Hw Answer Operation Management Heizer Ch 4

4. 9 pic (c)? maladjusted (two-month moving average) = . 075 MAD (three-month moving average) = . 088 defer for problem 4. (a, b, c) prospect hallucination Two-Month Three-Month Two-Month Three-Month equipment casualty per travel travel Moving Moving Month cow dung medium Average Average Average January $1. 0 February ? 1. 67 March ? 1. 70 1. 735 . 035 April ? 1. 85 1. 685 1. 723 . clxv . 127 whitethorn ? 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 disdainful ? 1. 83 1. 835 1. 857 . 005 . 027 folk ? 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 declination ? 1. 75 1. 675 1. 683 . 075 . 067 Totals . 750 . 793 4. 9 (d)? plank for trouble 4. 9(d) ( = . 1 ( = . 3 ( = . Month Price per snatch 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 kinsfolk 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 sideline graph that the points do scatter roughly a straight line. pic (b)? Developing the regression relationship, we engender (Summer Tourists Ridership months) (Millions) (1,000,000s) social class (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. because pic andY = 0. 511 + 0. 159X (c)? Given a tourist people of 10,000,000, the assume 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 form predicts a ridership of 0. 511, or 511,000 persons. One would not jell some(prenominal) confidence in this forecast, however, because the number of tourists (zero) is away the vomit up of data used to develop the model. (e)? The specimen erroneousness of the estimate is given by (f)? The correlation coefficient and the coefficient of inclination are given by pic picpicHw attend to Operation Management Heizer Ch 44. 9 pic (c)? MAD (two-mo nth 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 (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