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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 2 3 Surtido Cookies 23 Chapter 4 Moving Averages and Smoothing Methods 2 5 Problems 2 5 Cases: The Solar Alternative Company 43 Mr. Tux 45 Consumer Credit Counseling 4 6 Murphy Brothers Furniture 47 Five - year Revenue Projection for Downtown Radiology 4 9 Web Retailer 49 Southwest Medical Center 52 Surtido Cookies 54 Chapter 5 Time Series And Their Components 5 6 Problems 5 6 Cases: The Small Engine Doctor 76 Mr. Tux 79 Consumer Credit Counseling 8 0 Murphy Brothe rs Furniture 8 3 AAA Washington 8 4 Alomega Food Stores 87 Surtido Cookies 88 Southwest Medical Center 90 Chapter 6 Regression Analysis 94

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    Problems 94 Cases: Tiger Transport Compa ny 113 B utcher Products, Inc. 113 Ace Manufacturing 114 Mr. Tux 114 Consumer Credit Counseling 115 AAA Washington 117 Chapter 7 Mul tiple 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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    Cases: Golden Gardens Restaurant 224 Alom ega Food Stores 224 The Lydia E. Pinkha m Medicine Company 225 Chapter 11 Managing the Fore casting Process 226 Problems 226 Cases: Boundary Electronics 226 Busby Associates 227 Consumer Credit Counseling 228 Mr. Tux 228 Alomega Food Stores 228 Southwest Medical Center 229

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    PREFACE The goal of the ninth edition of Business Forecasting remains the same as t hat of the previous editions: T o 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 beg un 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. C omments for improvement are welcome. We can be reached at the following email addresses: John Hanke – john_hanke@msn.com Dean W ichern – 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. S 2 = 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 205625 X = 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 = = S X ( ) ( ) 91 . 4 76 . 10 30 / 71 . 13 96 . 1 76 . 10 / 96 . 1  =  =  n S X (5.85%, 15.67%) c. df = 30 − 1 = 29, t = 2.045

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    2 ( ) ( ) 12 . 5 76 . 10 30 / 71 . 13 045 . 2 76 . 10 / 045 . 2  =  =  n S X (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 . 102 41 . 23 = + = X 95% error margin: (102.59 − 23.41)/2 = 39.59 b. 1 −  = .90  Z = 1.645, 2 . 20 96 . 1 / 59 . 39 / , 63 = = = n S X ( ) 23 . 33 63 ) 2 . 20 ( 645 . 1 63 / 645 . 1  =  =  n S X (29.77, 96.23) 5. H 0 :  = 12.1 n = 100  = .05 H 1 :  > 12.1 S = 1.7 X = 13.5 Reject H 0 if Z > 1.645 Z = 100 7 . 1 1 . 12 5 . 13 − = 8.235 Reject H 0 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  1.96 49 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 = = S X 9 . 5 : 9 . 5 : 1 0  =   H H two - sided test,  = .05, critical value: | Z |= 1.96 Test statistic: 67 . 2 60 / 87 . 9 . 5 60 . 5 / 9 . 5 − = − = − = n S X Z Since | − 2.67| = 2.67 > 1.96, reject 0 H 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 against 0 H .

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    3 8 . df = n − 1 = 14 − 1 = 13, 52 . , 31 . 4 = = S X 4 : 4 : 1 0  =   H H one - sided test,  = .05, critical value: t = 1.771 Test statistic: 23 . 2 14 / 52 . 4 31 . 4 / 4 = − = − = n S X t Since 2.23 > 1.771, reject 0 H 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 against 0 H 9. H 0 :  = 700 n = 50  = .05 H 1 :   700 S = 50 X = 715 Reject H 0 if Z < - 1.96 or Z > 1.96 Z = 50 50 700 715 − = 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 against 0 H 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 Distributio n 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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    4 The null hypothesis that the mean is still 2.9 is true since the actual mean of the population of data is 2.91 with a standard deviation of 1.608; however, a few students may reject the null hypothesis, committing a Type I error. 11. a. b. Positive linear relationship c.  Y = 6058  Y 2 = 4,799,724  X = 59  X 2 = 513  XY = 48,665 r = .938

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    5 12. a. b. Positive linear relationship c.  Y = 2312  Y 2 = 515,878  X = 53.7  X 2 = 282.55  XY = 12,029.3 r = .95 Y ˆ = 32.5 + 36.4X Y ˆ = 32.5 + 3 6.4(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 H 0 : β 1 = 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 ea ch additional square foot of space. d. Size = 750  Rent = 275.5 + .518(750) = $664/month 15 . n = 175, 3 . 10 , 2 . 45 = = S X Point estimate: 2 . 45 = X 98% confidence interval: 1 −  = .98  Z = 2.33 ( ) ( ) 8 . 1 2 . 45 175 / 3 . 10 33 . 2 2 . 45 / 33 . 2  =  =  n S X  (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 44 2 . 45 / 44 = − = − = n S X Z Since | Z | = 1.54 < 2.33, do not reject 0 H 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 . 1 10 . 1 39 99 . 5 96 . 1 10 . 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, po sitive 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 re ly 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 horizon tal scale and the au tocorrelation coefficients, the correlations between Y t and Y t - k , are displayed as vertical bars at the appropriate time lags. The lengths and direction s (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, movi ng 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. M oving average s , 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 consumpt ion, 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 d ecomposition, census II, Winter s ’ 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 4 0 2004 2279 34 1991 2,371 - 7 2 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. r 1 = .895 H 0 : ρ 1 = 0          H 1 : ρ 1  0 

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    11  Reject if t < - 2.069 or t > 2.069 SE( k r ) = n r k i i  − = + 1 1 2 2 1 = ( ) 24 2 1 1 1 1 2 1  − = + i r = 24 1 = .204 ) SE(r k 1 1  − = r t = .204 0 895 . − = 4.39 Since the computed t (4.39) is greater than the critical t (2.069), reject the null. r 2 = .788     H 0 : ρ 2 = 0          H 1 : ρ 2  0   Reject if t < - 2.069 or t > 2.069     SE( k r ) = n r k i i  − = + 1 1 2 2 1  =  ( ) 24 895 . 2 1 1 2 1 2  − = + i =  2 6 24 . = .33      ) SE(r 1 1 1  − = 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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    12 The autocorrelation function follows.  18. a. r 1 = .376 b. The differenced data are stationary. See plot below.

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    13 The autocorrelation function follows. 19. Figure 3 - 18 - The d ata are nonstationary. (Trending data) Figure 3 - 19 - The data are random. Figure 3 - 20 - The data are seasonal. (Monthly data) Figure 3 - 21 - The data are stationary and have a pattern that could be modeled.

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    14 20. The data have a quarterly seasonal pattern as shown by the significant autocorrelation at time lag 4. First quarter earnings tend to be high, third quarter earnings tend to be low.

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    15 a. Time Data Forecast Error t Y t Y ˆ t e t e t e t 2 t t Y e t 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 .0 081 .2 143 - .2 143 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 .0 851 - .0 851 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 5.8 5 1.6 86 5 11.2227 - 2 . 1 988 b. MAD = 31 85 . 5 = .189 c. MSE = 31 6865 . 1 = .054 4 , RMSE = √.0544 = .2332

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    16 d. MAPE = 31 2227 . 11 = . 3620 or 36.2 % e. MPE = 31 1988 . 2 − = - .0 709 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 residuals Y Y e t t − = are listed below b. The residual autocorrelations follow Since, in this case, the residuals differ from the original observations by the constant 05 . 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 2 nd and 3 rd quarter earnings tend to be relatively large and 1 st and 4 th 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 m ethod 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, Winter s ’ 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. 22 12 2 1.0 0.8 0.6 0.4 0.2 0.0 - 0. 2 - 0.4 - 0.6 - 0.8 - 1.0 LBQ T Corr Lag LBQ T Corr Lag LBQ T Corr Lag LBQ T Corr Lag 165.14 156.84 153.39 152.33 148.66 146.40 144.86 144.37 138.70 136.27 134.55 127.70 1 21.68 106.87 100.72 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: 22 12 2 1.0 0.8 0.6 0.4 0.2 0.0 - 0.2 - 0.4 - 0.6 - 0.8 - 1.0 LBQ T Corr Lag LBQ T Corr Lag LBQ T Corr Lag LBQ T Corr Lag 47.00 42.09 41.93 40.02 38.98 38.95 38.93 34.85 29.52 29.32 27.72 26.2 0 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 sea sonal techniques such as Winter s ’ 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 l ook 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. T he 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. Winter s ’ three - parameter smoothing procedure 6. a. t Y t t Y ˆ e t e t e t 2 t t Y e t 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 2.40 25 .078 .078 7 19.51 19.98 - .47 .47 .2209 .024 - .024 8 20.63 19.51 1.12 1.12 1.254 4 .054 .054 9 19.78 20.63 - .85 .85 .7225 .043 - .043 10 21.25 19.78 1.47 1.47 2.16 09 .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 = .7 4 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: 4.631 9 MAD: 0.9422 MS E : 1.1728 The naïve approach is better. 8. a. See plot below. Y t 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 MS E : 64.6667 221.2 is forecast for period 9

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