Chap15_part2 - Solutions to End-of-Section and Chapter...

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Solutions to End-of-Section and Chapter Review Problems 163 15.31 (a) Coefficients Standard Error t Stat P-value Intercept 2.704449262 5.37548539 0.503107918 0.649503147 YLag1 0.89294783 0.424505135 2.103503014 0.126143957 YLag2 -0.708212436 0.602310389 -1.175826366 0.324481541 YLag3 0.802269914 0.491797033 1.631302876 0.20132482 Since the p -value = 0.20 > 0.05, do not reject H 0 that A 3 = 0. Third-order term can be deleted. (b) Coefficients Standard Error t Stat P-value Intercept 3.709551493 4.756100407 0.779956514 0.470706957 YLag1 0.949708028 0.440786085 2.154578061 0.083758887 YLag2 0.00080161 0.42035574 0.00190698 0.998552197 Since the p -value = 0.998 > 0.05, do not reject H 0 that A 2 = 0. Second-order term can be deleted. (c) Coefficients Standard Error t Stat P-value Intercept 4.824803547 3.189597232 1.51266859 0.17412552 YLag1 0.930752135 0.05809798 16.02038728 8.96802E-07 Since the p -value is virtually 0, reject H 0 that A 1 = 0. First-order autoregressive model is the appropriate model. (d) 1 ˆ 4.8248 0.9308 i i Y Y - = + Year Forecasts 2002 60.2705123 2003 60.92171152 2004 61.52781659 15.32 (a) 121 . 2 1 1 12 45 1 ) ˆ ( 1 2 = - - = - - - = = p n Y Y S n i i i YX . The standard error of the estimate is 2.121 billion constant 1995 dollars. (b) 5 . 1 12 18 ˆ 1 = = - = = n Y Y MAD n i i i . The mean absolute deviation is 1.5 billion constant 1995 dollars. 15.33 (a) 523 . 5 1 1 12 305 1 ) ˆ ( 1 2 = - - = - - - = = p n Y Y S n i i i YX . The standard error of the estimate is 5.523 billion constant 1995 dollars. (b) 167 . 3 12 38 ˆ 1 = = - = = n Y Y MAD n i i i . The mean absolute deviation is 3.167 billion constant 1995 dollars.
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164 Chapter 15: Time Series Forecasting and Index Numbers 15.34 (a) Residuals Plot -400 -300 -200 -100 0 100 200 300 400 500 600 1975 1977 1979 1981 1983 1985 1987 1989 1991 1993 1995 1997 1999 2001 Year Residuals (b) Excel output: Regression Statistics Multiple R 0.988517544 R Square 0.977166935 Adjusted R Square 0.976253613 Standard Error 238.738489 Observations 27 238.7385 YX S = (c) MAD = 5242.6761/27 = 194.1732 (d) The residuals in the linear trend model show strings of consecutive positive and negative values. An autoregressive model will probably do better. 15.35 (a) Residual Plots -200000 -150000 -100000 -50000 0 50000 100000 150000 Year Linear Quadratic Exponential AR1
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Solutions to End-of-Section and Chapter Review Problems 165 15.35 (b),(c) cont. Linear Quadratic Exponential AR1 Syx 71717.7 7 26532.99 80216.94 14815.12 MAD 57907.3 0 22068.16 67748.08 11641.89 (d) The residuals in the three trend models show strings of consecutive positive and negative values. The autoregressive model performs well for the historical data and has a fairly random pattern of residuals. The autoregressive model has the smallest values in MAD. Based on the principle of parsimony, the autoregressive model would probably be the best model for forecasting. 15.36
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This homework help was uploaded on 04/09/2008 for the course ENGR, STAT 320, 262, taught by Professor Harris during the Spring '08 term at Purdue.

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Chap15_part2 - Solutions to End-of-Section and Chapter...

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