Using the following data, determine the values of MAD and MSE. Which of these measurements of error seems to yield the best information about the forecasts?
Period | Value | Forecast |
1 | 19.4 | 16.6 |
2 | 23.6 | 19.1 |
3 | 24.0 | 22.0 |
4 | 26.8 | 24.8 |
5 | 29.2 | 25.9 |
6 | 35.5 | 28.6 |
A) MAD=3.58, MSE=15.77
B) MAD=3.58, MSE=94.59
C) MAD=21.5, MSE=15.77
D) MAD=21.5, MSE=94.59
Solution
Period | Value (A) | Forecast (F) | Deviation (A-F) | Absolute deviation |A-F| |
1 | 19.4 | 16.6 | 2.8 | 3.1 |
2 | 23.6 | 19.1 | 4.5 | 4 |
3 | 24 | 22 | 2 | 3 |
4 | 26.8 | 24.8 | 2 | 2 |
5 | 29.2 | 25.9 | 3.3 | 3.3 |
6 | 35.5 | 28.6 | 6.9 | 6.9 |
MAD = | A-F | / n
= 21.5 / 6
= 3.58
Period | Value (A) | Forecast (F) | A-F | (A-F)2 |
1 | 19.4 | 16.6 | 2.8 | 7.84 |
2 | 23.6 | 19.1 | 4.5 | 20.25 |
3 | 24 | 22 | 2 | 4 |
4 | 26.8 | 24.8 | 2 | 4 |
5 | 29.2 | 25.9 | 3.3 | 10.89 |
6 | 35.5 | 28.6 | 6.9 | 47.61 |
MSE = (A-F)2/ n
= 94.59
= 15.77
Option A
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