Lesson_10_11_Review_Problem_Solutions_F12

# Boats predictor coef se coef t p constant 16567 5215

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Unformatted text preview: in the "Fit" column of the printout. c. Perhaps boat owners do not buy a new trailer every time they buy a new boat. Alternatively, boat owners might make their own trailers or they may haul their boats in the back of a truck or on the roof of their car or truck. Applying the procedure described in textbook chapter 2, Computer Solutions 2.7, we obtain the corresponding Excel plot and equation shown below. C D E F G H I y = 0.5771x - 165.67 220 Trailers (thousands) A B 1 Boats Trailers 2 649 207 3 619 194 4 596 181 5 576 174 6 585 168 7 574 159 8 9 10 11 12 13 14 15 200 180 160 140 560 580 600 620 Boats (thousands) 640 660 PROBLEM # 10.16 a. Age 110 113 114 134 93 141 115 115 115 142 96 139 89 93 91 109 Repairs 327.67 376.68 392.52 443.14 342.62 476.16 324.74 338.98 433.45 526.37 362.42 448.76 335.27 350.94 291.81 467.80 3 138 83 100 137 474.48 354.15 420.11 416.04 a b1 = s xy s2 x = 936.82 = 2.47 b 0 = y − b1x = 395.21 – 2.47(113.35) = 115.24. 378.77 ˆ ˆ Regression line: y = 115.24 + 2.47x (Excel: y = 114.85 + 2.47x) b b1 = 2.47; for each additional month of age, repair costs increase on average by \$2.47. b 0 = 114.85 is the y ­intercept. c R 2 = s2 xy s2s2 xy = (936.82) 2 = .5659 (Excel: R 2 = .5659) 56.59% of the variation in repair (378.77)(4,094.79) costs s explained by the variation in ages. ȹ ȹ s 2 ȹ (936.82) 2 ȹ xy ȹ = 33,777 SSE = (n − 1)ȹ s 2 − 2 ȹ = (20 − 1)ȹ 4,094.79 − d y ȹ ȹ 378.77 ȹ s x ȹ ȹ Ⱥ ȹ Ⱥ sε = 33,777 SSE = 43.32 (Excel: s ε = 43.32). = 20 − 2 n−2 H 0 : β1 = 0 H1 : β1 ≠ 0 Rejection region: t > t α / 2, n − 2 = t .025,18 = 2.101or t < − t α / 2, n − 2 = − t.025,18 = −2.101 s b1 = t= sε (n − 1)s 2 x = 43.32 (20 − 1)(378.77) = .511 b 1 − β1 2.47 − 0 = = 4.84 (Excel: t = 4.84, p–value = .0001. There is enough evidence to s b1 .511 infer that repair costs and age are linearly related. ˆ e y = b0 + b1xg = 115.24 + 2.47(120) = 411.64 4 Prediction interval: y ± t α / 2,n −2 s ε 1 + ˆ = 411.64 ± 2.101(43.32) 1 + 2 1 (x g − x) (where t α / 2, n − 2 = t.025,18 = 2.101) + n (n − 1)s 2 x 1 (120 − 113.35) 2 + = 411.64 ± 93.54 20 (20 − 1)(378.77) Lower prediction limit = 318.1, upper prediction limit = 505.2 (Excel: 318.1, 505.2) PROBLEM # 10.17 The prediction interval for y gets wider as the x value on which the interval estimate is based gets farther away from the mean of x because there is less error in making interval estimates based on x values that are closer to the mean. This can be seen in the formula; the numerator of the fraction under the square root includes (x  ­ x )2 . This value, of course, increases as...
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## This note was uploaded on 09/30/2013 for the course COMM 215 taught by Professor Ghatri during the Fall '08 term at Concordia Canada.

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