Time-Series Forecasting 16-1 CHAPTER 16: TIME-SERIES FORECASTING 1. The effect of an unpredictable, rare event will be contained in the ___________ component. a) trend b) cyclical c) irregular d) seasonal ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: component factors, properties 2. The overall upward or downward pattern of the data in an annual time series will be contained in the ____________ component. a) trend b) cyclical c) irregular d) seasonal ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: component factors, properties 3. The fairly regular fluctuations that occur within each year would be contained in the _________________ component. a) trend b) cyclical c) irregular d) seasonal ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: component factors, properties 4. The annual multiplicative time-series model does not possess _______ component. a) a trend b) a cyclical c) an irregular d) a seasonal ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: component factors, properties Copyright ©2015 Pearson Education, Inc. 16-2 Time-Series Forecasting 5. Based on the following scatter plot, which of the time-series components is not present in this quarterly time series? 350 S to c k R e tur ns 300 250 200 150 100 50 0 0 10 20 30 40 Q u a rte rs a. b. c. d. Trend Seasonal Cyclical Irregular ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: component factors, properties Copyright ©2015 Pearson Education, Inc. 50 60 Time-Series Forecasting 16-3 6. After estimating a trend model for annual time-series data, you obtain the following residual plot against time, the problem with your model is that: a) The cyclical component has not been accounted for. b) The seasonal component has not been accounted for. c) The trend component has not been accounted for. d) The irregular component has not been accounted for. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: component factors 7. True or False: A trend is a persistent pattern in annual time-series data that has to be followed for several years. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: component factor 8. The method of moving averages is used a) to plot a series. b) to exponentiate a series. c) to smooth a series. d) in regression analysis. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: moving averages Copyright ©2015 Pearson Education, Inc. 16-4 Time-Series Forecasting 9. Which of the following methods should not be used for short-term forecasts into the future? a) Exponential smoothing b) Moving averages c) Linear trend model d) Autoregressive modeling ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: moving averages, properties 10. Which of the following statements about moving averages is not true? a) It can be used to smooth a series. b) It gives equal weight to all values in the computation. c) It is simpler than the method of exponential smoothing. d) It gives greater weight to more recent data. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: moving averages, properties 11. True or False: Given a data set with 15 yearly observations, a 3-year moving average will have fewer observations than a 5-year moving average. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: moving averages, properties 12. True or False: Given a data set with 15 yearly observations, there are only thirteen 3-year moving averages. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: moving averages, properties 13. True or False: Given a data set with 15 yearly observations, there are only seven 9-year moving averages. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: moving averages, properties Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting 16-5 14. Which of the following is not an advantage of exponential smoothing? a) It enables you to perform one-period ahead forecasting. b) It enables you to perform more than one-period ahead forecasting. c) It enables you to smooth out seasonal components. d) It enables you to smooth out cyclical components. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing, properties 15. When using the exponentially weighted moving average for purposes of forecasting rather than smoothing, a) the previous smoothed value becomes the forecast. b) the current smoothed value becomes the forecast. c) the next smoothed value becomes the forecast. d) None of the above. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing, properties 16. Which of the following statements about the method of exponential smoothing is not true? a) It gives greater weight to more recent data. b) It can be used for forecasting. c) It uses all earlier observations in each smoothing calculation. d) It gives greater weight to the earlier observations in the series. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing, properties 17. True or False: If a time series does not exhibit a long-term trend, the method of exponential smoothing may be used to obtain short-term predictions about the future. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: exponential smoothing Copyright ©2015 Pearson Education, Inc. 16-6 Time-Series Forecasting 18. A model that can be used to make predictions about long-term future values of a time series is a) linear trend. b) quadratic trend. c) exponential trend. d) All of the above. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: least-squares trend fitting, properties 19. You need to decide whether you should invest in a particular stock. You would like to invest if the price is likely to rise in the long run. You have data on the daily mean price of this stock over the past 12 months. Your best action is to a) compute moving averages b) perform exponential smoothing c) estimate a least square trend model d) compute the MAD statistic ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, properties 20. When a time series appears to be increasing at an increasing rate, such that percentage difference from value to value is constant, the appropriate model to fit is the a. linear trend. b. quadratic trend. c. exponential trend. d. None of the above. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: least-squares trend fitting, model selection 21. The method of least squares is used on time-series data for a) eliminating irregular movements. b) deseasonalizing the data. c) obtaining the trend equation. d) exponentially smoothing a series. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, properties Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting 16-7 22. True or False: A least squares linear trend line is just a simple regression line with the years recoded. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: least-squares trend fitting 23. True or False: The method of least squares may be used to estimate both linear and curvilinear trends. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting 24. Microsoft Excel was used to obtain the following quadratic trend equation: Sales = 100 – 10X + 15X2. The data used was from 2004 through 2013 coded 0 to 9. The forecast for 2014 is __________. ANSWER: 1,500 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, forecasting 25. The manager of a company believed that her company's profits were following an exponential trend. She used Microsoft Excel to obtain a prediction equation for the logarithm (base 10) of profits: log10(Profits) = 2 + 0.3X The data she used were from 2007 through 2012 coded 0 to 5. The forecast for 2013 profits is __________. ANSWER: 6,309.57 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, forecasting 26. A first-order autoregressive model for stock sales is: Salesi = 800 + 1.2(Sales)i-1. If sales in 2012 is 6,000, the forecast of sales for 2013 is __________. ANSWER: 8,000 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, forecasting Copyright ©2015 Pearson Education, Inc. 16-8 Time-Series Forecasting 27. A second-order autoregressive model for average mortgage rate is: Ratei = – 2.0 + 1.8(Rate)i-1 – 0.5 (Rate)i-2. If the average mortgage rate in 2012 was 7.0, and in 2011 was 6.4, the forecast for 2013 is __________. ANSWER: 7.4 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, forecasting 28. A second-order autoregressive model for average mortgage rate is: Ratei = – 2.0 + 1.8(Rate)i-1 – 0.5 (Rate)i-2. If the average mortgage rate in 2012 was 7.0, and in 2011 was 6.4, the forecast for 2014 is __________. ANSWER: 7.82 TYPE: FI DIFFICULTY: Difficult KEYWORDS: autoregressive model, forecasting 29. In selecting an appropriate forecasting model, the following approaches are suggested: a) Perform a residual analysis. b) Measure the size of the forecasting error. c) Use the principle of parsimony. d) All of the above. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: model selection 30. To assess the adequacy of a forecasting model, one measure that is often used is a) quadratic trend analysis. b) the MAD. c) exponential smoothing. d) moving averages. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: model selection 31. True or False: MAD is the summation of the residuals divided by the sample size. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: model selection Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting 16-9 32. True or False: The principle of parsimony indicates that the simplest model that gets the job done adequately should be used. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: model selection 33. True or False: In selecting a forecasting model, you should perform a residual analysis. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: model selection 34. True or False: Each forecast using the method of exponential smoothing depends on all the previous observations in the time series. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: exponential smoothing, properties 35. True or False: The MAD is a measure of the mean of the absolute discrepancies between the actual and the fitted values in a given time series. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: model selection SCENARIO 16-1 The number of cases of chardonnay wine sold by a Paso Robles winery in an 8-year period follows. Year Cases of Wine 2006 270 2007 356 2008 398 2009 456 2010 438 2011 478 2012 460 2013 480 Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting 36. Referring to Scenario 16-1, set up a scatter diagram (i.e., a time-series plot) with year on the horizontal X-axis. ANSWER: Scatter Plot Cases of Wine 16-10 600 500 400 300 200 100 0 2006 2007 2008 2009 2010 2011 2012 2013 Year TYPE: PR DIFFICULTY: Easy KEYWORDS: scatter plot 37. Referring to Scenario 16-1, does there appear to be a relationship between year and the number of cases of wine sold? a) No, there appears to be no relationship between the year and the number of cases of wine sold by the vintner. b) Yes, there appears to be a slight negative linear relationship between the year and the number of cases of wine sold by the vintner. c) Yes, there appears to be a slight positive relationship between the year and the number of cases of wine sold by the vintner. d) Yes, there appears to be a negative nonlinear relationship between the year and the number of cases of wine sold by the vintner. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: scatter plot, component factor Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting 16-11 38. After estimating a trend model for annual time-series data, you obtain the following residual plot against time, the problem with your model is that a) the cyclical component has not been accounted for. b) the seasonal component has not been accounted for. c) the trend component has not been accounted for. d) the irregular component has not been accounted for. 1 0.8 0.6 R esid u al s 0.4 0.2 0 -0.2 0 2 4 6 8 10 12 -0.4 -0.6 -0.8 -1 T im e (Ye a r) ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: component factor 39. The cyclical component of a time series a) represents periodic fluctuations which reoccur within 1 year. b) represents periodic fluctuations which usually occur in 2 or more years. c) is obtained by adjusting for the seasonal variation. d) is obtained by adjusting for calendar variation. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: component factor 40. Which of the following terms describes the overall long-term tendency of a time series? a) Trend b) Cyclical component c) Irregular component d) Seasonal component ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: component factor Copyright ©2015 Pearson Education, Inc. 16-12 Time-Series Forecasting 41. Which of the following terms describes the up and down movements of a time series that vary both in length and intensity? a) Trend b) Cyclical component c) Irregular component d) Seasonal component ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: component factor 42. The following is the list of MAD statistics for each of the models you have estimated from time-series data: Model MAD Linear Trend 1.38 Quadratic Trend 1.22 Exponential Trend 1.39 Second-order Autoregressive 0.71 Based on the MAD criterion, the most appropriate model is a) linear trend. b) quadratic trend. c) exponential trend. d) second-order autoregressive. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: model selection SCENARIO 16-2 The monthly advertising expenditures of a department store chain (in $1,000,000s) were collected over the last decade. The last 14 months of this time series follows: Month Expenditures ($) 1 1.4 2 1.8 3 1.6 4 1.5 5 1.8 6 1.7 7 1.9 8 2.2 9 1.9 10 1.9 11 2.1 12 2.4 13 2.8 14 3.1 Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting 16-13 43. Referring to Scenario 16-2, set up a scatter plot (i.e., time-series plot) with months on the horizontal X-axis. ANSWER: 3.5 Advertising Expenditures 3 2.5 2 1.5 1 0.5 0 0 2 4 6 8 Number of Months 10 12 14 TYPE: PR DIFFICULTY: Easy KEYWORDS: scatter plot 44. True or False: Referring to Scenario 16-2, advertising expenditures appear to be increasing in a linear rather than curvilinear manner over time. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: least-squares trend fitting SCENARIO 16-3 The following table contains the number of complaints received in a department store for the first 6 months of last year. Month Complaints January 36 February 45 March 81 April 90 May 108 June 144 Copyright ©2015 Pearson Education, Inc. 16-14 Time-Series Forecasting 45. Referring to Scenario 16-3, if a three-month moving average is used to smooth this series, what would be the second calculated value? a) 36 b) 40.5 c) 54 d) 72 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: moving averages 46. Referring to Scenario 16-3, if a three-month moving average is used to smooth this series, what would be the last calculated value? a) 72 b) 93 c) 114 d) 126 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: moving averages 47. Referring to Scenario 16-3, if a three-month moving average is used to smooth this series, how many values would it have? a) 2 b) 3 c) 4 d) 5 ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: moving averages, properties 48. Referring to Scenario 16-3, if this series is smoothed using exponential smoothing with a smoothing constant of 1/3, how many values would it have? a) 3 b) 4 c) 5 d) 6 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: exponential smoothing, properties Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting 16-15 49. Referring to Scenario 16-3, if this series is smoothed using exponential smoothing with a smoothing constant of 1/3, what would be the first value? a) 36 b) 39 c) 42 d) 45 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing 50. Referring to Scenario 16-3, if this series is smoothed using exponential smoothing with a smoothing constant of 1/3, what would be the second value? a) 39 b) 42 c) 45 d) 53 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing 51. Referring to Scenario 16-3, if this series is smoothed using exponential smoothing with a smoothing constant of 1/3, what would be the third value? a) 53 b) 65.33 c) 68 d) 81 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing 52. Referring to Scenario 16-3, suppose the last two smoothed values are 81 and 96 (Note: they are not). What would you forecast as the value of the time series for July? a) 81 b) 86 c) 91 d) 96 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing, forecasting Copyright ©2015 Pearson Education, Inc. 16-16 Time-Series Forecasting 53. Referring to Scenario 16-3, suppose the last two smoothed values are 81 and 96 (Note: they are not). What would you forecast as the value of the time series for September? a) 81 b) 86 c) 91 d) 96 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential smoothing, forecasting 54. If you want to recover the trend using exponential smoothing, you will choose a weight (W) that falls in the range a) 0, 0.2 b) c) d) 0.2, 0.4 0.6, 0.8 0.8,1.0 ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: exponential smoothing, forecasting, properties SCENARIO 16-4 The number of cases of merlot wine sold by a Paso Robles winery in an 8-year period follows. Year Cases of Wine 2005 270 2006 356 2007 398 2008 456 2009 358 2010 500 2011 410 2012 376 55. Referring to Scenario 16-4, a centered 3-year moving average is to be constructed for the wine sales. The result of this process will lead to a total of __________ moving averages. ANSWER: 6 TYPE: FI DIFFICULTY: Easy KEYWORDS: moving averages, properties Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting 16-17 56. Referring to Scenario 16-4, a centered 3-year moving average is to be constructed for the wine sales. The moving average for 2006 is __________. ANSWER: 341.33 TYPE: FI DIFFICULTY: Easy KEYWORDS: moving averages, properties 57. Referring to Scenario 16-4, a centered 3-year moving average is to be constructed for the wine sales. The moving average for 2009 is __________. ANSWER: 438 TYPE: FI DIFFICULTY: Easy KEYWORDS: moving averages, properties 58. Referring to Scenario 16-4, construct a centered 3-year moving average for the wine sales. ANSWER: Period Cases MA 1 270 * 2 356 341.333 3 398 403.333 4 456 404.000 5 358 438.000 6 500 422.667 7 410 428.667 8 376 * TYPE: PR DIFFICULTY: Moderate KEYWORDS: moving averages 59. Referring to Scenario 16-4, a centered 5-year moving average is to be constructed for the wine sales. The number of moving averages that will be calculated is __________. ANSWER: 4 TYPE: FI DIFFICULTY: Easy KEYWORDS: moving averages 60. Referring to Scenario 16-4, a centered 5-year moving average is to be constructed for the wine sales. The moving average for 2007 is __________. ANSWER: 367.6 TYPE: FI DIFFICULTY: Moderate KEYWORDS: moving averages Copyright ©2015 Pearson Education, Inc. 16-18 Time-Series Forecasting 61. Referring to Scenario 16-4, a centered 5-year moving average is to be constructed for the wine sales. The moving average for 2010 is __________. ANSWER: 420.0 TYPE: FI DIFFICULTY: Moderate KEYWORDS: moving averages 62. Referring to Scenario 16-4, construct a centered 5-year moving average for the wine sales. ANSWER: Period Cases MA 1 270 * 2 356 * 3 398 367.6 4 456 413.6 5 358 424.4 6 500 420.0 7 410 * 8 376 * TYPE: PR DIFFICULTY: Moderate KEYWORDS: moving averages 63. Referring to Scenario 16-4, exponential smoothing with a weight or smoothing constant of 0.2 will be used to smooth the wine sales. The value of E2, the smoothed value for 2006 is __________. ANSWER: 287.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing 64. Referring to Scenario 16-4, exponential smoothing with a weight or smoothing constant of 0.2 will be used to smooth the wine sales. The value of E4, the smoothed value for 2008 is __________. ANSWER: 338.7 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing 65. Referring to Scenario 16-4, exponential smoothing with a weight or smoothing constant of 0.2 will be used to forecast wine sales. The forecast for 2013 is __________. ANSWER: 380.2 TYPE: FI DIFFICULTY: Difficult KEYWORDS: exponential smoothing, forecasting Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting 16-19 66. Referring to Scenario 16-4, exponentially smooth the wine sales with a weight or smoothing constant of 0.2. ANSWER: Time CaseWine Smooth 1 270 270.000 2 356 287.200 3 398 309.360 4 456 338.688 5 358 342.550 6 500 374.040 7 410 381.232 8 376 380.186 TYPE: PR DIFFICULTY: Difficult KEYWORDS: exponential smoothing 67. Referring to Scenario 16-4, exponential smoothing with a weight or smoothing constant of 0.4 will be used to smooth the wine sales. The value of E2, the smoothed value for 2006 is __________. ANSWER: 304.4 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing 68. Referring to Scenario 16-4, exponential smoothing with a weight or smoothing constant of 0.4 will be used to smooth the wine sales. The value of E5, the smoothed value for 2009 is __________. ANSWER: 375.7 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing 69. Referring to Scenario 16-4, exponential smoothing with a weight or smoothing constant of 0.4 will be used to forecast wine sales. The forecast for 2013 is __________. ANSWER: 401.95 TYPE: FI DIFFICULTY: Difficult KEYWORDS: exponential smoothing, forecasting Copyright ©2015 Pearson Education, Inc. 16-20 Time-Series Forecasting 70. Referring to Scenario 16-4, exponentially smooth the wine sales with a weight or smoothing constant of 0.4. ANSWER: Time CaseWine Smooth 1 270 270.000 2 356 304.400 3 398 341.840 4 456 387.504 5 358 375.702 6 500 425.421 7 410 419.253 8 376 401.952 TYPE: PR DIFFICULTY: Difficult KEYWORDS: exponential smoothing SCENARIO 16-5 The number of passengers arriving at San Francisco on the Amtrak cross-country express on 6 successive Mondays were: 60, 72, 96, 84, 36, and 48. 71. Referring to Scenario 16-5, the number of arrivals will be smoothed with a 3-term moving average. There will be a total of __________ smoothed values. ANSWER: 4 TYPE: FI DIFFICULTY: Easy KEYWORDS: moving averages, properties 72. Referring to Scenario 16-5, the number of arrivals will be smoothed with a 3-term moving average. The first smoothed value will be __________. ANSWER: 76 TYPE: FI DIFFICULTY: Easy KEYWORDS: moving averages 73. Referring to Scenario 16-5, the number of arrivals will be smoothed with a 3-term moving average. The last smoothed value will be __________. ANSWER: 56 TYPE: FI DIFFICULTY: Easy KEYWORDS: moving averages Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting 16-21 74. Referring to Scenario 16-5, the number of arrivals will be smoothed with a 5-term moving average. The first smoothed value will be __________. ANSWER: 69.6 TYPE: FI DIFFICULTY: Moderate KEYWORDS: moving averages 75. Referring to Scenario 16-5, the number of arrivals will be smoothed with a 5-term moving average. The last smoothed value will be __________. ANSWER: 67.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: moving averages 76. Referring to Scenario 16-5, the number of arrivals will be exponentially smoothed with a smoothing constant of 0.1. The smoothed value for the second Monday will be __________. ANSWER: 61.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing 77. Referring to Scenario 16-5, the number of arrivals will be exponentially smoothed with a smoothing constant of 0.1. The smoothed value for the sixth Monday will be __________. ANSWER: 62.0 TYPE: FI DIFFICULTY: Difficult KEYWORDS: exponential smoothing 78. Referring to Scenario 16-5, the number of arrivals will be exponentially smoothed with a smoothing constant of 0.1. Then the forecast for the seventh Monday will be __________. ANSWER: 62.0 TYPE: FI DIFFICULTY: Difficult KEYWORDS: exponential smoothing, forecasting Copyright ©2015 Pearson Education, Inc. 16-22 Time-Series Forecasting 79. Referring to Scenario 16-5, exponentially smooth the number of arrivals using a smoothing constant of 0.1. ANSWER: Time Arrivals Smooth 1 60 60.0000 2 72 61.2000 3 96 64.6800 4 84 66.6120 5 36 63.5508 6 48 61.9957 TYPE: PR DIFFICULTY: Difficult KEYWORDS: exponential smoothing 80. Referring to Scenario 16-5, the number of arrivals will be exponentially smoothed with a smoothing constant of 0.25. The smoothed value for the second Monday will be __________. ANSWER: 63.0 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing 81. Referring to Scenario 16-5, the number of arrivals will be exponentially smoothed with a smoothing constant of 0.25. The smoothed value for the third Monday will be __________. ANSWER: 71.25 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential smoothing 82. Referring to Scenario 16-5, the number of arrivals will be exponentially smoothed with a smoothing constant of 0.25. The forecast of the number of arrivals on the seventh Monday will be __________. ANSWER: 60.6 TYPE: FI DIFFICULTY: Difficult KEYWORDS: exponential smoothing, forecasting Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting 16-23 83. Referring to Scenario 16-5, exponentially smooth the number of arrivals using a smoothing constant of 0.25. ANSWER: Time Arrivals Smooth 1 60 60.0000 2 72 63.0000 3 96 71.2500 4 84 74.4375 5 36 64.8281 6 48 60.6211 TYPE: PR DIFFICULTY: Difficult KEYWORDS: exponential smoothing SCENARIO 16-6 The president of a chain of department stores believes that her stores' total sales have been showing a linear trend since 1993. She uses Microsoft Excel to obtain the partial output below. The dependent variable is sales (in millions of dollars), while the independent variable is coded years, where 1993 is coded as 0, 1994 is coded as 1, etc. SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations Intercept Coded Year 0.604 0.365 0.316 4.800 17 Coefficients 31.2 0.78 84. Referring to Scenario 16-6, the fitted trend value (in millions of dollars) for 1993 is __________. ANSWER: 31.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, fitted value 85. Referring to Scenario 16-6, the fitted trend value (in millions of dollars) for 1998 is __________. ANSWER: 35.1 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, fitted value Copyright ©2015 Pearson Education, Inc. 16-24 Time-Series Forecasting 86. Referring to Scenario 16-6, the estimate of the amount by which sales (in millions of dollars) is increasing each year is __________. ANSWER: 0.78 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, slope, interpretation 87. Referring to Scenario 16-6, the forecast for sales (in millions of dollars) in 2013 is __________. ANSWER: 46.8 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, forecasting 88. Referring to Scenario 16-6, the forecast for sales (in millions of dollars) in 2015 is __________. ANSWER: 48.36 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, forecasting SCENARIO 16-7 The executive vice-president of a drug manufacturing firm believes that the demand for the firm's most popular drug has been evidencing an exponential trend since 1999. She uses Microsoft Excel to obtain the partial output below. The dependent variable is the log base 10 of the demand for the drug, while the independent variable is years, where 1999 is coded as 0, 2000 is coded as 1, etc. SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations Intercept Coded Year 0.996 0.992 0.991 0.02831 12 Coefficients 1.44 0.068 Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting 16-25 89. Referring to Scenario 16-7, the fitted trend value for 1999 is __________. ANSWER: 27.54 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, fitted value 90. Referring to Scenario 16-7, the fitted trend value for 2004 is __________. ANSWER: 60.26 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, fitted value 91. Referring to Scenario 16-7, the fitted exponential trend equation to predict Y is __________. ANSWER: 27.5423 1.1695 X TYPE: FI DIFFICULTY: Difficult KEYWORDS: least-squares trend fitting, fitted value 92. Referring to Scenario 16-7, the forecast for the demand in 2013 is __________. ANSWER: 246.6 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, forecasting 93. Referring to Scenario 16-7, the forecast for the demand in 2016 is __________. ANSWER: 394.46 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, forecasting Copyright ©2015 Pearson Education, Inc. 16-26 Time-Series Forecasting SCENARIO 16-8 The manager of a marketing consulting firm has been examining his company's yearly profits. He believes that these profits have been showing a quadratic trend since 1994. He uses Microsoft Excel to obtain the partial output below. The dependent variable is profit (in thousands of dollars), while the independent variables are coded years and squared of coded years, where 1994 is coded as 0, 1995 is coded as 1, etc. SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations Intercept Coded Year Year Squared 0.998 0.996 0.996 4.996 17 Coefficients 35.5 0.45 1.00 94. Referring to Scenario 16-8, the fitted value for 1994 is __________. ANSWER: 35.5 or $35,500 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, fitted value 95. Referring to Scenario 16-8, the fitted value for 1999 is __________. ANSWER: 62.75 or $62,750 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, fitted value 96. Referring to Scenario 16-8, the forecast for profits in 2014 is __________. ANSWER: 444.5 or $444,500 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, forecasting 97. Referring to Scenario 16-8, the forecast for profits in 2019 is __________. ANSWER: 671.75 or $671,750 TYPE: FI DIFFICULTY: Moderate KEYWORDS: least-squares trend fitting, forecasting Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting 16-27 SCENARIO 16-9 Given below are EXCEL outputs for various estimated autoregressive models for a company’s real operating revenues (in billions of dollars) from 1989 to 2012. From the data, you also know that the real operating revenues for 2010, 2011, and 2012 are 11.7909, 11.7757 and 11.5537, respectively. First-Order Autoregressive Model: Coefficients Standard Error Intercept XLag1 0.1802 1.0112 0.3980 0.0497 Second-Order Autoregressive Model: Coefficients Standard Error Intercept X Lag 1 X Lag 2 0.3005 1.1732 -0.1830 t Stat 0.4408 0.2347 0.2507 P-value 0.4528 20.3526 t Stat P-value 0.6817 4.9980 -0.7300 0.5036 0.0001 0.4743 Third-Order Autoregressive Model: Coefficients Standard Error t Stat P-value Intercept XLag1 XLag2 XLag3 0.6085 4.7617 -0.1860 -0.4330 0.5509 0.0002 0.8547 0.6704 0.3130 1.1737 -0.0694 -0.1221 0.5144 0.2465 0.3731 0.2820 0.6553 0.0000 98. Referring to Scenario 16-9 and using a 5% level of significance, what is the appropriate autoregressive model for the company’s real operating revenue? a) First-Order Autoregressive Model b) Second-Order Autoregressive c) Third-Order Autoregressive d) Any of the above ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: autoregressive model, t test on slope Copyright ©2015 Pearson Education, Inc. 16-28 Time-Series Forecasting 99. Referring to Scenario 16-9, if one decides to use the Third-Order Autoregressive model , what will the predicted real operating revenue for the company be in 2013? a) $11.55 billion b) $11.62 billion c) $11.84 billion d) $12.47 billion ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: autoregressive model, forecasting 100. Referring to Scenario 16-9, if one decides to use the Third-Order Autoregressive model , what will the predicted real operating revenue for the company be in 2014? a) $11.55 billion b) $11.62 billion c) $12.47 billion d) $12.57 billion ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: autoregressive model, forecasting 101. Referring to Scenario 16-9, if one decides to use the Third-Order Autoregressive model , what will the predicted real operating revenue for the company be in 2015? a) $11.59 billion b) $11.68 billion c) $11.84 billion d) $12.47 billion ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: autoregressive model, forecasting Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting 16-29 SCENARIO 16-10 Business closures in a city in the western U.S. from 2007 to 2012 were: 2007 10 2008 11 2009 13 2010 19 2011 24 2012 35 Microsoft Excel was used to fit both first-order and second-order autoregressive models, resulting in the following partial outputs: SUMMARY OUTPUT – 2nd Order Model Intercept X Variable 1 X Variable 2 Coefficients -5.77 0.80 1.14 SUMMARY OUTPUT – 1st Order Model Intercept X Variable 1 Coefficients -4.16 1.59 102. Referring to Scenario 16-10, the fitted values for the first-order autoregressive model are ________, ________, ________, ________, and ________. ANSWER: 11.74, 13.33, 16.51, 26.05, and 34.00 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, fitted value 103. Referring to Scenario 16-10, the residuals for the first-order autoregressive model are ________, ________, ________, ________, and ________. ANSWER: -.74, -.33, 2.49, -2.05, and 1.00 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, residual 104. Referring to Scenario 16-10, the value of the MAD for the first-order autoregressive model is ________. ANSWER: 1.322 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, model selection Copyright ©2015 Pearson Education, Inc. 16-30 Time-Series Forecasting 105. Referring to Scenario 16-10, the fitted values for the second-order autoregressive model are ________, ________, ________, and ________. ANSWER: 14.43, 17.17, 24.25, and 35.09 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, fitted value 106. Referring to Scenario 16-10, the residuals for the second-order autoregressive model are ________, ________, ________, and ________. ANSWER: – 1.43, 1.83, – 0.25, and – 0.09 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, residual 107. Referring to Scenario 16-10, the value of the MAD for the second-order autoregressive model is ________. ANSWER: 0.90 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, model selection 108. True or False: Referring to Scenario 16-10, the values of the MAD for the two models indicate that the first-order model should be used for forecasting. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: autoregressive model, model selection Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting 16-31 SCENARIO 16-11 The manager of a health club has recorded mean attendance in newly introduced step classes over the last 15 months: 32.1, 39.5, 40.3, 46.0, 65.2, 73.1, 83.7, 106.8, 118.0, 133.1, 163.3, 182.8, 205.6, 249.1, and 263.5. She then used Microsoft Excel to obtain the following partial output for both a first- and second-order autoregressive model. SUMMARY OUTPUT – 2nd Order Model Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations Intercept X Variable 1 X Variable 2 0.993 0.987 0.985 9.276 15 Coefficients 5.86 0.37 0.85 SUMMARY OUTPUT – 1st Order Model Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations Intercept X Variable 1 0.993 0.987 0.985 9.150 15 Coefficients 5.66 1.10 109. Referring to Scenario 16-11, using the first-order model, the forecast of mean attendance for month 16 is __________. ANSWER: 295.51 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, forecasting Copyright ©2015 Pearson Education, Inc. 16-32 Time-Series Forecasting 110. Referring to Scenario 16-11, using the first-order model, the forecast of mean attendance for month 17 is __________. ANSWER: 330.72 TYPE: FI DIFFICULTY: Moderate Y KEYWORDS: autoregressive model, forecasting 111. Referring to Scenario 16-11, using the second-order model, the forecast of mean attendance for month 16 is __________. ANSWER: 315.09 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, forecasting 112. Referring to Scenario 16-11, using the second-order model, the forecast of mean attendance for month 17 is __________. ANSWER: 346.42 TYPE: FI DIFFICULTY: Moderate KEYWORDS: autoregressive model, forecasting 113. True or False: Referring to Scenario 16-11, based on the parsimony principle, the secondorder model is the better model for making forecasts. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: autoregressive model, model selection SCENARIO 16-12 A local store developed a multiplicative time-series model to forecast its revenues in future quarters, using quarterly data on its revenues during the 5-year period from 2009 to 2013. The following is the resulting regression equation: log 10 Yˆ = 6.102 + 0.012 X – 0.129 Q1 – 0.054 Q2 + 0.098 Q3 where Yˆ is the estimated number of contracts in a quarter X is the coded quarterly value with X = 0 in the first quarter of 2008. Q1 is a dummy variable equal to 1 in the first quarter of a year and 0 otherwise. Q2 is a dummy variable equal to 1 in the second quarter of a year and 0 otherwise. Q3 is a dummy variable equal to 1 in the third quarter of a year and 0 otherwise. Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting 16-33 114. Referring to Scenario 16-12, the best interpretation of the constant 6.102 in the regression equation is: a) the fitted value for the first quarter of 2009, prior to seasonal adjustment, is log10(6.102). b) the fitted value for the first quarter of 2009, after to seasonal adjustment, is log10(6.102). c) the fitted value for the first quarter of 2009, prior to seasonal adjustment, is 106.102. d) the fitted value for the first quarter of 2009, after to seasonal adjustment, is 106.102. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, intercept, interpretation, seasonable data 115. Referring to Scenario 16-12, the best interpretation of the coefficient of X (0.012) in the regression equation is: a) the quarterly compound growth rate in revenues is around 2.8%. b) the annual growth rate in revenues is around 2.8%. c) the quarterly growth rate in revenues is around 1.2%. d) the annual growth rate in revenues is around 1.2%. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, slope, interpretation, seasonable data 116. Referring to Scenario 16-12, the estimated quarterly compound growth rate in revenues is around: a) 1.2%. b) 2.8%. c) 12%. d) 28%. ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, slope, interpretation, seasonable data Copyright ©2015 Pearson Education, Inc. 16-34 Time-Series Forecasting 117. Referring to Scenario 16-12, the best interpretation of the coefficient of Q2 (–0.054) in the regression equation is: a) the revenues in the second quarter of a year is approximately 5.4% lower than the average over all 4 quarters. b) the revenues in the second quarter of a year is approximately 5.4% lower than it would be during the fourth quarter. c) the revenues in the second quarter of a year is approximately 11.69% lower than the average over all 4 quarters. d) the revenues in the second quarter of a year is approximately 11.69% lower than it would be during the fourth quarter. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, slope, interpretation, seasonable data 118. Referring to Scenario 16-12, the best interpretation of the coefficient of Q3 (0.098) in the regression equation is: a) the revenues in the third quarter of a year is approximately 9.8% higher than the average over all 4 quarters. b) the revenues in the third quarter of a year is approximately 9.8% higher than it would be during the fourth quarter. c) the revenues in the third quarter of a year is approximately 25.31% higher than the average over all 4 quarters. d) the revenues in the third quarter of a year is approximately 25.31% higher than it would be during the fourth quarter. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, slope, interpretation, seasonable data 119. Referring to Scenario 16-12, to obtain the fitted value for the first quarter of 2013 using the model, which of the following sets of values should be used in the regression equation? a) X = 16, Q1 = 1, Q2 = 0, Q3 = 0 b) X = 16, Q1 = 0, Q2 = 1, Q3 = 0 c) X = 17, Q1 = 1, Q2 = 0, Q3 = 0 d) X = 17, Q1 = 0, Q2 = 1, Q3 = 0 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting, seasonable data Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting 16-35 120. Referring to Scenario 16-12, to obtain a fitted value for the fourth quarter of 2010 using the model, which of the following sets of values should be used in the regression equation? a) X = 7, Q1 = 0, Q2 = 0, Q3 = 0 b) X = 7, Q1 = 1, Q2 = 0, Q3 = 0 c) X = 8, Q1 = 0, Q2 = 0, Q3 = 0 d) X = 8, Q1 = 1, Q2 = 0, Q3 = 0 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting, seasonable data 121. Referring to Scenario 16-12, to obtain a forecast for the third quarter of 2014 using the model, which of the following sets of values should be used in the regression equation? a) X = 22, Q1 = 0, Q2 = 0, Q3 = 0 b) X = 22, Q1 = 0, Q2 = 0, Q3 = 1 c) X = 23, Q1 = 0, Q2 = 0, Q3 = 0 d) X = 23, Q1 = 0, Q2 = 0, Q3 = 1 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting, seasonable data 122. Referring to Scenario 16-12, using the regression equation, what is the forecast for the revenues in the third quarter of 2014? ANSWER: 2,910,717.12 TYPE: PR DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting, seasonable data 123. Referring to Scenario 16-12, using the regression equation, what is the forecast for the revenues in the first quarter of 2016? ANSWER: 2,037,042.08 TYPE: PR DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting, seasonable data 124. Referring to Scenario 16-12, using the regression equation, what is the forecast for the revenues in the fourth quarter of 2015? ANSWER: 2,666,858.67 TYPE: PR DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting, seasonable data Copyright ©2015 Pearson Education, Inc. 16-36 Time-Series Forecasting 125. Referring to Scenario 16-12, in testing the significance of the coefficient of X in the regression equation (0.012) which has a p-value of 0.0000. Which of the following is the best interpretation of this result? a) The quarterly growth rate in revenues is significantly different from 0% ( = 0.05). b) The quarterly growth rate in revenues is not significantly different from 0% ( = 0.05). c) The quarterly growth rate in revenues is significantly different from 1.2% ( = 0.05). d) The quarterly growth rate in revenues is not significantly different from 1.2% ( = 0.05). ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, t test on slope, decision, conclusion, interpretation, seasonable data 126. Referring to Scenario 16-12, in testing the significance of the coefficient for Q1 in the regression equation (– 0.129) which has a p-value of 0.492. Which of the following is the best interpretation of this result? a) The revenues in the first quarter of the year are significantly different from the revenues in an average quarter ( = 0.05). b) The revenues in the first quarter of the year are not significantly different from the revenues in an average quarter ( = 0.05). c) The revenues in the first quarter of the year are significantly different from the revenues in the fourth quarter ( = 0.05). d) The revenues in the first quarter of the year are not significantly different from the revenues in the fourth quarter ( = 0.05). ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, t test on slope, decision, conclusion, interpretation, seasonable data Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting 16-37 SCENARIO 16-13 Given below is the monthly time series data for U.S. retail sales of building materials over a specific year. Month 1 2 3 4 5 6 7 8 9 10 11 12 Retail Sales 6,594 6,610 8,174 9,513 10,595 10,415 9,949 9,810 9,637 9,732 9,214 9,201 The results of the linear trend, quadratic trend, exponential trend, first-order autoregressive, second-order autoregressive and third-order autoregressive model are presented below in which the coded month for the 1st month is 0: Linear trend model: Coefficients Standard Error t Stat P-value Intercept 7950.7564 617.6342 12.8729 0.0000 Coded Month 212.6503 95.1145 2.2357 0.0494 Standard Error t Stat Quadratic trend model: Coefficients P-value Intercept 6358.2473 417.2692 15.2378 0.0000 Coded Month 1168.1558 176.3526 6.6240 0.0001 Coded Month^2 -86.8641 15.4474 -5.6232 0.0003 Exponential trend model: Coefficients Standard Error t Stat P-value Intercept 3.8912 0.0315 123.3674 0.0000 Coded Month 0.0116 0.0049 2.3957 0.0376 First-order autoregressive: Coefficients Intercept YLag1 Standard Error t Stat P-value 3132.0951 1287.2899 2.4331 0.0378 0.6823 0.1398 4.8812 0.0009 Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting Second-order autoregressive:: Coefficients Intercept Standard Error t Stat P-value 4968.5789 766.9416 6.4784 0.0003 YLag1 0.9333 0.1547 6.0316 0.0005 YLag2 -0.4487 0.1238 -3.6235 0.0085 Third-order autoregressive:: Coefficients Intercept Standard Error t Stat P-value 6782.7567 2105.7115 3.2211 0.0234 YLag1 0.5481 0.3918 1.3990 0.2207 YLag2 0.0198 0.4034 0.0490 0.9628 YLag3 -0.2749 0.2234 -1.2308 0.2731 Below is the residual plot of the various models: Residual Plot 2500 2000 1500 1000 Residuals 16-38 Linear-trend Quadratic-trend 500 Exponential-trend 0 AR(1) 1 2 3 4 5 6 7 8 9 10 11 12 -500 AR(3) -1000 -1500 -2000 AR(2) Axis Title Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting 16-39 127. Referring to Scenario 16-13, construct a scatter plot (i.e., a time-series plot) with month on the horizontal X-axis. ANSWER: Time Series Plot 12,000 Retail Sales 10,000 8,000 6,000 4,000 2,000 0 0 2 4 6 8 10 12 14 Month TYPE: PR DIFFICULTY: Easy KEYWORDS: scatter plot 128. Referring to Scenario 16-13, if a five-month moving average is used to smooth this series, what would be the first calculated value? ANSWER: 8,297 TYPE: PR DIFFICULTY: Easy KEYWORDS: moving averages 129. Referring to Scenario 16-13, if a five-month moving average is used to smooth this series, what would be the last calculated value? ANSWER: 9,519 TYPE: PR DIFFICULTY: Easy KEYWORDS: moving averages 130. Referring to Scenario 16-13, if a five-month moving average is used to smooth this series, how many moving averages can you compute? ANSWER: 8 TYPE: PR DIFFICULTY: Easy KEYWORDS: moving averages Copyright ©2015 Pearson Education, Inc. 16-40 Time-Series Forecasting 131. Referring to Scenario 16-13, what is the exponentially smoothed value for the first month using a smoothing coefficient of W = 0.5? ANSWER: 6,594 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential smoothing 132. Referring to Scenario 16-13, what is the exponentially smoothed value for the second month using a smoothing coefficient of W = 0.5? ANSWER: 6,602 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential smoothing 133. Referring to Scenario 16-13, what is the exponentially smoothed value for the 12th month using a smoothing coefficient of W = 0.5 if the exponentially smooth value for the 10th and 11th month are 9,746.3672 and 9,480.1836, respectively? ANSWER: 9,340.5918 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential smoothing 134. Referring to Scenario 16-13, what is the exponentially smoothed forecast for the 13th month using a smoothing coefficient of W = 0.5 if the exponentially smooth value for the 10th and 11th month are 9,746.3672 and 9,480.1836, respectively? ANSWER: 9340.5918 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential smoothing, forecasting 135. Referring to Scenario 16-13, what is the exponentially smoothed value for the first month using a smoothing coefficient of W = 0.25? ANSWER: 6,594 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential smoothing 136. Referring to Scenario 16-13, what is the exponentially smoothed value for the second month using a smoothing coefficient of W = 0.25? ANSWER: 6,598 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential smoothing Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting 16-41 137. Referring to Scenario 16-13, what is the exponentially smoothed value for the 12th month using a smoothing coefficient of W = 0.25 if the exponentially smooth value for the 10th and 11th month are 9,477.7776 and 9,411.8332, respectively? ANSWER: 9,359.1249 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential smoothing 138. Referring to Scenario 16-13, what is the exponentially smoothed forecast for the 13th month using a smoothing coefficient of W = 0.25 if the exponentially smooth value for the 10th and 11th month are 9,477.7776 and 9,411.8332, respectively? ANSWER: 9359.1249 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential smoothing, forecasting, fitted value 139. Referring to Scenario 16-13, what is your forecast for the 13th month using the linear-trend model? ANSWER: 10,502.5606 TYPE: PR DIFFICULTY: Easy KEYWORDS: least squares trend fitting, forecasting, fitted value 140. Referring to Scenario 16-13, what is the p-value for the t test statistic for testing the significance of the quadratic term in the quadratic-trend model? ANSWER: 0.0003 TYPE: PR DIFFICULTY: Moderate KEYWORDS: least squares trend fitting, t test on slope 141. Referring to Scenario 16-13, what is the value of the t test statistic for testing the significance of the quadratic term in the quadratic-trend model? ANSWER: -5.6232 TYPE: PR DIFFICULTY: Moderate KEYWORDS: least squares trend fitting, t test on slope 142. True or False: Referring to Scenario 16-13, you can conclude that the quadratic term in the quadratic-trend model is statistically significant at the 5% level of significance. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: least squares trend fitting, t test on slope, decision, conclusion Copyright ©2015 Pearson Education, Inc. 16-42 Time-Series Forecasting 143. Referring to Scenario 16-13, what is your forecast for the 13th month using the quadratictrend model? ANSWER: 7,867.6818 TYPE: PR DIFFICULTY: Easy KEYWORDS: least squares trend fitting, forecasting, fitted value 144. Referring to Scenario 16-13, what is your forecast for the 13th month using the exponentialtrend model? ANSWER: 10,736.5937 TYPE: PR DIFFICULTY: Moderate KEYWORDS: least squares trend fitting, forecasting, fitted value 145. Referring to Scenario 16-13, what is your estimated annual compound growth rate using the exponential-trend model? ANSWER: 2.7157% TYPE: PR DIFFICULTY: Difficult KEYWORDS: least squares trend fitting, interpretation 146. Referring to Scenario 16-13, what is the value of the t test statistic for testing the appropriateness of the third-order autoregressive model? ANSWER: -1.2308 TYPE: PR DIFFICULTY: Easy KEYWORDS: autoregressive model, t test on slope 147. Referring to Scenario 16-13, what is the p-value of the t test statistic for testing the appropriateness of the third-order autoregressive model? ANSWER: 0.2731 TYPE: PR DIFFICULTY: Easy KEYWORDS: autoregressive model, t test on slope 148. True or False: Referring to Scenario 16-13, you can reject the null hypothesis for testing the appropriateness of the third-order autoregressive model at the 5% level of significance. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: autoregressive model, t test on slope, decision Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting 16-43 149. True or False: Referring to Scenario 16-13, you can conclude that the third-order autoregressive model is appropriate at the 5% level of significance. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: autoregressive model, t test on slope, conclusion 150. Referring to Scenario 16-13, what is the value of the t test statistic for testing the appropriateness of the second-order autoregressive model? ANSWER: -3.6235 TYPE: PR DIFFICULTY: Easy KEYWORDS: autoregressive model, t test on slope 151. Referring to Scenario 16-13, what is the p-value of the t test statistic for testing the appropriateness of the second-order autoregressive model? ANSWER: 0.0085 TYPE: PR DIFFICULTY: Easy KEYWORDS: autoregressive model, t test on slope 152. True or False: Referring to Scenario 16-13, you can reject the null hypothesis for testing the appropriateness of the second-order autoregressive model at the 5% level of significance. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: autoregressive model, t test on slope, decision 153. True or False: Referring to Scenario 16-13, you can conclude that the second-order autoregressive model is appropriate at the 5% level of significance. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: autoregressive model, t test on slope, conclusion Copyright ©2015 Pearson Education, Inc. 16-44 Time-Series Forecasting 154. Referring to Scenario 16-13, the best autoregressive model using the 5% level of significance is a) first-order b) second-order c) third-order d) none of the above ANSWER: b TYPE: MC DIFFICULTY: Difficult KEYWORDS: autoregressive model, t test on slope, model selection 155. Referring to Scenario 16-13, what is your forecast for the 13th month using the first-order autoregressive model? ANSWER: 9,410.0434 TYPE: PR DIFFICULTY: Easy KEYWORDS: least squares trend fitting, forecasting, fitted value 156. Referring to Scenario 16-13, what is your forecast for the 13th month using the secondorder autoregressive model? ANSWER: 9,421.1829 TYPE: PR DIFFICULTY: Easy KEYWORDS: least squares trend fitting, forecasting, fitted value 157. Referring to Scenario 16-13, what is your forecast for the 13th month using the third-order autoregressive model? ANSWER: 9,332.0326 TYPE: PR DIFFICULTY: Moderate KEYWORDS: least squares trend fitting, forecasting, fitted value 158. True or False: Referring to Scenario 16-13, the best model based on the residual plots is the linear-trend model. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: residual, model selection Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting 16-45 159. True or False: Referring to Scenario 16-13, the best model based on the residual plots is the quadratic-trend regression model. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: residual, model selection 160. True or False: Referring to Scenario 16-13, the best model based on the residual plots is the exponential-trend regression model. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: residual, model selection 161. True or False: Referring to Scenario 16-13, the best model based on the residual plots is the second-order autoregressive model. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: residual, model selection SCENARIO 16-14 A contractor developed a multiplicative time-series model to forecast the number of contracts in future quarters, using quarterly data on number of contracts during the 3-year period from 2011 to 2013. The following is the resulting regression equation: ln Yˆ = 3.37 + 0.117 X – 0.083 Q1 + 1.28 Q2 + 0.617 Q3 where Yˆ is the estimated number of contracts in a quarter X is the coded quarterly value with X = 0 in the first quarter of 2011. Q1 is a dummy variable equal to 1 in the first quarter of a year and 0 otherwise. Q2 is a dummy variable equal to 1 in the second quarter of a year and 0 otherwise. Q3 is a dummy variable equal to 1 in the third quarter of a year and 0 otherwise. 162. Referring to Scenario 16-14 , the best interpretation of the constant 3.37 in the regression equation is: a) the fitted value for the first quarter of 2011, prior to seasonal adjustment, is log10 3.37. b) the fitted value for the first quarter of 2011, after to seasonal adjustment, is log10 3.37. c) the fitted value for the first quarter of 2011, prior to seasonal adjustment, is 103.37. d) the fitted value for the first quarter of 2011, after to seasonal adjustment, is 103.37. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, intercept, interpretation, seasonable data Copyright ©2015 Pearson Education, Inc. 16-46 Time-Series Forecasting 163. Referring to Scenario 16-14, the best interpretation of the coefficient of X (0.117) in the regression equation is: a) the quarterly compound growth rate in contracts is around 30.92%. b) the annually compound growth rate in contracts is around 30.92%. c) the quarterly compound growth rate in contracts is around 11.7%. d) the annually compound growth rate in contracts is around 11.7%. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, slope, interpretation, seasonable data 164. Referring to Scenario 16-14, the best interpretation of the coefficient of Q3 (0.617) in the regression equation is: a) the number of contracts in the third quarter of a year is approximately 62% higher than the average over all 4 quarters. b) the number of contracts in the third quarter of a year is approximately 62% higher than it would be during the fourth quarter. c) the number of contracts in the third quarter of a year is approximately 314% higher than the average over all 4 quarters. d) the number of contracts in the third quarter of a year is approximately 314% higher than it would be during the fourth quarter. ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, slope, interpretation, seasonable data 165. Referring to Scenario 16-14, to obtain a forecast for the first quarter of 2014 using the model, which of the following sets of values should be used in the regression equation? a) X = 12, Q1 = 0, Q2 = 0, Q3 = 0 b) X = 12, Q1 = 1, Q2 = 0, Q3 = 0 c) X = 13, Q1 = 0, Q2 = 0, Q3 = 0 d) X = 13, Q1 = 1, Q2 = 0, Q3 = 0 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting, seasonable data Copyright ©2015 Pearson Education, Inc. Time-Series Forecasting 16-47 166. Referring to Scenario 16-14, to obtain a forecast for the fourth quarter of 2014 using the model, which of the following sets of values should be used in the regression equation? a) X = 15, Q1 = 0, Q2 = 0, Q3 = 0 b) X = 15, Q1 = 1, Q2 = 0, Q3 = 0 c) X = 16, Q1 = 0, Q2 = 0, Q3 = 0 d) X = 16, Q1 = 1, Q2 = 0, Q3 = 0 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting, seasonable data 167. Referring to Scenario 16-14, using the regression equation, which of the following values is the best forecast for the number of contracts in the third quarter of 2014? a) 49,091 b) 133,352 c) 421,697 d) 1,482,518 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting, seasonable data 168. Referring to Scenario 16-14, using the regression equation, which of the following values is the best forecast for the number of contracts in the second quarter of 2015? a) 144,212 b) 391,742 c) 1,238,797 d) 4,355,119 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, forecasting, seasonable data Copyright ©2015 Pearson Education, Inc. 16-48 Time-Series Forecasting 169. Referring to Scenario 16-14, in testing the coefficient of X in the regression equation (0.117) the results were a t-statistic of 9.08 and an associated p-value of 0.0000. Which of the following is the best interpretation of this result? a) The quarterly growth rate in the number of contracts is significantly different from 0% ( = 0.05). b) The quarterly growth rate in the number of contracts is not significantly different from 0% ( = 0.05). c) The quarterly growth rate in the number of contracts is significantly different from 100% ( = 0.05). d) The quarterly growth rate in the number of contracts is not significantly different from 100% ( = 0.05). ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, t test on slope, decision, conclusion, interpretation, seasonable data 170. Referring to Scenario 16-14, in testing the coefficient for Q1 in the regression equation (– 0.083), the results were a t-statistic of – 0.66 and an associated p-value of 0.530. Which of the following is the best interpretation of this result? a) The number of contracts in the first quarter of the year is significantly different from the number of contracts in an average quarter ( = 0.05). b) The number of contracts in the first quarter of the year is not significantly different from the number of contracts in an average quarter ( = 0.05). c) The number of contracts in the first quarter of the year is significantly different from the number of contracts in the fourth quarter for a given coded quarterly value of X ( = 0.05). d) The number of contracts in the first quarter of the year is not significantly different from the number of contracts in the fourth quarter for a given coded quarterly value of X ( = 0.05). ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential model, t test on slope, decision, conclusion, interpretation, seasonable data Copyright ©2015 Pearson Education, Inc.