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
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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.