Data collected on the yearly registrations for a Six Sigma seminar at the Qualit
ID: 3208711 • Letter: D
Question
Data collected on the yearly registrations for a Six Sigma seminar at the Quality College are shown in the following table:
Year
1
2
3
4
5
6
7
8
9
10
11
Registrations (000)
4.00
7.00
3.00
4.00
9.00
7.00
7.00
10.00
13.00
12.00
15.00
a) Calculate the forecasted registrations for years 2 through 12 using exponential smoothing, with a smoothing constant
(alpha) of 0.35 and a starting forecast of 5.00
for year 1 (round your responses to one decimal
place):
Year
1
2
3
4
5
6
7
8
9
10
11
12
Forecast (000)
5.00
nothing
nothing
nothing
nothing
nothing
nothing
nothing
nothing
nothing
nothing
nothing
b) Mean absolute deviation based on the forecast developed using the exponential smoothing method (with a smoothing constant
(alpha) = 0.35 and a starting forecast of Upper F1
= 5.00) is ?
registrations (round your response to one decimal place).
Year
1
2
3
4
5
6
7
8
9
10
11
Registrations (000)
4.00
7.00
3.00
4.00
9.00
7.00
7.00
10.00
13.00
12.00
15.00
Explanation / Answer
a. Applying exponential smoothing,
M1=5, M2=0.35Y2+0.65M1=0.35*7+0.65*5=5.7, M3=0.35*3+0.65*5.7=4.8, M4=0.35*4+0.65*4.8=4.5, M5=0.35*9+0.65*4.5=6.1, M6=0.35*7+0.65*6.1=6.4, M7=0.35*7+0.65*6.4=6.6, M8=0.35*10+0.65*6.6=6.7, M9=0.35*13+0.65*6.7=8.9, M10=0.35*12+0.65*8.9=10, M11=0.35*15+0.65*10=11.75
b. The Mean Absolute Deviation, MAD is as follows:
MAD=sigma |error|/n, where, error=absolute-forecast
={(5-4)+(7-5.7)+(3-4.8)+(4-4.5)+(9-6.1)+(7-6.4)+(7-6.6)+(10-6.7)+(13-8.9)+(12-10)+(15-11.75)}/12
=21.15/12
=1.8
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