Consider the following time series data.
Week | 1 | 2 | 3 | 4 | 5 | 6 |
Value | 18 | 13 | 16 | 11 | 17 | 14 |
Using the naïve method (most recent value) as the forecast for the next week, compute the following measures of forecast accuracy.
(a) | Mean absolute error |
If required, round your answer to one decimal place. | |
(b) | Mean squared error |
If required, round your answer to one decimal place. | |
(c) | Mean absolute percentage error |
If required, round your intermediate calculations and final answer to two decimal places. | |
(d) | What is the forecast for week 7? |
most recent value is used as forecast in naive method
period | demand | forcast | forecast error=demand value-forecast value | absolute forecast error | squared forcast error | Abs %error |
t | Dt | Ft | et=Dt-Ft | | et | | (et)² | | et/Dt | |
1 | 18 | |||||
2 | 13 | 18.000 | -5.00 | 5.00 | 25.00 | 38.46% |
3 | 16 | 13.000 | 3.00 | 3.00 | 9.00 | 18.75% |
4 | 11 | 16.000 | -5.00 | 5.00 | 25.00 | 45.45% |
4 | 17 | 11.000 | 6.00 | 6.00 | 36.00 | 35.29% |
6 | 14 | 17.000 | -3.00 | 3.00 | 9.00 | 21.43% |
forecast error=demand value-forecast value | absolute forecast error | squared forcast error | Abs %error | |
et=Dt-Ft | | et | | (et)² | | et/Dt | | |
total sum= | -4.00 | 22.00 | 104.00 | 159.39% |
n= | 5 | 5 | 5 | 5 |
average= | -0.80 | 4.40 | 20.80 | 31.88% |
a)
(a) Mean absolute error=Σ |et|/n =
4.4
(b) Mean squared error=Σ(et)²/n =
20.8
(c) Mean absolute percentage error = Σ | et/Dt |/n =
31.88%
(d) forecast for week 7 = 14
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