Download Chap. 13: Time Series: Descriptive Analyses, Models

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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
© 2011 Pearson Education, Inc
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.
© 2011 Pearson Education, Inc
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
© 2011 Pearson Education, Inc
Steps for Calculating
a 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 t 

100

 Time series value at base period 
© 2011 Pearson Education, Inc
Steps for Calculating
a Simple Index Number
Symbolically,
 Yt 
I t    100
 Y0 
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.
© 2011 Pearson Education, Inc
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.
© 2011 Pearson Education, Inc
Year
1990
1991
1992
1993
1994
1995
1996
1997
1998
1999
2000
2001
2002
2003
2004
2005
2006
$
1.299
1.098
1.087
1.067
1.075
1.111
1.224
1.199
1.03
1.136
1.484
1.42
1.345
1.561
1.852
2.27
2.572
Simple Index Number Solution
1990 Index Number (base period)
 1990price 
 1.299 

100  
100  100
 1.299 
 1990price 
1998 Index Number
 1998price 
 1.03 

100  
100  79.3
 1.299 
 1990price 
Indicates price had dropped by 20.7% (100 – 79.3)
between 1990 and 1998.
© 2011 Pearson Education, Inc
Simple Index Number Solution
2006 Index Number
 2006price 
 2.572 

100  
100  198
 1.299 
 1990price 
Indicates price had risen by 98% (100 – 198)
between 1990 and 2006.
© 2011 Pearson Education, Inc
Simple Index Numbers
1990–2006
© 2011 Pearson Education, Inc
Simple Index Numbers
1990–2006
Gasoline Price Simple Index
© 2011 Pearson Education, Inc
2006
2005
2004
2003
2002
2001
2000
1999
1998
1997
1996
1995
1994
1993
1992
1991
1990
250.0
200.0
150.0
100.0
50.0
0.0
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
© 2011 Pearson Education, Inc
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)
© 2011 Pearson Education, Inc
Simple Composite Index
Solution
First compute the total for
the three stocks for each
date.
© 2011 Pearson Education, Inc
Simple Composite Index
Solution
Now compute the
simple composite index
by dividing each total by
the January 2005 total.
For example, December
2006:
 12 / 06price 

100
 1/ 05price 
 99.64 

100
 95.49 
© 2011 Pearson Education, Inc  104.3
Simple Composite Index
Solution
© 2011 Pearson Education, Inc
Simple Composite Index
Solution
Simple Composite Index Numbers 2005 – 2006
120.0
100.0
80.0
60.0
40.0
20.0
© 2011 Pearson Education, Inc
-0
6
N
S06
J06
M
-0
6
M
-0
6
J06
-0
5
N
S05
J05
M
-0
5
M
-0
5
J05
0.0
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.
© 2011 Pearson Education, Inc
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
Q1t , Q2t ,K ,Qkt .
0
0
0
4. Calculate the weighted totals for each
time
k
period according to the formula  Qit Pit
© 2011 Pearson Education, Inc
i1
0
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,
k
It 
Q
i1
k
it0
Pit
it0
Pit
Q
i1
 100
0
© 2011 Pearson Education, Inc
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
100
500
200
1/31/2005 Price
45.51
13.17
36.81
12/29/2006 Price
61.41
7.51
30.72
Shares Purchased
© 2011 Pearson Education, Inc
Laspeyres Index Solution
Weighted total for base period (1/31/2005):
k
Q
i 1
P  100(45.51)  500(13.17)  200(36.81)
it0 it0
 18498
Weighted total for 12/29/2006:
k
Q
i 1
P  100(61.41)  500(7.51)  200(30.72)
it0 it
 16040
© 2011 Pearson Education, Inc
Laspeyres Index Solution
k
It 
Q
i 1
k
P
i ,1/ 31/ 05 i ,12 / 29 / 06
Q
i 1
100
P
i ,1/ 31/ 05 i ,1/ 31/ 05
16040

100
18498
 86.7
Indicates portfolio value had decreased by 13.3%
(100–86.7) between
1/31/2005
and
12/29/2006.
© 2011 Pearson Education, Inc
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
© 2011 Pearson Education, Inc
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
Q1t , Q2t ,K ,Qkt .
0
0
0
© 2011 Pearson Education, Inc
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.
k
Thus,
 Qit Pit
I t  i1
 100
k
 Qit Pit
0
i1
© 2011
Pearson Education, Inc
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
© 2011 Pearson Education, Inc
Paasche Index Solution
k
I1/ 31/ 05 
Q
P
Q
P
i 1
k
i 1
i ,1/ 31/ 05 i ,1/ 31/ 05
100
i ,1/ 31/ 05 i ,1/ 31/ 05
.8(45.51)  7(13.17)  5.6(36.81)

100
.8(45.51)  7(13.17)  5.6(36.81)
 100
© 2011 Pearson Education, Inc
Paasche Index Solution
k
I12 / 29 / 06 
Q
i 1
k
P
i12 / 29 / 06 i12 / 29 / 06
Q
i 1
100
P
i12 / 29 / 06 i1/ 31/ 05
.2(61.41)  10(7.51)  6.1(30.72)

100
.2(45.51)  10(13.17)  6.1(36.81)
274.774

 100  75.2
365.343
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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Education, Inc
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
© 2011 Pearson Education, Inc
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)
© 2011 Pearson Education, Inc
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.
© 2011 Pearson Education, Inc
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
© 2011 Pearson Education, Inc
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.
© 2011 Pearson Education, Inc
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
© 2011 Pearson Education, Inc
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
© 2011 Pearson Education, Inc
© 2011 Pearson Education, Inc
Dec-06
Nov-06
Oct-06
Sep-06
Aug-06
Jul-06
Jun-06
May-06
Apr-06
Mar-06
Feb-06
Jan-06
Dec-05
30
Nov-05
40
Oct-05
Sep-05
Aug-05
Jul-05
Jun-05
60
May-05
Apr-05
Mar-05
Feb-05
Jan-05
Exponential Smoothing
Solution
70
Actual Series
50
Smoothed Series
(w = .2)
20
10
0
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.
© 2011 Pearson Education, Inc
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
© 2011 Pearson Education, Inc
© 2011 Pearson Education, Inc
Dec-06
Nov-06
Oct-06
Sep-06
Aug-06
Jul-06
Jun-06
Smoothed Series
(w = .2)
May-06
Apr-06
Mar-06
Feb-06
Jan-06
Dec-05
Nov-05
Oct-05
30
Sep-05
40
Aug-05
Jul-05
Jun-05
60
May-05
Apr-05
Mar-05
Feb-05
Jan-05
Exponential Smoothing
Solution
70
Actual Series
50
Smoothed Series
(w = .8)
20
10
0
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
© 2011 Pearson Education, Inc
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)
© 2011 Pearson Education, Inc
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
© 2011 Pearson Education, Inc
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
© 2011 Pearson Education, Inc
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
© 2011 Pearson Education, Inc
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.
© 2011 Pearson Education, Inc
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
© 2011 Pearson Education, Inc
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
© 2011 Pearson Education, Inc
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.
© 2011 Pearson Education, Inc
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.
© 2011 Pearson Education, Inc
Steps for Calculating
Components of the Holt
Forecasting Model
…
3. 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(E
Et–1Education,
) + (1Inc – v)Tt–1
t –Pearson
© 2011
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.
© 2011 Pearson Education, Inc
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
© 2011 Pearson Education, Inc
Holt Solution
Completed series:
w = .8 v = .7
© 2011 Pearson Education, Inc
Holt Solution
Holt exponentially smoothed (w = .8 and v = .7)
65
Smoothed
60
50
45
40
Actual
35
l-0
Se 5
p05
N
ov
-0
5
Ja
n06
M
ar
-0
M 6
ay
-0
6
Ju
l-0
Se 6
p06
N
ov
-0
6
Ju
-0
5
5
ay
M
ar
-0
M
-0
5
30
Ja
n
Price
55
Date
© 2011 Pearson Education, Inc
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
© 2011 Pearson Education, Inc
Holt Forecasting Example
Use the Holt series to
forecast the closing price
of Daimler–Chrysler stock
on 1/31/2007 and
2/28/2007.
© 2011 Pearson Education, Inc
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
© 2011 Pearson Education, Inc
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.
© 2011 Pearson Education, Inc
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
© 2011 Pearson Education, Inc
Holt Solution
Completed series
Year
1995
1996
1997
1998
1999
2000
2001
2002
2003
2004
t
1
2
3
4
5
6
7
8
9
10
Tuition
$8,800
$9,206
$9,588
$10,076
$10,444
$10,818
$11,380
$12,014
$12,955
$13,743
w=.7
Et
v=.5
Tt
9206.0000
9595.2000
10051.0400
10454.1280
10833.3096
11335.1057
11945.1576
12810.9680
13672.7223
406.0000
397.6000
426.7200
414.9040
397.0428
449.4195
529.7356
697.7730
779.7637
© 2011 Pearson Education, Inc
Holt Solution
Holt–Winters exponentially smoothed (w = .7
and v = .5)
$15,000
$14,000
Tuition
$13,000
$12,000
$11,000
$10,000
Actual
Smoothed
$9,000
$8,000
1995
1996
1997
1998
1999
2000
Year
© 2011 Pearson Education, Inc
2001
2002
2003
2004
Holt Forecasting Thinking
Challenge
Use the Holt–Winters series to forecast tuition in
2005 and 2006
Year
1995
1996
1997
1998
1999
2000
2001
2002
2003
2004
t
1
2
3
4
5
6
7
8
9
10
w=.7
Et
Tuition
$8,800
$9,206
9206.0000
$9,588
9595.2000
$10,076
10051.0400
$10,444
10454.1280
$10,818
10833.3096
$11,380
11335.1057
$12,014
11945.1576
$12,955
12810.9680
© 2011
Pearson Education,
Inc
$13,743
13672.7223
v=.5
Tt
406.0000
397.6000
426.7200
414.9040
397.0428
449.4195
529.7356
697.7730
779.7637
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
© 2011 Pearson Education, Inc
13.6
Measuring Forecast Accuracy:
MAD and RMSE
© 2011 Pearson Education, Inc
Mean Absolute Deviation
• Mean absolute difference between the forecast
and actual values of the time series
nm
MAD 
 Y F
t
t
tn1
m
• where m = number of forecasts used
© 2011 Pearson Education, Inc
Mean Absolute Percentage
Error
• Mean of the absolute percentage of the
difference between the forecast and actual
values of the time series
nm

MAPE 
tn1
Y  F 
t
t
Yt
m
 100
• where m = number of forecasts used
© 2011 Pearson Education, Inc
Root Mean Squared Error
• Square root of the mean squared difference
between the forecast and actual values of the
time series
 Y  F 
nm
RMSE 
2
t
t
tn1
m
• where m = number of forecasts used
© 2011 Pearson Education, Inc
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)
© 2011 Pearson Education, Inc
Forecasting Accuracy
Example
Model I
MADI 
2.31  4.66  6.01  9.14
4
 5.53
 2.31   4.66    6.01  9.14 
MAPEI 
49.96
56.93
61.41
4
100  9.50
 2.31   4.66    6.01   9.14 
2
RMSEI 
58.28
2
2
4
© 2011 Pearson Education, Inc
2
 6.06
Forecasting Accuracy
Example
Model II
MADII 
2.82  4.15  5.50  8.63
4
 5.28
 2.82    4.15  5.50   8.63
MAPEII 
49.96
56.93
61.41
4
100  9.11
 2.82    4.15   5.50   8.63
2
RMSEII 
58.28
2
4
© 2011 Pearson Education, Inc
2
2
 5.70
Forecasting Accuracy
Example
Model III
MADIII 
3.45  2.42  2.67  4.71
4
 3.31
 3.45   2.42    2.67    4.71
MAPEIII 
49.96
56.93
61.41
4
100  5.85
 3.45   2.42    2.67    4.71
2
RMSEIII 
58.28
2
2
4
© 2011 Pearson Education, Inc
2
 3.44
13.7
Forecasting Trends:
Simple Linear Regression
© 2011 Pearson Education, Inc
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
© 2011 Pearson Education, Inc
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.
© 2011 Pearson Education, Inc
Simple Linear Regression
Solution
From Excel
Yˆt  7997.533
 528.158t
© 2011 Pearson Education, Inc
Simple Linear Regression
Solution
$15,000
Yˆt  7997.533  528.158t
$14,000
Tuition
$13,000
$12,000
$11,000
$10,000
$9,000
$8,000
1994
1995
1996
1997
1998
1999
2000
2001
Year
© 2011 Pearson Education, Inc
2002
2003
2004
2005
Simple Linear Regression
Solution
Forecast tuition for 2005 (t = 11):
Yˆ11  7997.533  528.158(11)  13807.27
95% prediction interval:
1 t p  t
yˆ  t / 2 s 1  
n
SStt
13807.27   2.306  286.84 

2
1 11  5.5 
1 
10
82.5
© 2011Pearson
Education,
Inc
13006.21
y11 
14608.33
2
13.8
Seasonal Regression Models
© 2011 Pearson Education, Inc
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
1
if quarter i
Qi  
if not
quarter
i
©02011 Pearson
Education, Inc
13.9
Autocorrelation and the
Durbin-Watson Test
© 2011 Pearson Education, Inc
Autocorrelation
• Time series data may have errors that are not
independent
• Time series residuals: Rˆt  Yt  Yˆt
• Correlation between residuals at different
points in time (autocorrelation)
• 1st order correlation: Correlation between
neighboring residuals (times t and t + 1)
© 2011 Pearson Education, Inc
Autocorrelation
Plot of residuals v. time for tuition data shows
residuals tend to group alternately into positive
and negative clusters
Residual v Time Plot
600
Residuals
400
200
0
-200
0
2
4
6
-400
© 2011 Pearson Education,
Inc
t
8
10
12
Durbin–Watson Test
• H0: No first–order autocorrelation of residuals
• Ha: Positive first–order autocorrelation of
residuals
• Test Statistic

n
d
t 2
Rˆt  Rˆt 1

2
n
2
ˆ
 Rt
t 1
© 2011 Pearson Education, Inc
Interpretation of DurbinWatson
d-Statistic
n
d
 Rφ  Rφ 
t
t2
n
2
φ
R
 t
t1
t1
Range of d : 0  d  4
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. © 2011 Pearson Education, Inc
Rejection Region for the Durbin–
Watson d Test
Rejection region:
evidence of
positive
autocorrelation
0
1
dL
dU
Possibly significant
autocorrelation
2
3
Nonrejection region:
insufficient evidence of
positive autocorrelation
© 2011 Pearson Education, Inc
4
d
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)
n
2
Test Statistic
ˆ ˆ
d
R  R 
t 2
t
n
t 1
2
ˆ
R

t
© 2011 Pearson Education,
Inc
t 1
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 value
dU, 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

n
d
t 2
Rˆt  Rˆt 1
n

2
2
ˆ
R

t
© 2011 Pearson Education,
Inc
t 1
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 value
dU, defines a “possibly significant” region
between dL, and dU,
© 2011 Pearson Education, Inc
Requirements for the Validity
of the d-Test
The residuals are normally distributed.
© 2011 Pearson Education, Inc
Durbin–Watson Test Example
Use the Durbin–Watson test to test for the
presence of autocorrelation in the tuition data.
Use α = .05.
© 2011 Pearson Education, Inc
Durbin–Watson Test Solution
• H0: No 1st–order
autocorrelation
• Ha: Positive 1st–order
autocorrelation
•   .05
n10=
1k
=
• Critical Value(s):
0
2
.88 1.32
4
d
© 2011 Pearson Education, Inc
Durbin–Watson Solution
Test Statistic

n
d
t 2
Rˆt  Rˆt 1

2
n
2
ˆ
R
 t
t 1
(152.1515  274.3091)2  (5.9939  152.1515)2  ...  (463.8909  204.0485) 2

(274.3091) 2  (152.1515)2  ...  (463.8909)2
 .51
© 2011 Pearson Education, Inc
Durbin–Watson Test Solution
• H0: No 1st–order
autocorrelation
Test Statistic:
d = .51
• Ha: Positive 1st–order
autocorrelation
•   .05
n10=
1k
=
• Critical Value(s):
0
2
.88 1.32
4
Decision:
Reject at  = .05
Conclusion:
There is evidence of
d positive autocorrelation
© 2011 Pearson Education, Inc
Key Ideas
Time Series Data
Data generated by processes over time.
© 2011 Pearson Education, Inc
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 number
3. Weighted composite number (Laspeyers
index or Pasche index)
© 2011 Pearson Education, Inc
Key Ideas
Time Series Components
1.
2.
3.
4.
Secular (long-term) trend
Cyclical effect
Seasonal effect
Residual effect
© 2011 Pearson Education, Inc
Key Ideas
Time Series Forecasting
Descriptive methods of forecasting with
smoothing:
1. Exponential smoothing
2. Holt’s method
© 2011 Pearson Education, Inc
Key Ideas
Time Series Forecasting
An Inferential forecasting method:
least squares regression
© 2011 Pearson Education, Inc
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)
© 2011 Pearson Education, Inc
Key Ideas
Time Series Forecasting
Problems with least squares regression
forecasting:
1. Prediction outside the experimental region
2. Regression errors are autocorrelated
© 2011 Pearson Education, Inc
Key Ideas
Autocorrelation
Correlation between time series residuals at
different points in time.
A test for first-order autocorrelation:
Durbin-Watson test
© 2011 Pearson Education, Inc