In: Statistics and Probability
Problem 15-07 (Algorithmic)
Refer to the gasoline sales time series data in the given table.
Week | Sales (1000s of gallons) |
1 | 18 |
2 | 21 |
3 | 18 |
4 | 24 |
5 | 18 |
6 | 15 |
7 | 21 |
8 | 19 |
9 | 22 |
10 | 21 |
11 | 15 |
12 | 23 |
Week |
Sales |
4-Week Moving Average |
5-Week Moving Average |
1 | 18 | ||
2 | 21 | ||
3 | 18 | ||
4 | 24 | ||
5 | 18 | ||
6 | 15 | ||
7 | 21 | ||
8 | 19 | ||
9 | 22 | ||
10 | 21 | ||
11 | 15 | ||
12 | 23 |
a.
Week | Sales | 4-Week Moving Average: 4w MA |
5-week Moving Average: 5w MA |
1 | 18 | ||
2 | 21 | ||
3 | 18 | ||
4 | 24 | ||
5 | 18 | (18+21+18+24)/4 =20.25 | |
6 | 15 | (21+18+24+18)/4 =20.25 | (18+21+18+24+18)/5 =19.8 |
7 | 21 | (18+24+18+15)/4 =18.75 | (21+18+24+18+15)/5 =19.2 |
8 | 19 | (24+18+15+21)/4 =19.5 | (18+24+18+15+21)/5 =19.2 |
9 | 22 | (18+15+21+19)/4 =18.25 | (24+18+15+21+19)/5 =19.4 |
10 | 21 | (15+21+19+22)/4 =19.25 | (18+15+21+19+22)/5 =19 |
11 | 15 | (21+19+22+21)/4 =20.75 | (15+21+19+22+21)/5 =19.6 |
12 | 23 | (19+22+21+15)/4 =19.25 | (21+19+22+21+15)/5 =19.6 |
b.
Actual: A | Forecast: F (4w MA) | Error square, E2 =(A - F)2 | Forecast:F (5w MA) | Error square, E2 =(A - F)2 |
18 | ||||
21 | ||||
18 | ||||
24 | ||||
18 | 20.25 | (18 - 20.25)2 =5.06 | ||
15 | 20.25 | 27.56 | 19.8 | (15 - 19.8)2 =23.04 |
21 | 18.75 | 5.06 | 19.2 | 3.24 |
19 | 19.5 | 0.25 | 19.2 | 0.04 |
22 | 18.25 | 14.06 | 19.4 | 6.76 |
21 | 19.25 | 3.06 | 19 | 4 |
15 | 20.75 | 33.06 | 19.6 | 21.16 |
23 | 19.25 | 14.06 | 19.6 | 11.56 |
Total | =102.17 | =69.8 |
MSE for four-week moving average = /n =102.17/8 =12.77
MSE for five-week moving average =/n =69.8/7 =9.97
c.
MSE for three-week moving average =13.69
MSE for four-week moving average =12.77
MSE for five-week moving average =9.97
Since MSE for five-week moving average is the smallest of the three, five weeks appear to be the best number of weeks of past data to use in the moving average computation.