In: Statistics and Probability
Problem 15-03 (Algorithmic) Consider the following time series data.
Week 1 2 3 4 5 6
Value 20 13 15 11 18 13
Using the naïve method (most recent value) as the forecast for the next week, compute the following measures of forecast accuracy: Mean absolute error (MAE) Mean squared error (MSE) Mean absolute percentage error (MAPE) Round your answers to two decimal places.
MAE =
MSE =
MAPE =
Using the average of all the historical data as a forecast for the next period, compute the same three values. Round your answers to two decimal places.
MAE =
MSE =
MAPE =
Which method appears to provide the more accurate forecasts for the historical data? Explain.
a)
Naïve method: |
Time period | Actual Value(A) | Moving avg. Forecast(F) | Forecast error E=|A-F| | Squared Forecast Error | |A-F|A |
1 | 20 | ||||
2 | 13 | 20 | 7 | 49 | 0.5385 |
3 | 15 | 13 | 2 | 4 | 0.1333 |
4 | 11 | 15 | 4 | 16 | 0.3636 |
5 | 18 | 11 | 7 | 49 | 0.3889 |
6 | 13 | 18 | 5 | 25 | 0.3846 |
7 | 13 | ||||
Total | 25 | 143 | 1.81 | ||
Average | 5.00 | 28.60 | 36.18% | ||
MAD | MSE | MAPE | |||
a) mean absolute error= | 5.00 | ||||
b) mean squared error = | 28.60 | ||||
c) mean absolute % error= | 36.18% | ||||
d) forecast for week 7 = | 13 |
2)
Historical data method: |
Time period | Actual Value(A) | historical Data Forecast(F) | Forecast error E=|A-F| | Squared Forecast Error | |A-F|A |
1 | 20 | ||||
2 | 13 | 20.00 | 7.00 | 49.00 | 0.54 |
3 | 15 | 16.50 | 1.50 | 2.25 | 0.10 |
4 | 11 | 16.00 | 5.00 | 25.00 | 0.45 |
5 | 18 | 14.75 | 3.25 | 10.56 | 0.18 |
6 | 13 | 15.40 | 2.40 | 5.76 | 0.18 |
7 | 15.00 | ||||
Total | 19.15 | 92.5725 | 1.46 | ||
Average | 3.83 | 18.51 | 29.16% | ||
MAD | MSE | MAPE |
a) mean absolute error= | 3.83 | |
b) mean squared error = | 18.51 | |
c) mean absolute % error= | 29.16% | |
d) forecast for week 7 = | 15.00 |
3)
average of all data method provide better forecast because of lower value of MAE, MSE and MAPE