In: Statistics and Probability
Consider the following gasoline sales time series data. Click on the datafile logo to reference the data.
Week |
Sales (1000s of gallons) |
1 |
16 |
2 |
21 |
3 |
19 |
4 |
24 |
5 |
18 |
6 |
16 |
7 |
19 |
8 |
17 |
9 |
23 |
10 |
20 |
11 |
15 |
12 |
22 |
a. Using a weight of 1/2 for the most recent observation, 1/3 for the second most recent observation, and 1/6 third the most recent observation, compute a three-week weighted moving average for the time series (to 2 decimals). Enter negative values as negative numbers.
|
|
Weighted Moving |
Forecast |
|
||
1 | ||||||
2 | ||||||
3 | ||||||
4 | ||||||
5 | ||||||
6 | ||||||
7 | ||||||
8 | ||||||
9 | ||||||
10 | ||||||
11 | ||||||
12 | ||||||
Total |
b. Compute the MSE for the weighted moving
average in part (a).
MSE =
Do you prefer this weighted moving average to the unweighted
moving average? Remember that the MSE for the unweighted moving
average is 13.69.
Prefer the unweighted moving average here; it has a
(greater/smaller) MSE.
c. Suppose you are allowed to choose any
weights as long as they sum to 1. Could you always find a set of
weights that would make the MSE at least as small for a weighted
moving average than for an unweighted moving average?
(Yes/No)
Answer a)
Now, the following table shows the calculations of the corresponding error metrics:
Total = ΣError2 = 23.3611+14.6944+17.3611+1+0.6944+30.25+0.1111+30.25+16 = 133.7221
Answer b)
MSE = ΣError2/n = 133.7221/9
MSE = 14.86
In this case, MSE for weighted moving average (14.86) is greater than MSE for unweighted moving average (13.69). Therefore, we prefer unweighted moving average.
Prefer the unweighted moving average here; it has a smaller MSE.
Answer c) Yes
Explanation
You could always find a weighted moving average at least as good as the unweighted one; actually the unweighted moving average is a special case of the weighted ones where the weights are equal.