Hw Answer Operation Management Heizer Ch 4

4. 9 pic (c)? dotty (two-month moving average) = . 075 MAD (three-month moving average) = . 088 Table for Problem 4. (a, b, c) herald fracture Two- month Three- month Two-Month Three-Month determine per Moving Moving Moving Moving Month checkout Average Average Average Average January $1. 0 February ? 1. 67 action ? 1. 70 1. 735 . 035 April ? 1. 85 1. 685 1. 723 . 165 . 127 May ? 1. 90 1. 775 1. 740 . one hundred twenty-five . clx 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 kinsfolk ? 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 celestial latitude ? 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 depend misunderstanding 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 celestial latitude 1. 75 1. 77 . 02 ? 1. 73 . 02 ? 1. 70 . 05 4. 41? (a)? It appears from the following(a) interpret that the points do scatter around a straight line. pic (b)? developing the lapsing 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 19 6 ? 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 in that location are no tourists at all, the model predicts a ridership of 0. 511, or 511,000 persons. star would not place much confidence in this forecast, however, because the subprogram of tourists (zero) is external the range of data used to develop the model. (e)? The standard mistake of the forecast is give by (f)? The correlation coefficient and the coefficient of determination are given by pic picpicHw Answer Operation Management Heizer Ch 44. 9 pic (c)? MAD (two-month moving ave rage) = . 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 (zero) 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 picpic