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Index Numbers
INDEX NUMBER: A number that measures the relative change in
price, quantity, value, or some other item of interest from one time
period to another.
Index numbers are statistical measures designed to show changes in a
variable or group of related variables with respect to time, geographic
location or other characteristics such as income, profession, etc.
A collection of index numbers for different years, locations, etc., is
sometimes called an index series.
Index numbers are meant to study the change in the effects of factors
which cannot be measured directly.
According to Bowley, “Index numbers are used to measure the
changes in some quantity which we cannot observe directly”.
For example, We may wish to compare the present agricultural
production or industrial production with that at the time of
independence or 1960/2000/2010, etc.
Types of Index Numbers
1. Price index Numbers: Price index numbers measure the relative
changes in prices of a commodities between two periods. Prices can
be either retail or wholesale.
e.g. Consumer Price Index (CPI)
2. Quantity Index Numbers: These index numbers are considered to
measure changes in the physical quantity of goods produced, consumed
or sold of an item or a group of items.
e.g. Number of cell phones produced annually
------------------------------------------------------------------------------Simple Index Number: A simple index number is a number that
measures a relative change in a single variable with respect to a base.
Composite Index Number: A composite index number is a number that
measures an average relative changes in a group of relative variables with
respect to a base.
Why compute indexes?
Easier to comprehend than actual numbers (percent change)
Ex. Production increased by 1,22,215 or by 20%
Provide convenient ways to express the change in the total of
a heterogeneous group of items
CPI
Ex. Consumer Price Index
Facilitate comparison of unlike series
Food
Rs. 1,000
Car
Rs.5,00,000
Cloth
Rs.5,000
Surgery
Rs.7,00,000
Indexes: Four classifications
Price
Measures the changes in
prices from a selected
base period to another
period.
Quantity
Measures the changes in
quantity consumed from the
base period to another period.
Special purpose
Combines and weights a
Value
heterogeneous group of series to
Measures the change in the value arrive at an overall index showing
of one or more items from the
the change in business activity
base period to the given period from the base period to the
(PxQ).
present.
Some Important Index number
• There are various uses of price index numbers.
• The wholesale price index number indicates the price changes in
wholesale markets.
• Consumer price index number or the cost of living index number
tells us about the changes in the prices faced by an individual
consumer. Its major application is in the calculation of dearness
allowance so that real wage does not decrease; or in comparing the
cost of living in, say, different regions. It is also used to measure
changes in purchasing power of money.
• The reciprocal of a general price index is known as purchasing
power of money with reference to the base period.
For example, if the price index number goes up to 150, it means
that the same amount of money will be able to purchase 0. 67 times
of the volume of goods being purchased in the base period.
Steps in Construction of Index number
• A. Selection of Base Period
• Since index numbers measure relative changes, they are expressed with one
Selected situation(i.e place, period, etc. ) as 100 . This is called the base or the
starting point of the index numbers.
• The base may be one day such as with index of retail prices, the average of year
or the average of a period.
While selecting a base period the following aspects should be taken into
consideration:
 The base date must be "normal" in the sense that the data chosen are not affected
by any irregular or abnormal situations such as natural calamities, war, etc. It is
desirable to restrict comparisons to stable periods for achieving accuracy.
 It should not be too back-dated as the patterns of trade, imports or consumer
preferences may change considerably if the time-span is too long. A ten to twenty
year interval is likely to be suitable for one base date, and after that the index
becomes more and more outdated. Greater accuracy is attained for moderate
short-run indices than for those covering greater span of time.
 For indices dealing with economic data, the base period should have some
economic significance.
B. Choice of a Suitable Average
• An index number is basically the result of averaging a series of data
(e.g., price relatives of several commodities). There are, however,
several ways of averaging a series: mean (i.e., arithmetic mean),
mode, median, geometric mean and harmonic mean.
• The question naturally arises as to which average to chose. The
mode has the merit of simplicity, but may be indefinite. The median
suffers from the same limitations. Moreover, neither of them takes
into account the size of the items at each end of a distribution.
• The harmonic mean has very little practical application to index
numbers. As a result, mode, median and harmonic mean are not
generally used in the calculation of index numbers.
• Thus, the arithmetic mean is most commonly used. However, the
geometric mean is sometimes used despite its slight difficulty in
calculation.
C. Selection of items and their Numbers
•
•
•
•
•
•
The number and kinds of commodities to be included in the
construction of an index number depend on the particular problem to
be dealt with, economy and ease of calculations.
Various practical considerations determine the number and kinds of
items to be taken into account. For a wholesale price index, the number
of commodities should be as large as possible.
On the other hand, for an index meant to serve as a predictor of price
movement rather than an indicator of changes over time, a much
smaller number of items may be adequate.
Care should, however, be taken to ensure that items chosen are not too
few which make the index unrepresentative of the general level.
A fixed set of commodities need not also be used for a very long period
as some items lose their importance with the passage of time and some
new items gain in significance.
In general, the commodities should be sensitive and representative of
the various elements in the price system.
D.
Collection of data
• As prices often vary from market to market, they should be collected at
regular intervals from various representative markets.
• It is desirable to select shops which are visited by a cross section of
customers.
• The reliability of the index depends greatly on the accuracy of the
quotations given for each constituent item.
Steps for Calculating
a Simple Index Number
1. Obtain the prices or quantities for the commodity
over the time period of interest.
2. Select a base period.
3. Calculate the index number for each period
according to the formula
Index number at time t
 Time series value at time t 

100

 Time series value at base period 
© 2011 Pearson Education, Inc
Steps for Calculating
a Simple Index Number
Symbolically,
 Yt 
I t    100
 Y0 
where It is the index number at time t, Yt is
the time series value at time t, and Y0 is
the time series value at the base period.
Simple Index Number Example
The table shows the price per quintal of
wheat for the years 1990 – 2006. Use 1990 as
the base year Calculate the simple index
number for 1990, 1998, and 2006.
Year
1990
1991
1992
1993
1994
1995
1996
1997
1998
1999
2000
2001
2002
2003
2004
2005
2006
Rs./Q
1299
1098
1087
1067
1075
1111
1224
1199
1030
1136
1484
1420
1345
1561
1852
2270
2572
Simple Index Number Example
1299 
1990 Index Number (base period) =  1990price 100   1.299
100  100
 1990price 
1998 Index Number =
2006 Index Number =
 1.299
1299 
1030 
 1998price 
 1.03

100  
100  79.3
1299 
 1.299
 1990price 
2572 
 2006price 
 2.572

100  
100  198
1299 
 1.299
 1990price 
•Indicates price had dropped by 20.7% (100 – 79.3) between
1990 and 1998.
•Indicates price had risen by 98% (198 – 100) between
1990 and 2006.
© 2011 Pearson Education, Inc
Simple Index Numbers
1990–2006
Wheat Price Simple Index
Gasoline
2006
2005
2004
2003
2002
2001
2000
1999
1998
1997
1996
1995
1994
1993
1992
1991
1990
250.0
200.0
150.0
100.0
50.0
0.0
Composite Index Number
Made up of two or more commodities
A simple index using the total price or total quantity
of all the series (commodities)
Disadvantage: Quantity of each commodity purchased
is not considered
Simple Composite Index Example
Suzuki
The table on the slide shows the
closing stock prices of 4wheellers on the last day of the
month for Suzuki, Ford, and
GM between 2005 and 2006.
Construct the simple
composite index using January
2005 as the base period.
Simple Composite Index example
Suzuki
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 / 06price 

 100
 1 / 05price 
 99.64 

 100
 95.49 
© 2011 Pearson Education, Inc
 104.3
Simple Composite Index Solution
Suzuki
© 2011 Pearson Education, Inc
Simple Composite Index Solution
Simple Composite Index Numbers 2005 – 2006
120.0
100.0
80.0
60.0
40.0
20.0
-0
6
N
S06
J06
M
-0
6
M
-0
6
J06
-0
5
N
S05
J05
M
-0
5
M
-0
5
J05
0.0
Weighted Composite Price Index
A weighted composite price index weights the prices by
quantities purchased prior to calculating totals for each time
period. The weighted totals are then used to compute the index in
the same way that the unweighted totals are used for simple
composite indexes.
• Laspeyres Index
• Paasche Index
• Fisher’s Ideal Index
Laspeyres Index
• Uses base period quantities as weights
– Appropriate when quantities remain approximately
constant over time period
• Example: Consumer Price Index (CPI)
Laspeyres Index
.
Advantages
Requires quantity data from only the base period. This allows a more
meaningful comparison over time. The changes in the index can be
attributed to changes in the price.
Disadvantages
Does not reflect changes in buying patterns over time. Also, it may
overweight goods whose prices increase.
Laspeyres Index Number Example
The table shows the prices(P) on 2014 and 2015 for Tur,
Mung and Udid. In 2005 an investor purchased the indicated
number of quantiy(Q). Construct the Laspeyres Index using
2005 as the base period.
Tur
Mung
Udid
Quantity Purchased(Qtl)
100
500
200
2014 Price(Rs)
90
80
110
2015 Price(Rs)
140
120
130
© 2011 Pearson Education, Inc
Laspeyres Index example
Tur
Mung
Udid
Quantity Purchased(Qtl)
100
500
200
2014 Price(Rs)
90
80
110
2015 Price(Rs)
140
120
130
Weighted total for base period (2014)
k
PQ0 QP0
i 1
it0
it0
 100(45.51)
 500(13.17)
 200(36.81)
100x90 + 500x80
+ 200x110
 18498
71,000
0- 2014
Weighted total for 2015:
k
Pt QP
0
Q
i 1
it0
it
t -2015
 100(61.41)  500(7.51)  200(30.72)
100x140 + 500x120 + 200x130
 16040
1,00,000
© 2011 Pearson Education, Inc
Laspeyres Index Solution
k
PI t 
 QP Q
P
 QP Q2014
P
i 1
k
i 1

0 05 i ,12 / 29 / 06
it ,1/ 31/
100
0 0
2014
i ,1/ 31/ 05 i ,1/
31/ 05
100000
16040
18498
71000
 86.7
140.84
100
Indicates the value had increased by 40.84% (140.84 100 = 40.84) between 2014 and 2015.
© 2011 Pearson Education, Inc
Paasche Index
.
Where
p is the price index
pt is the current price
p0 is the price of the base period
qt is the quantity used in the current period
q0 is the quantity used in the base period
Advantages
Because it uses quantities from the current period, it reflects current
buying habits. Appropriate when quantities change over time
Disadvantages
It requires quantity data for the current year. Because different quantities
are used each year, it is impossible to attribute changes in the index to
changes in price alone. It tends to overweight the goods whose prices
have declined. It requires the prices to be recomputed each year.
Paasche Index Number Example
The table shows the 2014 and 2015 prices(in Rs) and
production(in Qtl) for Brinjal, Potato and Onion. Calculate
the Paasche Index using 2014 as the base period.
Brinjal
Potato
Onion
Price
Prod.
Price
Prod.
Price
Prod.
2014
1700
10000
1600
15000
1600
30000
2015
2000
12000
1800
17000
2400
28000
Paasche Index
Potato
Brinjal
Onion
Price
Prod.
Price
Prod.
Price
Prod.
2014
1700
10000
1600
15000
1600
30000
2015
2000
12000
1800
17000
2400
28000
k
IP122015
/ 29 / 06 
 Q 2015
P
 Q 2015
P
i 1
k
i 1

/ 29 / 06
i12 / 29 / 06 i12
2015
100
i12 / 29 / 06 i1/ 31/ 05
2014
12000x2000
17000x1800
+ 28000x2400
 6.1(30.72)
.2(61.41) +10(7.51)
____
100 x 100
12000x1700
+ 17000x1600
+ 28000x1600
 6.1(36.81)
.2(45.51)  10(13.17)
121800000
274.774
____
= 131.81
75.2
 100x 100

92400000
365.343
2015 prices represent a 31.81% (131.81 – 100) increase from 2014
(assuming quantities were at 2014 levels for both periods)
Fisher’s Ideal Index
Fisher’s Ideal Index is obtained as geometric men of the
Laspeyres Index and Paasche Index, i.e. suquare root of the
product of Laspeyres Index and Paasche Index
• Laspeyres’ index tends to overweight goods whose
prices have increased.
• Paasche’s index, on the other hand, tends to overweight
goods whose prices have gone down.
• Fisher’s ideal index was developed in an attempt to
offset these shortcomings.
• Balances the negative effects of the Laspeyres’ and
Paasche’s indices.
Marshal-Edgeworth Index Number
In this index number, the average of the base year and current
year quantities are used as weights.
This index number is proposed by two English
economists Marshal and Edgeworth.
(∑Pt Qo + ∑Pt Qt )
P = ---------------------------- x 100
∑Po Qo + ∑Po Qt)
=
∑Pt (Qo+Qt)
----------------- x 100
∑Po(Qo+Qt)
Example:
Compute the weighted aggregative price index numbers
for 1981 with 1980 as base year using (1) Laspeyre’s Index
Number (2) Paashe’s Index Number (3) Fisher’s Ideal Index
Number (4) Marshal Edgeworth Index Number.
Commodity
A
B
C
D
Prices
1980
10
8
5
4
1981
12
8
6
4
Quantities
1980
1981
20
22
16
18
10
11
7
8
L = 112.32
P = 112.20
F = 112.26
M-E = 112.38
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
© 2011 Pearson Education, Inc
Exponential Smoothing
Constant
Exponential smoothing constant, 0 < w < 1
• w close to 0
– More weight given to previous values of time
series
– Smoother series
• w close to 1
– More weight given to current value of time series
– Series looks similar to original (more variable)
© 2011 Pearson Education, Inc
Steps for Calculating an
Exponentially Smoothed Series
1. Select an exponential smoothing constant, w,
between 0 and 1. Remember that small
values of w give less weight to the current
value of the series and yield a smoother
series. Larger choices of w assign more
weight to the current value of the series and
yield a more variable series.
© 2011 Pearson Education, Inc
Steps for Calculating an
Exponentially Smoothed Series
2. Calculate the exponentially smoothed series
Et from the original time series Yt as follows:
E1 = Y1
E2 = wY2 + (1 – w)E1
…
E3 = wY3 + (1 – w)E2
Et = wYt + (1 – w)Et–1
© 2011 Pearson Education, Inc
Exponential Smoothing
Example
The closing stock prices on the last
day of the month for Daimler–
Chrysler in 2005 and 2006 are
given in the table. Create an
exponentially smoothed series
using w = .2.
© 2011 Pearson Education, Inc
Exponential Smoothing
Solution
E1 = 45.51
E2 = .2(46.10) + .8(45.51) = 45.63
…
E3 = .2(44.72) + .8(45.63) = 45.45
E24 = .2(61.41) + .8(53.92) = 55.42
© 2011 Pearson Education, Inc
Exponential Smoothing
Solution
E1 = 45.51
E2 = .2(46.10) + .8(45.51) = 45.63
…
E3 = .2(44.72) + .8(45.63) = 45.45
E24 = .2(61.41) + .8(53.92) = 55.42
© 2011 Pearson Education, Inc
© 2011 Pearson Education, Inc
Dec-06
Nov-06
Oct-06
Sep-06
Aug-06
Jul-06
Jun-06
May-06
Apr-06
Mar-06
Feb-06
Jan-06
Dec-05
30
Nov-05
40
Oct-05
Sep-05
Aug-05
Jul-05
Jun-05
60
May-05
Apr-05
Mar-05
Feb-05
Jan-05
Exponential Smoothing
Solution
70
Actual Series
50
Smoothed Series
(w = .2)
20
10
0
Exponential Smoothing
Thinking Challenge
The closing stock prices on the last
day of the month for Daimler–
Chrysler in 2005 and 2006 are
given in the table. Create an
exponentially smoothed series
using w = .8.
© 2011 Pearson Education, Inc
Exponential Smoothing
Solution
E1 = 45.51
E2 = .8(46.10) + .2(45.51) = 45.98
…
E3 = .8(44.72) + .2(45.98) = 44.97
E24 = .8(61.41) + .2(57.75) = 60.68
© 2011 Pearson Education, Inc
© 2011 Pearson Education, Inc
Dec-06
Nov-06
Oct-06
Sep-06
Aug-06
Jul-06
Jun-06
Smoothed Series
(w = .2)
May-06
Apr-06
Mar-06
Feb-06
Jan-06
Dec-05
Nov-05
Oct-05
30
Sep-05
40
Aug-05
Jul-05
Jun-05
60
May-05
Apr-05
Mar-05
Feb-05
Jan-05
Exponential Smoothing
Solution
70
Actual Series
50
Smoothed Series
(w = .8)
20
10
0
13.3
Time Series Components
© 2011 Pearson Education, Inc
Descriptive v. Inferential
Analysis
• Descriptive Analysis
– Picture of the behavior of the time series
– e.g. Index numbers, exponential smoothing
– No measure of reliability
• Inferential Analysis
– Goal: Forecasting future values
– Measure of reliability
© 2011 Pearson Education, Inc
Time Series Components
Additive Time Series Model Yt = Tt + Ct + St + Rt
Tt = secular trend (describes long–term movements of Yt)
Ct = cyclical effect (describes fluctuations about the
secular trend attributable to business and economic
conditions)
St = seasonal effect (describes fluctuations that recur
during specific time periods)
Rt = residual effect (what remains after other components
have been removed)
© 2011 Pearson Education, Inc
13.4
Forecasting:
Exponential Smoothing
© 2011 Pearson Education, Inc
Exponentially Smoothed
Forecasts
• Assumes the trend and seasonal component are
relatively insignificant
• Exponentially smoothed forecast is constant for all
future values
• Ft+1 = Et
Ft+2 = Ft+1
Ft+3 = Ft+1
• Use for short–term forecasting only
© 2011 Pearson Education, Inc
Calculation of Exponentially
Smoothed Forecasts
1. Given the observed time series Y1, Y2, … , Yt,
first calculate the exponentially smoothed
values E1, E2, … , Et, using
E1 = Y1
E2 = wY2 + (1 – w)E1
Et = wYt + (1 – w)Et –1
© 2011 Pearson Education, Inc
Calculation of Exponentially
Smoothed Forecasts
2. Use the last smoothed value to forecast the
next time series value:
Ft +1 = Et
3. Assuming that Yt is relatively free of trend and
seasonal components, use the same forecast
for all future values of Yt:
Ft+2 = Ft+1
Ft+3 = Ft+1
© 2011 Pearson Education, Inc
Exponential Smoothing
Forecasting Example
The closing stock prices on the
last day of the month for
Daimler–Chrysler in 2005 and
2006 are given in the table
along with the exponentially
smoothed values using w = .2.
Forecast the closing price for
the January 31, 2007.
© 2011 Pearson Education, Inc
Exponential Smoothing
Forecasting Solution
F1/31/2007 = E12/29/2006 = 55.42
The actual closing price on 1/31/2007
for Daimler–Chrysler was 62.49.
Forecast Error = Y1/31/2007 – F1/31/2007
= 62.49 – 55.42
= 7.07
© 2011 Pearson Education, Inc
13.5
Forecasting Trends:
Holt’s Method
© 2011 Pearson Education, Inc
The Holt Forecasting Model
• Accounts for trends in time series
• Two components
– Exponentially smoothed component, Et
• Smoothing constant 0 < w < 1
– Trend component, Tt
• Smoothing constant 0 < v < 1
– Close to 0: More weight to past trend
– Close to 1: More weight to recent trend
© 2011 Pearson Education, Inc
Steps for Calculating
Components of the Holt
Forecasting Model
1. Select an exponential smoothing constant w
between 0 and 1. Small values of w give less
weight to the current values of the time series
and more weight to the past. Larger choices
assign more weight to the current value of the
series.
© 2011 Pearson Education, Inc
Steps for Calculating
Components of the Holt
Forecasting Model
2. Select a trend smoothing constant v between 0
and 1. Small values of v give less weight to the
current changes in the level of the series and
more weight to the past trend. Larger values
assign more weight to the most recent trend of
the series and less to past trends.
© 2011 Pearson Education, Inc
Steps for Calculating
Components of the Holt
Forecasting Model
…
3. Calculate the two components, Et and Tt, from
the time series Yt beginning at time t = 2 :
E2 = Y2 and T2 = Y2 – Y1
E3 = wY3 + (1 – w)(E2 + T2)
T3 = v(E3 – E2) + (1 – v)T2
Et = wYt + (1 – w)(Et–1 + Tt–1)
Tt = v(E
Et–1Education,
) + (1Inc – v)Tt–1
t –Pearson
© 2011
Holt Example
The closing stock prices on the
last day of the month for
Daimler–Chrysler in 2005 and
2006 are given in the table.
Calculate the Holt–Winters
components using w = .8 and
v = .7.
© 2011 Pearson Education, Inc
Holt Solution
w = .8 v = .7
E2 = Y2 and T2 = Y2 – Y1
E2 = 46.10 and T2 = 46.10 – 45.51 = .59
E3 = wY3 + (1 – w)(E2 + T2)
E3 = .8(44.72) + .2(46.10 + .59) = 45.114
T3 = v(E3 – E2) + (1 – v)T2
T3 = .7(45.114 – 46.10) + .3(.59) = –.5132
© 2011 Pearson Education, Inc
Holt Solution
Completed series:
w = .8 v = .7
© 2011 Pearson Education, Inc
Holt Solution
Holt exponentially smoothed (w = .8 and v = .7)
65
Smoothed
60
50
45
40
Actual
35
l-0
Se 5
p05
N
ov
-0
5
Ja
n06
M
ar
-0
M 6
ay
-0
6
Ju
l-0
Se 6
p06
N
ov
-0
6
Ju
-0
5
5
ay
M
ar
-0
M
-0
5
30
Ja
n
Price
55
Date
© 2011 Pearson Education, Inc
Holt’s Forecasting Methodology
1. Calculate the exponentially smoothed and
trend components, Et and Tt, for each
observed value of Yt (t ≥ 2) using the formulas
given in the previous box.
2. Calculate the one-step-ahead forecast using
Ft+1 = Et + Tt
3. Calculate the k-step-ahead forecast using
Ft+k = Et + kTt
© 2011 Pearson Education, Inc
Holt Forecasting Example
Use the Holt series to
forecast the closing price
of Daimler–Chrysler stock
on 1/31/2007 and
2/28/2007.
© 2011 Pearson Education, Inc
Holt Forecasting Solution
1/31/2007 is one–step–ahead:
F1/31/07 = E12/29/06 + T12/29/06
= 61.39 + 3.00 = 64.39
2/28/2007 is two–steps–ahead:
F2/28/07 = E12/29/06 + 2T12/29/06
= 61.39 + 2(3.00) = 67.39
© 2011 Pearson Education, Inc
Holt Thinking Challenge
The data shows the
average undergraduate
tuition at all 4–year
institutions for the years
1996–2004 (Source: U.S.
Dept. of Education).
Calculate the Holt–
Winters components
using w = .7 and v = .5.
© 2011 Pearson Education, Inc
Holt Solution
w = .7 v = .5
E2 = Y2 and T2 = Y2 – Y1
E2 = 9206 and T2 = 9206 – 8800 = 406
E3 = wY3 + (1 – w)(E2 + T2)
E3 = .7(9588) + .3(9206 + 406) = 9595.20
T3 = v(E3 – E2) + (1 – v)T2
T3 = .5(9595.20 – 9206) + .5(406) = 397.60
© 2011 Pearson Education, Inc
Holt Solution
Completed series
Year
1995
1996
1997
1998
1999
2000
2001
2002
2003
2004
t
1
2
3
4
5
6
7
8
9
10
Tuition
$8,800
$9,206
$9,588
$10,076
$10,444
$10,818
$11,380
$12,014
$12,955
$13,743
w=.7
Et
v=.5
Tt
9206.0000
9595.2000
10051.0400
10454.1280
10833.3096
11335.1057
11945.1576
12810.9680
13672.7223
406.0000
397.6000
426.7200
414.9040
397.0428
449.4195
529.7356
697.7730
779.7637
© 2011 Pearson Education, Inc
Holt Solution
Holt–Winters exponentially smoothed (w = .7
and v = .5)
$15,000
$14,000
Tuition
$13,000
$12,000
$11,000
$10,000
Actual
Smoothed
$9,000
$8,000
1995
1996
1997
1998
1999
2000
Year
© 2011 Pearson Education, Inc
2001
2002
2003
2004
Holt Forecasting Thinking
Challenge
Use the Holt–Winters series to forecast tuition in
2005 and 2006
Year
1995
1996
1997
1998
1999
2000
2001
2002
2003
2004
t
1
2
3
4
5
6
7
8
9
10
w=.7
Et
Tuition
$8,800
$9,206
9206.0000
$9,588
9595.2000
$10,076
10051.0400
$10,444
10454.1280
$10,818
10833.3096
$11,380
11335.1057
$12,014
11945.1576
$12,955
12810.9680
© 2011
Pearson Education,
Inc
$13,743
13672.7223
v=.5
Tt
406.0000
397.6000
426.7200
414.9040
397.0428
449.4195
529.7356
697.7730
779.7637
Holt Forecasting Solution
2005 is one–step–ahead: F11 = E10 + T10
13672.72 + 779.76 = $14,452.48
2006 is 2–steps–ahead: F12 = E10 + 2T10
=13672.72 +2(779.76) = $15,232.24
© 2011 Pearson Education, Inc
13.6
Measuring Forecast Accuracy:
MAD and RMSE
© 2011 Pearson Education, Inc
Mean Absolute Deviation
• Mean absolute difference between the forecast
and actual values of the time series
nm
MAD 
 Y F
t
t
tn1
m
• where m = number of forecasts used
© 2011 Pearson Education, Inc
Mean Absolute Percentage
Error
• Mean of the absolute percentage of the
difference between the forecast and actual
values of the time series
nm

MAPE 
tn1
Y  F 
t
t
Yt
m
 100
• where m = number of forecasts used
© 2011 Pearson Education, Inc
Root Mean Squared Error
• Square root of the mean squared difference
between the forecast and actual values of the
time series
 Y  F 
nm
RMSE 
2
t
t
tn1
m
• where m = number of forecasts used
© 2011 Pearson Education, Inc
Forecasting Accuracy
Example
Using the Daimler–Chrysler data from 1/31/2005 through
8/31/2006, three time series models were constructed and
forecasts made for the next four months.
• Model I: Exponential smoothing (w = .2)
• Model II: Exponential smoothing (w = .8)
• Model III: Holt–Winters (w = .8, v = .7)
© 2011 Pearson Education, Inc
Forecasting Accuracy
Example
Model I
MADI 
2.31  4.66  6.01  9.14
4
 5.53
 2.31   4.66    6.01  9.14 
MAPEI 
49.96
56.93
61.41
4
100  9.50
 2.31   4.66    6.01   9.14 
2
RMSEI 
58.28
2
2
4
© 2011 Pearson Education, Inc
2
 6.06
Forecasting Accuracy
Example
Model II
MADII 
2.82  4.15  5.50  8.63
4
 5.28
 2.82    4.15  5.50   8.63
MAPEII 
49.96
56.93
61.41
4
100  9.11
 2.82    4.15   5.50   8.63
2
RMSEII 
58.28
2
4
© 2011 Pearson Education, Inc
2
2
 5.70
Forecasting Accuracy
Example
Model III
MADIII 
3.45  2.42  2.67  4.71
4
 3.31
 3.45   2.42    2.67    4.71
MAPEIII 
49.96
56.93
61.41
4
100  5.85
 3.45   2.42    2.67    4.71
2
RMSEIII 
58.28
2
2
4
© 2011 Pearson Education, Inc
2
 3.44
13.7
Forecasting Trends:
Simple Linear Regression
© 2011 Pearson Education, Inc
Simple Linear Regression
• Model: E(Yt) = β0 + β1t
• Relates time series, Yt, to time, t
• Cautions
– Risky to extrapolate (forecast beyond observed
data)
– Does not account for cyclical effects
© 2011 Pearson Education, Inc
Simple Linear Regression
Example
The data shows the average
undergraduate tuition at all 4–
year institutions for the years
1996–2004 (Source: U.S.
Dept. of Education). Use least–
squares regression to fit a
linear model. Forecast the
tuition for 2005 (t = 11) and
compute a 95% prediction
interval for the forecast.
© 2011 Pearson Education, Inc
Simple Linear Regression
Solution
From Excel
Yˆt  7997.533
 528.158t
© 2011 Pearson Education, Inc
Simple Linear Regression
Solution
$15,000
Yˆt  7997.533  528.158t
$14,000
Tuition
$13,000
$12,000
$11,000
$10,000
$9,000
$8,000
1994
1995
1996
1997
1998
1999
2000
2001
Year
© 2011 Pearson Education, Inc
2002
2003
2004
2005
Simple Linear Regression
Solution
Forecast tuition for 2005 (t = 11):
Yˆ11  7997.533  528.158(11)  13807.27
95% prediction interval:
1 t p  t
yˆ  t / 2 s 1  
n
SStt
13807.27   2.306  286.84 

2
1 11  5.5 
1 
10
82.5
© 2011Pearson
Education,
Inc
13006.21
y11 
14608.33
2
13.8
Seasonal Regression Models
© 2011 Pearson Education, Inc
Seasonal Regression Models
• Takes into account secular trend and seasonal
effects (seasonal component)
• Uses multiple regression models
• Dummy variables to model seasonal
component
• E(Yt) = β0 + β1t + β2Q1 + β3Q2 + β4Q3
where
1
if quarter i
Qi  
if not
quarter
i
©02011 Pearson
Education, Inc
13.9
Autocorrelation and the
Durbin-Watson Test
© 2011 Pearson Education, Inc
Autocorrelation
• Time series data may have errors that are not
independent
• Time series residuals: Rˆt  Yt  Yˆt
• Correlation between residuals at different
points in time (autocorrelation)
• 1st order correlation: Correlation between
neighboring residuals (times t and t + 1)
© 2011 Pearson Education, Inc
Autocorrelation
Plot of residuals v. time for tuition data shows
residuals tend to group alternately into positive
and negative clusters
Residual v Time Plot
600
Residuals
400
200
0
-200
0
2
4
6
-400
© 2011 Pearson Education,
Inc
t
8
10
12
Durbin–Watson Test
• H0: No first–order autocorrelation of residuals
• Ha: Positive first–order autocorrelation of
residuals
• Test Statistic

n
d
t 2
Rˆt  Rˆt 1

2
n
2
ˆ
 Rt
t 1
© 2011 Pearson Education, Inc
Interpretation of DurbinWatson
d-Statistic
n
d
 Rö  Rö 
t
t2
n
2
ö
R
 t
t1
t1
Range of d : 0  d  4
1. If the residuals are uncorrelated, then d ≈ 2.
2. If the residuals are positively autocorrelated,
then d < 2, and if the autocorrelation is very
strong, d ≈ 2.
3. If the residuals are negatively autocorrelated,
then d >2, and if the autocorrelation is very
strong, d ≈ 4. © 2011 Pearson Education, Inc
Rejection Region for the Durbin–
Watson d Test
Rejection region:
evidence of
positive
autocorrelation
0
1
dL
dU
Possibly significant
autocorrelation
2
3
Nonrejection region:
insufficient evidence of
positive autocorrelation
© 2011 Pearson Education, Inc
4
d
Durbin–Watson d-Test for
Autocorrelation
One-tailed Test
H0: No first–order autocorrelation of residuals
Ha: Positive first–order autocorrelation of
residuals
(or Ha: Negative first–order autocorrelation)
n
2
Test Statistic
ˆ ˆ
d
R  R 
t 2
t
n
t 1
2
ˆ
R

t
© 2011 Pearson Education,
Inc
t 1
Durbin–Watson d-Test for
Autocorrelation
Rejection Region:
d < dL,
[or (4 – d) < dL,
If Ha : Negative first-order autocorrelation
where dL, is the lower tabled value
corresponding to k independent variables and n
observations. The corresponding upper value
dU, defines a “possibly significant” region
between dL, and dU,
© 2011 Pearson Education, Inc
Durbin–Watson d-Test for
Autocorrelation
Two-tailed Test
H0: No first–order autocorrelation of residuals
Ha: Positive or Negative first–order
autocorrelation of residuals
Test Statistic

n
d
t 2
Rˆt  Rˆt 1
n

2
2
ˆ
R

t
© 2011 Pearson Education,
Inc
t 1
Durbin–Watson d-Test for
Autocorrelation
Rejection Region:
d < dL, or (4 – d) < dL,
where dL, is the lower tabled value
corresponding to k independent variables and n
observations. The corresponding upper value
dU, defines a “possibly significant” region
between dL, and dU,
© 2011 Pearson Education, Inc
Requirements for the Validity
of the d-Test
The residuals are normally distributed.
© 2011 Pearson Education, Inc
Durbin–Watson Test Example
Use the Durbin–Watson test to test for the
presence of autocorrelation in the tuition data.
Use α = .05.
© 2011 Pearson Education, Inc
Durbin–Watson Test Solution
• H0: No 1st–order
autocorrelation
• Ha: Positive 1st–order
autocorrelation
•   .05
n10=
1k
=
• Critical Value(s):
0
2
.88 1.32
4
d
© 2011 Pearson Education, Inc
Durbin–Watson Solution
Test Statistic

n
d
t 2
Rˆt  Rˆt 1

2
n
2
ˆ
R
 t
t 1
(152.1515  274.3091)2  (5.9939  152.1515)2  ...  (463.8909  204.0485) 2

(274.3091) 2  (152.1515)2  ...  (463.8909)2
 .51
© 2011 Pearson Education, Inc
Durbin–Watson Test Solution
• H0: No 1st–order
autocorrelation
Test Statistic:
d = .51
• Ha: Positive 1st–order
autocorrelation
•   .05
n10=
1k
=
• Critical Value(s):
0
2
.88 1.32
4
Decision:
Reject at  = .05
Conclusion:
There is evidence of
d positive autocorrelation
© 2011 Pearson Education, Inc
Key Ideas
Time Series Data
Data generated by processes over time.
© 2011 Pearson Education, Inc
Key Ideas
Index Number
Measures the change in a variable over time
relative to a base period.
Types of Index numbers:
1. Simple index number
2. Simple composite index number
3. Weighted composite number (Laspeyers
index or Pasche index)
© 2011 Pearson Education, Inc
Key Ideas
Time Series Components
1.
2.
3.
4.
Secular (long-term) trend
Cyclical effect
Seasonal effect
Residual effect
© 2011 Pearson Education, Inc
Key Ideas
Time Series Forecasting
Descriptive methods of forecasting with
smoothing:
1. Exponential smoothing
2. Holt’s method
© 2011 Pearson Education, Inc
Key Ideas
Time Series Forecasting
An Inferential forecasting method:
least squares regression
© 2011 Pearson Education, Inc
Key Ideas
Time Series Forecasting
Measures of forecast accuracy:
1. mean absolute deviation (MAD)
2. mean absolute percentage error (MAPE)
3. root mean squared error (RMSE)
© 2011 Pearson Education, Inc
Key Ideas
Time Series Forecasting
Problems with least squares regression
forecasting:
1. Prediction outside the experimental region
2. Regression errors are autocorrelated
© 2011 Pearson Education, Inc
Key Ideas
Autocorrelation
Correlation between time series residuals at
different points in time.
A test for first-order autocorrelation:
Durbin-Watson test
© 2011 Pearson Education, Inc