Question
In: Statistics and Probability
These data show the number of gallons of gasoline sold by a gasoline distributor in Bennington,...
These data show the number of gallons of gasoline sold by a gasoline distributor in Bennington, Vermont, over the past 12 weeks.
| Week | Sales (1,000s of gallons) |
|---|---|
| 1 | 17 |
| 2 | 21 |
| 3 | 19 |
| 4 | 23 |
| 5 | 18 |
| 6 | 16 |
| 7 | 20 |
| 8 | 18 |
| 9 | 22 |
| 10 | 20 |
| 11 | 15 |
| 12 | 22 |
(a)
Compute four-week and five-week moving averages for the time series.
| Week | Time Series Value |
4-Week Moving Average Forecast |
5-Week Moving Average Forecast |
|---|---|---|---|
| 1 | 17 | ||
| 2 | 21 | ||
| 3 | 19 | ||
| 4 | 23 | ||
| 5 | 18 | 1 | |
| 6 | 16 | 2 | 3 |
| 7 | 20 | 4 | 5 |
| 8 | 18 | 6 | 7 |
| 9 | 22 | 8 | 9 |
| 10 | 20 | 10 | 11 |
| 11 | 15 | 12 | 13 |
| 12 | 22 | 14 | 15 |
(b)
Compute the MSE for the four-week moving average forecasts. (Round your answer to two decimal places.)
Compute the MSE for the five-week moving average forecasts. (Round your answer to two decimal places.)
(c)
What appears to be the best number of weeks of past data (three, four, or five) to use in the moving average computation? MSE for the three-week moving average is 10.22.
Three weeks appears to be best, because the three-week moving average provides the smallest MSE. Three weeks appears to be best, because the three-week moving average provides the largest MSE.
Four weeks appears to be best, because the four-week moving average provides the smallest MSE. Five weeks appears to be best, because the five-week moving average provides the smallest MSE. None appear better than the others, because they all provide the same MSE.
Solutions
Expert Solution
a)
Four-week moving average:
| Week | Sales | Four week moving Average | Error | Error² | |
| 1 | 17 | - | - | - | - |
| 2 | 21 | - | - | - | - |
| 3 | 19 | - | - | - | - |
| 4 | 23 | - | - | - | - |
| 5 | 18 | (17+21+19+23)/4 = | 20.000 | -2.000 | 4.000 |
| 6 | 16 | (21+19+23+18)/4 = | 20.250 | -4.250 | 18.063 |
| 7 | 20 | (19+23+18+16)/4 = | 19.000 | 1.000 | 1.000 |
| 8 | 18 | (23+18+16+20)/4 = | 19.250 | -1.250 | 1.563 |
| 9 | 22 | (18+16+20+18)/4 = | 18.000 | 4.000 | 16.000 |
| 10 | 20 | (16+20+18+22)/4 = | 19.000 | 1.000 | 1.000 |
| 11 | 15 | (20+18+22+20)/4 = | 20.000 | -5.000 | 25.000 |
| 12 | 22 | (18+22+20+15)/4 = | 18.750 | 3.250 | 10.563 |
| Total | -3.250 | 77.188 | |||
Five-week moving average:
| Week | Sales | Five week moving Average | Error | Error² | |
| 1 | 17 | - | - | - | - |
| 2 | 21 | - | - | - | - |
| 3 | 19 | - | - | - | - |
| 4 | 23 | - | - | - | - |
| 5 | 18 | - | - | - | - |
| 6 | 16 | (17+21+19+23+18)/5 = | 19.600 | -3.600 | 12.960 |
| 7 | 20 | (21+19+23+18+16)/5 = | 19.400 | 0.600 | 0.360 |
| 8 | 18 | (19+23+18+16+20)/5 = | 19.200 | -1.200 | 1.440 |
| 9 | 22 | (23+18+16+20+18)/5 = | 19.000 | 3.000 | 9.000 |
| 10 | 20 | (18+16+20+18+22)/5 = | 18.800 | 1.200 | 1.440 |
| 11 | 15 | (16+20+18+22+20)/5 = | 19.200 | -4.200 | 17.640 |
| 12 | 22 | (20+18+22+20+15)/5 = | 19.000 | 3.000 | 9.000 |
| Total | -1.200 | 51.840 | |||
b)
MSE for the four-week moving average forecasts = ∑(A-F)²/n = 77.188/8 = 9.65
MSE for the five-week moving average forecasts = ∑(A-F)²/n = 51.84/7 = 7.41
c)
Five weeks appears to be best, because the five-week moving average provides the smallest MSE.
orchestra answered 1 year agoRelated Solutions
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Suppose these data show the number of gallons of gasoline sold
by a gasoline distributor in Bennington, Vermont, over the past 12
weeks.
Week
Sales (1,000s
of gallons)
1
17
2
21
3
19
4
23
5
18
6
16
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21
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19
9
23
10
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17
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20
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