review for exam 2_Forecasting_Smith Insurance Agency
xlsx
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School
Tufts University *
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Course
361A
Subject
Aerospace Engineering
Date
Dec 6, 2023
Type
xlsx
Pages
14
Uploaded by sontran450
Period
Usage
1
17
2
24
3
21
4
20
5
17
6
21
7
21
8
19
9
17
10
18
11
20
12
21
Monthly water usage at Smith Insurance Agency for the past year
has been provided.
a) graph this time series. Does a stationary time series seem appr
b) Using regression, show that a stationary time series is reasonab
c) Using a four-month moving average, forecast water usage for b
January and the entire upcoming year.
d) What is the MAE of using four-month moving average forecasti
e) Determine the forecast for the upcoming January if exponentia
smoothing constant of 0.1.
f) Repeat part e with smoothing constant of 0.6.
g) Which of the methods of part d, e, and f perform better based
r, as measured in 100s of cubic feet,
ropriate from the graph?
ble.
both the upcoming month of
ting method?
al smoothing is used with a
on MAE?
Period
Usage
1
17
2
24
3
21
4
20
5
17
6
21
7
21
8
19
9
17
10
18
11
20
12
21
Questions
a) graph this time series. Does a stationary time series seem appropriate from the
b) Using regression, show that a stationary time series is reasonable.
c) Using a four-month moving average, forecast water usage for both the upcomin
month of January and the entire upcoming year.
d) What is the MAE of using four-month moving average forecasting method?
e) Determine the forecast for the upcoming January if exponential smoothing is us
a smoothing constant of 0.1.
f) Repeat part e with smoothing constant of 0.6.
g) Which of the methods of part d, e, and f perform better based on MAE?
1
2
3
4
5
6
7
8
9
10
11
0
5
10
15
20
25
30
Usage
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a. Yes, looking at the time series, there is not much "predictable" variation in data, so a station
stationary model seems appropriate
b. Run regression of usage with respect to month # (period).
From regression output, p-value of month # is p = .689.
A stationary model seems appropriat
signifcant, so we cannot reject H0:
_1 = 0 which means that we can assume that
_1=0 (slop
?
?
c.
forecast for January= (17+18+20+21)/4=19.
since the model is stationary and we don't hav
forecast for each month in the next year is 19 as well. So total water usage forecast for next ye
the upcoming months would be the same value of 19 given that this is a stationary time series
d. We need to calculate MAE. We use the performance measure template to get MAE value.
For 4-period moving average, start from month 5 (this is the frst month we have 4 months of
of 4 data points before month 5 as a forecast for month 5. Then drag the formula down to calc
months. The MAE will appear in the template . MAD is 1.72. See next sheet.
e,f. Since we don’t have any initial forecast, we have to come up with an initial forecast:
_2
?
=
smoothing mehtod formula to arrive at the forecast for period 3.
_3
_2+(1
)
_2=0.1 2
?
=??
−? ?
∗
If you do this in Excel, you can just drag down the formula to calculate forecast for other perio
reference to the cell you put your alpha value in, so that Excel treats it as a fxed cell).
Please
For part f, replace the cell for smoothing constant to 0.6. Py attention that the formulas in calc
cell for smoothing constant. With smoothing constant of 0.1, the forecast for each month of n
constant of 0.6, the forecast for each month of next year is 20.28.
g.
Incorporate the forecasts into a new forecasting accuracy template.
With smoothing constant of 0.1, MAE = 2.31
With smoothing constant of 0.6, MAE = 2.13
Comparing the methods, it seems 4-month moving average is best in terms of MAE.
e graph?
ng
sed with
12
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.12923
R Square
0.0167
Adjusted R
-0.08163
Standard E 2.232051
Observatio
12
ANOVA
df
SS
MS
F
Regression
1 0.846154 0.846154
0.16984
Residual
10 49.82051 4.982051
Total
11 50.66667
Coefficients
tandard Erro
t Stat
P-value
Intercept
20.16667 1.373732
14.6802
4.3E-08
Period
-0.076923 0.186653 -0.412117 0.688949
smoothing constant
Period
Usage
forecasts
0.6
1
17
2
24
17
3
21
21.2
4
20
21.08
5
17
20.432
6
21
18.3728
7
21
19.94912
8
19
20.57965
9
17
19.63186
10
18
18.05274
11
20
18.0211
12
21
19.20844
13
20.28338
nary model seems appropriate Yes, a
te since the linear term, Period, is not
pe of the regression line is 0).
ve data beyond the current year, the
ear is 19*12=228 as forecast for each of
s.
data before it). Calculate the average
culate the forecast for all other
_1=17. Then apply the exponential
=?
24+(1−0.1) 17=17.7.
∗
ods. (remember: press F4 on the
see the solution spreadsheet.
culation of forecasts are referring to the
next year is 18.86. With smoothing
ignificance F
0.688949
Lower 95%Upper 95%
Lower 95.0%
Upper 95.0%
17.1058 23.22753
17.1058 23.22753
-0.492813 0.338967 -0.492813 0.338967
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INPUTS
Number of Periods of Data Collected =
12
Performance measures for forecasted values below
MSE =
3.87
MAE =
1.72
MAPE =
9.26%
Absolute
Error
Absolute
Period
Usage
Forecasted values
Error
Error
Squared
% Error
1
17
2
24
3
21
4
20
5
17
20.5
-3.5
3.5
12.25
0.205882
6
21
20.5
0.5
0.5
0.25
0.02381
7
21
19.75
1.25
1.25
1.5625
0.059524
8
19
19.75
-0.75
0.75
0.5625
0.039474
9
17
19.5
-2.5
2.5
6.25
0.147059
10
18
19.5
-1.5
1.5
2.25
0.083333
11
20
18.75
1.25
1.25
1.5625
0.0625
12
21
18.5
2.5
2.5
6.25
0.119048
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INPUTS
Number of Periods of Data Collected =
12
Performance measures for forecasted values below
MSE =
7.87
MAE =
2.13
MAPE =
10.52%
smoothing constant
0.6
Absolute
Error
Absolute
Period
Usage
Forecasted values
Error
Error
Squared
% Error
1
17
2
24
17
7
7
49
0.291667
3
21
21.2
-0.2
0.2
0.04
0.009524
4
20
21.08
-1.08
1.08
1.1664
0.054
5
17
20.432
-3.432
3.432
11.77862
0.201882
6
21
18.3728
2.6272
2.6272
6.90218
0.125105
7
21
19.94912
1.05088
1.05088
1.104349
0.050042
8
19
20.579648
-1.579648 1.579648
2.495288
0.083139
9
17
19.6318592
-2.631859 2.631859
6.926683
0.154815
10
18
18.05274368
-0.052744 0.052744
0.002782
0.00293
11
20
18.021097472
1.978903
1.978903
3.916055
0.098945
12
21
19.2084389888
1.791561
1.791561
3.209691
0.085312
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