# ch6 - Chapter 6 Estimation 6.1 Sample Characteristics...

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Chapter 6 Estimation 6.1 Sample Characteristics 6.1–2 (a) x = 4 3 = 1 . 333; (b) s 2 = 88 69 = 1 . 275. 6.1–4 (a) x = 1 . 711, s = 0 . 486; (b) and (d) graphs. 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7 1 2 3 4 5 6 Figure 6.1–4: (b) Female underwater weights, (d) Female (above) and male underwater weights (c) The fit in the left tail is not the best. (d) Females generally weigh much less than males underwater. 73

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74 Chapter 6 6.1–6 (a) Stems Leaves Frequency Depths 3 * 10 1 1 3 t 25 1 2 3 f 0 2 3 s 70 70 70 70 4 6 3 81 83 90 90 90 90 90 90 92 9 15 4 * 00 00 00 00 00 02 02 05 10 10 10 15 15 15 14 (14) 4 t 20 20 20 20 20 25 30 30 8 16 4 f 40 50 2 8 4 s 60 63 70 75 75 5 6 4 80 1 1 Table 6.1–6: Ordered stem-and-leaf diagram of Barry Bonds’s home run distances (b) min = 310, e q 1 = 390, e m = 405, e q 3 = 422 . 5, max = 480; (c) 300 320 340 360 380 400 420 440 460 480 500 Figure 6.1–6: Distances of Barry Bonds’s homeruns (d) The interquartile range is IQR = 32 . 5. Inner fences could be drawn at 356.25 and 453.75. Outer fences could be drawn at 307.5 and 502.5.
Estimation 75 6.1–8 (a) The order statistics are: 5, 5, 5, 7, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 11, 11, 11, 11, 13, 13, 13, 13, 15, 15, 17, 19, 19, 19, 19, 19, 21, 21, 23, 23, 25, 25, 25, 27, 27, 33, 33, 35, 35, 37, 39, 41, 43, 43, 45, 47, 49, 49, 51, 53, 57, 57, 61, 61, 63, 63, 63, 63, 65, 65, 65, 65, 67, 71, 71, 75, 75, 75, 83, 83, 85, 87, 89, 91, 93, 95, 95, 101, 109, 127, 131, 131, 135, 177, 213, 247, 307, 413, 443, 471, 507, 515, 615, 703, 877, 1815; (b) y 1 = 5, e q 1 = 35 / 2, e m = 48, e q 3 = 173 / 2, y 100 = 1 , 815; (c) 0 500 1000 1500 0 500 1000 1500 Figure 6.1–8: (c) Roulette data (d) showing fences and outliers (d) Inner fence at 173 / 2 + 1 . 5 * 69 = 190, outer fence at 173 / 2 + 3 * 69 = 293 . 5; (e) x = 112 . 12; (f) The mean is influenced greatly by the outliers. 6.1–10 (a) Stems Leaves Frequency Depths 11 805 1 1 12 380 523 527 590 690 837 930 7 8 13 008 143 172 217 300 325 342 343 350 425 698 710 728 852 980 15 23 14 087 285 375 505 507 548 655 667 807 875 893 977 12 35 15 010 022 062 082 085 110 143 225 260 290 555 702 958 970 980 992 997 17 (17) 16 133 217 360 545 623 810 860 917 993 9 24 17 088 120 323 422 857 883 6 15 18 045 308 607 648 977 5 9 19 252 788 980 3 4 20 392 1 1 (Multiply numbers by 10 - 2 .) Table 6.1–10: Ordered stem-and-leaf diagram of race times for 125 male runners (b) min = 118 . 05, e q 1 = 137 . 01, e m = 150 . 72, e q 3 = 167 . 6325, max = 203 . 92;

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76 Chapter 6 (c) 100 120 140 160 180 200 Figure 6.1–10: Race times for women (above) and men. 6.1–12 (a) Stems Leaves Frequency Depths 101 7 1 1 102 0 0 0 3 4 103 0 4 104 0 4 105 8 9 2 6 106 1 3 3 6 6 7 7 8 8 9 (9) 107 3 7 9 3 10 108 8 1 7 109 1 3 9 3 6 110 0 2 2 3 3 (Multiply numbers by 10 - 1 .) Table 6.1–12: Ordered stem-and-leaf diagram of weights of indicator housings (b) 102 104 106 108 110 Figure 6.1–12: Weights of indicator housings min = 101 . 7, e q 1 = 106 . 0, e m = 106 . 7, e q 3 = 108 . 95, max = 110 . 2; (c) The interquartile range in IQR = 108 . 95 - 106 . 0 = 2 . 95 . The inner fence is located at 106 . 7 - 1 . 5(2 . 95) = 102 . 275 so there are four suspected outliers.
Estimation 77 6.2 Point Estimation 6.2–2 The likelihood function is L ( θ ) = 1 2 πθ n/ 2 exp " - n X i =1 ( x i - μ ) 2 / 2 θ # , 0 < θ < . The logarithm of the likelihood function is ln L ( θ ) = - n 2 (ln 2 π ) - n 2 (ln θ ) - 1 2 θ n X i =1 ( x i - μ ) 2 . Setting the first derivative equal to zero and solving for θ yields d ln L ( θ ) = - n 2 θ + 1 2 θ 2 n X i =1 ( x i - μ ) 2 = 0 θ = 1 n n X i =1 ( x i - μ ) 2 . Thus b θ = 1 n n X i =1 ( X i - μ ) 2 . To see that b θ is an unbiased estimator of θ , note that E ( b θ ) = E σ 2 n n X i =1 ( X i - μ ) 2 σ 2 !

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