# Chap Time Series Descriptive Analyses Models Forecasting

Document Sample

```					Statistics for Business and
Economics

Chapter 13
Time Series:
Descriptive Analyses, Models, &
Forecasting
Learning Objectives
1. Describe Time Series
2. Explain Descriptive Analyses
3. Define Time Series Components
4. Explain Forecasting
5. Describe Measures of Accuracy
6. Define Autocorrelation
7. Explain Durbin–Watson Test
Time Series

•   Data generated by processes over time
•   Describe and predict output of processes
•   Descriptive analysis
–   Understanding patterns
•   Inferential analysis
–   Forecast future values
Descriptive Analysis:
Index Numbers
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
Simple Index Number
Based on price/quantity of a single commodity

 Yt 
I t   100
 Y0 
where
Yt = value at time t
Y0 = value at time 0 (base period)
Simple Index Number Example
Year      \$
1990   1.299
The table shows the price per       1991   1.098
1992   1.087
gallon of regular gasoline in the   1993   1.067
U.S for the years 1990 – 2006.      1994   1.075
1995   1.111
Use 1990 as the base year (prior    1996   1.224
1997   1.199
to the Gulf War). Calculate the     1998   1.03
simple index number for 1990,       1999   1.136
2000   1.484
1998, and 2006.                     2001   1.42
2002   1.345
2003   1.561
2004   1.852
2005   2.27
2006   2.572
Simple Index Number Solution

1990 Index Number (base period)
 1990price        1.299 
           100         100  100
 1990price        1.299 

1998 Index Number
 1998price        1.03 
           100         100  79.3
 1990price        1.299 

Indicates price had dropped by 20.7% (100 – 79.3)
between 1990 and 1998.
Simple Index Number Solution

2006 Index Number
 2006price        2.572 
           100         100  198
 1990price        1.299 

Indicates price had risen by 98% (100 – 198)
between 1990 and 2006.
Simple Index Numbers
1990–2006
Simple Index Numbers
1990–2006

Gasoline Price Simple Index

250.0
200.0
150.0
100.0
50.0
0.0
1990
1991
1992
1993
1994
1995
1996
1997
1998
1999
2000
2001
2002
2003
2004
2005
2006
Composite Index Number
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
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)
Simple Composite Index
Solution
First compute the total for
the three stocks for each
date.
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 
 104.3
Simple Composite Index
Solution
Simple Composite Index
Solution
Simple Composite Index Numbers 2005 – 2006
120.0
100.0
80.0
60.0
40.0
20.0
0.0
5

6
5

5

6

6
05

06
05

05

06

06
-0

-0
-0

-0

-0

-0
J-

J-

J-

J-
S-

S-
N

N
M

M

M

M
Weighted Composite Index
Number
Weighted Composite Price
Index
• Weights prices by quantities purchased before
computing totals
• Weighted totals used to compute composite
index
• Laspeyres Index
– Uses base period quantities as weights
• Paasche Index
– Uses quantities from each period as weights
Laspeyres Index
• Uses base period quantities as weights
– Appropriate when quantities remain approximately
constant over time period
• Example: Consumer Price Index (CPI)
Calculating a Laspeyres Index
weighted total for period t
It                                 100
weighted total for base period
k

Q     it0   Pit
   i 1
k
100
Q
i 1
it0   Pit0

where
Pit= price for each commodity at time t
Qit= quantity of each commodity at time t
t0 = base period
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
Laspeyres Index Solution
Weighted total for base period (1/31/2005):
k

Q
i 1
it0   Pit0  100(45.51)  500(13.17)  200(36.81)

 18498

Weighted total for 12/29/2006:
k

Q
i 1
it0   Pit  100(61.41)  500(7.51)  200(30.72)

 16040
Laspeyres Index Solution
k

Q                  P
i ,1/ 31/ 05 i ,12 / 29 / 06
It    i 1
k
100
Q
i 1
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.
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
– Must know purchase quantities for each time
period
– Difficult to interpret a change in index when base
period is not used
Calculating a Paasche Index
weighted total for period t
It                                 100
weighted total for base period
k

Q P   it it           Weights are
   i 1
k
100   quantities for
Q P
i 1
it it0
time period t

where
Pit= price for each commodity at time t
Qit= quantity of each commodity at time t
t0 = base period
Laspeyres 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
Paasche Index Solution
k

Q                P
i ,1/ 31/ 05 i ,1/ 31/ 05
I1/ 31/ 05    i 1
k
100
Q
i 1
P
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
Paasche Index Solution
k

Q                  P
i12 / 29 / 06 i12 / 29 / 06
I12 / 29 / 06    i 1
k
100
Q
i 1
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)
Exponential Smoothing
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
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)
Calculating an Exponential
Smoothed Series
E1 = Y1      (same as original series)
E2 = wY2 + (1 – w)E1
E3 = wY3 + (1 – w)E2
…

Et = wYt + (1 – w)Et–1

Weight given       Weight given to
to current         previous “smoothed”
series value       value
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.
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
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
0
10
20
30
40
50
60
70
Jan-05
Feb-05
Mar-05
Apr-05
May-05
Jun-05
Jul-05
Aug-05
Sep-05
Oct-05
Actual Series

Nov-05
Dec-05
(w = .2)

Jan-06
Feb-06
Mar-06
Solution

Smoothed Series

Apr-06
May-06
Jun-06
Jul-06
Aug-06
Sep-06
Exponential Smoothing

Oct-06
Nov-06
Dec-06
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.
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) = 60.08
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) = 60.08
0
10
20
30
40
50
60
70
Jan-05
Feb-05
Mar-05
Apr-05
May-05
Jun-05
Jul-05
Aug-05
Sep-05
Oct-05
Actual Series

(w = .2)
Nov-05
Dec-05
Jan-06
Feb-06
Smoothed Series

Mar-06
Solution*

Apr-06
May-06
Jun-06
Jul-06
(w = .8)

Aug-06
Sep-06
Oct-06
Exponential Smoothing

Nov-06
Smoothed Series

Dec-06
Inferential Analysis
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
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)
Forecasting: Exponential
Smoothing
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
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.
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
Forecasting Trends: The
Holt–Winters Forecasting
Model
The Holt–Winters 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
Calculating The Holt–Winters
Model
• Select values for the exponential smoothing
constant w and the trend smoothing constant v
• 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
Holt–Winters 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.
Holt–Winters 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
Holt–Winters Solution
Completed series:
w = .8 v = .7
Holt–Winters Solution
Holt–Winters exponentially smoothed (w = .8 and v = .7)
65

60                              Smoothed
55

50
Price

45

40

35
Actual
30
5

6
5

6
5

M 6
05

06
Se 5

Se 6
05

06
-0

-0
-0

-0
-0

-0
l-0

l-0
n-

n-
p-

p-
ay

ay
ov

ov
ar

ar
Ju

Ju
Ja

Ja
M

M
M

N

N
Date
Holt–Winters Forecasting
– Ft+1 = Et + Tt

– Ft+k = Et + kTt
Holt–Winters Forecasting
Example
Use the Holt–Winters
series to forecast the
closing price of Daimler–
Chrysler stock on
1/31/2007 and 2/28/2007.
Holt–Winters Forecasting
Solution*
F1/31/07 = E12/29/06 + T12/29/06
= 61.39 + 3.00 = 64.39

F2/28/07 = E12/29/06 + 2T12/29/06
= 61.39 + 2(3.00) = 67.39
Holt–Winters Thinking Challenge
The data shows the
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.
Holt–Winters 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
Holt–Winters Solution*
Completed series
w=.7        v=.5
Year    t   Tuition      Et          Tt
1995   1     \$8,800
1996   2     \$9,206    9206.0000   406.0000
1997   3     \$9,588    9595.2000   397.6000
1998   4    \$10,076   10051.0400   426.7200
1999   5    \$10,444   10454.1280   414.9040
2000   6    \$10,818   10833.3096   397.0428
2001   7    \$11,380   11335.1057   449.4195
2002   8    \$12,014   11945.1576   529.7356
2003   9    \$12,955   12810.9680   697.7730
2004   10   \$13,743   13672.7223   779.7637
Holt–Winters Solution*
Holt–Winters exponentially smoothed (w = .7
and v = .5)
\$15,000
\$14,000

\$13,000
Tuition

\$12,000
\$11,000

\$10,000   Actual                                       Smoothed
\$9,000

\$8,000
1995   1996   1997   1998   1999          2000   2001   2002   2003   2004
Year
Holt–Winters Forecasting
Thinking Challenge
Use the Holt–Winters series to forecast tuition in
2005 and 2006
w=.7        v=.5
Year    t   Tuition      Et          Tt
1995   1     \$8,800
1996   2     \$9,206    9206.0000   406.0000
1997   3     \$9,588    9595.2000   397.6000
1998   4    \$10,076   10051.0400   426.7200
1999   5    \$10,444   10454.1280   414.9040
2000   6    \$10,818   10833.3096   397.0428
2001   7    \$11,380   11335.1057   449.4195
2002   8    \$12,014   11945.1576   529.7356
2003   9    \$12,955   12810.9680   697.7730
2004   10   \$13,743   13672.7223   779.7637
Holt–Winters 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
Measuring Forecast
Accuracy
Mean Absolute Deviation
• Mean absolute difference between the forecast
and actual values of the time series
m

Y F  t   t
MAD    t 1
m

• where m = number of forecasts used
Mean Absolute Percentage
Error
• Mean of the absolute percentage of the
difference between the forecast and actual
values of the time series
m
Yt  Ft 
t 1      Yt
MAPE                         100
m

• where m = number of forecasts used
Root Mean Squared Error
• Square root of the mean squared difference
between the forecast and actual values of the
time series
m

 Y  F 
2
t   t
RMSE      t 1
m
• where m = number of forecasts used
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)
Forecasting Accuracy
Example
Model I
2.31  4.66  6.01  9.14
MADI                                            5.53
4

 2.31   4.66    6.01  9.14 
49.96       56.93           58.28       61.41
MAPEI                                                      100  9.50
4

 2.31   4.66    6.01  9.14 
2           2               2           2

RMSEI                                                          6.06
4
Forecasting Accuracy
Example
Model II
2.82  4.15  5.50  8.63
MADII                                            5.28
4

 2.82    4.15  5.50   8.63
49.96       56.93           58.28       61.41
MAPEII                                                      100  9.11
4

 2.82    4.15   5.50   8.63
2           2               2           2

RMSEII                                                          5.70
4
Forecasting Accuracy
Example
Model III
3.45  2.42  2.67  4.71
MADIII                                             3.31
4

 3.45   2.42    2.67    4.71
49.96        56.93           58.28       61.41
MAPEIII                                                       100  5.85
4

 3.45   2.42    2.67    4.71
2           2               2           2

RMSEIII                                                           3.44
4
Forecasting Trends: Simple
Linear Regression
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
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.
Simple Linear Regression
Solution
From Excel

ˆ
Yt  7997.533  528.158t
Simple Linear Regression
Solution
\$15,000

\$14,000
ˆ
Yt  7997.533  528.158t
\$13,000
Tuition

\$12,000

\$11,000

\$10,000

\$9,000

\$8,000
1994   1995   1996   1997   1998   1999     2000   2001   2002   2003   2004   2005
Year
Simple Linear Regression
Solution
Forecast tuition for 2005 (t = 11):
ˆ
Y11  7997.533  528.158(11)  13807.27
95% prediction interval:
1 t p  t   
2

y  t / 2 s 1  
ˆ
n     SStt

1 11  5.5 
2

13807.27   2.306  286.84    1 
10    82.5

13006.21  y11  14608.33
Seasonal Regression
Models
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 
0 if not quarter i
Autocorrelation and The
Durbin–Watson Test
Autocorrelation
• Time series data may have errors that are not
independent
ˆ         ˆ
• Time series residuals: Rt  Yt  Yt
• Correlation between residuals at different
points in time (autocorrelation)
• 1st order correlation: Correlation between
neighboring residuals (times t and t + 1)
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

400
Residuals

200

0
0   2     4      6         8   10   12
-200

-400
t
Durbin–Watson Test
• Ho: No first–order autocorrelation of residuals
• Ha: Positive first–order autocorrelation of
residuals
• Test Statistic

                  
n                      2
ˆ ˆ
Rt  Rt 1
d   t 2
n

 Rt
ˆ2
t 1
Interpretation of d Statistic
• 0≤d≤4
• If residuals uncorrelated, then d ≈ 2
• If residuals positively autocorrelated, then
d<2
• If residuals negatively autocorrelated, then
d >2
Rejection Region for the Durbin–
Watson d Test

Rejection region:
evidence of
positive
autocorrelation

d
0           1     dL   dU   2                  3           4
Possibly significant        Nonrejection region:
autocorrelation             insufficient evidence of
positive autocorrelation
Durbin–Watson Test Example
Use the Durbin–Watson test to test for the
presence of autocorrelation in the tuition data.
Use α = .05.
Durbin–Watson Test Solution
• H0: No 1st–order               Test Statistic:
autocorrelation
• Ha: Positive 1st–order
autocorrelation
•   .05 n = 10 k = 1
Decision:
• Critical Value(s):

Conclusion:
d
0           2            4
.88 1.32
Durbin–Watson Solution
Test Statistic

                  
n                      2
ˆ ˆ
Rt  Rt 1
d   t 2
n

ˆ
Rt 2
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
Durbin–Watson Test Solution
• H0: No 1st–order           Test Statistic:
autocorrelation              d = .51
• Ha: Positive 1st–order
autocorrelation
•   .05 n = 10 k = 1
Decision:
• Critical Value(s):         Reject at  = .05
Conclusion:
d There is evidence of
0           2            4 positive autocorrelation
.88 1.32
Conclusion
1. Described Time Series
2. Explained Descriptive Analyses
3. Defined Time Series Components
4. Explained Forecasting
5. Described Measures of Accuracy
6. Defined Autocorrelation
7. Explained Durbin–Watson Test

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 5 posted: 8/16/2012 language: English pages: 92