (a) p(0)=0.42, p(1)=0.46, p(2)=0.12 3.2: E(Y)=200 dollars och Var(Y)=110000 dollars^2 3.4: E(X)=3.5 och Var(X)=35/12 3.8: E(X)=1.5 och Var(X)=1.25 3.9: 1/16 3.10: 0.28 3.11: (a) p(x,y)=1/36 om x=y, 1/18 om xy (b) beroende (c) p(1)=11/36, p(2)=9/36, p(3)=7/36, p(4)=5/36, p(5)=3/36, p(6)=1/36 (d) 2/9 3.12: (a) beroende (b) beroende 3.13: (a) p(0)=0.2, p(2)=0.7, p(4)=0.1 (b) p(-2)=0.3, p(0)=0.4, p(2)=0.3 (c) p(0)=0.8, p(1)=0.1, p(4)=0.1 3.16: (a) beroende (b) 0.6 3.19: (a) E(100X)=0 och Var(100X)=40000 (b) E(100Y)=0 och Var(100Y)=40000 (c) E(50X+50Y)=0 och Var(50X+50Y)=20000 Le3 3.5: (a) 19/33 (b) 5/3 3.20: (a) 0.0596 (b) 0.986 3.21: 0.245 3.23: (a) 0.005 (b) 0.00314 3.24: (a) 0.0328 (b) 0.4096 3.25: (a) 0.1111 (b) 0.1937 3.27: (a) 0.945 (b) 0.061 3.28: 3.29: 0.0923 3.30: 0.0166 3.31: (a) 0.2061 (b) 1.25 3.35: 9/13 3.37: 0.827 Le4 4.2: (a) 0.003 (b) 0.217 4.6: 55/6 4.7: 0.875 4.8: 0.264 4.9: (a) 0.269 (b) 0.666 4.10: 0.4764 4.11: (a) 0.6 (b) 15 4.14: (a) 4 och 0.2 (b) 0.353 4.16: (a) 0.8944 (b) 0.8944 (c) 0.1056 (d) 0.7888 (e) 1 (f) 0 (g) 0.84 4.17: (a) 0.1611 (b) 0.1611 (c) 0.8389 (d) 0.3222 (e) 1 (f) 0 (g) 1.28 4.18: (a) 0.9965 (b) 0.3802 (c) 0.6233 (d) 0.2321 (e)1 f) 0.8413, (g) -2.12 4.19: (a) 0.9974 (b) 1 4.29: 0.567 4.30: 0.8693 4.32: (a) 4 (b) 6 Le5 4.23: 0.9918 4.24: 0.2033 4.25: (a) 0.4544 (b) 0.9997 4.26: 0.3226 4.27: 0.6318 4.28: 0.1151 4.31: (a) 0 (b) 38 5.1: 0.01 5.2: (a) 0.3693 (b) 1 (c) 5 (d) 2 (e) 0.7351 (f) −0.5902 (g) 4.0026 5.6: Le6 6.1: (a) p00 = p01 = p11 = p12 = p21 = p22 = 0.5 anc p02 = p10 = p20 = 0 (c) 0.125 6.3: (a) P = ((0.6 0.4), (0.2 0.8)) row-wise (b) 0.28 6.9: (a) 0.0579 (b) 0.0001 6.12: (a) 3 seconds (b) 0.000027 (c) 0.9699 6.14: 0.2810, 20 seconds, 17.32 seconds 6.17: (a) 0.945 (b) 0.061 6.18: 0.0162 6.19: (a) 3.75 seconds (b) 0.4129, i.e. observing 50 or more messages is not unlikely 6.20: 0.884 6.21: 0.215 6.22: (a) 0.735 (b) 7500 dollars and 3354 dollars 6.23: (a) 0.003 (b) 0.6 minutes and 0.12 minutes^2 6.24: 0.544 Le7 8.1: (b) min(X) = 37, Q1_hat = 43, M_hat = 50, Q3_hat = 56, max(X) = 60, and min(X) = 21, Q1_hat = 35, M_hat = 39, Q3_hat = 46, max(X) = 53. 8.2: (a) X_bar = 17.9540, s^2 = 9.9682, s= 3.1573 (b) 0.4465 8.3: 8.4: 0.9930 8.5: pop<-matrix(data=c(seq(1790,2010,10),3.9,5.3,7.2,9.6,12.9,17.1,23.2,31.4,38.6,50.2,63.0, 76.2,92.2,106.0,123.2,132.2,151.3,179.3,203.3,226.5,248.7,281.4,308.7), ncol = 2,nrow = 23) plot(pop,xlim=c(1790,2010)) 8.6: popinc<-matrix(data=NA,ncol=2,nrow=22) for(i in 1:22){ popinc[i,1]<-pop[i+1] popinc[i,2]<-pop[i+1,2]-pop[i,2] } plot(popinc,xlim=c(1800,2010)) mean(popinc[,2]) median(popinc[,2]) var(popinc[,2]) 8.7: poprel<-matrix(data=NA,ncol=2,nrow=22) for(i in 1:22){ poprel[i,1]<-pop[i+1] poprel[i,2]<-(pop[i+1,2]-pop[i,2])/pop[i,2] } plot(poprel,xlim=c(1800,2010)) mean(poprel[,2]) median(poprel[,2]) var(poprel[,2]) 8.9: (a) X_bar = 49.6, M_hat = 47.5, Q1_hat = 43, Q3_hat = 52, and s = 23.5 (b) 105 is an outlier (c) X_bar = 43.4, M_hat = 45, Q1_hat = 43, Q3_hat = 51, and s = 13.9 (d) note the change in sample mean and standard deviation whereas the quartiles do not change that much (robust). Le8 9.1: (b) 5/8 (c) 0.171 9.2: (b) 5/20 9.3: (a) a_hat = min(x_i) and b_hat = max(x_i) (b) lambda_hat = 1/X_bar (c) mu_hat = X_bar (d) sigma_hat = sqrt(sum_i (x_i - mu)^2/n) (e) mu_hat = X_bar and sigma_hat = sqrt(sum_i (x_i - X_hat)^2/n) 9.4: 2.1766 Le9 9.7: (a) 37.7 ± 1.5 or [36.2, 39.2] (b) H0 : μ = 35 vs HA : μ > 35. Z = 2.9348 ≥ z_0.01 = 2.326. Therefore, reject H0 in favor of HA. 9.8: (a) 42± 1.225 or [40.775, 43.225] (b) 0.1790 9.9: (a) 50± 33.7 or [16.3, 83.7] (b) H0 : μ = 80 vs HA : μ ̸= 80. t = −2.598 belongs to the acceptance region (−2.920; 2.920), therefore, H0 is not rejected. (c) [11.6, 89.4] (thousand dollars) 9.10: (a) 0.12 ± 0.047 or [0.073, 0.167] (b) H0 : p ≤ 0.1 (or H0 : p = 0.1) vs HA : p > 0.1. Z = 0.9428. Therefore H0 is not rejected in either case. 9.12: (a) 0.62 ± 0.054 or [0.566, 0.674] (b) 0.2358 9.13: 0.0062 9.16: (a) −0.02 ± 0.043 or [−0.063, 0.023] (b) H0 : p1 = p2 is not rejected because the 98% confidence interval contains 0 9.17: 0.0325 or 3.25% for pHat1 = 45%, 0.0312 or 3.12% for pHat2 = 35%, 0.0450 or 4.50% for pHat1 − pHat2 = 10% Le10 10.31: (a) 4.5 and 2.25 10.32: (b) 4/13 10.33: (b) 0.4 and 0.02, (c) 0.22, (d) reject H0 10.34: (a) 17.6611, (b) [16.7657,18.5565], (c) reject H0 10.35: (a) 17.9540, (b) [17.0234, 18.8846], (c) reject H0 10.36: (a) N(5.5, 0.255), (b) N(5.575, 0.245), 5.575 and 0.060, (c) [5.095, 6.055] 10.37: 0.5145 10.39: 10.40: (a) 0.9 and 9/1100, (b) [0.7743, 1], (c) reject H0 Le11 11.1: y_hat = 0.009 + 0.000246x and 0.107 seconds 11.3: (a) y_hat = 24.77 − 0.0000395x 11.5: (a) b0 = 13.3586, b1 = 2.3569, b2 = 4.0926 (2nd edition), b0 = 1.5741, b1 = 2.3569, b2 = 4.0926 (3rd edition) 11.8: (a) y_hat = 19.50 + 17.36x, (b) 88.94 11.10: (a) y_hat = −32.1+ 1.36x, (c) 260.4 and 267.2 million people 11.11: (a) b0 = 5.8700, b1 = 0.0057, and b2 = 0.0068, (b) 320.2 and 335.0 million people 11.12: (a) the quadratic model is more appropriate, (b) b0 = 7.716, b1 = 1.312, and b2 = −0.024, (d) 25.53 and 28.41 lbs Le12 11.2: (c) [0.4547, 0.6703]. This interval does not contain 0, therefore, the slope is significant (significantly different from 0) at a 1% level of significance. 11.3: (b) R2 = r2 = 0.0289. This model can explain only 2.89% of the total variation of the number of miles per gallon. This is a rather poor fit. The number of miles per gallon should probably depend on many factors other than mileage, (c) [16.74, 30.04] and [22.78, 24.00] 11.4: (b) there is significant evidence against the null hypothesis H0 : β1 = 1.8 (1000 dollars), (c) [50.40; 67.02], (d) It is wrong to say that the actual investment will be between 50,400 and 67,020 with probability 0.95. Rather, 95% refers to the long run proportion of prediction intervals constructed by the same procedure as we used in (c). That is, about 95% of prediction intervals computed from a large number of independent samples will contain the actual 2017 investment 11.5: (b) 52.8047 thousand dollars, (c) Our prediction will decrease by b2 = 4.0926 thousand dollars