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Statistics for Business and Economics Chapter 13 Time Series: Descriptive Analyses, Models, & Forecasting Learning Objectives 1. Describe Time Series 2. Explain Descriptive Analyses 3. Define Time Series Components 4. Explain Forecasting 5. Describe Measures of Accuracy 6. Define Autocorrelation 7. Explain Durbin–Watson Test Time Series • Data generated by processes over time • Describe and predict output of processes • Descriptive analysis – Understanding patterns • Inferential analysis – Forecast future values Descriptive Analysis: Index Numbers Index Number • Measures change over time relative to a base period • Price Index measures changes in price – e.g. Consumer Price Index (CPI) • Quantity Index measures changes in quantity – e.g. Number of cell phones produced annually Simple Index Number Based on price/quantity of a single commodity Yt I t 100 Y0 where Yt = value at time t Y0 = value at time 0 (base period) Simple Index Number Example Year $ 1990 1.299 The table shows the price per 1991 1.098 1992 1.087 gallon of regular gasoline in the 1993 1.067 U.S for the years 1990 – 2006. 1994 1.075 1995 1.111 Use 1990 as the base year (prior 1996 1.224 1997 1.199 to the Gulf War). Calculate the 1998 1.03 simple index number for 1990, 1999 1.136 2000 1.484 1998, and 2006. 2001 1.42 2002 1.345 2003 1.561 2004 1.852 2005 2.27 2006 2.572 Simple Index Number Solution 1990 Index Number (base period) 1990price 1.299 100 100 100 1990price 1.299 1998 Index Number 1998price 1.03 100 100 79.3 1990price 1.299 Indicates price had dropped by 20.7% (100 – 79.3) between 1990 and 1998. Simple Index Number Solution 2006 Index Number 2006price 2.572 100 100 198 1990price 1.299 Indicates price had risen by 98% (100 – 198) between 1990 and 2006. Simple Index Numbers 1990–2006 Simple Index Numbers 1990–2006 Gasoline Price Simple Index 250.0 200.0 150.0 100.0 50.0 0.0 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 Composite Index Number Composite Index Number • Made up of two or more commodities • A simple index using the total price or total quantity of all the series (commodities) • Disadvantage: Quantity of each commodity purchased is not considered Composite Index Number Example The table on the next slide shows the closing stock prices on the last day of the month for Daimler–Chrysler, Ford, and GM between 2005 and 2006. Construct the simple composite index using January 2005 as the base period. (Source: Nasdaq.com) Simple Composite Index Solution First compute the total for the three stocks for each date. Simple Composite Index Solution Now compute the simple composite index by dividing each total by the January 2005 total. For example, December 2006: 12 / 06price 100 1/ 05price 99.64 100 95.49 104.3 Simple Composite Index Solution Simple Composite Index Solution Simple Composite Index Numbers 2005 – 2006 120.0 100.0 80.0 60.0 40.0 20.0 0.0 5 6 5 5 6 6 05 06 05 05 06 06 -0 -0 -0 -0 -0 -0 J- J- J- J- S- S- N N M M M M Weighted Composite Index Number Weighted Composite Price Index • Weights prices by quantities purchased before computing totals • Weighted totals used to compute composite index • Laspeyres Index – Uses base period quantities as weights • Paasche Index – Uses quantities from each period as weights Laspeyres Index • Uses base period quantities as weights – Appropriate when quantities remain approximately constant over time period • Example: Consumer Price Index (CPI) Calculating a Laspeyres Index weighted total for period t It 100 weighted total for base period k Q it0 Pit i 1 k 100 Q i 1 it0 Pit0 where Pit= price for each commodity at time t Qit= quantity of each commodity at time t t0 = base period Laspeyres Index Number Example The table shows the closing stock prices on 1/31/2005 and 12/29/2006 for Daimler– Chrysler, Ford, and GM. On 1/31/2005 an investor purchased the indicated number of shares of each stock. Construct the Laspeyres Index using 1/31/2005 as the base period. Daimler–Chrysler GM Ford Shares Purchased 100 500 200 1/31/2005 Price 45.51 13.17 36.81 12/29/2006 Price 61.41 7.51 30.72 Laspeyres Index Solution Weighted total for base period (1/31/2005): k Q i 1 it0 Pit0 100(45.51) 500(13.17) 200(36.81) 18498 Weighted total for 12/29/2006: k Q i 1 it0 Pit 100(61.41) 500(7.51) 200(30.72) 16040 Laspeyres Index Solution k Q P i ,1/ 31/ 05 i ,12 / 29 / 06 It i 1 k 100 Q i 1 P i ,1/ 31/ 05 i ,1/ 31/ 05 16040 100 18498 86.7 Indicates portfolio value had decreased by 13.3% (100–86.7) between 1/31/2005 and 12/29/2006. Paasche Index • Uses quantities for each period as weights – Appropriate when quantities change over time • Compare current prices to base period prices at current purchase levels • Disadvantages – Must know purchase quantities for each time period – Difficult to interpret a change in index when base period is not used Calculating a Paasche Index weighted total for period t It 100 weighted total for base period k Q P it it Weights are i 1 k 100 quantities for Q P i 1 it it0 time period t where Pit= price for each commodity at time t Qit= quantity of each commodity at time t t0 = base period Laspeyres Index Number Example The table shows the 1/31/2005 and 12/29/2006 prices and volumes in millions of shares for Daimler–Chrysler, Ford, and GM. Calculate the Paasche Index using 1/31/2005 as the base period. (Source: Nasdaq.com) Daimler–Chrysler Ford GM Price Volume Price Volume Price Volume 1/31/2005 45.51 .8 13.17 7.0 36.81 5.6 12/29/2006 61.41 .2 7.51 10.0 30.72 6.1 Paasche Index Solution k Q P i ,1/ 31/ 05 i ,1/ 31/ 05 I1/ 31/ 05 i 1 k 100 Q i 1 P i ,1/ 31/ 05 i ,1/ 31/ 05 .8(45.51) 7(13.17) 5.6(36.81) 100 .8(45.51) 7(13.17) 5.6(36.81) 100 Paasche Index Solution k Q P i12 / 29 / 06 i12 / 29 / 06 I12 / 29 / 06 i 1 k 100 Q i 1 P i12 / 29 / 06 i1/ 31/ 05 .2(61.41) 10(7.51) 6.1(30.72) 100 .2(45.51) 10(13.17) 6.1(36.81) 274.774 100 75.2 365.343 12/29/2006 prices represent a 24.8% (100 – 75.2) decrease from 1/31/2005 (assuming quantities were at 12/29/2006 levels for both periods) Exponential Smoothing Exponential Smoothing • Type of weighted average • Removes rapid fluctuations in time series (less sensitive to short–term changes in prices) • Allows overall trend to be identified • Used for forecasting future values • Exponential smoothing constant (w) affects “smoothness” of series Exponential Smoothing Constant Exponential smoothing constant, 0 < w < 1 • w close to 0 – More weight given to previous values of time series – Smoother series • w close to 1 – More weight given to current value of time series – Series looks similar to original (more variable) Calculating an Exponential Smoothed Series E1 = Y1 (same as original series) E2 = wY2 + (1 – w)E1 E3 = wY3 + (1 – w)E2 … Et = wYt + (1 – w)Et–1 Weight given Weight given to to current previous “smoothed” series value value Exponential Smoothing Example The closing stock prices on the last day of the month for Daimler– Chrysler in 2005 and 2006 are given in the table. Create an exponentially smoothed series using w = .2. Exponential Smoothing Solution E1 = 45.51 E2 = .2(46.10) + .8(45.51) = 45.63 E3 = .2(44.72) + .8(45.63) = 45.45 … E24 = .2(61.41) + .8(53.92) = 55.42 Exponential Smoothing Solution E1 = 45.51 E2 = .2(46.10) + .8(45.51) = 45.63 E3 = .2(44.72) + .8(45.63) = 45.45 … E24 = .2(61.41) + .8(53.92) = 55.42 0 10 20 30 40 50 60 70 Jan-05 Feb-05 Mar-05 Apr-05 May-05 Jun-05 Jul-05 Aug-05 Sep-05 Oct-05 Actual Series Nov-05 Dec-05 (w = .2) Jan-06 Feb-06 Mar-06 Solution Smoothed Series Apr-06 May-06 Jun-06 Jul-06 Aug-06 Sep-06 Exponential Smoothing Oct-06 Nov-06 Dec-06 Exponential Smoothing Thinking Challenge The closing stock prices on the last day of the month for Daimler– Chrysler in 2005 and 2006 are given in the table. Create an exponentially smoothed series using w = .8. Exponential Smoothing Solution* E1 = 45.51 E2 = .8(46.10) + .2(45.51) = 45.98 E3 = .8(44.72) + .2(45.98) = 44.97 … E24 = .8(61.41) + .2(57.75) = 60.08 Exponential Smoothing Solution* E1 = 45.51 E2 = .8(46.10) + .2(45.51) = 45.98 E3 = .8(44.72) + .2(45.98) = 44.97 … E24 = .8(61.41) + .2(57.75) = 60.08 0 10 20 30 40 50 60 70 Jan-05 Feb-05 Mar-05 Apr-05 May-05 Jun-05 Jul-05 Aug-05 Sep-05 Oct-05 Actual Series (w = .2) Nov-05 Dec-05 Jan-06 Feb-06 Smoothed Series Mar-06 Solution* Apr-06 May-06 Jun-06 Jul-06 (w = .8) Aug-06 Sep-06 Oct-06 Exponential Smoothing Nov-06 Smoothed Series Dec-06 Inferential Analysis Descriptive v. Inferential Analysis • Descriptive Analysis – Picture of the behavior of the time series – e.g. Index numbers, exponential smoothing – No measure of reliability • Inferential Analysis – Goal: Forecasting future values – Measure of reliability Time Series Components Additive Time Series Model Yt = Tt + Ct + St + Rt Tt = secular trend (describes long–term movements of Yt) Ct = cyclical effect (describes fluctuations about the secular trend attributable to business and economic conditions) St = seasonal effect (describes fluctuations that recur during specific time periods) Rt = residual effect (what remains after other components have been removed) Forecasting: Exponential Smoothing Exponentially Smoothed Forecasts • Assumes the trend and seasonal component are relatively insignificant • Exponentially smoothed forecast is constant for all future values • Ft+1 = Et Ft+2 = Ft+1 Ft+3 = Ft+1 • Use for short–term forecasting only Exponential Smoothing Forecasting Example The closing stock prices on the last day of the month for Daimler–Chrysler in 2005 and 2006 are given in the table along with the exponentially smoothed values using w = .2. Forecast the closing price for the January 31, 2007. Exponential Smoothing Forecasting Solution F1/31/2007 = E12/29/2006 = 55.42 The actual closing price on 1/31/2007 for Daimler–Chrysler was 62.49. Forecast Error = Y1/31/2007 – F1/31/2007 = 62.49 – 55.42 = 7.07 Forecasting Trends: The Holt–Winters Forecasting Model The Holt–Winters Forecasting Model • Accounts for trends in time series • Two components – Exponentially smoothed component, Et • Smoothing constant 0 < w < 1 – Trend component, Tt • Smoothing constant 0 < v < 1 – Close to 0: More weight to past trend – Close to 1: More weight to recent trend Calculating The Holt–Winters Model • Select values for the exponential smoothing constant w and the trend smoothing constant v • E2 = Y2 and T2 = Y2 – Y1 • E3 = wY3 + (1 – w)(E2 + T2) T3 = v(E3 – E2) + (1 – v)T2 … • Et = wYt + (1 – w)(Et–1 + Tt–1) Tt = v(Et – Et–1) + (1 – v)Tt–1 Holt–Winters Example The closing stock prices on the last day of the month for Daimler–Chrysler in 2005 and 2006 are given in the table. Calculate the Holt–Winters components using w = .8 and v = .7. Holt–Winters Solution w = .8 v = .7 E2 = Y2 and T2 = Y2 – Y1 E2 = 46.10 and T2 = 46.10 – 45.51 = .59 E3 = wY3 + (1 – w)(E2 + T2) E3 = .8(44.72) + .2(46.10 + .59) = 45.114 T3 = v(E3 – E2) + (1 – v)T2 T3 = .7(45.114 – 46.10) + .3(.59) = –.5132 Holt–Winters Solution Completed series: w = .8 v = .7 Holt–Winters Solution Holt–Winters exponentially smoothed (w = .8 and v = .7) 65 60 Smoothed 55 50 Price 45 40 35 Actual 30 5 6 5 6 5 M 6 05 06 Se 5 Se 6 05 06 -0 -0 -0 -0 -0 -0 l-0 l-0 n- n- p- p- ay ay ov ov ar ar Ju Ju Ja Ja M M M N N Date Holt–Winters Forecasting • One–step–ahead forecast – Ft+1 = Et + Tt • k–step–ahead forecasting – Ft+k = Et + kTt Holt–Winters Forecasting Example Use the Holt–Winters series to forecast the closing price of Daimler– Chrysler stock on 1/31/2007 and 2/28/2007. Holt–Winters Forecasting Solution* 1/31/2007 is one–step–ahead: F1/31/07 = E12/29/06 + T12/29/06 = 61.39 + 3.00 = 64.39 2/28/2007 is two–steps–ahead: F2/28/07 = E12/29/06 + 2T12/29/06 = 61.39 + 2(3.00) = 67.39 Holt–Winters Thinking Challenge The data shows the average undergraduate tuition at all 4–year institutions for the years 1996–2004 (Source: U.S. Dept. of Education). Calculate the Holt– Winters components using w = .7 and v = .5. Holt–Winters Solution* w = .7 v = .5 E2 = Y2 and T2 = Y2 – Y1 E2 = 9206 and T2 = 9206 – 8800 = 406 E3 = wY3 + (1 – w)(E2 + T2) E3 = .7(9588) + .3(9206 + 406) = 9595.20 T3 = v(E3 – E2) + (1 – v)T2 T3 = .5(9595.20 – 9206) + .5(406) = 397.60 Holt–Winters Solution* Completed series w=.7 v=.5 Year t Tuition Et Tt 1995 1 $8,800 1996 2 $9,206 9206.0000 406.0000 1997 3 $9,588 9595.2000 397.6000 1998 4 $10,076 10051.0400 426.7200 1999 5 $10,444 10454.1280 414.9040 2000 6 $10,818 10833.3096 397.0428 2001 7 $11,380 11335.1057 449.4195 2002 8 $12,014 11945.1576 529.7356 2003 9 $12,955 12810.9680 697.7730 2004 10 $13,743 13672.7223 779.7637 Holt–Winters Solution* Holt–Winters exponentially smoothed (w = .7 and v = .5) $15,000 $14,000 $13,000 Tuition $12,000 $11,000 $10,000 Actual Smoothed $9,000 $8,000 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 Year Holt–Winters Forecasting Thinking Challenge Use the Holt–Winters series to forecast tuition in 2005 and 2006 w=.7 v=.5 Year t Tuition Et Tt 1995 1 $8,800 1996 2 $9,206 9206.0000 406.0000 1997 3 $9,588 9595.2000 397.6000 1998 4 $10,076 10051.0400 426.7200 1999 5 $10,444 10454.1280 414.9040 2000 6 $10,818 10833.3096 397.0428 2001 7 $11,380 11335.1057 449.4195 2002 8 $12,014 11945.1576 529.7356 2003 9 $12,955 12810.9680 697.7730 2004 10 $13,743 13672.7223 779.7637 Holt–Winters Forecasting Solution* 2005 is one–step–ahead: F11 = E10 + T10 13672.72 + 779.76 = $14,452.48 2006 is 2–steps–ahead: F12 = E10 + 2T10 =13672.72 +2(779.76) = $15,232.24 Measuring Forecast Accuracy Mean Absolute Deviation • Mean absolute difference between the forecast and actual values of the time series m Y F t t MAD t 1 m • where m = number of forecasts used Mean Absolute Percentage Error • Mean of the absolute percentage of the difference between the forecast and actual values of the time series m Yt Ft t 1 Yt MAPE 100 m • where m = number of forecasts used Root Mean Squared Error • Square root of the mean squared difference between the forecast and actual values of the time series m Y F 2 t t RMSE t 1 m • where m = number of forecasts used Forecasting Accuracy Example Using the Daimler–Chrysler data from 1/31/2005 through 8/31/2006, three time series models were constructed and forecasts made for the next four months. • Model I: Exponential smoothing (w = .2) • Model II: Exponential smoothing (w = .8) • Model III: Holt–Winters (w = .8, v = .7) Forecasting Accuracy Example Model I 2.31 4.66 6.01 9.14 MADI 5.53 4 2.31 4.66 6.01 9.14 49.96 56.93 58.28 61.41 MAPEI 100 9.50 4 2.31 4.66 6.01 9.14 2 2 2 2 RMSEI 6.06 4 Forecasting Accuracy Example Model II 2.82 4.15 5.50 8.63 MADII 5.28 4 2.82 4.15 5.50 8.63 49.96 56.93 58.28 61.41 MAPEII 100 9.11 4 2.82 4.15 5.50 8.63 2 2 2 2 RMSEII 5.70 4 Forecasting Accuracy Example Model III 3.45 2.42 2.67 4.71 MADIII 3.31 4 3.45 2.42 2.67 4.71 49.96 56.93 58.28 61.41 MAPEIII 100 5.85 4 3.45 2.42 2.67 4.71 2 2 2 2 RMSEIII 3.44 4 Forecasting Trends: Simple Linear Regression Simple Linear Regression • Model: E(Yt) = β0 + β1t • Relates time series, Yt, to time, t • Cautions – Risky to extrapolate (forecast beyond observed data) – Does not account for cyclical effects Simple Linear Regression Example The data shows the average undergraduate tuition at all 4– year institutions for the years 1996–2004 (Source: U.S. Dept. of Education). Use least– squares regression to fit a linear model. Forecast the tuition for 2005 (t = 11) and compute a 95% prediction interval for the forecast. Simple Linear Regression Solution From Excel ˆ Yt 7997.533 528.158t Simple Linear Regression Solution $15,000 $14,000 ˆ Yt 7997.533 528.158t $13,000 Tuition $12,000 $11,000 $10,000 $9,000 $8,000 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 Year Simple Linear Regression Solution Forecast tuition for 2005 (t = 11): ˆ Y11 7997.533 528.158(11) 13807.27 95% prediction interval: 1 t p t 2 y t / 2 s 1 ˆ n SStt 1 11 5.5 2 13807.27 2.306 286.84 1 10 82.5 13006.21 y11 14608.33 Seasonal Regression Models Seasonal Regression Models • Takes into account secular trend and seasonal effects (seasonal component) • Uses multiple regression models • Dummy variables to model seasonal component • E(Yt) = β0 + β1t + β2Q1 + β3Q2 + β4Q3 where 1 if quarter i Qi 0 if not quarter i Autocorrelation and The Durbin–Watson Test Autocorrelation • Time series data may have errors that are not independent ˆ ˆ • Time series residuals: Rt Yt Yt • Correlation between residuals at different points in time (autocorrelation) • 1st order correlation: Correlation between neighboring residuals (times t and t + 1) Autocorrelation Plot of residuals v. time for tuition data shows residuals tend to group alternately into positive and negative clusters Residual v Time Plot 600 400 Residuals 200 0 0 2 4 6 8 10 12 -200 -400 t Durbin–Watson Test • Ho: No first–order autocorrelation of residuals • Ha: Positive first–order autocorrelation of residuals • Test Statistic n 2 ˆ ˆ Rt Rt 1 d t 2 n Rt ˆ2 t 1 Interpretation of d Statistic • 0≤d≤4 • If residuals uncorrelated, then d ≈ 2 • If residuals positively autocorrelated, then d<2 • If residuals negatively autocorrelated, then d >2 Rejection Region for the Durbin– Watson d Test Rejection region: evidence of positive autocorrelation d 0 1 dL dU 2 3 4 Possibly significant Nonrejection region: autocorrelation insufficient evidence of positive autocorrelation Durbin–Watson Test Example Use the Durbin–Watson test to test for the presence of autocorrelation in the tuition data. Use α = .05. Durbin–Watson Test Solution • H0: No 1st–order Test Statistic: autocorrelation • Ha: Positive 1st–order autocorrelation • .05 n = 10 k = 1 Decision: • Critical Value(s): Conclusion: d 0 2 4 .88 1.32 Durbin–Watson Solution Test Statistic n 2 ˆ ˆ Rt Rt 1 d t 2 n ˆ Rt 2 t 1 (152.1515 274.3091) 2 (5.9939 152.1515) 2 ... (463.8909 204.0485) 2 (274.3091) 2 (152.1515) 2 ... (463.8909) 2 .51 Durbin–Watson Test Solution • H0: No 1st–order Test Statistic: autocorrelation d = .51 • Ha: Positive 1st–order autocorrelation • .05 n = 10 k = 1 Decision: • Critical Value(s): Reject at = .05 Conclusion: d There is evidence of 0 2 4 positive autocorrelation .88 1.32 Conclusion 1. Described Time Series 2. Explained Descriptive Analyses 3. Defined Time Series Components 4. Explained Forecasting 5. Described Measures of Accuracy 6. Defined Autocorrelation 7. Explained Durbin–Watson Test

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