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

0.950

0.960

0.970

0.980

0.990

1.000

1.010

1.020

1.030

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

cl x ir

cl

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) ?

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.

Yes!

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

No differences compared to the multiplicative case.

-80.000

-60.000

-40.000

-20.000

0.000

20.000

40.000

60.000

80.000

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

cl x ir

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) ?

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

Yes!

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

0

100

200

300

400

500

600

700

800

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

y

Time series graph from Excel Workbook:

Seasonal variation increases with level Support for the multiplicative model

Prediction intervals much wider with additive model Strong support for the multiplicative model