Msb11e ppt ch13

© 2011 Pearson Education, Inc


Statistics for Business and Economics

Chapter 13

Time Series:Descriptive Analyses, Models, &

Forecasting


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


Content

13.7 Forecasting Trends: Simple Linear Regression

13.8 Seasonal Regression Models

13.9 Autocorrelation and the Durbin-Watson Test


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.


Time Series

• Data generated by processes over time

• Describe and predict output of processes

• Descriptive analysis

– Understanding patterns

• Inferential analysis

– Forecast future values


13.1

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


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


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


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


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.


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.


Simple Index Numbers 1990–2006


Simple Index Numbers 1990–2006

Gasoline Price Simple Index

0.0

50.0

100.0

150.0

200.0

250.0

19901991199219931994199519961997199819992000200120022003200420052006


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.



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

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞=⎜ ⎟⎝ ⎠

=





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


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.


Laspeyres Index

• Uses base period quantities as weights– Appropriate when quantities remain approximately

constant over time period

• Example: Consumer Price Index (CPI)


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

∑


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


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

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:


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.


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


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.


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


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


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

=

=

= ×

+ += ×

+ +=

∑

∑


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)


13.2

Descriptive Analysis: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)


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.


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

…


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

…



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-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)


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.



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

…


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




13.3

Time Series Components


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



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)


13.4

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


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


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


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


13.5

Forecasting Trends:

Holt’s Method


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


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.



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.



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

…


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.


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


Holt Solution

Completed series:

w = .8 v = .7


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


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


Holt Forecasting Example

Use the Holt series to forecast the closing price of Daimler–Chrysler stock on 1/31/2007 and 2/28/2007.


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


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.


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


Holt Solution

Completed series


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


Holt Forecasting Thinking Challenge

Use the Holt–Winters series to forecast tuition in 2005 and 2006


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


13.6

Measuring Forecast Accuracy:

MAD and RMSE


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


Mean Absolute Percentage Error

• Mean of the absolute percentage of the difference between the forecast and actual values of the time series


MAPE =

Yt −Ft( )Ytt=n+1

n+m

∑m

×100


Root Mean Squared Error

• Square root of the mean squared difference between the forecast and actual values of the time series


RMSE =

Yt −Ft( )2

t=n+1

n+m

∑m


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)



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

− + + += =



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

− + + += =



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

− + + += =


13.7

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 7997.533 528.158tY t= +



$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= +



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:


13.8

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

Q

i=

1 if quarter i0 if notquarter i

⎧⎨⎩


13.9

Autocorrelation and theDurbin-Watson Test


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= −


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


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

−=

=

−=

∑

∑


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


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


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

−=

=

−=

∑

∑



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,



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

−=

=

−=

∑

∑



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,


Requirements for the Validity of the d-Test

The residuals are normally distributed.


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:

• Ha:

• n = k =

• Critical Value(s):

No 1st–orderautocorrelation

Positive 1st–orderautocorrelation

.05 10 1

2 40 d.88 1.32


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

−=

=

−=

− + − + + −=

+ + +

=

∑

∑


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


Key Ideas

Time Series Data

Data generated by processes over time.


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)


Key Ideas


1. Secular (long-term) trend

2. Cyclical effect

3. Seasonal effect

4. Residual effect


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Time Series Forecasting

Descriptive methods of forecasting with smoothing:

1. Exponential smoothing

2. Holt’s method


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An Inferential forecasting method:

least squares regression


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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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Problems with least squares regression forecasting:

1. Prediction outside the experimental region

2. Regression errors are autocorrelated


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Autocorrelation

Correlation between time series residuals at different points in time.

A test for first-order autocorrelation:

Durbin-Watson test

Date post:	29-May-2015
Category:	Technology
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Msb11e ppt ch13

Technology