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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 nm MAD Y F t t tn1 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 nm MAPE tn1 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 nm RMSE 2 t t tn1 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 © 2011Pearson 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 t2 n 2 ö R t t1 t1 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