Statistics for Business and Economics chapter 18 Time Series Analysis and Forecasting

Chapter 18
Time Series Analysis and Forecasting

Learning Objectives
1.

Be able to construct a time series plot and identify the underlying pattern in the data.

2.

Understand how to measure forecast accuracy.

3.

Be able to use smoothing techniques such as moving averages and exponential smoothing to
forecast a time series with a horizontal pattern.

4.

Know how simple linear regression and Holt’s linear exponential smoothing can be used to forecast
a time series with a linear trend.

5.

Be able to develop a quadratic trend equation and an exponential trend equation to forecast a time
series with a curvilinear or nonlinear trend.

6.

Know how to develop forecasts for a time series that has a seasonal pattern.

7.

Know how time series decomposition can be used to separate or decompose a time series into
season, trend, and irregular components.

8.

Be able to deseasonalize a time series.

9.

Know the definition of the following terms:
time series
time series plot
horizontal pattern
stationary time series
trend pattern
seasonal pattern
cyclical pattern
mean absolute error

mean squared error
mean absolute percentage error
moving average
weighted moving average
smoothing constant
time series decomposition
additive model
multiplicative model

18 - 1

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Chapter 18

Solutions:
1.

The following table shows the calculations for parts (a), (b), and (c).

Week
1
2
3
4
5
6
a.

MAE = 22/5 = 4.4

b.

MSE = 104/5 = 20.8

c.

MAPE = 159.38/5 = 31.88

d.

Forecast for week 7 is 14

2.

Squared
Forecast
Error

5
3
5
6
3
22

Percentage
Error

25
9
25
36
9
104

-38.46

18.75
-45.45
35.29
-21.43
-51.30

Absolute Value
of Percentage
Error
38.46
18.75
45.45
35.29
21.43
159.38

The following table shows the calculations for parts (a), (b), and (c).

Week
1
2
3
4
5
6

3.

Time Series
Forecast

Value
Forecast
Error
18
13
18
-5
16
13
3
11
16
-5
17
11
6
14
17
-3
Totals

Absolute
Value of
Forecast
Error

Time Series
Forecast
Value
Forecast

Error
18
13
18.00
-5.00
16
15.50
0.50
11
15.67
-4.67
17
14.50
2.50
14
15.00
-1.00
Totals

Absolute
Value of
Forecast
Error

Squared
Forecast
Error

5.00
0.50

4.67
2.50
1.00
13.67

Percentage
Error

25.00
0.25
21.81
6.25
1.00
54.31

a.

MAE = 13.67/5 = 2.73

b.

MSE = 54.31/5 = 10.86

c.

MAPE = 105.89/5 = 21.18

d.

Forecast for week 7 is (18 + 13 + 16 + 11 + 17 + 14) / 6 = 14.83

-38.46
3.13
-42.45
14.71
-7.14
-70.21

The following table shows the measures of forecast error for both methods.

MAE
MSE
MAP
E

Exercise
1
4.40
20.80
31.88

18 ­ 2

Exercise
2
2.73
10.86
21.18

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Absolute Value
of Percentage
Error
38.46
3.13
42.45
14.71
7.14
105.86


Time Series Analysis and Forecasting
For each measure of forecast accuracy the average of all the historical data provided more
accurate forecasts than simply using the most recent value.
4.

a.
Month
1
2
3
4
5
6
7

Squared
Time Series

Forecast
Forecast
Value
Forecast
Error
Error
24
13
24
-11
121
20
13
7
49
12
20
-8
64
19
12
7
49
23
19
4
16
15
23
-8

64
Total
363

MSE = 363/6 = 60.5
Forecast for month 8 = 15
b.
Week
1
2
3
4
5
6
7

Squared
Time Series
Forecast
Forecast
Value
Forecast
Error
Error
24
13
24.00
-11.00
121.00
20

18.50
1.50
2.25
12
19.00
-7.00
49.00
19
17.25
1.75
3.06
23
17.60
5.40
29.16
15
18.50
-3.50
12.25
Total
216.72

MSE = 216.72/6 = 36.12
Forecast for month 8 = (24 + 13 + 20 + 12 + 19 + 23 + 15) / 7 = 18
c.

5.

The average of all the previous values is better because MSE is smaller.

a.

18 ­ 3

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Chapter 18

The data appear to follow a horizontal pattern.
b.

Three-week moving average.
Week
1
2
3
4
5
6

Squared
Time Series
Forecast
Forecast
Value
Forecast
Error
Error

18
13
16
11
15.67
-4.67
21.78
17
13.33
3.67
13.44
14
14.67
-0.67
0.44
Total
35.67

MSE = 35.67/3 = 11.89.
The forecast for week 7 = (11 + 17 + 14) / 3 = 14
c.

Smoothing constant = .2
Week
1
2
3
4
5
6

Squared
Time Series
Forecast
Forecast
Value
Forecast
Error
Error
18
13
18.00
-5.00
25.00
16
17.00
-1.00
1.00
11
16.80
-5.80
33.64
17
15.64
1.36
1.85
14
15.91
-1.91
3.66

Total
65.15

MSE = 65.15/5 = 13.03
The forecast for week 7 is .2(14) + (1 - .2)15.91 = 15.53
d.
e.

The three-week moving average provides a better forecast since it has a smaller MSE.
Smoothing constant = .4

18 ­ 4

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Time Series Analysis and Forecasting

Week
1
2
3
4
5
6

Squared
Time Series
Forecast

Forecast
Value
Forecast
Error
Error
18
13
18.00
-5.00
25.00
16
16.00
0.00
0.00
11
16.00
-5.00
25.00
17
14.00
3.00
9.00
14
15.20
-1.20
1.44
Total
60.44

MSE = 60.44/5 = 12.09

The exponential smoothing forecast using α = .4 provides a better forecast than the exponential
smoothing forecast using α = .2 since it has a smaller MSE.
6.

a.

The data appear to follow a horizontal pattern.
Three-week moving average.
Week
1
2
3
4
5
6
7

Squared
Time Series
Forecast
Forecast
Value
Forecast
Error
Error
24
13
20
12
19.00

-7.00
49.00
19
15.00
4.00
16.00
23
17.00
6.00
36.00
15
18.00
-3.00
9.00
Total
110.00

MSE = 110/4 = 27.5.
The forecast for week 8 = (19 + 23 + 15) / 3 = 19

18 ­ 5

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Chapter 18
b.

Smoothing constant = .2

Week
1
2
3
4
5
6
7

Squared
Time Series
Forecast
Forecast
Value
Forecast
Error
Error
24
13
24.00
-11.00
121.00
20
21.80
-1.80
3.24
12
21.44
-9.44
89.11

19
19.55
-0.55
0.30
23
19.44
3.56
12.66
15
20.15
-5.15
26.56
Total
252.87

MSE = 252.87/6 = 42.15
The forecast for week 8 is .2(15) + (1 - .2)20.15 = 19.12
c.

The three-week moving average provides a better forecast since it has a smaller MSE.

d.

Smoothing constant = .4
Time Series
Forecast
Value
Forecast
Error
24

13
24.00
-11.00
20
19.60
0.40
12
19.76
-7.76
19
16.66
2.34
23
17.59
5.41
15
19.76
-4.76
Total
MSE = 238.72/6 = 39.79
Week
1
2
3
4
5
6
7

7.

a.

121.00
0.16
60.22
5.49
29.23
22.62
238.72

The exponential smoothing forecast using α = .4 provides a better forecast than the exponential
smoothing forecast using α = .2 since it has a smaller MSE.

Week
1
2
3
4
5
6
7
8
9
10
11
12
b.

Squared

Value of
Forecast
Error

Time-Series
Value
17
21
19
23
18
16
20
18
22
20
15
22

4-Week
Moving
Average
Forecast

(Error)2

20.00
20.25
19.00
19.25

18.00
19.00
20.00
18.75
Totals

4.00
18.06
1.00
1.56
16.00
1.00
25.00
10.56
77.18

5-Week
Moving
Average
Forecast

(Error)2

19.60
19.40
19.20
19.00
18.80
19.20
19.00

MSE(4-Week) = 77.18 / 8 = 9.65

18 ­ 6

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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

12.96
0.36
1.44
9.00
1.44
17.64
9.00
51.84


Time Series Analysis and Forecasting
MSE(5-Week) = 51.84 / 7 = 7.41
c.
8.

For the limited data provided, the 5-week moving average provides the smallest MSE.

a.
Week
1
2
3

4
5
6
7
8
9
10
11
12
b.

Time-Series
Value
17
21
19
23
18
16
20
18
22
20
15
22

Weighted Moving
Average Forecast

19.33

21.33
19.83
17.83
18.33
18.33
20.33
20.33
17.83

Forecast
Error

(Error)2

3.67
-3.33
-3.83
2.17
-0.33
3.67
-0.33
-5.33
4.17
Total

13.47
11.09
14.67
4.71
0.11

13.47
0.11
28.41
17.39
103.43

MSE = 103.43 / 9 = 11.49
Prefer the unweighted moving average here; it has a smaller MSE.

c.

You could always find a weighted moving average at least as good as the unweighted one.
Actually the unweighted moving average is a special case of the weighted ones where the
weights are equal.
The following tables show the calculations for = .1 

9.

Week
1
2
3
4
5
6
7
8
9
10
11

12

Time Series
Value
Forecast
17
21
17.00
19
17.40
23
17.56
18
18.10
16
18.09
20
17.88
18
18.10
22
18.09
20
18.48
15
18.63
22
18.27

Absolute

Value of
Forecast
Error

Forecast
Error
4.00
1.60
5.44
-0.10
-2.09
2.12
-0.10
3.91
1.52
-3.63
3.73
Totals

4.00
1.60
5.44
0.10
2.09
2.12
0.10
3.91
1.52
3.63
3.73

28.24

Squared
Forecast
Error
16.00
2.56
29.59
0.01
4.37
4.49
0.01
15.29
2.31
13.18
13.91
101.72

Absolute Value
of Percentage
Error

Percentage
Error
19.05
8.42
23.65
-0.56
-13.06
10.60

-0.56
17.77
7.60
-24.20
16.95
65.67

19.05
8.42
23.65
0.56
13.06
10.60
0.56
17.77
7.60
24.20
16.95
142.42

The following tables show the calculations for = .2
Week

Time Series Forecast
Value

Forecast
Error

Absolute

Value of
Forecast

18 ­ 7

Squared
Forecast
Error

Percentage
Error

© 2010 Cengage Learning. All Rights Reserved.
May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Absolute Value
of Percentage
Error


Chapter 18
Error
1
2
3
4
5
6
7
8

9
10
11
12

17
21
19
23
18
16
20
18
22
20
15
22

17.00
17.80
18.04
19.03
18.83
18.26
18.61
18.49
19.19
19.35
18.48

4.00
1.20
4.96
-1.03
-2.83
1.74
-0.61
3.51
0.81
-4.35
3.52
Totals

4.00
1.20
4.96
1.03
2.83
1.74
0.61
3.51
0.81
4.35
3.52
28.56

16.00
1.44
24.60
1.06

8.01
3.03
0.37
12.32
0.66
18.92
12.39
98.80

19.05
6.32
21.57
-5.72
-17.69
8.70
-3.39
15.95
4.05
-29.00
16.00
35.84

19.05
6.32
21.57
5.72
17.69
8.70
3.39
15.95

4.05
29.00
16.00
147.44

a. MSE for = .1 = 101.72/11 = 9.25
MSE for = .2 = 98.80/11 = 8.98

= .2 provides more accurate forecasts based upon MSE
b.  MAE for = .1 = 28.24/11 = 2.57
MAE for = .2 = 28.56/11 = 2.60
= .1 provides more accurate forecasts based upon MAE; but, they are very close.
c.

MAPE for = .1 = 142.42/11 = 12.95%
MAPE for = .2 = 147.44/11 = 13.40%

= .1 provides more accurate forecasts based upon MAPE
10. a.

F13 = .2Y12 + .16Y11 + .64(.2Y10 + .8F10) = .2Y12 + .16Y11 + .128Y10 + .512F10
F13 = .2Y12 + .16Y11 + .128Y10 + .512(.2Y9 + .8F9) = .2Y12 + .16Y11 + .128Y10 + .1024Y9 + .4096F9
F13 = .2Y12 + .16Y11 + .128Y10 + .1024Y9 + .4096(.2Y8 + .8F8) = .2Y12 + .16Y11 + .128Y10 + .1024Y9 + .
08192Y8 + .32768F8

b.

The more recent data receives the greater weight or importance in determining the forecast. The moving
averages method weights the last n data values equally in determining the forecast.

11. a.

18 ­ 8

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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Time Series Analysis and Forecasting

The first two time series values may be an indication that the time series has shifted to a
new higher level, as shown by the remainig 10 values. But, overall, the time series plot
exhibits a horizontal pattern.
b.
Month
1
2
3
4
5
6
7
8
9
10
11
12

Yt
80

82
84
83
83
84
85
84
82
83
84
83

3-Month Moving
Averages
Forecast

82.00
83.00
83.33
83.33
84.00
84.33
83.67
83.00
83.00
Totals

(Error)2

1.00

0.00
0.45
2.79
0.00
5.43
0.45
1.00
0.00
11.12

α=2
Forecast
80.00
80.40
81.12
81.50
81.80
82.24
82.79
83.03
82.83
82.86
83.09

MSE(3-Month) = 11.12 / 9 = 1.24
MSE(α = .2) = 39.06 / 11 = 3.55
A 3-month moving average provides the most accurate forecast using MSE
c.

3-month moving average forecast = (83 + 84 + 83) / 3 = 83.3

12. a.

18 ­ 9

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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(Error)2
4.00
12.96
3.53
2.25
4.84
7.62
1.46
1.06
0.03
1.30
0.01
39.06


Chapter 18

The data appear to follow a horizontal pattern.
b.
Month
1
2

3
4
5
6
7
8
9
10
11
12

Time-Series
Value
9.5
9.3
9.4
9.6
9.8
9.7
9.8
10.5
9.9
9.7
9.6
9.6

3-Month Moving
Average Forecast

9.40

9.43
9.60
9.70
9.77
10.00
10.07
10.03
9.73

Totals

(Error)2

0.04
0.14
0.01
0.01
0.53
0.01
0.14
0.18
0.02
1.08

4-Month Moving
Average Forecast

9.45
9.53
9.63

9.73
9.95
9.98
9.97
9.92

MSE(3-Month) = 1.08 / 9 = .12
MSE(4-Month) = 1.09 / 8 = .14
Use 3-Month moving averages.
c.

Forecast = (9.7 + 9.6 + 9.6) / 3 = 9.63

13. a.

18 ­ 10

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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(Error)2

0.12
0.03
0.03
0.59
0.00
0.08
0.14
0.10

1.09


Time Series Analysis and Forecasting

The data appear to follow a horizontal pattern.
b.
Month
1
2
3
4
5
6
7
8
9
10
11
12

Time-Series
Value
240
350
230
260
280
320
220

310
240
310
240
230

3-Month Moving
Average Forecast

273.33
280.00
256.67
286.67
273.33
283.33
256.67
286.67
263.33
Totals

(Error)2

177.69
0.00
4010.69
4444.89
1344.69
1877.49
2844.09
2178.09

1110.89
17,988.52

α = .2
Forecast
240.00
262.00
255.60
256.48
261.18
272.95
262.36
271.89
265.51
274.41
267.53

(Error)2
12100.00
1024.00
19.36
553.19
3459.79
2803.70
2269.57
1016.97
1979.36
1184.05
1408.50
27,818.49

MSE(3-Month) = 17,988.52 / 9 = 1998.72
MSE(α = .2) = 27,818.49 / 11 = 2528.95
Based on the above MSE values, the 3-month moving averages appears better. However,
exponential smoothing was penalized by including month 2 which was difficult for any
method to forecast. Using only the errors for months 4 to 12, the MSE for exponential
smoothing is:
MSE(α = .2) = 14,694.49 / 9 = 1632.72
Thus, exponential smoothing was better considering months 4 to 12.
c.

Using exponential smoothing,
F13 = α Y12 + (1 - α)F12 = .20(230) + .80(267.53) = 260

14. a.

18 ­ 11

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Chapter 18

The data appear to follow a horizontal pattern.
b.

Smoothing constant = .3.
Month t
1

2
3
4
5
6
7
8
9
10
11
12

Time-Series Value
Yt
105
135
120
105
90
120
145
140
100
80
100
110

Forecast Ft Forecast Error Squared Error
Y t - Ft
(Yt - Ft)2

105.00
114.00
115.80
112.56
105.79
110.05
120.54
126.38
118.46
106.92
104.85

30.00
6.00
-10.80
-22.56
14.21
34.95
19.46
-26.38
-38.46
-6.92
5.15

Total

900.00
36.00
116.64
508.95

201.92
1221.50
378.69
695.90
1479.17
47.89
26.52
5613.18

MSE = 5613.18 / 11 = 510.29
Forecast for month 13: F13 = .3(110) + .7(104.85) = 106.4

c. Smoothing constant = .5
Time-Series Value

Forecast Ft
18 ­ 12

Forecast Error

Squared Error

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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Time Series Analysis and Forecasting
Month t
1
2

3
4
5
6
7
8
9
10
11
12

Yt
105
135
120
105
90
120
145
140
100
80
100
110

Y t - Ft
105
120
120
112.50

101.25
110.63
127.81
133.91
116.95
98.48
99.24

30.00
0.00
-15.00
-22.50
18.75
34.37
12.19
-33.91
-36.95
1.52
10.76

(Yt - Ft)2
900.00
0.00
225.00
506.25
351.56
1181.30
148.60
1149.89
1365.30

2.31
115.78
5945.99

MSE = 5945.99 / 11 = 540.55
Forecast for month 13: F13 = .5(110) + .5(99.24) = 104.62
Conclusion: a smoothing constant of .3 is better than a smoothing constant of .5 since the MSE is less for 0.3.
15. a.

You might think the time series plot shown above exhibits some trend. But, this is simply due to
the fact that the smallest value on the vertical axis is 7.1, as shown by the following version of the
plot.

18 ­ 13

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Chapter 18

In other words, the time series plot shows an underlying horizontal pattern.
b/c.
Week
1
2
3
4
5
6

7
8
9
10
d.

Time-Series
Value
7.35
7.40
7.55
7.56
7.60
7.52
7.52
7.70
7.62
7.55

α = .2
Forecast
7.35
7.36
7.40
7.43
7.46
7.48
7.48
7.53
7.55

(Error)2
.0025
.0361
.0256
.0289
.0036
.0016
.0484
.0081
.0000
.1548

α = .3
Forecast
7.35
7.36
7.42
7.46
7.50
7.51
7.51
7.57
7.58

MSE(α = .2) = .1548 / 9 = .0172
MSE(α = .3) = .1178 / 9 = .0131
Use α = .3.
F11 = .3Y10 + .7F10 = .3(7.55) + .7(7.58) = 7.57

16. a.

18 ­ 14

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(Error)2
.0025
.0361
.0196
.0196
.0004
.0001
.0361
.0025
.0009
.1178


Time Series Analysis and Forecasting

The time series plot indicates a possible linear trend in the data. This could be due to decreasing
viewer interest in watching the Masters. But, closer inspection of the data indicates that the two
highest ratings correspond to years 1997 and 2001, years in which Tiger Woods won the
tournament. In fact, four of the five highest ratings occurred when Tiger Woods has won the
tournament. So, instead of an underlying linear trend in the time series, the pattern observed may
be simply due to the effect Tiger Woods has on ratings and not necessarily on any long-term
decrease in viewer interest.
b.

The methods disucssed in this section are only applicable for a time series that has a horizontal
pattern. So, if there is really a long-term linear trend in the data, the methods disucssed in this
section are not appropriate.

c.

The following time series plot shows the ratings for years 2002 – 2008.

The time series plot for the data for years 2002 – 2008 exhibits a horizontal pattern. It seems
reasonable to conclude that the extreme values observed in 1997 and 2001 are more attributable to
viewer interest in the performance of Tiger Woods. Basing the forecast on years 2002 – 2008 does
seem reasonable. But, because of the injury that Tiger Woods experienced in the 2008 season, if
he is able to play in the 2009 Masters then the rating for 2009 may be significantly higher than
suggested by the data for years 2002 – 2008. These types of issues are what make forecasting in
18 ­ 15

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Chapter 18
practice so difficult. For the methods to work, we have to be able to assume that the pattern in the
past is appropriate for the future. But, because of the great influence Tiger Woods has on viewer
interest, making this assumption for this time series may not be appropriate.
17. a.

The time series plot shows a linear trend.
n

b.

t 

�t
t 1

n

n



15
3
5

Y 

(t  t )(Yt  Y )  21

�Y
t 1

n

t



55
 11
5

( t  t ) 2  10

n

b1 

�(t  t )(Y  Y )
t

t 1

n

�(t  t )



2

21
 2.1
10

t 1

b0  Y  b1 t  11  (2.1)(3)  4.7

Tt  4.7  2.1t
c.

18.

T6  4.7  2.1(6)  17.3

Holt’s linear exponential smoothing using α = .3 and β = .5
t

Value

Estimated
Level

Estimated
Trend
18 ­ 16

Forecast

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Time Series Analysis and Forecasting
1
2
3

4
5

6
11
9
14
15

6.00
11.00
13.90
16.70
18.55

5.00
5.00
3.95
3.37
2.61

11.00
16.00
17.85
20.07

Forecast for week 6 = 18.55 + 2.61 = 21.16
19. a.

The time series plot shows a linear trend.

n

b.

t 

�t
t 1

n

n



28
4
7

Y 

(t  t )(Yt  Y )  138

�Y
t 1

n

t



700
 100
7

(t  t ) 2  28

n

b1 

�(t  t )(Y  Y )
t

t 1

n

�(t  t )



2

138
 4.9286
28

t 1

b0  Y  b1 t  100  ( 4.9286)(4)  119.714

Tt  119.714  4.9286t
c.

T8  119.714  4.9286(8)  80.29

20. a.

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Chapter 18

The time series plot exhibits a curvilinear trend.
b.

Using Minitab, the linear trend equation is Tt =107.857 -28.9881 t +2.65476 t 2

c.

Tt =107.857 -28.9881(8) +2.65476 (8) 2 = 45.86

21. a.

The time series plot shows a linear trend

n

b.

t 

�t
t 1

n

n



45
5
9

Y 

(t  t )(Yt  Y )  87.4

�Y
t 1

n

t



108
 12
9

(t  t ) 2  60

18 ­ 18

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Time Series Analysis and Forecasting

n

b1 

�(t  t )(Y  Y )
t

t 1

n

�(t  t )



2

87.4
 1.4567
60

t 1

b0  Y  b1t  12  (1.4567)(5)  4.7165

Tt  4.7165  1.4567t
c.

T10  4.7165  1.4567(10)  19.28

22. a.

The time series plot shows a downward linear trend
n

b.

t 

�t
t 1

n

n



28
4
7

Y 

(t  t )(Yt  Y )  19.6

�Y
t 1

n

t



77
 11
7

(t  t ) 2  28

n

b1 

�(t  t )(Y  Y )
t

t 1

n

�(t  t )

2



19.6
 .7
28

t 1

b0  Y  b1 t  11  ( .7)(4)  13.8

Tt  13.8  .7t
c.

2010 corresponds to time period t = 8. T8  13.8  .7(8)  8.2

d.

If SCF can continue to decrease the percentage of funds spent on administrative and fund-raising

by .7% per year, the forecast of expenses for 2015 is 4.70%.
18 ­ 19

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Chapter 18
23. a.

The time series plot shows an upward linear trend
n

b.

t 

�t
t 1

n

n



36
 4.5
8

Y 

(t  t )(Yt  Y )  74.5

�Y
t 1

t

n



223.8
 27.98
8

(t  t ) 2  42

n

b1 

�(t  t )(Y  Y )
t

t 1

n

�(t  t )

2



74.5
 1.7738
42

t 1

b0  Y  b1t  27.98  (1.7738)(4.5)  19.9979

Tt  19.9979  1.7738t
c.

The average cost/unit has been increasing by approximately $1.77 per year.

d.

T9  19.9979  1.7738(9)  35.96

18 ­ 20

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Time Series Analysis and Forecasting

24. a.

The time series plot shows a linear trend.
b.

Using Minitab, the linear trend equation is Tt  7.5623  .07541t

c.

A forecast for August corresponds to t = 11.

T11  7.5623  .07541(11)  6.7328
d.

Given the uncertainty in global market conditions, making a prediction for December using only
time is not recommended.

25. a.

A linear trend is not appropriate.
b.

The following output shows the results of using Minitab’s Time Series – Trend Analysis
procedure to fit a quadratic trend to the time series.
18 ­ 21

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Chapter 18

The quadratic trend equation is Tt  472.7  62.9t  5.303t 2
c.

T11  472.7  62.9(11)  5.303(11) 2  422

d.

Sales appear to have bottomed-out in year 6 and appear to be increasing in a linear fashion for
years 6 – 10. So, another alternative would be to use the data for years 6 – 10 (or 5 -10) to
develop a linear trend equation to forecast sales in year 11.

26. a.

A linear trend is not appropriate.

b.

The following output shows the results of using Minitab’s Time Series – Trend Analysis
procedure to fit a quadratic trend to the time series.

18 ­ 22

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Time Series Analysis and Forecasting

The quadratic trend equation is Tt  5.702  2.889t  .1618t 2
c.

T11  5.702  2.889(11)  .1618(11) 2  17.90

27. a.

The time series plot indicates a slight curvature in the data.

b.

The following output shows the results of using Minitab’s Time Series – Trend Analysis
procedure to fit a quadratic trend equation to the time series.

18 ­ 23

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Chapter 18

c.

The following output shows the results of using Minitab’s Time Series – Trend Analysis
procedure to fit an exponential trend equation to the time series.

d.

The following output shows the results of using Minitab’s Time Series – Trend Analysis

procedure to fit a linear trend equation to the time series.

18 ­ 24

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Time Series Analysis and Forecasting

e. We recommend using the quadratic trend equation because it provides the best fit (smallest
MSE).
f.

Using the quadratic trend equation the estimate of value for 2009 (t = 12) is
T12  217.2  20.10(12)  5.196(12) 2  1207

28. a.

The time series plot shows a horizontal pattern. But, there is a seasonal pattern in the data. For
instance, in each year the lowest value occurs in quarter 2 and the highest value occurs in quarter
4.
b.

A portion of the Minitab regression output is shown below.
The regression equation is
Value = 77.0 - 10.0 Qtr1 - 30.0 Qtr2 - 20.0 Qtr3

c.

The quarterly forecasts for next year are as follows:
18 ­ 25

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