4-19: Summer-month bus and subway ridership in Washington, DC, is believed to be tied heavily to the number of tourists visiting the city. During the past 12 years, the following data have been obtained:
- a) Plot these data and decide if a linear model is reasonable.
As the number of tourists (x-axis) increases, the ridership appears to increase: a linear model is reasonable. (1 mark)
- b) Develop a regression relationship.
| Year | Tourists (x) | Ridership (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 |
| 6 | 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 |
| Sx = 132 | Sy = 27.1 | Sx2 = 1796 | Sy2 = 71.59 | Sxy = 352.9 |
= Sx / n = 132 / 12 = 11
= Sy / n = 27.1 / 12 = 2.26
\ The relationship is y = 0.511 + 0.159x (2 marks)
- c) What is expected ridership if 10 million tourists visit the city in a year?
Y = 0.511 + 0.159(10) = 2.101 or 2,101,000 persons. (2 marks)
- d) Explain the predicted ridership if there are no tourists at all.
If there are no tourists at all, the model predicts a ridership of 0.511 or 511,000 persons
Consulting income at Dr. Thomas W. Jones Associates for the period February to July has been as follows. Use trend-adjusted exponential smoothing to forecast August’s income. Assume that the initial forecast for February is $65,000 and the initial trend adjustment is 0. The smoothing constants selected are a = .1 and b = .2.
Forecast Ft = a(At-1) + (1 – a)(Ft-1 + Tt-1), Trend Tt = b(Ft – Ft-1) + (1 – b)T t-1
| Mo | Inc | Forecast | Trend | FIT | Err2 |
| Feb | 70.0 | 65.0 | 0 | 65+0 = 65 | 25.0 |
| Mar | 68.5 | 0.1(70)+0.9(65) = 65.5 | 0.2(65.5–65)+(0.8)0 = 0.1 | 65.5+0.1=65.6 | 8.4 |
| Apr | 64.8 | 0.1(68.5)+0.9(65.6) = 65.89 | 0.2(65.89-65.5)+(0.8)0.1 = 0.16 | 65.89+0.16=66.05 | 1.6 |
| May | 71.7 | 0.1(64.8)+0.9(66.05) = 65.93 | 0.2(65.92-65.89)+(0.8)0.16 = 0.13 | 65.93+0.13=66.06 | 31.9 |
| Jun | 71.3 | 0.1(71.7)+0.9(66.06) = 66.62 | 0.2(66.62-65.93)+(0.8)0.13 = 0.25 | 66.62+0.25=66.87 | 19.7 |
| Jul | 72.8 | 0.1(71.3)+0.9(66.87) = 67.31 | 0.2(67.31-66.62)+(0.8)0.25 = 0.33 | 67.31+0.33=67.64 | 26.6 |
| Aug | 0.1(72.8)+0.9(67.64) = 68.16 | 0.2(68.16-67.31)+(0.8)0.33 = 0.43 | 68.16+0.43=68.60 | ||
| Σ = | 113.2 |
MSE = 113.2 / 6 = 18.87
Resolve problem 4.19 with a = .1 and b = .8. Using MSE, which smoothing constants provide a better forecast?
| Mo | Inc | Forecast | Trend | FIT | Err2 |
| Feb | 70.0 | 65.0 | 0 | 65+0 = 65 | 25.0 |
| Mar | 68.5 | 0.1(70)+0.9(65) = 65.5 | 0.8(65.5–65)+(0.2)0 = 0.4 | 65.5+0.4=65.9 | 6.76 |
| Apr | 64.8 | 0.1(68.5)+0.9(65.9) = 66.16 | 0.8(66.16-65.5)+(0.2)0.4 = 0.61 | 66.16+0.61=66.77 | 3.87 |
| May | 71.7 | 0.1(64.8)+0.9(66.77) = 66.57 | 0.8(66.57-66.16)+(0.2)0.61 = 0.45 | 66.57+0.45=67.02 | 21.89 |
| Jun | 71.3 | 0.1(71.7)+0.9(67.02) = 67.49 | 0.8(67.49-66.57)+(0.2)0.45 = 0.82 | 67.49+0.82=68.31 | 8.91 |
| Jul | 72.8 | 0.1(71.3)+0.9(68.31) = 68.61 | 0.8(68.61-67.49)+(0.2)0.82 = 1.06 | 68.61+1.06=69.68 | 9.76 |
| Σ = | 76.19 |
MSE = 76.19 / 6 = 12.70
Based upon the MSE criterion, exponential smoothing with a = .1 and b = .8 provides a better forecast.
The accountant at Tick Wing Coal Distributors, Inc., in San Francisco notes that the demand for coal seems to be tied to an index of weather severity developed by the U.S. Weather Bureau. When weather was extremely cold in the U.S. over the past five years (and the index was thus high), coal sales were high. The accountant proposes that one good forecast of next year’s coal demand could be made by developing a regression equation and then consulting the Farmer’s Almanac to see how severe next year’s winter would be.
For the data in the following table, derive a least squares regression and compute the coefficient of correlation of the data. Also compute the standard error of the estimate.
Coal Sales, y(in millions of tons) |
4 | 1 | 4 | 6 | 5 |
| Weather Index, x | 2 | 1 | 4 | 5 | 3 |
| X | Y | X2 | Y2 | XY |
| 2 | 4 | 4 | 16 | 8 |
| 1 | 1 | 1 | 1 | 1 |
| 4 | 4 | 16 | 16 | 16 |
| 5 | 6 | 25 | 36 | 30 |
| 3 | 5 | 9 | 25 | 15 |
| Σ = 15 | Σ = 20 | Σ = 55 | Σ = 94 | Σ = 70 |
15 / 5 = 3 20 / 5 = 4
= = 1.0
= 4 – 1(3) = 1.0
Our equation is then Y = a + bX, ie Y = 1.0 + 1.0X
Correlation Coefficient:
=
r = = = = 0.845
Standard Error of the Estimate:
= = = 1.15
Given the following data, use exponential smoothing (a = 0.2) to develop a demand forecast. Assume the forecast for the initial period is 5.
Exponential Smoothing Forecast Ft = Ft-1 + a(At-1 – Ft-1)
ie F2 = F1 + a(A1 – F1) = 5 + 0.2 (7 – 5) = 5.4. Carrying this through to week 7 gives:
| Period |
Demand |
Exponentially Smoothed Forecast |
| 1 | 7 | 5 |
| 2 | 9 | 5 + 0.2 (7 – 5) = 5.4 |
| 3 | 5 | 5.4 + 0.2 (9 – 5.4) = 6.12 |
| 4 | 9 | 6.12 + 0.2 (5 – 6.12) = 5.90 |
| 5 | 13 | 5.90 + 0.2 (9 – 5.90) = 6.52 |
| 6 | 8 | 6.52 + 0.2 (13 – 6.52) = 7.82 |
| 7 | 7.82 + 0.2 (8 – 7.82) = 7.86 |
The director of the Riley County, Kansas, library system would like to forecast evening patron usage for next week. Below are the data for the past 4 weeks:
- a) Calculate a seasonal index for each day of the week.
| Day | Week 1 | Week 2 | Week 3 | Week 4 | Avg |
| Monday | 210 | 215 | 220 | 225 | 217.5 |
| Tuesday | 178 | 180 | 176 | 178 | 178 |
| Wednesday | 250 | 250 | 260 | 260 | 255 |
| Thursday | 215 | 213 | 220 | 225 | 218.3 |
| Friday | 160 | 165 | 175 | 176 | 169 |
| Saturday | 180 | 185 | 190 | 190 | 186.3 |
| S = 1224.1 |
Average Daily Demand = S Average Demand / 6 Days = 1224.1 / 6 = 204
Seasonal Index = Average Demand / Average Daily Demand
Seasonal Index for Monday = 217.5 / 204 = 1.066
Seasonal Index for Tuesday = 178 / 204 = 0.873
Seasonal Index for Wednesday = 255 / 204 = 1.25
Seasonal Index for Thursday = 218.3 / 204 = 1.07
Seasonal Index for Friday = 169 / 204 = 0.828
Seasonal Index for Saturday = 186.3 / 204 = 0.913
- b) If the trend equation for this problem is y = 201.74 + 0.18x, what is the forecast for each day of week 5? Round your forecast to the nearest whole number.
Note that each day is one period along the x-axis. So in week five, Monday is x = 25, Tuesday is x = 26, etc.
| Day | x | y | Seasonal Index | Forecast (Rounded) |
| Monday | 25 | 201.74 + 0.18(25) = 206.24 | 1.066 | 206.24 (1.066) = 220 |
| Tuesday | 26 | 201.74 + 0.18(26) = 206.42 | 0.873 | 206.42 (0.873) = 180 |
| Wednesday | 27 | 201.74 + 0.18(27) = 206.6 | 1.25 | 206.6 (1.25) = 258 |
| Thursday | 28 | 201.74 + 0.18(28) = 206.78 | 1.07 | 206.78 (1.07) = 221 |
| Friday | 29 | 201.74 + 0.18(29) = 206.96 | 0.828 | 206.96 (0.828) = 171 |
| Saturday | 30 | 201.74 + 0.18(30) = 207.14 | 0.913 | 207.14 (0.913) = 189 |
5-31 and 5-32
Forecast Ft = a(At-1) + (1 – a)(Ft-1 + Tt-1), Trend Tt = b(Ft – Ft-1) + (1 – b)T t-1
| Mo | Inc | Forecast | Trend | FIT | Err2 |
| Feb | 70.0 | 65.0 | 0 | 65+0 = 65 | 25.0 |
| Mar | 68.5 | 0.1(70)+0.9(65) = 65.5 | 0.2(65.5–65)+(0.8)0 = 0.1 | 65.5+0.1=65.6 | 8.4 |
| Apr | 64.8 | 0.1(68.5)+0.9(65.6) = 65.89 | 0.2(65.89-65.5)+(0.8)0.1 = 0.16 | 65.89+0.16=66.05 | 1.6 |
| May | 71.7 | 0.1(64.8)+0.9(66.05) = 65.93 | 0.2(65.92-65.89)+(0.8)0.16 = 0.13 | 65.93+0.13=66.06 | 31.9 |
| Jun | 71.3 | 0.1(71.7)+0.9(66.06) = 66.62 | 0.2(66.62-65.93)+(0.8)0.13 = 0.25 | 66.62+0.25=66.87 | 19.7 |
| Jul | 72.8 | 0.1(71.3)+0.9(66.87) = 67.31 | 0.2(67.31-66.62)+(0.8)0.25 = 0.33 | 67.31+0.33=67.64 | 26.6 |
| Aug | 0.1(72.8)+0.9(67.64) = 68.16 | 0.2(68.16-67.31)+(0.8)0.33 = 0.43 | 68.16+0.43=68.60 | ||
| Σ = | 113.2 |
MSE = 113.2 / 6 = 18.87
5-33
a) Quarter Data MA CMA Percentage1 218 2 247 3 243 250 250.875 96.860986554 292 251.75 252.625 115.58634341 225 253.5 255 88.235294122 254 256.5 257.375 98.688683833 255 258.25 259.375 98.313253014 299 260.5 261.875 114.1766111 234 263.25 264.375 88.51063832 265 265.5 269 98.513011153 264 272.5 274.5 96.174863394 327 276.5 278.75 117.3094171 250 281 284.125 87.989441272 283 287.25 290.875 97.292651483 289 294.5 4 356 The percentage column is simply the data column, divided by the CMA, and multiplied by 100. Now, we average the percentages for each quarter. We get: Quarter 1: 88.24Quarter 2: 98.16Quarter 3: 97.11Quarter 4: 115.69 Since the average of these four number is 99.8, we multiply them by 100/99.8=1.001 in order to normalize them, thus getting: Quarter 1: 88.41Quarter 2: 98.35Quarter 3: 97.3Quarter 4: 115.91 So these are the seasonal indices. b) We get: Quarter Data Desasonalized1 218 246.55560452 247 251.12523263 243 249.72553314 292 251.9029011 225 254.47252762 254 258.2421423 255 262.05765824 299 257.94166911 234 264.65142872 265 269.42585693 264 271.30675214 327 282.09674181 250 282.74725292 283 287.72648113 289 296.99867934 356 307.1144957 As you can see, the spikes in every 4th quarter are smoothed. Y X246.5556045 1251.1252326 2249.7255331 3251.902901 4254.4725276 5258.242142 6262.0576582 7257.9416691 8264.6514287 9269.4258569 10271.3067521 11282.0967418 12282.7472529 13287.7264811 14 296.9986793 15307.1144957 16 So, we have to find the coefficients ‘a’ and ‘b’ in the following regression: Y = a + bX The formula for a and b can be found in slide 6 in the Powerpoint presentation I provided above. We get that these values are: a = 237.28b = 3.65 So the trend line would then be… Time Trend1 240.932 244.583 248.234 251.885 255.536 259.187 262.838 266.489 270.1310 273.7811 277.4312 281.0813 284.7314 288.3815 292.0316 295.68 Each value was calculated using the coefficients ‘a’ and ‘b’. For example, the first value is simply 237.28+3.65*1, the next one is 237.28+3.65*2 and so on. c) 17 299.3318 302.9819 306.6320 310.28 d) Forecasts deseasonalized final forecast17 299.33 264.144131318 302.98 297.41965719 306.63 297.787918120 310.28 358.9653847
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