Solution Manual for Business Forecasting, 9th Edition

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CONTENTS
Preface
Chapter 2 A Review of Basic Statistical Concepts 1
Problems 1
Cases:
Alcam Electronics 8
Mr. Tux 8
Alomega Food Stores 8
Chapter 3 Exploring Data Patterns and Choosing a Forecasting Technique 9
Problems 9
Cases:
Murphy Brothers Furniture 20
Mr. Tux 20
Consumer Credit Counseling 21
Alomega Food Stores 23
Surtido Cookies 23
Chapter 4 Moving Averages and Smoothing Methods 25
Problems 25
Cases:
The Solar Alternative Company 43
Mr. Tux 45
Consumer Credit Counseling 46
Murphy Brothers Furniture 47
Five-year Revenue Projection for Downtown Radiology 49
Web Retailer 49
Southwest Medical Center 52
Surtido Cookies 54
Chapter 5 Time Series And Their Components 56
Problems 56
Cases:
The Small Engine Doctor 76
Mr. Tux 79
Consumer Credit Counseling 80
Murphy Brothers Furniture 83
AAA Washington 84
Alomega Food Stores 87
Surtido Cookies 88
Southwest Medical Center 90
Chapter 6 Regression Analysis 94

Problems 94
Cases:
Tiger Transport Company 113
Butcher Products, Inc. 113
Ace Manufacturing 114
Mr. Tux 114
Consumer Credit Counseling 115
AAA Washington 117
Chapter 7 Multiple Regression 120
Problems 120
Cases:
The Bond Market 140
AAA Washington 141
Fantasy Baseball (A) 143
Fantasy Baseball (B) 145
Chapter 8 Regression With Time Series Data 146
Problems 146
Cases:
Business Activity Index for Spokane County 164
Restaurant Sales 165
Mr. Tux 165
Consumer Credit Counseling 166
AAA Washington 168
Alomega Food Stores 169
Surtido Cookies 169
Southwest Medical Center 171
Chapter 9 Box-Jenkins (ARIMA) Methodology 173
Problems 173
Cases:
Restaurant Sales 202
Mr. Tux 203
Consumer Credit Counseling 205
The Lydia E. Pinkham Medicine Company 207
City of College Station 209
UPS Air Finance Division 210
AAA Washington 212
Web Retailer 213
Surtido Cookies 215
Southwest Medical Center 218
Chapter 10 Judgmental Elements in Forecasting 223
Problems 223

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PREFACE
The goal of the ninth edition of Business Forecasting remains the same as that
of the previous editions: To present the basic statistical techniques that are useful for
preparing individual business forecasts and long-range plans. This instructor’s manual
contains answers to chapter-end problems and comments on the case studies that
appear at the end of every chapter.
Our work in forecasting over many years has taught us that intuition and good
judgment are essential components of a good forecasting process, a point we stress in
Chapters 1 and 11. This is a difficult concept to put across in the remaining chapters,
which deal with important forecasting techniques involving data analysis. We hope
that the instructor can bring a measure of real-world common sense to the study of
forecasting to supplement the quantitative material with which this instructor’s manual
is concerned.
We also hope that students can gain practical experience through hands on use
of computer programs in their study of forecasting. Our solutions to problems and
cases here rely heavily on Minitab software. Forecasters have just begun to tap the
potential offered by resources on the Internet.
Data sets that appear in the text are available in several formats on the CD
included with the book and on our website maintained by Prentice Hall at
www.prenhall.com/Hanke.
Finally, we hope you find the ninth edition of Business Forecasting useful.
Comments for improvement are welcome. We can be reached at the following email
addresses:
John Hanke – john_hanke@msn.com
Dean Wichern – d-wichern@tamu.edu

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1
CHAPTER 2
A REVIEW OF BASIC STATISTICAL CONCEPTS
ANSWERS TO PROBLEMS AND CASES
1. Descriptive Statistics
Variable N Mean Median StDev SE Mean
Orders 28 21.32 17.00 13.37 2.53
Variable Min Max Q1 Q3
Orders 5.00 54.00 11.25 28.75
a.X = 21.32
b. S = 13.37
c. S2 = 178.76
d. If the policy is successful, smaller orders will be eliminated and the mean will
increase.
e. If the change causes all customers to consolidate a number of small orders into
large orders, the standard deviation will probably decrease. Otherwise, it is very
difficult to tell how the standard deviation will be affected.
f. The best forecast over the long-term is the mean of 21.32.
2. Descriptive Statistics
Variable N Mean Median StDev SE Mean
Prices 12 176654 180000 39440 11385
Variable Min Max Q1 Q3
Prices 121450 253000 138325 205625X
= 176,654 and S = 39,440
3. a. Point estimate:%76.10=X
b. 1− = .95  Z = 1.96, n = 30,71.13,76.10 == SX( ) ( )
91.476.1030/71.1396.176.10/96.1 == nSX
(5.85%, 15.67%)
c. df = 30−1 = 29, t = 2.045

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2( ) ( )
12.576.1030/71.13045.276.10/045.2 == nSX
(5.64%, 15.88%)
d. We see that the 95% confidence intervals in b and c are not much different
because the multipliers 1.96 and 2.045 are nearly the same magnitude.
This explains why a sample of size n = 30 is often taken as the cutoff between
large and small samples.
4. a. Point estimate:63
2
59.10241.23 =
+
=X
95% error margin: (102.59 − 23.41)/2 = 39.59
b. 1− = .90  Z = 1.645,2.2096.1/59.39/,63 === nSX( )
23.3363)2.20(645.163/645.1 == nSX
(29.77, 96.23)
5. H0:  = 12.1 n = 100  = .05
H1:  > 12.1 S = 1.7X = 13.5
Reject H0 if Z > 1.645
Z =100
7.1
1.125.13 − = 8.235
Reject H0 since the computed Z (8.235) is greater than the critical Z (1.645). The mean has
increased.
6. point estimate: 8.1 seats
interval estimate: 8.1  1.9649
7.5  6.5 to 9.7 seats
Forecast 8.1 empty seats per flight; very likely the mean number of empty seats will lie
between 6.5 and 9.7.
7. n = 60,87.,60.5 == SX9.5:
9.5:
1
0

=


H
H
two-sided test,  = .05, critical value: |Z|= 1.96
Test statistic:67.2
60/87.
9.560.5
/
9.5 −=
−
=
−
= nS
X
Z
Since |−2.67| = 2.67 > 1.96, reject0H at the 5% level. The mean satisfaction rating is
different from 5.9.
p-value: P(Z < − 2.67 or Z > 2.67) = 2 P(Z > 2.67) = 2(.0038) = .0076, very strong
evidence against0H .

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3
8. df = n −1 = 14 −1 = 13,52.,31.4 == SX4:
4:
1
0

=


H
H
one-sided test,  = .05, critical value: t = 1.771
Test statistic:23.2
14/52.
431.4
/
4 =
−
=
−
= nS
X
t
Since 2.23 > 1.771, reject0H at the 5% level. The medium-size serving contains an
average of more than 4 ounces of yogurt.
p-value: P(t > 2.23) = .022, strong evidence against0H
9. H0:  = 700 n = 50  = .05
H1:   700 S = 50X = 715
Reject H0 if Z < -1.96 or Z > 1.96
Z =50
50
700715 − = 2.12
Since the calculated Z is greater than the critical Z (2.12 > 1.96), reject the null hypothesis.
The forecast does not appear to be reasonable.
p-value: P(Z < − 2.12 or Z > 2.12) = 2 P(Z > 2.12) = 2(.017) = .034, strong evidence
against0H
10. This problem can be used to illustrate how a random sample is selected with Minitab. In
order to generate 30 random numbers from a population of 200 click the following menus:
  Calc>Random Data>Integer

The Integer Distribution dialog box shown in the figure below appears. The number of
random digits desired, 30, is entered in the Number of rows of data to generate space. C1
is entered for Store in column(s) and 1 and 200 are entered as the Minimum and
Maximum values. OK is clicked and the 30 random numbers appear in Column 1 of the
worksheet.

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5
12. a.
b. Positive linear relationship
c. Y = 2312 Y2 = 515,878 X = 53.7
X2 = 282.55 XY = 12,029.3 r = .95Yˆ
= 32.5 + 36.4XYˆ
= 32.5 + 36.4(5.2) = 222
13. This is a good population for showing how random samples are taken. If three-digit
random numbers are generated from Minitab as demonstrated in Problem 10, the selected
items for the sample can be easily found. In this population,  = 0.06 so most
students will get a sample correlation coefficient r close to 0. The least squares line will,
in
most cases, have a slope coefficient close to 0, and students will not be able to reject the
null hypothesis H0: β1 = 0 (or, equivalently, ρ = 0) if they carry out the hypothesis test.
14. a.

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6
b. Rent = 275.5 + .518 Size
c. Slope coefficient = .518  Increase of $.518/month for each additional square
foot of space.
d. Size = 750  Rent = 275.5 + .518(750) = $664/month
15. n = 175,3.10,2.45 == SX
Point estimate:2.45=X
98% confidence interval: 1− = .98  Z = 2.33( ) ( )
8.12.45175/3.1033.22.45/33.2 == nSX
 (43.4, 47.0)
Hypothesis test:44:
44:
1
0

=


H
H
two-sided test,  = .02, critical value: |Z|= 2.33
Test statistic:54.1
175/3.10
442.45
/
44 =
−
=
−
= nS
X
Z
Since |Z| = 1.54 < 2.33, do not reject0H at the 2% level.
As expected, the results of the hypothesis test are consistent with the confidence
interval for ;  = 44 is not ruled out by either procedure.

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7
16. a.700,63:
700,63:
1
0

=


H
H
b.3.4:
3.4:
1
0

=


H
H
c.1300:
1300:
1
0

=


H
H
17. Large sample 95% confidence interval for mean monthly return μ:)78.,98.2(88.110.1
39
99.5
96.110.1 −−=−
μ = .94 (%) is not a realistic value for mean monthly return of client’s
account since it falls outside the 95% confidence interval. Client may have a
case.
18. a.
b. r = .581, positive linear association between wages and length of service.
Other variables affecting wages may be size of bank and previous experience.
c. WAGES = 324.3 + 1.006 LOS
WAGES = 324.3 + 1.006 (80) = 405

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8
CASE 2-1: ALCAM ELECTRONICS
In our consulting work, business people sometimes tell us that business schools teach a risk-
taking attitude that is too conservative. This is often reflected, we are told, in students choosing too
low a significance level: such a choice requires extreme evidence to move one from the status quo.
This case can be used to generate a discussion on this point as David chooses  = .01 and ends up
"accepting" the null hypothesis that the mean lifetime is 5000 hours.
Alice's point is valid: the company may be put in a bad position if it insists on very dramatic
evidence before abandoning the notion that its components last 5000 hours. In fact, the indifference
 (p-value) is about .0375; at any higher level the null hypothesis of 5000 hours is rejected.
CASE 2-2: MR. TUX
In this case, John Mosby tries some primitive ways of forecasting his monthly sales. The
things he tries make some sort of sense, at least for a first cut, given that he has had no formal
training in forecasting methods. Students should have no trouble finding flaws in his efforts, such
as:
1. The mean value for each year, if projected into the future, is of little value since
month-to-month variability is missing.
2. His free-hand method of fitting a regression line through his data can be improved
upon using the least squares method, a technique now found on inexpensive hand
calculators. The large standard deviation for his monthly data suggests
considerable month-to-month variability and, perhaps, a strong
seasonal effect, a factor not accounted for when the values for a year are averaged.
Both the hand-fit regression line and John's interest in dealing with the monthly seasonal
factor suggest techniques to be studied in later chapters. His efforts also point out the value of
learning about well-established formal forecasting methods rather than relying on intuition and
very simple methods in the absence of knowledge about forecasting. We hope students will begin
to appreciate the value of formal forecasting methods after learning about John's initial efforts.
CASE 2-3: ALOMEGA FOOD STORES
Julie’s initial look at her data using regression analysis is a good start. She found that the
r-squared value of 36% is not very high. Using more predictor variables, along with examining
their significance in the equation, seems like a good next step. The case suggests that other
techniques may prove even more valuable, techniques to be discussed in the chapters that follow.
Examining the residuals of her equation might prove useful. About how large are these
errors? Are forecast errors in this range acceptable to her? Do the residuals seem to remain in
the same range over time, or do they increase over time? Are a string of negative residuals
followed by a string of positive residuals or vice versa? These questions involve a deeper
understanding of forecasting using historical values and these matters will be discussed more
fully in later chapters.
CHAPTER 3

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9
EXPLORING DATA PATTERNS AND
CHOOSING A FORECASTING TECHNIQUE
ANSWERS TO PROBLEMS AND CASES
1. Qualitative forecasting techniques rely on human judgment and intuition. Quantitative
forecasting techniques rely more on manipulation of historical data.
2. A time series consists of data that are collected, recorded, or observed over successive
increments of time.
3. The secular trend of a time series is the long-term component that represents the growth or
decline in the series over an extended period of time. The cyclical component is the wave-
like fluctuation around the trend. The seasonal component is a pattern of change that
repeats itself year after year. The irregular component is that part of the time
series remaining after the other components have been removed.
4. Autocorrelation is the correlation between a variable, lagged one or more period, and itself.
5. The autocorrelation coefficient measures the correlation between a variable, lagged one or
more periods, and itself.
6. The correlogram is a useful graphical tool for displaying the autocorrelations for various
lags of a time series. Typically, the time lags are shown on a horizontal scale and the
autocorrelation coefficients, the correlations between Yt and Yt-k, are displayed as vertical
bars at the appropriate time lags. The lengths and directions (from 0) of the bars indicate
the magnitude and sign of the of the autocorrelation coefficients. The lags at which
significant autocorrelations occur provide information about the nature of the time series.
7. a. nonstationary series
b. stationary series
c. nonstationary series
d. stationary series
8. a. stationary series
b. random series
c. trending or nonstationary series
d. seasonal series
e. stationary series
f. trending or nonstationary series
9. Naive methods, simple averaging methods, moving averages, and Box-Jenkins methods.
Examples are: the number of breakdowns per week on an assembly line having a uniform
production rate; the unit sales of a product or service in the maturation stage of its life

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10
cycle; and the number of sales resulting from a constant level of effort.
10. Moving averages, simple exponential smoothing, Holt's linear exponential smoothing,
simple regression, growth curves, and Box-Jenkins methods. Examples are: sales
revenues of consumer goods, demand for energy consumption, and use of raw materials.
Other examples include: salaries, production costs, and prices, the growth period of the
life cycle of a new product.
11. Classical decomposition, census II, Winters’ exponential smoothing, time series multiple
regression, and Box-Jenkins methods. Examples are: electrical consumption,
summer/winter activities (sports like skiing), clothing, and agricultural growing seasons,
retail sales influenced by holidays, three-day weekends, and school calendars.
12. Classical decomposition, economic indicators, econometric models, multiple regression,
and Box-Jenkins methods. Examples are: fashions, music, and food.
13. 1985 2,413 - 1999 2358 114
1986 2,407 -6 2000 2329 -29
1987 2,403 -4 2001 2345 16
1988 2,396 -7 2002 2254 -91
1989 2,403 7 2003 2245 -9
1990 2,443 40 2004 2279 34
1991 2,371 -72
1992 2,362 -9
1993 2,334 -28
1994 2,362 28
1995 2,336 -26
1996 2,344 8
1997 2,384 40
1998 2,244 -140
Yes! The original series has a decreasing trend.
14. 0  1.96 (1 80 ) = 0  1.96 (.1118) = 0  .219
15. a. MPE
b. MAPE
c. MSE or RMSE
16. All four statements are true.
17. a. r1 = .895
H0: ρ1 = 0H1: ρ1  0

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11

Reject if t < -2.069 or t > 2.069
SE(kr ) =n
r
k
i
i
−
=
+
1
1
2
21 =( )
24
21
11
1
2
1
−
=
+ i
r =24
1 = .204)SE(rk
11

−
= r
t
=.204
0895. − = 4.39
Since the computed t (4.39) is greater than the critical t (2.069), reject the null.
r2 = .788
 
H0: ρ2 = 0H1: ρ2  0

Reject if t < -2.069 or t > 2.069
   SE(kr ) =n
r
k
i
i
−
=
+
1
1
2
21 =( )
24
895.21
12
1
2

−
=
+ i =2 6
24
. = .33
  )SE(r1
11

−
= r
t= −.
.
788 0
33 = 2.39
Since the computed t (4.39) is greater than the critical t (2.069), reject the null.
b. The data are nonstationary. See plot below.

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15
a. Time Data Forecast Error
t YtYˆ t e tet e t
2t
t
Y
et
t
Y
e
1 .40 - - - - - -
2 .29 .40 -.11 .11 .0121 .3793 -.3793
3 .24 .29 -.05 .05 .0025 .2083 -.2083
4 .32 .24 .08 .08 .0064 .2500 .2500
5 .47 .32 .15 .15 .0225 .3191 .3191
6 .34 .47 -.13 .13 .0169 .3824 -.3824
7 .30 .34 -.04 .04 .0016 .1333 -.1333
8 .39 .30 .09 .09 .0081 .2308 .2308
9 .63 .39 .24 .24 .0576 .3810 .3810
10 .43 .63 -.20 .20 .0400 .4651 -.4651
11 .38 .43 -.05 .05 .0025 .1316 -.1316
12 .49 .38 .11 .11 .0121 .2245 .2245
13 .76 .49 .27 .27 .0729 .3553 .3553
14 .51 .76 -.25 .25 .0625 .4902 -.4902
15 .42 .51 -.09 .09 .0081 .2143 -.2143
16 .61 .42 .19 .19 .0361 .3115 .3115
17 .86 .61 .25 .25 .0625 .2907 .2907
18 .51 .86 -.35 .35 .1225 .6863 -.6863
19 .47 .51 -.04 .04 .0016 .0851 -.0851
20 .63 .47 .16 .16 .0256 .2540 .2540
21 .94 .63 .31 .31 .0961 .3298 .3298
22 .56 .94 -.38 .38 .1444 .6786 -.6786
23 .50 .56 -.06 .06 .0036 .1200 -.1200
24 .65 .50 .15 .15 .0225 .2308 .2308
25 .95 .65 .30 .30 .0900 .3158 .3158
26 .42 .95 -.53 .53 .2809 1.2619 -1.2619
27 .57 .42 .15 .15 .0225 .2632 .2632
28 .60 .57 .03 .03 .0009 .0500 .0500
29 .93 .60 .33 .33 .1089 .3548 .3548
30 .38 .93 -.55 .55 .3025 1.4474 -1.4474
31 .37 .38 -.01 .01 .0001 .0270 -.0270
32 .57 .37 .20 .20 .0400 .3509 .3509
5.85 1.6865 11.2227 -2.1988
b. MAD =31
85.5 = .189
c. MSE =31
6865.1 = .0544 , RMSE = √.0544 = .2332

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16
d. MAPE =31
2227.11 = .3620 or 36.2%
e. MPE =31
1988.2− = -.0709
21. a. Time series plot follows
b. The sales time series appears to vary about a fixed level so it is stationary.
c. The sample autocorrelation function for the sales series follows:
The sample autocorrelations die out rapidly. This behavior is consistent with a
stationary series. Note that the sales data are not random. Sales in adjacent
weeks tend to be positively correlated.

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17
22. a. The residualsYYe tt −= are listed below
b. The residual autocorrelations follow
Since, in this case, the residuals differ from the original observations by the
constant05.2460=Y , the residual autocorrelations will be the same as the
autocorrelations for the sales numbers. There is significant residual
autocorrelation at lag 1 and the autocorrelations die out in an exponential fashion.
The random model is not adequate for these data.
23. a. & b. Time series plot follows.

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18
Since this series is trending upward, it is nonstationary. There is also a seasonal
pattern since 2nd and 3rd quarter earnings tend to be relatively large and 1st and 4th
quarter earnings tend to be relatively small.
c. The autocorrelation function for the first 10 lags follows.
The autocorrelations are consistent with choice in part b. The autocorrelations fail
to die out rapidly consistent with nonstationary behavior. In addition, there are
relatively large autocorrelations at lags 4 and 8, indicating a quarterly seasonal
pattern.

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19
24. a. & b. Time series plot of fourth differences follows.
The time series of fourth differences appears to be stationary as it varies
about a fixed level.
25. a. 98/99Inc 98/99For 98/99Err 98/99AbsErr 98/99Err^2 98/99AbE/Inc
70.01 50.87 19.14 19.14 366.34 0.273390
133.39 93.83 39.56 39.56 1564.99 0.296574
129.64 92.51 37.13 37.13 1378.64 0.286409
100.38 80.55 19.83 19.83 393.23 0.197549
95.85 70.01 25.84 25.84 667.71 0.269588
157.76 133.39 24.37 24.37 593.90 0.154475
126.98 129.64 -2.66 2.66 7.08 0.020948
93.80 100.38 -6.58 6.58 43.30 0.070149
Sum 175.11 5015.17 1.5691
b. MAD = 175.11/8 = 21.89, RMSE = √5015.17 = 70.82, MAPE = 1.5691/8 = .196
or 19.6%
c. Naïve forecasting method of part a assumes fourth differences are random.
Autocorrelation function for fourth differences suggests they are not random.
Error measures suggest naïve method not very accurate. In particular, on average,
there is about a 20% error. However, naïve method does pretty well for 1999.
Hard to think of another naïve method that will do better.

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20
CASE 3-1A: MURPHY BROTHERS FURNITURE
1. The retail sales series has a trend and a monthly seasonal pattern.
2. Yes! Julie has determined that her data have a trend and should be first differenced. She has
also found out that the first differenced data are seasonal.
3. Techniques that she should consider include classical decomposition, Winters’
exponential smoothing, time series multiple regression, and Box-Jenkins methods.
4. She will know which technique works best by comparing error measurements such as MAD,
MSE or RMSE, MAPE, and MPE.
CASE 3-1B: MURPHY BROTHERS FURNITURE
1. The retail sales series has a trend and a monthly seasonal pattern.
2. The patterns appear to be somewhat similar. More actual data is needed in order to reach a
definitive conclusion.
3. This question should create a lively discussion. There are good reasons to use either set of
data. The retail sales series should probably be used until more actual sales data is available.
CASE 3-2: MR. TUX
1. This case affords students an opportunity to learn about the use of autocorrelation functions,
and to continue following John Mosby's quest to find a good forecasting method for his
data.
With the use of Minitab, the concept of first differencing data is also illustrated. The
summary should conclude that the sales data have both a trend and a seasonal component.
2. The trend is upward. Since there are significant autocorrelation coefficients at time lags 12
and 24, the data have a monthly seasonal pattern.
3. There is a 49% random component. That is, about half the variability in John’s monthly
sales is not accounted for by trend and seasonal factors. John, and the students analyzing
these results, should realize that finding an accurate method of forecasting these data could
be very difficult.
4. Yes, the first differences have a seasonal component. Given the autocorrelations at lags 12
and 24, the monthly changes are related 12, 24, … months apart. This information should be
used in developing a forecasting model for changes in monthly sales.

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21
CASE 3-3: CONSUMER CREDIT COUNSELING
1. First, Dorothy used Minitab to compute the autocorrelation function for the number of new
clients. The results are shown below.22122
1.0
0.8
0.6
0.4
0.2
0.0
-0.2
-0.4
-0.6
-0.8
-1.0
LBQTCorrLagLBQTCorrLagLBQTCorrLagLBQTCorrLag
165.14
156.84
153.39
152.33
148.66
146.40
144.86
144.37
138.70
136.27
134.55
127.70
121.68
106.87
100.72
96.90
93.71
87.87
81.61
75.60
67.18
55.51
42.86
24.08
1.26
0.83
0.46
0.87
0.69
0.58
0.33
1.14
0.75
0.64
1.30
1.25
2.05
1.35
1.09
1.01
1.40
1.49
1.50
1.85
2.30
2.56
3.50
4.83
0.25
0.16
0.09
0.17
0.13
0.11
0.06
0.22
0.14
0.12
0.24
0.23
0.36
0.23
0.18
0.17
0.23
0.24
0.24
0.28
0.33
0.35
0.43
0.49
24
23
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
Autocorrelation Function for Clients
Since the autocorrelations failed to die out rapidly, Dorothy concluded her series was
trending or nonstationary. She then decided to difference her time series.

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22
The autocorrelations for the first differenced series are:22122
1.0
0.8
0.6
0.4
0.2
0.0
-0.2
-0.4
-0.6
-0.8
-1.0
LBQTCorrLagLBQTCorrLagLBQTCorrLagLBQTCorrLag
47.00
42.09
41.93
40.02
38.98
38.95
38.93
34.85
29.52
29.32
27.72
26.20
24.07
19.67
18.91
18.91
18.49
18.39
18.34
17.87
17.83
17.82
17.66
17.43
1.44
-0.26
-0.92
0.69
-0.11
0.09
-1.41
1.67
-0.32
-0.93
0.92
-1.11
1.65
-0.69
0.02
-0.52
0.26
0.17
-0.57
0.17
0.10
-0.33
0.41
-4.11
0.19
-0.03
-0.12
0.09
-0.02
0.01
-0.18
0.21
-0.04
-0.12
0.11
-0.14
0.20
-0.08
0.00
-0.06
0.03
0.02
-0.07
0.02
0.01
-0.04
0.05
-0.42
24
23
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
Autocorrelations for Differenced Data
2. The differences appear to be stationary and are correlated in consecutive time periods.
Given
the somewhat large autocorrelations at lags 12 and 24, a monthly seasonal pattern should be
considered.
3. Dorothy would recommend that various seasonal techniques such as Winters’ method of
exponential smoothing (Chapter 4), classical decomposition (Chapter 5), time series
multiple regression (Chapter 8) and Box-Jenkins methods (ARIMA models in Chapter 9) be
considered.

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23
CASE 3-4: ALOMEGA FOOD STORES
The sales data from Chapter 1 for the Alomega Food Stores case are reprinted in Case
3-4. The case suggests that Julie look at the data pattern for her sales data.
The autocorrelation function for sales follows.
Autocorrelations suggest an up and down pattern that is very regular. If one month is
relatively high, next month tends to be relatively low and so forth. Very regular
pattern is suggested by persistence of autocorrelations at relatively large lags.
The changing of the sign of the autocorrelations from one lag to the next is consistent with
an up and down pattern in the time series. If high sales tend to be followed by low sales or
low sales by high sales, autocorrelations at odd lags will be negative and autocorrelations at
even lags positive.
The relatively large autocorrelation at lag 12, 0.53, suggests there may also be a seasonal
pattern. This issue is explored in Case 5-6.
CASE 3-5: SURTIDO COOKIES
1. A time series plot and the autocorrelation function for Surtido Cookies sales follow.

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24
The graphical evidence above suggests Surtido Cookies sales vary about a fixed level with
a strong monthly seasonal component. Sales are typically high near the end of the year and
low during the beginning of the year.
2. 03Sales NaiveFor Err AbsErr AbsE/03Sales MAD = 678369/5 = 135674
1072617 681117 391500 391500 0.364995 MAPE = .816833/5 = .163 or 16.3%
510005 549689 -39684 39684 0.077811
579541 497059 82482 82482 0.142323
771350 652449 118901 118901 0.154147
590556 636358 -45802 45802 0.077557
Sum 678369 0.816833
MAD appears large because of the big numbers for sales. MAPE is fairly large but
perhaps tolerable. In any event, Jame is convinced he can do better.

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25
CHAPTER 4
MOVING AVERAGES AND SMOOTHING METHODS
ANSWERS TO PROBLEMS AND CASES
1. Exponential smoothing
2. Naive
3. Moving average
4. Holt's two-parameter smoothing procedure
5. Winters’ three-parameter smoothing procedure
6. a.
t YttYˆ e tet e t
2t
t
Y
et
t
Y
e
1 19.39 19.00 .39 .39 .1521 .020 .020
2 18.96 19.39 - .43 .43 .1849 .023 -.023
3 18.20 18.96 - .76 .76 .5776 .042 -.042
4 17.89 18.20 - .31 .31 .0961 .017 -.017
5 18.43 17.89 .54 .54 .2916 .029 .029
6 19.98 18.43 1.55 1.55 2.4025 .078 .078
7 19.51 19.98 - .47 .47 .2209 .024 -.024
8 20.63 19.51 1.12 1.12 1.2544 .054 .054
9 19.78 20.63 - .85 .85 .7225 .043 -.043
10 21.25 19.78 1.47 1.47 2.1609 .069 .069
11 21.18 21.25 - .07 .07 .0049 .003 -.003
12 22.14 21.18 .96 .96 .9216 .043 .043
8.92 8.990 .445 .141
b. MAD =12
92.8 = .74
c. MSE =8 99
12
. = .75

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26
d. MAPE =.445
12 = .0371
e. MPE =.141
12 = .0118
f. 22.14
7.
Price AVER1 FITS1 RESI1
19.39 * * *
18.96 * * *
18.20 18.8500 * *
17.89 18.3500 18.8500 -0.96000
18.43 18.1733 18.3500 0.08000
19.98 18.7667 18.1733 1.80667
19.51 19.3067 18.7667 0.74333
20.63 20.0400 19.3067 1.32333
19.78 19.9733 20.0400 -0.26000
21.25 20.5533 19.9733 1.27667
21.18 20.7367 20.5533 0.62667
22.14 21.5233 20.7367 1.40333
Accuracy Measures
MAPE: 4.6319 MAD: 0.9422 MSE: 1.1728
The naïve approach is better.
8. a. See plot below.
Yt Avg Fits Res
200 * * *
210 * * *
215 * * *
216 * * *
219 212 * *
220 216 212 8
225 219 216 9
226 221.2 219 7
221.2
Accuracy Measures
MAPE: 3.5779 MAD: 8.0000 MSE: 64.6667
221.2 is forecast for period 9

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