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• Index number for base year will always equal 100
• Index number for base year will always equal 100
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• Model III (Holt–Winters w=.8 and v=.7) has the lowest MAD, MAPE, and RMSE of all three models, thus it yields the most accurate predictions.
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• Msb11e ppt ch13

1. 1. © 2011 Pearson Education, Inc
2. 2. Statistics for Business and Economics Chapter 13 Time Series: Descriptive Analyses, Models, & Forecasting © 2011 Pearson Education, Inc
3. 3. 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 © 2011 Pearson Education, Inc
4. 4. Content 13.7 Forecasting Trends: Simple Linear Regression 13.8 Seasonal Regression Models 13.9 Autocorrelation and the Durbin-Watson Test © 2011 Pearson Education, Inc
5. 5. 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. © 2011 Pearson Education, Inc
6. 6. Time Series • Data generated by processes over time • Describe and predict output of processes • Descriptive analysis – • Understanding patterns Inferential analysis – Forecast future values © 2011 Pearson Education, Inc
7. 7. 13.1 Descriptive Analysis: Index Numbers © 2011 Pearson Education, Inc
8. 8. 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 © 2011 Pearson Education, Inc
9. 9. 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  © 2011 Pearson Education, Inc
10. 10. Steps for Calculating 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. © 2011 Pearson Education, Inc
11. 11. 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
12. 12. 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. © 2011 Pearson Education, Inc
13. 13. 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. © 2011 Pearson Education, Inc
14. 14. Simple Index Numbers 1990–2006 © 2011 Pearson Education, Inc
15. 15. 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 © 2011 Pearson Education, Inc
16. 16. 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 © 2011 Pearson Education, Inc
17. 17. 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) © 2011 Pearson Education, Inc
18. 18. Simple Composite Index Solution First compute the total for the three stocks for each date. © 2011 Pearson Education, Inc
19. 19. 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 © 2011 Pearson Education, Inc
20. 20. Simple Composite Index Solution © 2011 Pearson Education, Inc
21. 21. Simple Composite Index 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 © 2011 Pearson Education, Inc N-06
22. 22. 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. © 2011 Pearson Education, Inc
23. 23. Laspeyres Index • Uses base period quantities as weights – Appropriate when quantities remain approximately constant over time period • Example: Consumer Price Index (CPI) © 2011 Pearson Education, Inc
24. 24. 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 © 2011 Pearson Education, Inc i=1
25. 25. Steps for Calculating a Laspeyres Index 5. Calculate the Laspeyres index, It, at time t by taking the ratio of the weighted total at time t to the base period weighted total and multiplying by 100–that is, κ It = ∑Θ ιτ0 Π ιτ ∑Θ ιτ0 Π ιτ ι=1 κ ι=1 × 100 0 © 2011 Pearson Education, Inc
26. 26. 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 © 2011 Pearson Education, Inc
27. 27. 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 © 2011 Pearson Education, Inc
28. 28. 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 © 2011 Pearson Education, Inc 12/29/2006.
29. 29. 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 © 2011 Pearson Education, Inc
30. 30. Steps for Calculating a Paasche Index 1. Collect price information for each of the k price series to be used in the composite index. Denote these series by P1t, P2t, …, Pkt . 2. Select a base period. Call this time period t0. 3. Collect purchase quantity information for the base period. Denote the k quantities by Q1t , Q2t ,K ,Qkt . 0 0 0 © 2011 Pearson Education, Inc
31. 31. Steps for Calculating a Paasche Index 4. Calculate the Paasche index for time t by multiplying the ratio of the weighted total at time t to the weighted total at time t0 (base period) by 100, where the weights used are the purchase quantities for time period t. κ Thus, ∑ Θιτ Π ιτ =1 I t = ικ × 100 ∑ Θιτ Π ιτ ι=1 0 © 2011 Pearson Education, Inc
32. 32. 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 © 2011 Pearson Education, Inc
33. 33. 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 © 2011 Pearson Education, Inc
34. 34. 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) ©
35. 35. 13.2 Descriptive Analysis: Exponential Smoothing © 2011 Pearson Education, Inc
36. 36. 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
37. 37. 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
38. 38. 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
39. 39. 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
40. 40. 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
41. 41. 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
42. 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 © 2011 Pearson Education, Inc
43. 43. Exponential Smoothing 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 © 2011 Pearson Education, Inc
44. 44. 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
45. 45. 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
46. 46. Exponential Smoothing 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 © 2011 Pearson Education, Inc
47. 47. 13.3 Time Series Components © 2011 Pearson Education, Inc
48. 48. 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
49. 49. 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
50. 50. 13.4 Forecasting: Exponential Smoothing © 2011 Pearson Education, Inc
51. 51. 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
52. 52. 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 © 2011 Pearson Education, Inc
53. 53. 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 © 2011 Pearson Education, Inc
54. 54. 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
55. 55. 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
56. 56. 13.5 Forecasting Trends: Holt’s Method © 2011 Pearson Education, Inc
57. 57. 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
58. 58. 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
59. 59. 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
60. 60. 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 = 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
61. 61. 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
62. 62. 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
63. 63. Holt Solution Completed series: w = .8 v = .7 © 2011 Pearson Education, Inc
64. 64. 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 © 2011 Pearson Education, Inc
65. 65. 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
66. 66. 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
67. 67. 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
68. 68. 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
69. 69. 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
70. 70. Holt Solution Completed series © 2011 Pearson Education, Inc
71. 71. 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 © 2011 Pearson Education, Inc 2003 2004
72. 72. Holt Forecasting Thinking Challenge Use the Holt–Winters series to forecast tuition in 2005 and 2006 © 2011 Pearson Education, Inc
73. 73. 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
74. 74. 13.6 Measuring Forecast Accuracy: MAD and RMSE © 2011 Pearson Education, Inc
75. 75. 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 © 2011 Pearson Education, Inc
76. 76. 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 • where m = number of forecasts used © 2011 Pearson Education, Inc
77. 77. 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 • where m = number of forecasts used © 2011 Pearson Education, Inc
78. 78. 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
79. 79. 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
80. 80. 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 2 © 2011 Pearson Education, Inc 2 = 5.70
81. 81. 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
82. 82. 13.7 Forecasting Trends: Simple Linear Regression © 2011 Pearson Education, Inc
83. 83. 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
84. 84. 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
85. 85. Simple Linear Regression Solution From Excel ˆ Yt = 7997.533 + 528.158t © 2011 Pearson Education, Inc
86. 86. Simple Linear Regression 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 © 2011 Pearson Education, Inc 2004 2005
87. 87. 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 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
88. 88. 13.8 Seasonal Regression Models © 2011 Pearson Education, Inc
89. 89. 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 © 2011 Pearson Education, Inc
90. 90. 13.9 Autocorrelation and the Durbin-Watson Test © 2011 Pearson Education, Inc
91. 91. 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) © 2011 Pearson Education, Inc
92. 92. 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 © 2011 Pearson Education, Inc 12
93. 93. 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 © 2011 Pearson Education, Inc
94. 94. 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
95. 95. 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
96. 96. 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
97. 97. Durbin–Watson d-Test for 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 © 2011 Pearson Education, Inc
98. 98. 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= ˆ ˆ Rt − Rt −1 t =2 ) 2 n ˆ2 ∑ Rt © 2011tPearson Education, Inc =1
99. 99. Durbin–Watson d-Test for 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 © 2011 Pearson Education, Inc
100. 100. Requirements for the Validity of the d-Test The residuals are normally distributed. © 2011 Pearson Education, Inc
101. 101. 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
102. 102. 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 © 2011 Pearson Education, Inc
103. 103. 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 © 2011 Pearson Education, Inc
104. 104. 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 © 2011 Pearson Education, Inc
105. 105. Key Ideas Time Series Data Data generated by processes over time. © 2011 Pearson Education, Inc
106. 106. 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
107. 107. Key Ideas Time Series Components 1. 2. 3. 4. Secular (long-term) trend Cyclical effect Seasonal effect Residual effect © 2011 Pearson Education, Inc
108. 108. Key Ideas Time Series Forecasting Descriptive methods of forecasting with smoothing: 1. Exponential smoothing 2. Holt’s method © 2011 Pearson Education, Inc
109. 109. Key Ideas Time Series Forecasting An Inferential forecasting method: least squares regression © 2011 Pearson Education, Inc
110. 110. 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
111. 111. 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
112. 112. 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