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Chap Time Series Descriptive Analyses Models Forecasting

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Chap Time Series Descriptive Analyses Models Forecasting Powered By Docstoc
					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
• Disadvantages
  – 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
• One–step–ahead forecast
  – Ft+1 = Et + Tt


• k–step–ahead forecasting
  – 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*
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
Holt–Winters 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.
     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

				
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