msb11e ppt ch13
TRANSCRIPT
© 2011 Pearson Education, Inc
© 2011 Pearson Education, Inc
Statistics for Business and Economics
Chapter 13
Time Series:Descriptive Analyses, Models, &
Forecasting
© 2011 Pearson Education, Inc
Content
13.1 Descriptive Analysis: Index Numbers
13.2 Descriptive Analysis: Exponential Smoothing
13.3 Time Series Components
13.4 Forecasting: Exponential Smoothing
13.5 Forecasting Trends: Holt’s Method
13.6 Measuring Forecast Accuracy: MAD and RMSE
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Content
13.7 Forecasting Trends: Simple Linear Regression
13.8 Seasonal Regression Models
13.9 Autocorrelation and the Durbin-Watson Test
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Learning Objectives
• Focus on methods for analyzing data generated by a process over time (i.e., time series data).
• Present descriptive methods for characterizing time series data.
• Present inferential methods for forecasting future values of time series data.
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Time Series
• Data generated by processes over time
• Describe and predict output of processes
• Descriptive analysis
– Understanding patterns
• Inferential analysis
– Forecast future values
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13.1
Descriptive Analysis:Index Numbers
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Index Number
• Measures change over time relative to a base period
• Price Index measures changes in price
– e.g. Consumer Price Index (CPI)
• Quantity Index measures changes in quantity
– e.g. Number of cell phones produced annually
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Steps for Calculatinga Simple Index Number
1. Obtain the prices or quantities for the commodity over the time period of interest.
2. Select a base period.
3. Calculate the index number for each period according to the formula
Index number at time t
=
Time series value at time tTime series value at base period
⎛
⎝⎜⎞
⎠⎟100
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Steps for Calculatinga Simple Index Number
Symbolically,
where It is the index number at time t, Yt is the time series value at time t, and Y0 is the time series value at the base period.
I
t=
Yt
Y0
⎛
⎝⎜⎞
⎠⎟100
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Simple Index Number Example
The table shows the price per gallon of regular gasoline in the U.S for the years 1990 – 2006. Use 1990 as the base year (prior to the Gulf War). Calculate the simple index number for 1990, 1998, and 2006.
Year $ 1990 1.2991991 1.0981992 1.0871993 1.0671994 1.0751995 1.1111996 1.2241997 1.1991998 1.031999 1.1362000 1.4842001 1.422002 1.3452003 1.5612004 1.8522005 2.272006 2.572
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Simple Index Number Solution
1990 Index Number (base period)
1998price 1.03100 100 79.3
1990price 1.299
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
1998 Index Number
1990price 1.299100 100 100
1990price 1.299
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
Indicates price had dropped by 20.7% (100 – 79.3) between 1990 and 1998.
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Simple Index Number Solution
2006 Index Number2006price 2.572
100 100 1981990price 1.299
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
Indicates price had risen by 98% (100 – 198) between 1990 and 2006.
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Simple Index Numbers 1990–2006
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Simple Index Numbers 1990–2006
Gasoline Price Simple Index
0.0
50.0
100.0
150.0
200.0
250.0
19901991199219931994199519961997199819992000200120022003200420052006
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Composite Index Number
• Made up of two or more commodities
• A simple index using the total price or total quantity of all the series (commodities)
• Disadvantage: Quantity of each commodity purchased is not considered
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Composite Index Number Example
The table on the next slide shows the closing stock prices on the last day of the month for Daimler–Chrysler, Ford, and GM between 2005 and 2006. Construct the simple composite index using January 2005 as the base period. (Source: Nasdaq.com)
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Simple Composite Index Solution
First compute the total for the three stocks for each date.
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Simple Composite Index Solution
Now compute the simple composite index by dividing each total by the January 2005 total. For example, December 2006:
12 / 06price100
1/ 05price
99.64100
95.49
104.3
⎛ ⎞⎜ ⎟⎝ ⎠
⎛ ⎞=⎜ ⎟⎝ ⎠
=
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Simple Composite Index Solution
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Simple Composite Index Solution
Simple Composite Index Numbers 2005 – 2006
0.0
20.0
40.0
60.0
80.0
100.0
120.0
J-05 M-05 M-05 J-05 S-05 N-05 J-06 M-06 M-06 J-06 S-06 N-06
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Weighted Composite Price Index
A weighted composite price index weights the prices by quantities purchased prior to calculating totals for each time period. The weighted totals are then used to compute the index in the same way that the unweighted totals are used for simple composite indexes.
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Laspeyres Index
• Uses base period quantities as weights– Appropriate when quantities remain approximately
constant over time period
• Example: Consumer Price Index (CPI)
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Steps for Calculating a Laspeyres Index
1. Collect price information for each of the k price series to be used in the composite index. Denote these series by P1t, P2t, …, Pkt .
2. Select a base period. Call this time period t0.
3. Collect purchase quantity information for the base period. Denote the k quantities by
4. Calculate the weighted totals for each time
period according to the formula
Q
1t0, Q
2t0,K ,Q
kt0.
Q
it0P
iti=1
k
∑
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Steps for Calculating a Laspeyres Index
5. Calculate the Laspeyres index, It, at time t by taking the ratio of the weighted total at time t to the base period weighted total and multiplying by 100–that is,
It=
Qit0Pit
i=1
k
∑
Qit0Pit0
i=1
k
∑×100
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Laspeyres Index Number Example
The table shows the closing stock prices on 1/31/2005 and 12/29/2006 for Daimler–Chrysler, Ford, and GM. On 1/31/2005 an investor purchased the indicated number of shares of each stock. Construct the Laspeyres Index using 1/31/2005 as the base period.
Daimler–Chrysler GM Ford
Shares Purchased 100 500 200
1/31/2005 Price 45.51 13.17 36.81
12/29/2006 Price 61.41 7.51 30.72
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Laspeyres Index Solution
0 01
100(45.51) 500(13.17) 200(36.81)
18498
k
it iti
Q P=
= + +
=
∑
01
100(61.41) 500(7.51) 200(30.72)
16040
k
it iti
Q P=
= + +
=
∑
Weighted total for base period (1/31/2005):
Weighted total for 12/29/2006:
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Laspeyres Index Solution
,1/31/ 05 ,12 / 29 / 061
,1/31/ 05 ,1/31/ 051
100
16040100
1849886.7
k
i ii
t k
i ii
Q PI
Q P
=
=
= ×
= ×
=
∑
∑
Indicates portfolio value had decreased by 13.3% (100–86.7) between 1/31/2005 and 12/29/2006.
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Paasche Index
• Uses quantities for each period as weights– Appropriate when quantities change over time
• Compare current prices to base period prices at current purchase levels
• Disadvantages– Must know purchase quantities for each time
period– Difficult to interpret a change in index when base
period is not used
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Steps for Calculating a Paasche Index
1. Collect price information for each of the k price series to be used in the composite index. Denote these series by P1t, P2t, …, Pkt .
2. Select a base period. Call this time period t0.
3. Collect purchase quantity information for the base period. Denote the k quantities by
Q
1t0, Q
2t0,K ,Q
kt0.
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Steps for Calculating a Paasche Index
4. Calculate the Paasche index for time t by multiplying the ratio of the weighted total at time t to the weighted total at time t0 (base period) by 100, where the weights used are the purchase quantities for time period t. Thus,
It=
QitPiti=1
k
∑
QitPit0i=1
k
∑×100
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Paasche Index Number Example
The table shows the 1/31/2005 and 12/29/2006 prices and volumes in millions of shares for Daimler–Chrysler, Ford, and GM. Calculate the Paasche Index using 1/31/2005 as the base period. (Source: Nasdaq.com)
Daimler–Chrysler Ford GM
Price Volume Price Volume Price Volume
1/31/2005 45.51 .8 13.17 7.0 36.81 5.6
12/29/2006 61.41 .2 7.51 10.0 30.72 6.1
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Paasche Index Solution
,1/31/ 05 ,1/31/ 051
1/31/ 05
,1/31/ 05 ,1/31/ 051
100
.8(45.51) 7(13.17) 5.6(36.81)100
.8(45.51) 7(13.17) 5.6(36.81)
100
k
i iik
i ii
Q PI
Q P
=
=
= ×
+ += ×
+ +=
∑
∑
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Paasche Index Solution
12/ 29 / 06 12/ 29 / 061
12/ 29 / 06
12/ 29 / 06 1/31/ 051
100
.2(61.41) 10(7.51) 6.1(30.72)100
.2(45.51) 10(13.17) 6.1(36.81)
274.774100 75.2
365.343
k
i ii
k
i ii
Q PI
Q P
=
=
= ×
+ += ×
+ +
= × =
∑
∑
12/29/2006 prices represent a 24.8% (100 – 75.2) decrease from 1/31/2005 (assuming quantities were at 12/29/2006 levels for both periods)
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13.2
Descriptive Analysis:Exponential Smoothing
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Exponential Smoothing
• Type of weighted average
• Removes rapid fluctuations in time series (less sensitive to short–term changes in prices)
• Allows overall trend to be identified
• Used for forecasting future values
• Exponential smoothing constant (w) affects “smoothness” of series
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Exponential Smoothing Constant
Exponential smoothing constant, 0 < w < 1
• w close to 0– More weight given to previous values of time
series– Smoother series
• w close to 1– More weight given to current value of time series– Series looks similar to original (more variable)
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Steps for Calculating an Exponentially Smoothed Series
1. Select an exponential smoothing constant, w, between 0 and 1. Remember that small values of w give less weight to the current value of the series and yield a smoother series. Larger choices of w assign more weight to the current value of the series and yield a more variable series.
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Steps for Calculating an Exponentially Smoothed Series
2. Calculate the exponentially smoothed series Et from the original time series Yt as follows:
E1 = Y1
E2 = wY2 + (1 – w)E1
E3 = wY3 + (1 – w)E2
Et = wYt + (1 – w)Et–1
…
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Exponential Smoothing Example
The closing stock prices on the last day of the month for Daimler–Chrysler in 2005 and 2006 are given in the table. Create an exponentially smoothed series using w = .2.
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Exponential Smoothing Solution
E1 = 45.51
E2 = .2(46.10) + .8(45.51) = 45.63
E3 = .2(44.72) + .8(45.63) = 45.45
E24 = .2(61.41) + .8(53.92) = 55.42
…
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Exponential Smoothing Solution
E1 = 45.51
E2 = .2(46.10) + .8(45.51) = 45.63
E3 = .2(44.72) + .8(45.63) = 45.45
E24 = .2(61.41) + .8(53.92) = 55.42
…
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Exponential Smoothing Solution
0
10
20
30
40
50
60
70
Jan-05Feb-05Mar-05Apr-05May-05Jun-05Jul-05Aug-05Sep-05Oct-05Nov-05Dec-05Jan-06Feb-06Mar-06Apr-06May-06Jun-06Jul-06Aug-06Sep-06Oct-06Nov-06Dec-06
Actual Series
Smoothed Series (w = .2)
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Exponential Smoothing Thinking Challenge
The closing stock prices on the last day of the month for Daimler–Chrysler in 2005 and 2006 are given in the table. Create an exponentially smoothed series using w = .8.
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Exponential Smoothing Solution
E1 = 45.51
E2 = .8(46.10) + .2(45.51) = 45.98
E3 = .8(44.72) + .2(45.98) = 44.97
E24 = .8(61.41) + .2(57.75) = 60.68
…
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0
10
20
30
40
50
60
70
Jan-05Feb-05Mar-05Apr-05May-05Jun-05Jul-05Aug-05Sep-05Oct-05Nov-05Dec-05Jan-06Feb-06Mar-06Apr-06May-06Jun-06Jul-06Aug-06Sep-06Oct-06Nov-06Dec-06
Exponential Smoothing Solution
Actual Series
Smoothed Series (w = .2)
Smoothed Series (w = .8)
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13.3
Time Series Components
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Descriptive v. Inferential Analysis
• Descriptive Analysis– Picture of the behavior of the time series– e.g. Index numbers, exponential smoothing– No measure of reliability
• Inferential Analysis– Goal: Forecasting future values– Measure of reliability
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Time Series Components
Additive Time Series Model Yt = Tt + Ct + St + Rt
Tt = secular trend (describes long–term movements of Yt)
Ct = cyclical effect (describes fluctuations about the secular trend attributable to business and economic conditions)
St = seasonal effect (describes fluctuations that recur during specific time periods)
Rt = residual effect (what remains after other components have been removed)
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13.4
Forecasting:Exponential Smoothing
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Exponentially Smoothed Forecasts
• Assumes the trend and seasonal component are relatively insignificant
• Exponentially smoothed forecast is constant for all future values
• Ft+1 = Et Ft+2 = Ft+1
Ft+3 = Ft+1
• Use for short–term forecasting only
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Calculation of Exponentially Smoothed Forecasts
1. Given the observed time series Y1, Y2, … , Yt, first calculate the exponentially smoothed values E1, E2, … , Et, using
E1 = Y1 E2 = wY2 + (1 – w)E1
Et = wYt + (1 – w)Et –1
M
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Calculation of Exponentially Smoothed Forecasts
2. Use the last smoothed value to forecast the next time series value:
Ft +1 = Et
3. Assuming that Yt is relatively free of trend and seasonal components, use the same forecast for all future values of Yt:
Ft+2 = Ft+1
Ft+3 = Ft+1 M
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Exponential Smoothing Forecasting Example
The closing stock prices on the last day of the month for Daimler–Chrysler in 2005 and 2006 are given in the table along with the exponentially smoothed values using w = .2. Forecast the closing price for the January 31, 2007.
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Exponential Smoothing Forecasting Solution
F1/31/2007 = E12/29/2006 = 55.42
The actual closing price on 1/31/2007 for Daimler–Chrysler was 62.49.
Forecast Error = Y1/31/2007 – F1/31/2007
= 62.49 – 55.42 = 7.07
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13.5
Forecasting Trends:
Holt’s Method
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The Holt Forecasting Model
• Accounts for trends in time series• Two components
– Exponentially smoothed component, Et
• Smoothing constant 0 < w < 1
– Trend component, Tt
• Smoothing constant 0 < v < 1– Close to 0: More weight to past trend– Close to 1: More weight to recent trend
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Steps for Calculating Components of the Holt
Forecasting Model
1. Select an exponential smoothing constant w between 0 and 1. Small values of w give less weight to the current values of the time series and more weight to the past. Larger choices assign more weight to the current value of the series.
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Steps for Calculating Components of the Holt
Forecasting Model
2. Select a trend smoothing constant v between 0 and 1. Small values of v give less weight to the current changes in the level of the series and more weight to the past trend. Larger values assign more weight to the most recent trend of the series and less to past trends.
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Steps for Calculating Components of the Holt
Forecasting Model3. Calculate the two components, Et and Tt, from the
time series Yt beginning at time t = 2 :
E2 = Y2 and T2 = Y2 – Y1
E3 = wY3 + (1 – w)(E2 + T2)T3 = v(E3 – E2) + (1 – v)T2
Et = wYt + (1 – w)(Et–1 + Tt–1)Tt = v(Et – Et–1) + (1 – v)Tt–1
…
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Holt Example
The closing stock prices on the last day of the month for Daimler–Chrysler in 2005 and 2006 are given in the table. Calculate the Holt–Winters components using w = .8 and v = .7.
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Holt Solution
w = .8 v = .7
E2 = Y2 and T2 = Y2 – Y1
E2 = 46.10 and T2 = 46.10 – 45.51 = .59
E3 = wY3 + (1 – w)(E2 + T2)E3 = .8(44.72) + .2(46.10 + .59) = 45.114
T3 = v(E3 – E2) + (1 – v)T2
T3 = .7(45.114 – 46.10) + .3(.59) = –.5132
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Holt Solution
Completed series:
w = .8 v = .7
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30
35
40
45
50
55
60
65
Jan-05Mar-05May-05Jul-05Sep-05Nov-05Jan-06Mar-06May-06Jul-06Sep-06Nov-06
Date
Price
Holt Solution
Holt exponentially smoothed (w = .8 and v = .7)
Actual
Smoothed
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Holt’s Forecasting Methodology
1. Calculate the exponentially smoothed and trend components, Et and Tt, for each observed value of Yt (t ≥ 2) using the formulas given in the previous box.
2. Calculate the one-step-ahead forecast using
Ft+1 = Et + Tt
3. Calculate the k-step-ahead forecast using
Ft+k = Et + kTt
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Holt Forecasting Example
Use the Holt series to forecast the closing price of Daimler–Chrysler stock on 1/31/2007 and 2/28/2007.
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Holt Forecasting Solution
1/31/2007 is one–step–ahead:
F1/31/07 = E12/29/06 + T12/29/06
= 61.39 + 3.00 = 64.39
2/28/2007 is two–steps–ahead:
F2/28/07 = E12/29/06 + 2T12/29/06
= 61.39 + 2(3.00) = 67.39
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Holt Thinking Challenge
The data shows the average undergraduate tuition at all 4–year institutions for the years 1996–2004 (Source: U.S. Dept. of Education). Calculate the Holt–Winters components using w = .7 and v = .5.
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Holt Solution
w = .7 v = .5
E2 = Y2 and T2 = Y2 – Y1
E2 = 9206 and T2 = 9206 – 8800 = 406
E3 = wY3 + (1 – w)(E2 + T2)E3 = .7(9588) + .3(9206 + 406) = 9595.20
T3 = v(E3 – E2) + (1 – v)T2
T3 = .5(9595.20 – 9206) + .5(406) = 397.60
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Holt Solution
Completed series
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Holt Solution
$8,000
$9,000
$10,000
$11,000
$12,000
$13,000
$14,000
$15,000
1995 1996 1997 1998 1999 2000 2001 2002 2003 2004
Year
Actual Smoothed
Holt–Winters exponentially smoothed (w = .7 and v = .5)
Tu
itio
n
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Holt Forecasting Thinking Challenge
Use the Holt–Winters series to forecast tuition in 2005 and 2006
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Holt Forecasting Solution
2005 is one–step–ahead: F11 = E10 + T10
13672.72 + 779.76 = $14,452.48
2006 is 2–steps–ahead: F12 = E10 + 2T10 =13672.72 +2(779.76) = $15,232.24
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13.6
Measuring Forecast Accuracy:
MAD and RMSE
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Mean Absolute Deviation
• Mean absolute difference between the forecast and actual values of the time series
• where m = number of forecasts used
MAD =
Yt −Ftt=n+1
n+m
∑m
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Mean Absolute Percentage Error
• Mean of the absolute percentage of the difference between the forecast and actual values of the time series
• where m = number of forecasts used
MAPE =
Yt −Ft( )Ytt=n+1
n+m
∑m
×100
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Root Mean Squared Error
• Square root of the mean squared difference between the forecast and actual values of the time series
• where m = number of forecasts used
RMSE =
Yt −Ft( )2
t=n+1
n+m
∑m
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Forecasting Accuracy Example
Using the Daimler–Chrysler data from 1/31/2005 through 8/31/2006, three time series models were constructed and forecasts made for the next four months.• Model I: Exponential smoothing (w = .2)• Model II: Exponential smoothing (w = .8)• Model III: Holt–Winters (w = .8, v = .7)
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Forecasting Accuracy Example
Model I
2.31 4.66 6.01 9.145.53
4IMAD− + + +
= =
( ) ( ) ( ) ( )2.31 4.66 6.01 9.14
49.96 56.93 58.28 61.41100 9.50
4IMAPE
−+ + +
= × =
( ) ( ) ( ) ( )2 2 2 22.31 4.66 6.01 9.14
6.064IRMSE
− + + += =
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Forecasting Accuracy Example
Model II
2.82 4.15 5.50 8.635.28
4IIMAD− + + +
= =
( ) ( ) ( ) ( )2.82 4.15 5.50 8.63
49.96 56.93 58.28 61.41100 9.11
4IIMAPE
−+ + +
= × =
( ) ( ) ( ) ( )2 2 2 22.82 4.15 5.50 8.63
5.704IIRMSE
− + + += =
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Forecasting Accuracy Example
Model III
3.45 2.42 2.67 4.713.31
4IIIMAD− + + +
= =
( ) ( ) ( ) ( )3.45 2.42 2.67 4.71
49.96 56.93 58.28 61.41100 5.85
4IIIMAPE
−+ + +
= × =
( ) ( ) ( ) ( )2 2 2 23.45 2.42 2.67 4.71
3.444IIIRMSE
− + + += =
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13.7
Forecasting Trends:
Simple Linear Regression
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Simple Linear Regression
• Model: E(Yt) = β0 + β1t
• Relates time series, Yt, to time, t
• Cautions– Risky to extrapolate (forecast beyond observed
data)– Does not account for cyclical effects
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Simple Linear Regression Example
The data shows the average undergraduate tuition at all 4–year institutions for the years 1996–2004 (Source: U.S. Dept. of Education). Use least–squares regression to fit a linear model. Forecast the tuition for 2005 (t = 11) and compute a 95% prediction interval for the forecast.
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Simple Linear Regression Solution
From Excel
ˆ 7997.533 528.158tY t= +
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Simple Linear Regression Solution
$8,000
$9,000
$10,000
$11,000
$12,000
$13,000
$14,000
$15,000
1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005
Year
Tuition
ˆ 7997.533 528.158tY t= +
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Simple Linear Regression Solution
Forecast tuition for 2005 (t = 11):
11ˆ 7997.533 528.158(11) 13807.27Y = + =
( )
( )( ) ( )
2
/ 2
2
11
1ˆ 1
11 5.5113807.27 2.306 286.84 1
10 82.5
13006.21 14608.33
p
tt
t ty t s
n SS
y
α
−± + +
−± + +
≤ ≤
95% prediction interval:
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13.8
Seasonal Regression Models
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Seasonal Regression Models
• Takes into account secular trend and seasonal effects (seasonal component)
• Uses multiple regression models
• Dummy variables to model seasonal component
• E(Yt) = β0 + β1t + β2Q1 + β3Q2 + β4Q3 where
Q
i=
1 if quarter i0 if notquarter i
⎧⎨⎩
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13.9
Autocorrelation and theDurbin-Watson Test
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Autocorrelation
• Time series data may have errors that are not independent
• Time series residuals:
• Correlation between residuals at different points in time (autocorrelation)
• 1st order correlation: Correlation between neighboring residuals (times t and t + 1)
ˆ ˆt t tR Y Y= −
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Autocorrelation
Plot of residuals v. time for tuition data shows residuals tend to group alternately into positive and negative clusters
Residual v Time Plot
-400
-200
0
200
400
600
0 2 4 6 8 10 12
t
Residuals
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Durbin–Watson Test
• H0: No first–order autocorrelation of residuals
• Ha: Positive first–order autocorrelation of residuals
• Test Statistic
( )212
2
1
ˆ ˆ
ˆ
n
t tt
n
tt
R Rd
R
−=
=
−=
∑
∑
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Interpretation of Durbin-Watson d-Statistic
1. If the residuals are uncorrelated, then d ≈ 2.2. If the residuals are positively autocorrelated,
then d < 2, and if the autocorrelation is very strong, d ≈ 2.
3. If the residuals are negatively autocorrelated, then d >2, and if the autocorrelation is very strong, d ≈ 4.
d =R̂t −R̂t−1( )
t=2
n
∑
R̂t2
t=1
n
∑ Range of d : 0 ≤d≤4
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Rejection Region for the Durbin–Watson d Test
32dL dU 40 1d
Rejection region: evidence of positive autocorrelation
Nonrejection region: insufficient evidence of positive autocorrelation
Possibly significant autocorrelation
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Durbin–Watson d-Test for Autocorrelation
One-tailed Test
H0: No first–order autocorrelation of residuals
Ha: Positive first–order autocorrelation of residuals
(or Ha: Negative first–order autocorrelation)
Test Statistic ( )212
2
1
ˆ ˆ
ˆ
n
t tt
n
tt
R Rd
R
−=
=
−=
∑
∑
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Durbin–Watson d-Test for Autocorrelation
Rejection Region:
d < dL,
[or (4 – d) < dL,If Ha : Negative first-order autocorrelation
where dL, is the lower tabled value corresponding to k independent variables and n observations. The corresponding upper valuedU, defines a “possibly significant” region between dL, and dU,
© 2011 Pearson Education, Inc
Durbin–Watson d-Test for Autocorrelation
Two-tailed Test
H0: No first–order autocorrelation of residuals
Ha: Positive or Negative first–order autocorrelation of residuals
Test Statistic
( )212
2
1
ˆ ˆ
ˆ
n
t tt
n
tt
R Rd
R
−=
=
−=
∑
∑
© 2011 Pearson Education, Inc
Durbin–Watson d-Test for Autocorrelation
Rejection Region:
d < dL, or (4 – d) < dL,
where dL, is the lower tabled value corresponding to k independent variables and n observations. The corresponding upper valuedU, defines a “possibly significant” region between dL, and dU,
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Requirements for the Validity of the d-Test
The residuals are normally distributed.
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Durbin–Watson Test Example
Use the Durbin–Watson test to test for the presence of autocorrelation in the tuition data. Use α = .05.
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Durbin–Watson Test Solution
• H0:
• Ha:
• n = k =
• Critical Value(s):
No 1st–orderautocorrelation
Positive 1st–orderautocorrelation
.05 10 1
2 40 d.88 1.32
© 2011 Pearson Education, Inc
Durbin–Watson Solution
Test Statistic
( )212
2
1
2 2 2
2 2 2
ˆ ˆ
ˆ
(152.1515 274.3091) (5.9939 152.1515) ... (463.8909 204.0485)
(274.3091) (152.1515) ... (463.8909)
.51
n
t tt
n
tt
R Rd
R
−=
=
−=
− + − + + −=
+ + +
=
∑
∑
© 2011 Pearson Education, Inc
Durbin–Watson Test Solution
• H0:
• Ha:
• n = k =
• Critical Value(s):
Test Statistic:
Decision:
Conclusion:
No 1st–orderautocorrelation
Positive 1st–orderautocorrelation
.05 10 1
2 40 d.88 1.32
d = .51
Reject at = .05
There is evidence of positive autocorrelation
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Key Ideas
Time Series Data
Data generated by processes over time.
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Key Ideas
Index Number
Measures the change in a variable over time relative to a base period.
Types of Index numbers:1. Simple index number 2. Simple composite index number3. Weighted composite number (Laspeyers index or Pasche index)
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Key Ideas
Time Series Components
1. Secular (long-term) trend
2. Cyclical effect
3. Seasonal effect
4. Residual effect
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Key Ideas
Time Series Forecasting
Descriptive methods of forecasting with smoothing:
1. Exponential smoothing
2. Holt’s method
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Key Ideas
Time Series Forecasting
An Inferential forecasting method:
least squares regression
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Key Ideas
Time Series Forecasting
Measures of forecast accuracy:
1. mean absolute deviation (MAD)
2. mean absolute percentage error (MAPE)
3. root mean squared error (RMSE)
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Key Ideas
Time Series Forecasting
Problems with least squares regression forecasting:
1. Prediction outside the experimental region
2. Regression errors are autocorrelated
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Key Ideas
Autocorrelation
Correlation between time series residuals at different points in time.
A test for first-order autocorrelation:
Durbin-Watson test