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Chapter 16 time series and forecasting

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• http://www.lmvp.org/boxplots.htm Example 1 is what is known as an even distribution (Figure 9). The individual values are; 2 3 4 5 6 7 8 and 9. The average of these values is 5.5 which also happens to be the median value (remember when dealing with an even number of values the median equals the average of the two middle values, in this case 5 and 6). The minimum and maximum values are 2 and 9 respectively, giving us a range of 7. This is considered an even distribution because the mean and median are located in the middle of the range. In other words the minimum and maximum value are equal distances from the median value. The box plot for this data has the median line in the middle of the box and the minimum and maximum lines extend equal distances from the box.   Example 2 contains the following values; 2 3 4 5 6 7 10 and 13 (Figure 9). The average equals 6.25 and the median is 5.5. Minimum and maximum values are 2 and 13, giving us a range of 11. Having an average that is larger than the median suggests that the data is not balanced around the median. Review of the data shows that the two high values (10 and 13) deviate from the median more than the two low values (2 and 3). This leads to data referred to as skewed. When we compare the box plot for Example 2 to the box plot for Example 1 we see two indications that the data was skewed: 1) the vertical line identifying the maximum is longer than the minimum line, and 2) the box extends higher above the median line than it does below.       In Example 3 the values are 2 3 4 4.25 4.75 7 8 and 9 (Figure 9). The average is 5.25 and the median is 4.5. The range is 7 with minimum and maximum values of 2 and 9 respectively. We again have skewed data as the median and average are not equal. This time instead of extreme high values, the skewing is caused by the values of 4, 4.25 and 4.75 being clumped close together. Comparison of this box plot to Example 1 shows that the plots are the same with the exception of the median line. In Example 3 we are tipped off to the skewness in the data by the fact that the area of box above the median line is larger than the area below.   Example 4 consist of the following values; 2 3 3.25 3.50 3.75 4 5 and 9 (Figure 9). The average for these data equals 4.19 and the median is 3.375. The minimum, maximum and range are the same as seen in Examples 1 and 3. Again we have skewed data caused by a clumping of low values (similar to Example 3). This time the number of values clumped together is greater, causing the box to be smaller. Clues that the data are skewed are that the median line is not in the center of the box and the maximum line extends farther away from the box than the minimum line. For more about Box Plots, try this page: http://www.lmvp.org/introduction/understanding.htm
• More explanation: http://www.ncsu.edu/labwrite/res/gt/gt-reg-home.html
• More explanation: http://www.ncsu.edu/labwrite/res/gt/gt-reg-home.html
• More explanation: http://www.ncsu.edu/labwrite/res/gt/gt-reg-home.html
• 006

1. 1. IBS Statistics Year 1 Dr. Ning DING
2. 2. Table of content <ul><li>Review </li></ul><ul><li>Learning Goals </li></ul><ul><li>Chapter 16: Time Series and Forecasting </li></ul><ul><li>Exercises </li></ul>
3. 3. Chapter 3: Describing Data Review Chapter 3 Why Dispersion? Central Tendency?
4. 4. Chapter 3: Describing Data Review Chapter 3 Dispersion Range Variation Standard Deviation
5. 5. Chapter 3: Describing Data Review Chapter 3 Dispersion
6. 6. Chapter 4: Describing Data P42 Example Ch2 The distribution is skewed to __________ because the mean is __________the median. the right larger than Mean =23.06 Review Chapter 4 Interquartile Range
7. 7. Review Chapter 4 Chapter 4: Describing Data 2 3 4 5 6 7 8 9 2 3 4 5 6 7 10 13 2 3 4 4.25 4.75 7 8 9 2 3 3.25 3.50 3.75 4 5 9 Mean= 5.5 6.25 5.25 4.19 Median= 5.5 5.5 4.5 3.38
8. 8. Review Chapter 4 Chapter 4: Describing Data Mean= 5.5 6.25 5.25 4.19 Median= 5.5 5.5 4.5 3.38 Most skewed? http://qudata.com/online/statcalc/
9. 9. Chapter 12: Sim Reg & Corr Sample Exam P.4 Ŷ = a + b X a = -1.8181 Review Chapter 12
10. 10. Chapter 12: Sim Reg & Corr Review Chapter 12 Negative Correlation Positive Correlation
11. 11. Exercise Chapter 12: Sim Reg & Corr Sample Exam P.4 Ŷ = -1.8182 + 0.1329X a = -1.8181
12. 12. Applicable when time series follows fairly linear trend that have definite rhythmic pattern Chapter 16: Time Series & Forecasting Review Chapter 16
13. 13. 1+2+3+4+5+4+3= 22 / 7 = 3.143 Seven-Year Moving Total Moving Average 2+3+4+5+4+3+2= 23 / 7 = 3.286 3+4+5+4+3+2+3= 24 / 7 = 3.429
14. 14. Learning Goals <ul><li>Chapter 16: </li></ul><ul><ul><li>Define the components of a time series </li></ul></ul><ul><ul><li>Compute a moving average </li></ul></ul><ul><ul><li>Determine a linear trend equation </li></ul></ul><ul><ul><li>Compute a trend equation for a nonlinear trend </li></ul></ul><ul><ul><li>Use a trend equation to forecast future time periods and to develop seasonally adjusted forecasts </li></ul></ul><ul><ul><li>Determine and interpret a set of seasonal indexes </li></ul></ul><ul><ul><li>Desearsonalize data using a seasonal index </li></ul></ul><ul><ul><li>Test for autocorrelation </li></ul></ul>
15. 15. Exercise Chapter 16: Time Series & Forecasting Ŷ = a + b t = 1.73 a = 22.67 -1.73*4 = 15.75 P152 N6 Ch16 a = Y - b X
16. 16. 3. Linear Trend Chapter 16: Time Series & Forecasting Ŷ = a + b t = 1.73 Ŷ = 22.67 + 1.73 t a = 22.67 = 22.67 a = Y - b X
17. 17. 3. Linear Trend Chapter 16: Time Series & Forecasting Ŷ = a + b t Odd-numbered Even-numbered a = Y - b X
18. 18. 4. Seasonal Variation Understanding seasonal fluctuations help plan for sufficient goods and materials on hand to meet varying seasonal demand Chapter 16: Time Series & Forecasting
19. 19. 4. Seasonal Variation Chapter 16: Time Series & Forecasting Seasonal variations are fluctuations that coincide with certain seasons and are repeated year after year
20. 20. 4. Seasonal Variation Chapter 16: Time Series & Forecasting Seasonal Index: A number, usually expressed in percent, that expresses the relative value of a season with respect to the average for the year (100%) Sales for December are 26.8% above an average month. Sales for July are 14% below an average month.
21. 21. 4. Seasonal Variation Chapter 16: Time Series & Forecasting 2005 2006 2007 2008 2009 2010 Sales Report: in \$ millions
22. 22. 4. Seasonal Variation Chapter 16: Time Series & Forecasting Step 1: Re-organize the data 2005 2006 2007 2008 2009 2010
23. 23. 6.7+4.6+10.0+12.7=34 /4=8.50 4.6+10.0+12.7+6.5=33.8 /4=8.45 Step 2: Moving Average
24. 24. Step 3: Centered Moving Average
25. 25. Step 4: Specific Seasonal Index
26. 26. 10/8.475=1.180 12.7/8.45=1.503 6.5/8.425=0.772 Step 4: Specific Seasonal Index
27. 27. + + + = Step 5: Typical Quarterly Index 2005 2006 2007 2008 2009 2010 *(0.9978) *(0.9978) *(0.9978) *(0.9978)
28. 28. Step 6: Interpret Sales for the Fall are 51.9% above the typical quarter. Sales for the Winter are 23.5% below the typical quarter. 2005 2006 2007 2008 2009 2010
29. 29. Appliance Center sells a variety of electronic equipment and home appliances. For the last four years the following quarterly sales (in \$ millions) were reported. Determine a typical seasonal index for each of the four quarters. Exercise Chapter 16: Time Series & Forecasting P161 No.10 Ch16
30. 30. Exercise Chapter 16: Time Series & Forecasting P161 No.10 Ch16 Step 1: Reorganize the data Step 2: Moving Average Step 3: Centered Moving Average Step 4: Specific Seasonal Index
31. 31. Exercise Chapter 16: Time Series & Forecasting P161 No.10 Ch16 Step 5: Reorganize the data Step 6: Calculate the mean for each quarter Step 7: Sum up the four means Step 8: Divide 4 by Total of four means to get Correction Factor Step 9: Mean * Correction Factor
32. 32. 5. Deseasonalizing Data Chapter 16: Time Series & Forecasting To remove the seasonal fluctuations so that the trend and cycle can be studied. Ŷ = a + b X Ŷ = a + b t
33. 33. 5. Deseasonalizing Data Chapter 16: Time Series & Forecasting 76.5 57.5 114.1 151.9 / 0.765 = 8.759 / 0.575 = 8.004 / 1.141 = 8.761 / 1.519 = 8.361 / 1.519 / 0.765 / 0.575 / 1.141 / 0.765 / 0.575 / 1.141 / 1.519 = 8.498 = 8.004 = 8.586 = 8.953 = 9.021 = 8.700 = 9.112 = 9.283
34. 34. Ŷ = a + b t Ŷ = 8.1096 + 0.0899 t Sale increased at a rate of 0.0899 (\$ millions) per quarter. Ŷ = 8.1096 + 0.0899 * 25 = 10.3571 \$ millions 10.3571*0.765 = 7.9232 \$ millions Chapter 16: Time Series & Forecasting 76.5 57.5 114.1 151.9
35. 35. Exercise Chapter 16: Time Series & Forecasting 1. Calculate the seasonal indices for each quarter, express them as a ratio and not as a %. You may round to 4 dec. places. 2. Interpret the seasonal index quarter II. 3. Deseasonalized the original revenue for 2008 quarter I. 4. For 2011 quarter II the forecasted revenue from the trend line was 55. Calculate the seasonalized revenue for 2011 quarter II. Friday Oct 22, 2010 Pigeon hole Ning Ding
36. 36. Summary <ul><li>Chapter 16: </li></ul><ul><ul><li>A seasonal factor can be estimated using the ratio-to-moving-average method. </li></ul></ul><ul><ul><li>The six-step procedure yields a seasonal index for each period. </li></ul></ul><ul><ul><li>The seasonal factor is used to adjust forecasts, taking into account the effects of the season. </li></ul></ul>Chapter 16: Time Series & Forecasting
37. 37. Step 1: Reorganize the data Step 2: Moving Average Step 3: Centered Moving Average Step 4: Specific Seasonal Index Step 5: Reorganize the data Step 6: Calculate the mean for each quarter Step 7: Sum up the four means Step 8: Divide 4 by Total of four means to get Correction Factor Step 9: Mean * Correction Factor Hint