Msb11e ppt ch13

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 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
 Τιµ ε σ ιεσϖ υε ατ τιµ ε τ 
ερ
αλ
=
 100
ερ
αλ
ε ιοδ
 Τιµ ε σ ιεσϖ υε ατ βασ περ 

a Simple Index Number
Symbolically,
 Ψ
I t =  τ  100
Ψ
0
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.

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,

Year
1990
1991
1992
1993
1994
1995
1996
1997
1998
1999
2000
2001
2002
2003
2004
2005
Inc
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.

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.


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 1992
1991
1993 1995 1997
1994 1996
1998 2000 2002
1999 2001 2003
2004 2006
2005


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.


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

Solution


Solution
Simple Composite Index Numbers 2005 – 2006

120.0
100.0
80.0
60.0
40.0
20.0
0.0
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
Q1t , Q2t ,K ,Qkt .
0

0

0

4. Calculate the weighted totals for each time
κ
period according to the formula ∑ Qit0 Pit
i=1

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,
κ

It =

∑Θ

ιτ0

Π
ιτ

∑Θ

ιτ0

Π
ιτ

ι=1
κ

ι=1

× 100

0


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


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

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.

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

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


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.
κ
Thus,
∑ Θιτ Π
ιτ
=1
I t = ικ
× 100
∑ Θιτ Π
ιτ
ι=1

0


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
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


Paasche Index Solution
P
∑Q
k

I12 / 29 / 06 =

i =1
k

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 for2011 Pearson Education, Inc
both periods)
©

13.2
Descriptive Analysis:
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

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

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.


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


Solution
70
60

Actual Series

50
40
30

Smoothed Series
(w = .2)

20
10
0
Jan-05 Mar-05 May-05 Jul-05 Sep-05 Nov-05 Jan-06 Mar-06 May-06 Jul-06 Sep-06 Nov-06
Feb-05 Apr-05 Jun-05 Aug-05 Oct-05 Dec-05 Feb-06 Apr-06 Jun-06 Aug-06 Oct-06 Dec-06


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.


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


Solution
70
60

Actual Series

50
40
30

Smoothed Series
(w = .2)

Smoothed Series
(w = .8)

20
10
0
Jan-05 Mar-05 May-05 Jul-05 Sep-05 Nov-05 Jan-06 Mar-06 May-06 Jul-06 Sep-06 Nov-06
Feb-05 Apr-05 Jun-05 Aug-05 Oct-05 Dec-05 Feb-06 Apr-06 Jun-06 Aug-06 Oct-06 Dec-06


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:


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
M
Et = wYt + (1 – w)Et –1

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 =MFt+1


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.

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

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 Model
3. Calculate the two components, Et and Tt, from
the time series Yt beginning at time t = 2 :
E2 = Y 2

and T2 = Y2 – Y1

…

E3 = wY3 + (1 – w)(E2 + T2)
T3 = v(E3 – E2) + (1 – v)T2
Et = wY2011(1 – w)(Et–1 + Tt–1)
© t + Pearson Education, Inc

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


Holt Solution
Holt exponentially smoothed (w = .8 and v = .7)
65
60

Smoothed

55
50
Price

45
40
35
30

Actual

Jan-05
Jan-06
Mar-05 ay-05 Jul-05
Nov-05
Mar-06 ay-06 Jul-06
Nov-06
M
Sep-05
M
Sep-06
D a te


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
Holt–Winters exponentially smoothed (w = .7
and v = .5)
$15,000
$14,000

Tuition

$13,000
$12,000
$11,000
$10,000
$9,000
$8,000

Actual
1995

1996

Smoothed
1997

1998

1999

2000

2001

2002

Ye ar


2003

2004

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
ν+ µ

MAD =

∑ Ψ− Φ

τ= ν+1

τ

τ

µ

• 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
(Ψ − Φ )
τ
τ
∑ Ψ
τ= ν+1
τ
ν+ µ

MAPE =

µ

× 100


Root Mean Squared Error
• Square root of the mean squared difference
between the forecast and actual values of the
time series
ν+ µ

RMSE =

∑ (Ψ − Φ )

τ= ν+1

τ

2

τ

µ


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)


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


2

= 6.06

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

2


2

= 5.70

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


2

= 3.44

13.7
Forecasting Trends:
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


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.

Solution
From Excel

ˆ
Yt = 7997.533 + 528.158t


Solution
$15,000
$14,000

ˆ
Yt = 7997.533 + 528.158t

$13,000
$12,000
$11,000
Tuition
$10,000
$9,000
$8,000
1994

1995

1996

1997

1998

1999

2000

2001

2002

2003

Year


2004

2005

Solution
Forecast tuition for 2005 (t = 11):
ˆ
Y11 = 7997.533 + 528.158(11) = 13807.27
95% prediction interval:
1 (t p − t
ˆ
y ± tα / 2 s 1 + +
n
SStt
13807.27 ± (2.306 )(286.84 )

)

2

1 (11 − 5.5 )
1+ +
10
82.5

13006.21 ≤ Pearson Education, Inc
© 2011 y11 ≤ 14608.33

2

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
1
ιφ θυαρ ι
τερ
Qi = 
τερ
0 ιφ νοτθυαρ ι

13.9
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
200
0
Residuals
0
-200

2

4

6

8

10

-400
t


12

Durbin–Watson Test
• H0: No first–order autocorrelation of residuals
• Ha: Positive first–order autocorrelation of
residuals
• Test Statistic

∑(
n

d=

ˆ ˆ
Rt − Rt −1

t =2

)

2

n

ˆ2
∑ Rt
t =1


Interpretation of DurbinWatson d-Statistic
ν
d=

ˆ
∑( Ρ
τ=2

ˆ
− Ρτ−1 )
τ

ν

ˆ
Ρτ2
∑
τ=1

Ρ ανγε οφ δ : 0 ≤ δ ≤ 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

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

Test Statistic

d=

ˆ ˆ
Rt − Rt −1

t =2

)

2

n

ˆ2
∑ Rt

© 2011tPearson Education, Inc
=1

Autocorrelation
Rejection Region:
d < dL,α
[or (4 – d) < dL,α]
If Ha : Negative first-order autocorrelation
where dL,α is the lower tabled value corresponding
to k independent variables and n observations.
The corresponding upper value
dU,α defines a “possibly significant” region
between dL,α and dU,α

Autocorrelation
Two-tailed Test
H0: No first–order autocorrelation of residuals
Ha: Positive or Negative first–order
autocorrelation of residuals
Test Statistic

∑(
n

d=

ˆ ˆ
Rt − Rt −1

t =2

)

2

n

ˆ2
∑ Rt

© 2011tPearson Education, Inc
=1

Autocorrelation
Rejection Region:
d < dL,α/2 or (4 – d) < dL,α/2
where dL,α/2 is the lower tabled value
corresponding to k independent variables and n
observations. The corresponding upper value
dU,α/2 defines a “possibly significant” region
between dL,α/2 and dU,α/2

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: No 1st–order
autocorrelation
• Ha:

Positive 1st–order
autocorrelation
.05

10

• α=
n=
k=
• Critical Value(s):

0

2
.88 1.32

1

4

d


Durbin–Watson Solution
Test Statistic

∑(
n

d=

ˆ ˆ
Rt − Rt −1

t =2

)

2

n

Rt 2
∑ˆ
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
autocorrelation
• Ha:

d = .51

Positive 1st–order
autocorrelation
.05

10

• α=
n=
k=
• Critical Value(s):

0

Test Statistic:

2
.88 1.32

1

4

Decision:
Reject at α = .05
Conclusion:
There is evidence of
d 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 number
3. Weighted composite number (Laspeyers
index or Pasche index)

Key Ideas
1.
2.
3.
4.

Secular (long-term) trend
Cyclical effect
Seasonal effect
Residual effect


Key Ideas
Time Series Forecasting
Descriptive methods of forecasting with
smoothing:
1. Exponential smoothing
2. Holt’s method


Key Ideas
An Inferential forecasting method:
least squares regression


Key Ideas
Measures of forecast accuracy:
1. mean absolute deviation (MAD)
2. mean absolute percentage error (MAPE)
3. root mean squared error (RMSE)


Key Ideas
Problems with least squares regression
forecasting:
1. Prediction outside the experimental region
2. Regression errors are autocorrelated


Key Ideas
Autocorrelation
Correlation between time series residuals at
different points in time.
A test for first-order autocorrelation:
Durbin-Watson test


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