Hw Answer Operation Management Heizer Ch 4

4. 9 [pic] (c)? MAD (two-month moving average) = . 075 MAD (three-month moving average) = . 088 |Table for Problem 4. (a, b, c) | | | | |Forecast | ||Error| | | | |Two-Month |Three-Month |Two-Month |Three-Month | | |Price per |Moving |Moving |Moving |Moving | |Month |Chip |Average |Average |Average |Average | |January |$1. 0 | | | | | |February |? 1. 67 | | | | | |March |? 1. 70 |1. 735 | |. 035 | | |April |? 1. 85 |1. 685 |1. 723 |. 165 |. 127 | |May |? 1. 90 |1. 775 |1. 740 |. 125 |. 160 | |June |? 1. 87 |1. 75 |1. 817 |. 005 |. 053 | |July |? 1. 80 |1. 885 |1. 873 |. 085 |. 073 | |August |? 1. 83 |1. 835 |1. 857 |. 005 |. 027 | |September |? 1. 70 |1. 815 |1. 833 |. 115 |. 133 | |October |? 1. 65 |1. 765 |1. 777 |. 115 |. 127 | |November |? 1. 70 |1. 675 |1. 27 |. 025 |. 027 | |December |? 1. 75 |1. 675 |1. 683 |. 075 |. 067 | | | | |Totals |. 750| |. 793| | | | | | | | | | | |4. 9 |(d)? Table for Problem 4. 9(d): | | | | |( = . 1 | | |( = . 3 | | |( = . | |Month |Price per Chip |Forecast ||Error| |Forecast ||Error| |Forecast ||Error| | |January |$1. 80 |$1. 80 |$. 00 |$1. 80 |$. 00 |$1. 80 |$. 00 | |February |1. 67 |1. 80 |. 13 |? 1. 80 | . 13 |? 1. 80 |. 13 | |March |1. 70 |1. 79 |. 09 |? 1. 76 | . 06 |? 1. 74 |. 04 | |April |1. 85 |1. 78 |. 07 |? 1. 74 | . 11 |? 1. 72 |. 13 | |May |1. 0 |1. 79 |. 11 |? 1. 77 | . 13 |? 1. 78 |. 12 | |June |1. 87 |1. 80 |. 07 |? 1. 81 | . 06 |? 1. 84 |. 03 | |July |1. 80 |1. 80 |. 00 |? 1. 83 | . 03 |? 1. 86 |. 06 | |August |1. 83 |1. 80 |. 03 |? 1. 82 | . 01 |? 1. 83 |. 00 | |September |1. 70 |1. 81 |. 11 |? 1. 82 | . 12 |? 1. 83 |. 13 | |October |1. 65 |1. 80 |. 5 |? 1. 79 | . 14 |? 1. 76 |. 11 | |November |1. 70 |1. 78 |. 08 |? 1. 75 | . 05 |? 1. 71 |. 01 | |December |1. 75 |1. 77 |. 02 |? 1. 73 | . 02 |? 1. 70 |. 05 | | | | | 4. 41? (a)? It appears from the following graph that the points do scatter around a straight line. [pic] (b)? Developing the regression relationship, we have: (Summer |Tourists |Ridership | | | | |months) |(Millions) |(1,000,000s) | | | | |Year |(X) |(Y) |X2 |Y2 |XY | |? 1 |? 7 |1. 5 |? 49 |? 2. 25 |10. 5 | |? 2 |? 2 |1. 0 | 4 |? 1. 00 |? 2. 0 | |? 3 |? 6 |1. 3 |? 36 |? 1. 69 |? 7. 8 | |? 4 |? 4 |1. 5 |? 16 |? 2. 25 |? 6. 0 | |? 5 |14 |2. 5 |196 |? 6. 25 |35. 0 | |? |15 |2. 7 |225 |? 7. 29 |40. 5 | |? 7 |16 |2. 4 |256 |? 5. 76 |38. 4 | |? 8 |12 |2. 0 |144 |? 4. 00 |24. 0 | |? 9 |14 |2. 7 |196 |? 7. 29 |37. 8 | |10 |20 |4. 4 |400 |19. 36 |88. 0 | |11 |15 |3. 4 |225 |11. 56 |51. 0 | |12 |? 7 |1. 7 |? 49 |? 2. 89 |11. 9 | and (X = 132, (Y = 27. 1, (XY = 352. 9, (X2 = 1796, (Y2 = 71. 59, [pic] = 11, [pic]= 2. 6. Then: [pic] andY = 0. 511 + 0. 159X (c)? Given a tourist population of 10,000,000, the model predicts a ridership of: Y = 0. 511 + 0. 159 ( 10 = 2. 101, or 2,101,000 persons. (d)? If there are no tourists at all, the model predicts a ridership of 0. 511, or 511,000 persons. One would not place much confidence in this forecast, however, because the number of tourists (ze ro) is outside the range of data used to develop the model. (e)? The standard error of the estimate is given by: (f)? The correlation coefficient and the coefficient of determination are given by: [pic] ———————– [pic][pic] Hw Answer Operation Management Heizer Ch 4 4. 9 [pic] (c)? MAD (two-month moving average) = . 075 MAD (three-month moving average) = . 088 |Table for Problem 4. (a, b, c) | | | | |Forecast | ||Error| | | | |Two-Month |Three-Month |Two-Month |Three-Month | | |Price per |Moving |Moving |Moving |Moving | |Month |Chip |Average |Average |Average |Average | |January |$1. 0 | | | | | |February |? 1. 67 | | | | | |March |? 1. 70 |1. 735 | |. 035 | | |April |? 1. 85 |1. 685 |1. 723 |. 165 |. 127 | |May |? 1. 90 |1. 775 |1. 740 |. 125 |. 160 | |June |? 1. 87 |1. 75 |1. 817 |. 005 |. 053 | |July |? 1. 80 |1. 885 |1. 873 |. 085 |. 073 | |August |? 1. 83 |1. 835 |1. 857 |. 005 |. 027 | |September |? 1. 70 |1. 815 |1. 833 |. 115 |. 133 | |October |? 1. 65 |1. 765 |1. 777 |. 115 |. 127 | |November |? 1. 70 |1. 675 |1. 27 |. 025 |. 027 | |December |? 1. 75 |1. 675 |1. 683 |. 075 |. 067 | | | | |Totals |. 750| |. 793| | | | | | | | | | | |4. 9 |(d)? Table for Problem 4. 9(d): | | | | |( = . 1 | | |( = . 3 | | |( = . | |Month |Price per Chip |Forecast ||Error| |Forecast ||Error| |Forecast ||Error| | |January |$1. 80 |$1. 80 |$. 00 |$1. 80 |$. 00 |$1. 80 |$. 00 | |February |1. 67 |1. 80 |. 13 |? 1. 80 | . 13 |? 1. 80 |. 13 | |March |1. 70 |1. 79 |. 09 |? 1. 76 | . 06 |? 1. 74 |. 04 | |April |1. 85 |1. 78 |. 07 |? 1. 74 | . 11 |? 1. 72 |. 13 | |May |1. 0 |1. 79 |. 11 |? 1. 77 | . 13 |? 1. 78 |. 12 | |June |1. 87 |1. 80 |. 07 |? 1. 81 | . 06 |? 1. 84 |. 03 | |July |1. 80 |1. 80 |. 00 |? 1. 83 | . 03 |? 1. 86 |. 06 | |August |1. 83 |1. 80 |. 03 |? 1. 82 | . 01 |? 1. 83 |. 00 | |September |1. 70 |1. 81 |. 11 |? 1. 82 | . 12 |? 1. 83 |. 13 | |October |1. 65 |1. 80 |. 5 |? 1. 79 | . 14 |? 1. 76 |. 11 | |November |1. 70 |1. 78 |. 08 |? 1. 75 | . 05 |? 1. 71 |. 01 | |December |1. 75 |1. 77 |. 02 |? 1. 73 | . 02 |? 1. 70 |. 05 | | | | | 4. 41? (a)? It appears from the following graph that the points do scatter around a straight line. [pic] (b)? Developing the regression relationship, we have: (Summer |Tourists |Ridership | | | | |months) |(Millions) |(1,000,000s) | | | | |Year |(X) |(Y) |X2 |Y2 |XY | |? 1 |? 7 |1. 5 |? 49 |? 2. 25 |10. 5 | |? 2 |? 2 |1. 0 | 4 |? 1. 00 |? 2. 0 | |? 3 |? 6 |1. 3 |? 36 |? 1. 69 |? 7. 8 | |? 4 |? 4 |1. 5 |? 16 |? 2. 25 |? 6. 0 | |? 5 |14 |2. 5 |196 |? 6. 25 |35. 0 | |? |15 |2. 7 |225 |? 7. 29 |40. 5 | |? 7 |16 |2. 4 |256 |? 5. 76 |38. 4 | |? 8 |12 |2. 0 |144 |? 4. 00 |24. 0 | |? 9 |14 |2. 7 |196 |? 7. 29 |37. 8 | |10 |20 |4. 4 |400 |19. 36 |88. 0 | |11 |15 |3. 4 |225 |11. 56 |51. 0 | |12 |? 7 |1. 7 |? 49 |? 2. 89 |11. 9 | and (X = 132, (Y = 27. 1, (XY = 352. 9, (X2 = 1796, (Y2 = 71. 59, [pic] = 11, [pic]= 2. 6. Then: [pic] andY = 0. 511 + 0. 159X (c)? Given a tourist population of 10,000,000, the model predicts a ridership of: Y = 0. 511 + 0. 159 ( 10 = 2. 101, or 2,101,000 persons. (d)? If there are no tourists at all, the model predicts a ridership of 0. 511, or 511,000 persons. One would not place much confidence in this forecast, however, because the number of tourists (ze ro) is outside the range of data used to develop the model. (e)? The standard error of the estimate is given by: (f)? The correlation coefficient and the coefficient of determination are given by: [pic] ———————– [pic][pic]