Exercise 7.1
a. Find and report the four seasonal factors for quarter 1, 2, 3 and 4
sn1 = 1.191, sn2 = 1.521, sn3 = 0.804, sn4 = 0.484
b. What is the equation of the estimated trend that has been calculated using the
deaseasonalized data?
trt = 220.53893 + 19.949897  t
c. Compute (using estimated trend and seasonal factors) ŷ17, ŷ18, ŷ19, and ŷ20
ŷ17 = (220.53893 + 19.949897  17)  1.191  666.59  667
ŷ18 = (220.53893 + 19.949897  18)  1.521  881.63  882
ŷ19 = (220.53893 + 19.949897  19)  0.804  482.07  482
ŷ20 = (220.53893 + 19.949897  20)  0.484  299.86  300
d. Compute a point forecast of the total tractor sales for the next year (year 5)
ŷ17 + ŷ18 + ŷ19 + ŷ20  2330 (With rounded forecasts  2331 )
e. Do the cyclical factors determine a well-defined cycle for these data? Explain your
answer.
Make a plot of cl together with cl  ir
1.030
1.020
1.010
1.000
cl x ir
0.990
cl
0.980
0.970
0.960
0.950
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
In general the cycles would cover longer periods than a year, but this
estimated cycle seems to have a period corresponding with four quarters.
 The answer is no.
f. In figure 7.5, find and report point forecasts of the tractor sales (based on trend
and seasonal factors) for each of the quarters of next year (year 5). Do the values
agree with your answers to part (c) ?
Yes!
g. (This is the task of assignment 2 )
h. Figure 7.6 (page 342) presents the MINITAB output of the 95% prediction
interval forecasts of the deseasonalized tractor sales for each of the four quarters of
the next year (year 5). use the results in this output to compute approximate 95%
prediction interval forecasts of tractor sales for each of the quarters of the next year.
Quarter 17:
95% P.I. for y17* = d17 :
(545.64, 573.76)
See p. 339 in the textbook
Error bound: (573.76 – 545.64)/2 = 14.06
 Approx. 95% P.I. for y17 : (666.59 – 14.06, 666.59 + 14.06 ) 
(653, 681)
Analogously for quarters 18-20:
Error bound of P:I: for d18 :
(594.00 – 565.31)/2 = 14.345
 Approx 95% P.I. for y18:
(867, 896)
Error bound of P:I: for d19 :
(614.26 – 584.94)/2 = 14.66
 Approx 95% P.I. for y19:
(467, 498)
Error bound of P:I: for d20 :
(634.55 – 604.55)/2 = 15
 Approx 95% P.I. for y20:
(285, 315)
Exercise 7.3 (Refer to Excel Worksheet, proceed as with 7.1)
a. Find and report the four seasonal factors for quarter 1, 2, 3 and 4
sn1 = 70.59, sn2 = 210.76, sn3 = –76.95, sn4 = –204.41
b. What is the equation of the estimated trend that has been calculated using the
deaseasonalized data?
trt = 214.8895833 + 19.75563725  t
c. Compute (using estimated trend and seasonal factors) ŷ17, ŷ18, ŷ19, and ŷ20
ŷ17 = (214.8895833 + 19.75563725  17) +70.59  621.33  621
ŷ18 = (214.8895833 + 19.75563725  18) + 210.76  781.25  781
ŷ19 = (214.8895833 + 19.75563725  19) – 76.95  513.30  513
ŷ20 = (214.8895833 + 19.75563725  20) – 204.41  405.59  406
d. Compute a point forecast of the total tractor sales for the next year (year 5)
ŷ17 + ŷ18 + ŷ19 + ŷ20  2321 (With rounded forecasts  2321 )
e. Do the cyclical factors determine a well-defined cycle for these data? Explain your
answer.
Make a plot of cl together with cl  ir
80.000
60.000
40.000
20.000
cl x ir
0.000
1
2
3
4
5
6
7
8
9
10
11
-20.000
-40.000
-60.000
-80.000
No differences compared to the multiplicative case.
12
13
14
15
16
cl
f. In figure 7.5, find and report point forecasts of the tractor sales (based on trend
and seasonal factors) for each of the quarters of next year (year 5). Do the values
agree with your answers to part (c) ?
Yes!
g. (This is the task of assignment 2 )
h. Figure 7.6 (page 342) presents the MINITAB output of the 95% prediction
interval forecasts of the deseasonalized tractor sales for each of the four quarters of
the next year (year 5). use the results in this output to compute approximate 95%
prediction interval forecasts of tractor sales for each of the quarters of the next year.
To be able to do the corresponding for the additive mode, we need a new
Minitab (or SPSS or Eviews) analysis
The regression equation is
d = 215 + 19.8 t
Predictor
Coef
SE Coef
T
P
Constant
214.89
20.08
10.70
0.000
t
19.756
2.076
9.52
0.000
S = 38.2841
R-Sq = 86.6%
R-Sq(adj) = 85.7%
Predicted Values for New Observations
New
Obs
Fit
SE Fit
95% CI
95% PI
1
550.74
20.08
(507.68, 593.79)
(458.02, 643.45)
2
570.49
21.92
(523.47, 617.51)
(475.87, 665.11)
3
590.25
23.81
(539.18, 641.31)
(493.55, 686.94)
4
610.00
25.72
(554.83, 665.17)
(511.08, 708.93)
Quarter 17:
95% P.I. for y17* = d17 :
(458.02, 643.45)
See p. 339 in the textbook
Error bound: (643.45 – 458.02)/2 = 92.715
 Approx. 95% P.I. for y17 : (621.33 – 92.715, 621.33 + 92.715 ) 
(529, 714)
Analogously for quarters 18-20:
Error bound of P:I: for d18 :
(665.11 – 475.87)/2 = 94.62
 Approx 95% P.I. for y18:
(687, 876)
Error bound of P:I: for d19 :
(686.94 – 493.55)/2 = 96.695
 Approx 95% P.I. for y19:
(417, 610)
Error bound of P:I: for d20 :
(708.93 – 511.08)/2 = 98.925
 Approx 95% P.I. for y20:
(307, 505)
Does this method seem more appropriate for these data than the multiplicative
method?
y
800
Time series graph from
Excel Workbook:
700
600
500
400
y
300
200
100
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Seasonal variation increases with level  Support for the multiplicative model
Prediction intervals much wider with additive model  Strong support for the
multiplicative model