Consider the following time series data. Develop a three-week moving average for
ID: 452995 • Letter: C
Question
Consider the following time series data.
Develop a three-week moving average for this time series. Compute MSE and a forecast for week 7. Round your answers to two decimal places.
MSE:
The forecast for week 7:
Use a= 0.2 to compute the exponential smoothing values for the time series. Compute MSE and a forecast for week 7. Round your answers to two decimal places.
MSE:
The forecast for week 7:
Compare the three-week moving average forecast with the exponential smoothing forecast using a= 0.2. Which appears to provide the better forecast based on MSE?
Three-week moving average
Use trial and error to find a value of the exponential smoothing coefficient that results in a smaller MSE than what you calculated for a= 0.2. Find a value of for the smallest MSE. Round your answer to three decimal places.
a=
Explanation / Answer
Develop a three-week moving average for this time series. Compute MSE and a forecast for week 7. Round your answers to two decimal places.
Use a= 0.2 to compute the exponential smoothing values for the time series. Compute MSE and a forecast for week 7. Round your answers to two decimal places.
Answer :
Using exponential smoothing, list all terms for the time series {18,13,16,12,18,15,15.78} using = 0.2
t = 1
s1 = x0
s1 = 18
t = 2
st = xt - 1 + (1 - )st - 1
s2 = x2 - 1 + (1 - )s2 - 1
s2 = 0.2(x1) + (1 - 0.2)s1
s2 = 0.2(x1) + (0.8)s1
s2 = 0.2(13) + (0.8)18
s2 = 2.6 + 14.4
s2 = 17
t = 3
st = xt - 1 + (1 - )st - 1
s3 = x3 - 1 + (1 - )s3 - 1
s3 = 0.2(x2) + (1 - 0.2)s2
s3 = 0.2(x2) + (0.8)s2
s3 = 0.2(16) + (0.8)17
s3 = 3.2 + 13.6
s3 = 16.8
t = 4
st = xt - 1 + (1 - )st - 1
s4 = x4 - 1 + (1 - )s4 - 1
s4 = 0.2(x3) + (1 - 0.2)s3
s4 = 0.2(x3) + (0.8)s3
s4 = 0.2(12) + (0.8)16.8
s4 = 2.4 + 13.44
s4 = 15.84
t = 5
st = xt - 1 + (1 - )st - 1
s5 = x5 - 1 + (1 - )s5 - 1
s5 = 0.2(x4) + (1 - 0.2)s4
s5 = 0.2(x4) + (0.8)s4
s5 = 0.2(18) + (0.8)15.84
s5 = 3.6 + 12.672
s5 = 16.272
t = 6
st = xt - 1 + (1 - )st - 1
s6 = x6 - 1 + (1 - )s6 - 1
s6 = 0.2(x5) + (1 - 0.2)s5
s6 = 0.2(x5) + (0.8)s5
s6 = 0.2(15) + (0.8)16.272
s6 = 3 + 13.0176
s6 = 16.0176
t = 7
st = xt - 1 + (1 - )st - 1
s7 = x7 - 1 + (1 - )s7 - 1
s7 = 0.2(x6) + (1 - 0.2)s6
s7 = 0.2(x6) + (0.8)s6
s7 = 0.2(15.78) + (0.8)16.0176
s7 = 3.156 + 12.81408
s7 = 15.97008
Our 7 smoothed values are {18, 17, 16.8, 15.84, 16.272, 16.0176, 15.97008}
Use trial and error to find a value of the exponential smoothing coefficient that results in a smaller MSE than what you calculated for a= 0.2. Find a value of for the smallest MSE. Round your answer to three decimal places.
Answer :
0.4 is the minimum.
Using exponential smoothing, list all terms for the time series {18,13,16,12,18,15,15.78} using = 0.4
t = 1
s1 = x0
s1 = 18
t = 2
st = xt - 1 + (1 - )st - 1
s2 = x2 - 1 + (1 - )s2 - 1
s2 = 0.4(x1) + (1 - 0.4)s1
s2 = 0.4(x1) + (0.6)s1
s2 = 0.4(13) + (0.6)18
s2 = 5.2 + 10.8
s2 = 16
t = 3
st = xt - 1 + (1 - )st - 1
s3 = x3 - 1 + (1 - )s3 - 1
s3 = 0.4(x2) + (1 - 0.4)s2
s3 = 0.4(x2) + (0.6)s2
s3 = 0.4(16) + (0.6)16
s3 = 6.4 + 9.6
s3 = 16
t = 4
st = xt - 1 + (1 - )st - 1
s4 = x4 - 1 + (1 - )s4 - 1
s4 = 0.4(x3) + (1 - 0.4)s3
s4 = 0.4(x3) + (0.6)s3
s4 = 0.4(12) + (0.6)16
s4 = 4.8 + 9.6
s4 = 14.4
t = 5
st = xt - 1 + (1 - )st - 1
s5 = x5 - 1 + (1 - )s5 - 1
s5 = 0.4(x4) + (1 - 0.4)s4
s5 = 0.4(x4) + (0.6)s4
s5 = 0.4(18) + (0.6)14.4
s5 = 7.2 + 8.64
s5 = 15.84
t = 6
st = xt - 1 + (1 - )st - 1
s6 = x6 - 1 + (1 - )s6 - 1
s6 = 0.4(x5) + (1 - 0.4)s5
s6 = 0.4(x5) + (0.6)s5
s6 = 0.4(15) + (0.6)15.84
s6 = 6 + 9.504
s6 = 15.504
t = 7
st = xt - 1 + (1 - )st - 1
s7 = x7 - 1 + (1 - )s7 - 1
s7 = 0.4(x6) + (1 - 0.4)s6
s7 = 0.4(x6) + (0.6)s6
s7 = 0.4(15.78) + (0.6)15.504
s7 = 6.312 + 9.3024
s7 = 15.6144
Our 7 smoothed values are {18, 16, 16, 14.4, 15.84, 15.504, 15.6144}