Business Forecasting 9th Edition Hanke Solutions Manual

Business Forecasting 9th Edition Hanke Solutions Manual

richard@qwconsultancy.com

1|Pa ge

CONTENTS

Preface Chapter 2

Chapter 3

Chapter 4

Chapter 5

Chapter 6

A Review of Basic Statistical Concepts Problems Cases: Alcam Electronics Mr. Tux Alomega Food Stores

1

Exploring Data Patterns and Choosing a Forecasting Technique Problems Cases: Murphy Brothers Furniture Mr. Tux Consumer Credit Counseling Alomega Food Stores Surtido Cookies

9

Moving Averages and Smoothing Methods Problems Cases: The Solar Alternative Company Mr. Tux Consumer Credit Counseling Murphy Brothers Furniture Five-year Revenue Projection for Downtown Radiology Web Retailer Southwest Medical Center Surtido Cookies

25

Time Series And Their Components Problems Cases: The Small Engine Doctor Mr. Tux Consumer Credit Counseling Murphy Brothers Furniture AAA Washington Alomega Food Stores Surtido Cookies Southwest Medical Center Regression Analysis

56

1 8 8 8

9 20 20 21 23 23

25 43 45 46 47 49 49 52 54

56 76 79 80 83 84 87 88 90 94

Problems Cases: Tiger Transport Company Butcher Products, Inc. Ace Manufacturing Mr. Tux Consumer Credit Counseling AAA Washington Chapter 7

Chapter 8

Chapter 9

Chapter 10

94 113 113 114 114 115 117

Multiple Regression Problems Cases: The Bond Market AAA Washington Fantasy Baseball (A) Fantasy Baseball (B)

120

Regression With Time Series Data Problems Cases: Business Activity Index for Spokane County Restaurant Sales Mr. Tux Consumer Credit Counseling AAA Washington Alomega Food Stores Surtido Cookies Southwest Medical Center

146

Box-Jenkins (ARIMA) Methodology Problems Cases: Restaurant Sales Mr. Tux Consumer Credit Counseling The Lydia E. Pinkham Medicine Company City of College Station UPS Air Finance Division AAA Washington Web Retailer Surtido Cookies Southwest Medical Center

173

Judgmental Elements in Forecasting Problems

223

120 140 141 143 145

146 164 165 165 166 168 169 169 171

173 202 203 205 207 209 210 212 213 215 218

223

Cases: Golden Gardens Restaurant Alomega Food Stores The Lydia E. Pinkham Medicine Company Chapter 11

Managing the Forecasting Process Problems Cases: Boundary Electronics Busby Associates Consumer Credit Counseling Mr. Tux Alomega Food Stores Southwest Medical Center

224 224 225 226 226 226 227 228 228 228 229

CHAPTER 2 A REVIEW OF BASIC STATISTICAL CONCEPTS ANSWERS TO PROBLEMS AND CASES 1.

2.

Descriptive Statistics Variable Orders

N 28

Mean Median StDev SE Mean 21.32 17.00 13.37 2.53

Variable Orders

Min 5.00

Max 54.00

Q1 Q3 11.25 28.75

a.

X = 21.32

b.

S = 13.37

c.

S = 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 Prices

N Mean Median StDev SE Mean 12 176654 180000 39440 11385

Variable Min Max Q1 Q3 Prices 121450 253000 138325 205625

X = 176,654 and 3.

S = 39,440

a.

Point estimate: X = 10.76%

b.

1− = .95  Z = 1.96, n = 30, X = 10 .76 , S = 13 .71

(

)

(

)

X  1.96 S / n = 10.76  1.96 13.71/ 30 = 10.76  4.91 (5.85%, 15.67%)

c.

df = 30−1 = 29, t = 2.045 1

(

)

(

)

X  2.045 S / n = 10.76  2.045 13.71/ 30 = 10.76  5.12 (5.64%, 15.88%)

d.

4.

a.

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.

Point estimate: X = 23.41 + 102.59 = 63 2

95% error margin: (102.59 − 23.41)/2 = 39.59 b.

1− = .90  Z = 1.645, X = 63, S / n = 39.59 / 1.96 = 20.2

(

)

X  1.645 S / n = 63  1.645(20.2) = 63  33.23 (29.77, 96.23)

5.

H0:  = 12.1 H1:  > 12.1

 = .05 X = 13.5

n = 100 S = 1.7

Reject H0 if Z > 1.645 Z=

13.5 − 12.1 = 8.235 1.7 100

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  1.96 5.7

 6.5 to 9.7 seats 49 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, X = 5.60 , S = .87 H 0 :  = 5.9 two-sided test,  = .05, critical value: |Z|= 1.96 H 1 :   5.9 Test statistic: Z =

X − 5.9 S/ n

=

5.60 − 5.9 .87 / 60

= −2.67

Since |−2.67| = 2.67 > 1.96, reject H 0 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 H 0 .

2

8.

df = n −1 = 14 −1 = 13, X = 4.31, S = .52 H0 :  = 4 one-sided test,  = .05, critical value: t = 1.771 H1 :   4 Test statistic: t =

X −4 S/ n

=

4.31 − 4 .52 / 14

= 2.23

Since 2.23 > 1.771, reject H 0 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 H 0 9.

H0:  = 700 H1:   700

n = 50 S = 50

 = .05 X = 715

Reject H0 if Z < -1.96 or Z > 1.96 Z=

715 − 700 = 2.12 50 50

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 H 0 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.

3

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 2 X = 513

2

Y = 4,799,724 XY = 48,665

X = 59 r = .938 4

12.

a.

b.

Positive linear relationship

c.

Y = 2312 Y2 = 515,878 X2 = 282.55 XY = 12,029.3

X = 53.7 r = .95

Ŷ = 32.5 + 36.4X Ŷ = 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.

5

15.

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

n = 175, X = 45 .2, S = 10 .3 Point estimate: X = 45.2 98% confidence interval: 1− = .98  Z = 2.33 X  2.33 S / n = 45.2  2.33 10.3 / 175 = 45.2  1.8  (43.4, 47.0)

(

)

(

)

Hypothesis test: H 0 :  = 44 two-sided test,  = .02, critical value: |Z|= 2.33 H 1 :   44 Test statistic: Z =

X − 44 S/ n

=

45.2 − 44 10.3 / 175

= 1.54

Since |Z| = 1.54 < 2.33, do not reject H 0 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.

6

16.

a.

b.

c.

17.

H 0 :  = 63,700 H1 :   63,700 H 0 :  = 4.3 H1 :   4.3 H 0 :  = 1300 H1 :   1300

Large sample 95% confidence interval for mean monthly return μ: − 1.10  1.96

5.99 = −1.10  1.88  39

(−2.98, .78 )

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 7

CASE 2-1: ALCAM ELECTRONICS In our consulting work, business people sometimes tell us that business schools teach a risktaking 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 8

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 wavelike 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. b. c. d.

nonstationary series stationary series nonstationary series stationary series

8.

a. b. c. d. e. f.

stationary series random series trending or nonstationary series seasonal series stationary series 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 9

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 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998

2,413 2,407 2,403 2,396 2,403 2,443 2,371 2,362 2,334 2,362 2,336 2,344 2,384 2,244

-6 -4 -7 7 40 -72 -9 -28 28 -26 8 40 -140

1999 2358 114 2000 2329 -29 2001 2345 16 2002 2254 -91 2003 2245 -9 2004 2279 34

Yes! The original series has a decreasing trend.

14.

0  1.96 ( 1

15.

a. b. c.

16.

All four statements are true.

17.

a.

80

) = 0  1.96 (.1118) = 0  .219

MPE MAPE MSE or RMSE

r1 = .895 H0: ρ1 = 0H1: ρ1  0 10



Reject if t < -2.069 or t > 2.069 k −1

1−1

1 + 2 (r1 )

1 + 2 ri 2 SE( r k ) =

i =1

n

2

i =1

=

=

24

1 = .204 24

.895 − 0 r − 1 = = 4.39 t= 1 .204 SE(rk)

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

2 −1

k −1

1 + 2 (.895)

1 + 2 ri 2 i =1



 

SE( r k ) =



 

r − 1 .788− 0 = = 2.39 t= 1 .33 SE(r1)

n

=

i =1

24

2

=

2.6 = .33 24

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.

11

The autocorrelation function follows.

 18.

a.

r1 = .376

b.

The differenced data are stationary. See plot below.

12

The autocorrelation function follows.

19.

Figure 3-18 - The data 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.

13

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.

14

a. Time Data Forecast Error t

Yt

Ŷ t

et

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

.40 .29 .24 .32 .47 .34 .30 .39 .63 .43 .38 .49 .76 .51 .42 .61 .86 .51 .47 .63 .94 .56 .50 .65 .95 .42 .57 .60 .93 .38 .37 .57

.40 .29 .24 .32 .47 .34 .30 .39 .63 .43 .38 .49 .76 .51 .42 .61 .86 .51 .47 .63 .94 .56 .50 .65 .95 .42 .57 .60 .93 .38 .37

-.11 -.05 .08 .15 -.13 -.04 .09 .24 -.20 -.05 .11 .27 -.25 -.09 .19 .25 -.35 -.04 .16 .31 -.38 -.06 .15 .30 -.53 .15 .03 .33 -.55 -.01 .20

et

et

et

2

Yt

et Yt

.11 .0121 .3793 -.3793 .05 .0025 .2083 -.2083 .08 .0064 .2500 .2500 .15 .0225 .3191 .3191 .13 .0169 .3824 -.3824 .04 .0016 .1333 -.1333 .09 .0081 .2308 .2308 .24 .0576 .3810 .3810 .20 .0400 .4651 -.4651 .05 .0025 .1316 -.1316 .11 .0121 .2245 .2245 .27 .0729 .3553 .3553 .25 .0625 .4902 -.4902 .09 .0081 .2143 -.2143 .19 .0361 .3115 .3115 .25 .0625 .2907 .2907 .35 .1225 .6863 -.6863 .04 .0016 .0851 -.0851 .16 .0256 .2540 .2540 .31 .0961 .3298 .3298 .38 .1444 .6786 -.6786 .06 .0036 .1200 -.1200 .15 .0225 .2308 .2308 .30 .0900 .3158 .3158 .53 .2809 1.2619 -1.2619 .15 .0225 .2632 .2632 .03 .0009 .0500 .0500 .33 .1089 .3548 .3548 .55 .3025 1.4474 -1.4474 .01 .0001 .0270 -.0270 .20 .0400 .3509 .3509 5.85 1.6865 11.2227 -2.1988

b.

MAD =

5.85 = .189 31

c.

MSE =

1.6865 = .0544 , 31

RMSE = √.0544 = .2332

15

21.

11.2227 = .3620 or 36.2% 31

d.

MAPE =

e.

MPE =

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:

− 2.1988 = -.0709 31

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.

16

22.

a.

The residuals et = Yt − Y are listed below

b.

The residual autocorrelations follow

Since, in this case, the residuals differ from the original observations by the constant Y = 2460.05 , 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.

17

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.

18

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 70.01 133.39 129.64 100.38 95.85 157.76 126.98 93.80

98/99For 50.87 93.83 92.51 80.55 70.01 133.39 129.64 100.38

98/99Err 19.14 39.56 37.13 19.83 25.84 24.37 -2.66 -6.58

98/99AbsErr 19.14 39.56 37.13 19.83 25.84 24.37 2.66 6.58

Sum

175.11

98/99Err^2 366.34 1564.99 1378.64 393.23 667.71 593.90 7.08 43.30

98/99AbE/Inc 0.273390 0.296574 0.286409 0.197549 0.269588 0.154475 0.020948 0.070149

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.

19

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.

20

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.

Autocorrelation Function for Clients 1.0 0.8 0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6 -0.8 -1.0

2

12

22

Lag

Corr

T

LBQ

Lag

Corr

T

LBQ

Lag

Corr

1

0.49

4.83

24.08

8

0.23

1.40

93.71

15

2

0.43

3.50

42.86

9

0.17

1.01

96.90

16

3

0.35

2.56

55.51

10

0.18

1.09 100.72

4

0.33

2.30

67.18

11

0.23

1.35 106.87

5

0.28

1.85

75.60

12

0.36

6

0.24

1.50

81.61

13

0.23

7

0.24

1.49

87.87

14

0.24

T

LBQ

Lag

Corr

0.12

0.64 136.27

22

0.09

0.46 153.39

0.14

0.75 138.70

23

0.16

0.83 156.84

17

0.22

1.14 144.37

24

0.25

1.26 165.14

18

0.06

0.33 144.86

2.05 121.68

19

0.11

0.58 146.40

1.25 127.70

20

0.13

0.69 148.66

1.30 134.55

21

0.17

0.87 152.33

T

LBQ

Since the autocorrelations failed to die out rapidly, Dorothy concluded her series was trending or nonstationary. She then decided to difference her time series.

21

The autocorrelations for the first differenced series are: Autocorrelations for Differenced Data 1.0 0.8 0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6 -0.8 -1.0

2

Lag

Corr

12

T

LBQ

1 -0.42 -4.11 2

Lag

22

Corr

T

LBQ

Lag

T

LBQ

Lag

T

LBQ

17.43

8 0.03

0.26

18.49

15 -0.12 -0.93

29.32

22 -0.12 -0.92

41.93

0.41

17.66

9 -0.06 -0.52

18.91

16 -0.04 -0.32

29.52

23 -0.03 -0.26

42.09

3 -0.04 -0.33

17.82

10 0.00

0.02

18.91

17

1.67

34.85

24

47.00

4

0.01

0.10

17.83

11 -0.08 -0.69

19.67

18 -0.18 -1.41

38.93

5

0.02

0.17

17.87

12 0.20

1.65

24.07

19

0.09

38.95

6 -0.07 -0.57

18.34

13 -0.14 -1.11

26.20

20 -0.02 -0.11

38.98

7

18.39

14 0.11

27.72

21

40.02

0.05

0.02

0.17

0.92

Corr

0.21

0.01

0.09

0.69

Corr

0.19

1.44

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.

22

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.

23

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 1072617 510005 579541 771350 590556

NaiveFor 681117 549689 497059 652449 636358

Err AbsErr 391500 391500 -39684 39684 82482 82482 118901 118901 -45802 45802 Sum 678369

AbsE/03Sales 0.364995 0.077811 0.142323 0.154147 0.077557 0.816833

MAD = 678369/5 = 135674 MAPE = .816833/5 = .163 or 16.3%

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.

24

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

Yt

Ŷt

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

19.39 18.96 18.20 17.89 18.43 19.98 19.51 20.63 19.78 21.25 21.18 22.14

19.00 19.39 18.96 18.20 17.89 18.43 19.98 19.51 20.63 19.78 21.25 21.18

et

et

et

et

2

Yt

.39 .39 .1521 .020 - .43 .43 .1849 .023 - .76 .76 .5776 .042 - .31 .31 .0961 .017 .54 .54 .2916 .029 1.55 1.55 2.4025 .078 - .47 .47 .2209 .024 1.12 1.12 1.2544 .054 - .85 .85 .7225 .043 1.47 1.47 2.1609 .069 - .07 .07 .0049 .003 .96 .96 .9216 .043 8.92

b. MAD =

8.92 = .74 12

c. MSE =

8.99 = .75 12

8.990

25

.445

et Yt

.020 -.023 -.042 -.017 .029 .078 -.024 .054 -.043 .069 -.003 .043 .141

d. MAPE =

e. MPE =

.445 = .0371 12

.141 = .0118 12

f. 22.14 7.

Price 19.39 18.96 18.20 17.89 18.43 19.98 19.51 20.63 19.78 21.25 21.18 22.14

AVER1 * * 18.8500 18.3500 18.1733 18.7667 19.3067 20.0400 19.9733 20.5533 20.7367 21.5233

FITS1 * * * 18.8500 18.3500 18.1733 18.7667 19.3067 20.0400 19.9733 20.5533 20.7367

Accuracy Measures MAPE: 4.6319 MAD: 0.9422

RESI1 * * * -0.96000 0.08000 1.80667 0.74333 1.32333 -0.26000 1.27667 0.62667 1.40333

MSE: 1.1728

The naïve approach is better.

8.

a. See plot below.

Yt 200 210 215 216 219 220 225 226

Avg * * * * 212 216 219 221.2

Fits * * * * * 212 216 219 221.2

Accuracy Measures MAPE: 3.5779

Res * * * * * 8 9 7

MAD: 8.0000

MSE: 64.6667

221.2 is forecast for period 9 26

b. & c. See plot below.

Yt 200 210 215 216 219 220 225 226

Smoothed 200.000 204.000 208.400 211.440 214.464 216.678 220.007 222.404

Accuracy Measures MAPE: 3.2144

Forecast 200.000 200.000 204.000 208.400 211.440 214.646 216.678 220.007 222.404

MAD: 7.0013

MSE: 58.9657

Caution: If Minitab is used, the final result depends on how many values are averaged for the initial value. If 1 value is averaged, so in this case the initial value is 200, the forecast for period 4 is 208.4.

The forecast error for time period 3 is 11.

27

9.

a. & c, d, e, f

3-month moving-average (See plot below.)

Month Yield 1 9.29 2 9.99 3 10.16 4 10.25 5 10.61 6 11.07 7 11.52 8 11.09 9 10.80 10 10.50 11 10.86 12 9.97

MA Forecast * * * * 9.813 * 10.133 9.813 10.340 10.133 10.643 10.340 11.067 10.643 11.227 11.067 11.137 11.227 10.797 11.137 10.720 10.797 10.443 10.720

Accuracy Measures MAPE: 4.5875

Error * * * 0.437 0.477 0.730 0.877 0.023 -0.427 -0.637 0.063 -0.750

MAD: 0.4911

MSE: 0.3193

Forecast for month 13 (Jan.) is 10.443

28

MPE: .6904

b. & c, d, e, f

5-month moving-average (See plot below.)

Month Yield MA Forecast Error 1 9.29 * * * 2 9.99 * * * 3 10.16 * * * 4 10.25 * * * 5 10.61 10.060 * * 6 11.07 10.416 10.060 1.010 7 11.52 10.722 10.416 1.104 8 11.09 10.908 10.722 0.368 9 10.80 11.018 10.908 -0.108 10 10.50 10.996 11.018 -0.518 11 10.86 10.954 10.996 -0.136 12 9.97 10.644 10.954 -0.984 Accuracy Measures MAPE: 5.5830 MAD: 0.6040 Forecast for month 13 (Jan.) is 10.644

29

MSE: 0.5202

MPE: .7100

g.

10.

Use 3-month moving average forecast: 10.4433

Accuracy Measures (See plot below.)

MAPE: 5.8926

MAD: 0.6300

MSE: 0.5568

MPE: 5.0588

Forecast for month 13 (Jan. 2007) is 10.4996 No! The accuracy measures favor the three-month moving average procedure, but the values of the forecasts are not much different.

30

11.

See plot below.

Month Demand Smooth Forecast 1 2 3 4 5 6 7 8 9 10 11 12

205 251 304 284 352 300 241 284 312 289 385 256

205.000 228.000 266.000 275.000 313.500 306.750 273.875 278.938 295.469 292.234 338.617 297.309

Accuracy Measures MAPE: 14.67

Error

205.000 0.0000 205.000 46.0000 228.000 76.0000 266.000 18.0000 275.000 77.0000 313.500 -13.5000 306.750 -65.7500 273.875 10.1250 278.938 33.0625 295.469 -6.4688 292.234 92.7656 338.617 -82.6172

MAD:

43.44

MSE: 2943.24

Forecast for month 13 (Jan. 2007) is 297.309

12.

Naïve method - Forecast for 1996 Q2: 25.68 (Actual: 26.47) MAPE = 8.622 MAD = 1.916 MSE = 5.852 5-month moving average - Forecast for 1996 Q2: 24.244 (Actual: 26.47) MAPE: 9.791 MAD: 2.249 MSE: 7.402 31

Exponential smoothing with a Smoothing Constant of Alpha: 0.696 MAPE: 8.425 MAD: 1.894 MSE: 5.462 Forecast for 1996 Q2: 25.227

(Actual: 26.47)

Based on the error measures and the forecast for Q2 of 1996, the naïve method and simple exponential smoothing are comparable. Either method could be used.

13.

a.

 = .4 Accuracy Measures MAPE: 14.05 MAD: 24.02

MSE: 1174.50

Forecast for Q1 2000: 326.367 b.

 = .6 Accuracy Measures MAPE: 14.68 MAD: 24.56

MSE: 1080.21

Forecast for Q1 2000: 334.070 c.

Looking at the error measures, there is not much difference between the two choices of smoothing constant. The error measures for α = .4 are slightly better.

The forecasts for the two choices of smoothing constant are also not much different.

32

d.

14.

The residual autocorrelations for α = .4 are shown below. The residual autocorrelations for α = .6 are similar. There are significant residual autocorrelations at lags 1, 4 and (very nearly) 8. A forecasting method that yields no significant residual autocorrelations would be desirable.

None of the techniques do much better than the naïve method. Simple exponential Smoothing with α close to 1, say .95, is essentially the naïve method.

Accuracy Measures for Naïve Method MAPE: 42.57 MAD: 1.685

MSD: 4.935

Using the naïve method, the forecast for 2000 would be 6.85.

15.

A time series plot of quarterly Revenues and the autocorrelation function show that the data are seasonal with a trend. After some experimentation, Winters’ multiplicative smoothing with smoothing constants α (level) = 0.8, β (trend) = 0.1 and γ (seasonal) = 0.1 is used to forecast future Revenues. See plot below.

Accuracy Measures MAPE 3.8 MAD 69.1 MSE 11146.4

Forecasts 33

Quarter Forecast 71 2444.63 72 1987.98 73 2237.98 74 1887.74 75 2456.18 76 1997.36

Lower Upper 2275.34 2613.92 1773.84 2202.12 1969.23 2506.72 1559.46 2216.01 2065.70 2846.65 1543.10 2451.62

An examination of the autocorrelation coefficients for the residuals from Winters’ multiplicative smoothing shown below indicates that none of them are significantly different from zero.

16.

a.

An Excel time series plot for Sales follows 34

The data appear to be seasonal with relatively large sales in August, September, October and November, and relatively small sales in July and December. b. & c. The Excel spreadsheet for calculating MAPE for the naïve forecasts and the simple exponential smoothing forecasts is shown below.

Note: MAPE2 for simple exponential smoothing in the Excel spreadsheet 35

is calculated with a divisor of 23 (since the first smoothed value is set equal to the first observation). Using a divisor of 24 gives MAPE2 = 7.69%, the value reported by Minitab. d.

Neither the naïve model nor simple exponential smoothing is likely to generate accurate forecasts of future monthly sales since neither model allows for seasonality.

e.

The results of Winters’ multiplicative smoothing with  =  =  = .5 is shown in the Minitab plot below.

The forecast for January, 2003 is 467. f.

From part e, MAPE = 2.45%. Prefer Winters’ multiplicative smoothing since it allows for seasonality and has the smallest MAPE of the three models considered.

g.

The residual autocorrelations from Winters’ multiplicative smoothing are shown below. The residual autocorrelations suggest Winters’ method works well for these data since they are all insignificantly different from 0.

36

17.

a.

The four-week moving average seems to represent the data a little better.

Compare the error measures for the four-week moving average in the figure

below with the five-week moving average results in Figure 4-4.

b.

Simple exponential smoothing with a smoothing constant of α = .7 does a better job of smoothing the data than a four-week moving average as judged by the uniformly smaller error measures shown in the plot below.

37

18.

a.

As the order of the moving average increases, the smoothed data become more wavelike. Looking at the results for orders k =10 and k = 15, and counting the number of years from one peak to the next, it appears as if the number of severe earthquakes is on about a 30 year cycle.

b.

The results of simple exponential smoothing with a smoothing constant of α = .4 are shown in the plot below. The forecast for the number of severe earthquakes for 2000 is 20.

38

The residual autocorrelation function is shown below. There are no significant residual autocorrelations. Simple exponential smoothing seems to provide a good fit to the earthquake data.

19.

c.

There can be no seasonal component in the number of severe earthquakes since these data are recorded on an annual basis.

a.

The results of Holt’s smoothing with α (level) = .9 and β (trend) = .1 for

39

Southwest Airline’s quarterly income are shown below. A plot of the residual autocorrelation function follows. It appears as if Holt’s procedure represents the data well but the residual autocorrelations have significant spikes at the seasonal lags of 4 and 8 suggesting a seasonal component is not captured by Holt’s method.

b.

Winters’ multiplicative smoothing with α = β = γ =.2 was applied to the quarterly income data and the results are shown in the plot below. The forecasts for the four quarters of 2000 are: 40

Quarter 49 50 51 52

Forecast 88.960 184.811 181.464 117.985

The forecasts seem reasonable but the residual autocorrelation function below has a significant spike at lag 1. So although Winters’ procedure captures the trend and seasonality, there is still some association in consecutive observations not accounted for by Winters’ method.

41

20.

A time series plot of The Gap quarterly sales is shown below.

This time series is trending upward and has a seasonal pattern with third and fourth quarter Gap sales relatively large. Moreover the variability in this series is increasing with the level suggesting a multiplicative Winters’ smoothing procedure or a transformation of the data (say logarithms of sales) to stabilize the variability.

The results of Winters’ multiplicative smoothing with smoothing constants 42

α = β = γ =.2 are shown in the plot below.

The forecasts for the four quarters of 2005 are: Forecasts Quarter 101 102 103 104

Forecast 3644.18 3775.78 4269.27 5267.82

Lower 3423.79 3551.94 4041.58 5035.90

Upper 3864.57 3999.62 4496.96 5499.74

The forecasts seem reasonable, however, the residuals autocorrelations shown below indicate there is still some autocorrelation at low lags, including the seasonal lag S = 4, that is not accounted for by Winters’ method. A better model is needed. This issue is explored in later chapters of the text.

43

CASE 4-1: THE SOLAR ALTERNATIVE COMPANY This case provides the student with an opportunity to deal with a frequent real world problem: small data sets. A plot of the two years of data shows both an upward trend and seasonal pattern. The forecasting model that is selected must do an accurate job for at least three months into the future.

Averaging methods are not appropriate for this data set because they do not work when data has a trend, seasonality, or some other systematic pattern. Moving average models tend to smooth out the seasonal pattern of the data instead of making use of it to forecast.

A naive model that takes into account both the trend and the seasonality of the data might work. Since the seasonal pattern appears to be strong, a good forecast might take the same value it did in the corresponding month one year ago or Yt+1 = Yt-11.

However, as it stands, this forecast ignores the trend. One approach to estimate trend is to calculate the increase from each month in 2005 to the same month in 2006. As an example, the increase from January, 2005 to January, 2006 is equal to (Y13 - Y1) = (17 - 5) = 12.

After the increases for all 12 months are calculated, they can be summed and then divided by 12. The forecast for each month of 2007 could then be calculated as the value for the same month in 2006 plus the average increase for each of the 12 months from 2005 to the same month in 2006.

Consequently, the forecast for January, 2007 is Y25 = 17 + [(17 - 5) + (14 - 6) + (20 - 10) + (23 - 13) + (30 - 18) + (38 - 15) + (44 - 23) + (41 - 26) + (33 - 21) + (23 - 15) + (26 - 12) + (17 - 14)]/12

44

Y25 = 17 +

148 = 17 + 12 = 29 12

The forecasts for 2007 are:

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

29 26 32 35 42 50 56 53 45 35 38 29

Winters’ multiplicative method with smoothing constants α = .1, β = .1, γ = .3 seems to represent the data fairly well (see plot below) and produces the forecasts: Month Forecast Jan/2007 19.8 Feb/2007 18.0 Mar/2007 26.8 Apr/2007 32.0 May/2007 42.4 Jun/2007 45.8 Jul/2007 58.4 Aug/2007 58.9 Sep/2007 47.6 Oct/2007 33.7 Nov/2007 33.5 Dec/2007 28.0

45

The naïve forecasts are not unreasonable but the Winters’ forecasts seem to have captured the seasonal pattern a little better, particularly for the first 3 months of the year. Notice that if the trend and seasonal pattern are strong, Winters’ smoothing procedure can work well even with only two years of monthly data.

CASE 4-2: MR TUX This case shows how several exponential smoothing methods can be applied to the Mr. Tux data. John Mosby tries simple exponential smoothing and exponential smoothing with adjustments for trend and seasonal factors, along with a three-month moving average.

Students can begin to see that several forecasting methods are typically tried when an important variable must be forecast. Some method of comparing them must be used, such as the three accuracy methods discussed in this case. Students should be asked their opinions of John's progress in his forecasting efforts given these accuracy values. It should be apparent to most that the degree of accuracy achieved is not sufficient and that further study is needed. Students should be reminded that they are looking at actual data, and that the problems faced by John Mosby really occurred.

1.

Of the methods attempted, Winters’ multiplicative smoothing was the best method John found. Each forecast was typically off by about 25,825. The error in each forecast was about 22% of the value of the variable being forecast.

2.

There are other choices for the smoothing constants that lead to smaller error measures.

For example, with α = β = γ = .1, MAD = 22,634 and MAPE = 20.

3.

John should examine plots of the residuals and the residual autocorrelations. If Winters’ procedure is adequate, the residuals should appear to be random. In addition, John can 46

examine the forecasts for the next 12 months to see if they appear to be reasonable.

4.

The ideal value for MPE is 0. If MPE is negative, then, on average, the predicted values are too high (larger than the actual values).

CASE 4-3: CONSUMER CREDIT COUNSELING 1.

Students should realize immediately that simply using the basic naive approach of using last period to predict this period will not allow for forecasts for the rest of 1993. Since the autocorrelation coefficients presented in Case 3-3 indicate some seasonality, a naive model using April 1992 to predict April 1993, May 1992 to predict May 1993 and so forth might be tried. This approach produces the error measures MAD = 23.39

MSE = 861.34

MAPE = 18.95

over the data region, and are not particularly attractive given the magnitudes of the new client numbers.

2.

A moving average model of any order cannot be defended since any moving average will produce flat line forecasts for the rest of 1993. That is, the forecasts will lie along a horizontal line whose level is the last value for the moving average. The seasonal pattern will be ignored.

3.

Since the data have a seasonal component, Winters’ multiplicative smoothing procedure with smoothing constants α = β = γ =.2 was tried. For these choices: MAD = 19.29, MSE = 545.41 and MAPE = 16.74. For smoothing constants α = .5, β = γ = .1, MAD = 16.94, MSE = 451.26 and MAPE = 14.30.

4.

Of the methods attempted, the Winters’ multiplicative smoothing procedure with smoothing constants α = .5, β = γ = .1 is best based on MAD, MSD, and MAPE.

5.

Using Winters’ procedure in 4, the forecasts for the remainder of 1993 are: Month Apr/1993 May/1993 Jun/1993 Jul/1993 Aug/1993 Sep/1993 Oct/1993 Nov/1993 Dec/1993

Forecast 148 141 148 141 143 136 159 146 126

47

6.

There are no significant residual autocorrelations (see plot below). Winters’ multiplicative smoothing seems adequate.

CASE 4-4: MURPHY BROTHERS FURNITURE 1.

No adequate smoothing model was found! A Winters’ multiplicative model using α = .3, β = .2 and γ = .1 was deemed the best but there was still some significant residual autocorrelation.

48

2.

Yes. A Winters’ multiplicative smoothing procedure with α = .8, β = .1 and γ = .1 was adequate. Also, a naïve model that combined seasonal and trend estimates (similar to Equation 4.5) was found to be adequate. The trend and seasonal pattern in actual Murphy Brother’s sales are consistent and pronounced so a naïve model is likely to work well.

3.

Based on the forecasting methods tested, actual Murphy Brother’s sales data should be used. A plot of the results for the best Winters’ procedure follows.

An examination of the autocorrelation coefficients for the residuals from this Winters’ model shown below indicates that none of them are significantly different from zero.

However, Julie decided to use the naïve model because it was very simple and she could 49

explain it to her father.

CASE 4-5: FIVE-YEAR REVENUE PROJECTION FOR DOWNTOWN RADIOLOGY This case is designed to emphasize the use of subjective probability estimates in a forecasting situation. The methodology used to generate revenue forecasts is both appropriate and accurately employed. The key to answering the question concerning the accuracy of the projections hinges on the accuracy of the assumptions made and estimates used. Examination of the report indicates that the analysts were conservative each time they made an assumption or computed an estimate. This is probably one of the major reasons why the Professional Marketing Associates’ (PMA) forecast is considerably lower. Since we do not know how the accountant projected the number of procedures, it is difficult to determine why his revenue projections were higher. However, it is reasonable to assume that his forecast of the number of cases for each type of procedure was not nearly as sophisticated or thorough as PMAs. Therefore, the recommendation to management should indicate that the PMA forecast, while probably on the conservative side, is more likely to be accurate.

Downtown Radiology evidently agreed with PMA's forecast. They decided not to purchase a 9,800 series CT scanner. They also decided to purchase a less expensive MRI. Finally, they decided to obtain outside funding and did not resort to any type of public offering.

They built their new imaging center, purchased an MRI and have created a very successful imaging center.

CASE 4-6: WEB RETAILER 1.

The time series plot for Orders shows a slight upward trend and a seasonal pattern with peaks in December. Because of the relatively small data set, the autocorrelations are only computed for a limited number of lags, 6 in this case. Consequently with monthly data, the seasonality does not show up in the autocorrelation function. There is significant positive autocorrelation at lag 1, so Orders in consecutive months are correlated.

The time series plot for CPO shows a downward trend but a seasonal component is not readily apparent. There is significant positive autocorrelation at lag 1 and the autocorrelations die out relatively slowly. The CPO series is nonstationary and observations in consecutive time periods are correlated.

2.

Winters’ multiplicative smoothing with α = β = γ = .1 works well (see plot below).

Forecasts for the next 4 months follow. Residual autocorrelation function below has no significant autocorrelations.

Month Forecast Lower Upper Jul/2003 3524720 3072265 3977174 Aug/2003 3885780 3431589 4339972 Sep/2003 3656581 3200544 4112618 Oct/2003 4141277 3683287 4599266

50

3.

Simple exponential smoothing with α = .77 (the optimal α in Minitab) represents the the CPO data well but, like any “averaging” procedure, produces flat-line forecasts.

Forecasts of CPO for the next 4 months are: Month Forecast Lower Upper Jul/2003 0.1045 0.0787 0.1303 Aug/2003 0.1045 0.0787 0.1303 Sep/2003 0.1045 0.0787 0.1303 Oct/2003 0.1045 0.0787 0.1303 The results for simple exponential smoothing are pictured below. There are no significant residual autocorrelations (see plot below).

51

4.

Multiplying the Orders forecasts in 2 by the CPO forecasts in 3 gives the Contacts forecasts:

Month

Forecast 52

Jul/2003 Aug/2003 Sep/2003 Oct/2003

368333 406064 382113 432763

5.

It seems reasonable to forecast Contacts directly if the data are available.

Multiplying a forecast of Orders by a forecast of CPO to get a forecast of Contacts has the potential for introducing additional error (uncertainty) into the process.

6.

It may or may not be better to focus on the number of units and contacts per unit to get a forecast of contacts. It depends on the nature of the data (ease of modeling) and the amount of relevant data available.

CASE 4-7: SOUTHWEST MEDICAL CENTER 1.

Autocorrelation function for total visits suggests time series is nonstationary (since autocorrelations slow to die out) and seasonal (relatively large autocorrelation at lag 12).

2.

There is no adequate smoothing method to represent Mary’s data. Winters’ multiplicative smoothing with α = β = .5 and γ = .2 seems to do as well as any smoothing procedure (see error measures in plot below). Forecasts for the remainder of FY2003-04 generated by Winters’ procedure follow.

Month Forecast Mar/2004 1465.8 Apr/2004 1490.5 May/2004 1453.7 Jun/2004 1465.4 Jul/2004 1568.7 Aug/2004 1552.7

Lower Upper 1249.9 1681.7 1252.6 1728.4 1189.3 1718.1 1171.2 1759.6 1242.3 1895.1 1192.4 1913.0

Forecasts seem high. Residual autocorrelation function pictured below indicates some remaining significant autocorrelation not captured by Winters’ method.

53

3.

If another forecasting method can adequately account for the autocorrelation in the Total Visits data, it is likely to produce “better” forecasts. This issue is explored in subsequent cases.

4.

The forecasts from Winters’ smoothing show an upward trend. If they are to be believed, perhaps additional medical staff are required to handle the expected increased demand. At this point however, further study is required.

54

CASE 4-8: SURTIDO COOKIES 1.

Jame learned that Surtido Cookie sales have a strong seasonal pattern (sales are relatively high during the last two months of the year, low during the spring) with very little, if any, trend (see Case 3-5).

2.

The autocorrelation function for sales (see Case 3-5) is consistent with the time series plot. The autocorrelations die out (consistent with no trend) and have a spike at the seasonal lag 12 (consistent with a seasonal component).

3.

Winters’ multiplicative smoothing with α = β = γ = .2 seems to represent the data fairly well and produce reasonable forecasts (see plot below). However, there is still some significant residual autocorrelation at low lags.

Month Forecast Lower Upper Jun/2003 653254 91351 1215157 Jul/2003 712159 141453 1282865 Aug/2003 655889 75368 1236411 Sep/2003 1532946 941647 2124245 Oct/2003 1710520 1107533 2313507 Nov/2003 2133888 1518354 2749421 Dec/2003 1903589 1274702 2532476

55

4.

Karin’s forecasts follow.

Month Forecast Jun/2003 618914 Jul/2003 685615 Aug/2003 622795 Sep/2003 1447864 Oct/2003 1630271 Nov/2003 2038257 Dec/2003 1817989 These forecasts have the same pattern as the forecasts generated by Winters’ method but are uniformly lower. Winters’ forecasts seem more consistent with recent history.

56

CHAPTER 5 TIME SERIES AND THEIR COMPONENTS ANSWERS TO PROBLEMS AND CASES 1.

The purpose of decomposing a time series variable is to observe its various elements in isolation. By doing so, insights into the causes of the variability of the series are frequently gained. A second important reason for isolating time series components is to facilitate the forecasting process.

2.

The multiplicative components model works best when the variability of the time series increases with the level. That is, the values of the series spread out as the trend increases, and the set of observations have the appearance of a megaphone or funnel.

3.

The basic forces that affect and help explain the trend-cycle of a series are population growth, price inflation, technological change, and productivity increases.

4.

a.

Exponential

b.

Growth curve (Gompertz)

c.

Linear

5.

Weather and the calendar year such as holidays affect the seasonal component.

6.

a. & b.

c.

23.89 billion 57

7.

d.

648.5 billion

e.

The trend estimate is below Value Line’s estimate of 680 billion.

f.

Inflation, population growth, and new technology affect the trend of capital spending.

a. & b.

c.

Y = 9310 + 1795(19) = 43,415

d.

Cyclical component might be indicated because of wavelike behavior of observations about fitted straight line. However, if there is a cyclical affect, it is very slight.

8.

Median equals (109 .8 + 105 .9) 2 = 107.85

9.

Ŷ = TS = 850(1.12) = $952

10.

(80.0 + 85.4)/2 = 82.7 Ŷ = TS = 900(.827) = $744.3

11.

All of the statements are correct except d.

12.

Month

Sales ($ Thousands)

Seasonal Index (%)

Deseasonalized Data 58

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

125 113 189 201 206 241 230 245 271 291 320 419

51 50 87 93 95 99 96 89 103 120 131 189

245 226 217 216 217 243 240 275 263 243 244 222

The statement is not true. When the data are deseasonalized, it shows that business is about the same.

13.

a. & b. Would use both the trend and seasonal indices to forecast although seasonal component is not strong in this example (see plot and seasonal indices below).

Fitted Trend Equation Yt = 2268.0 + 22.1*t

Seasonal Indices Period

Index 59

1 2 3 4

0.969 1.026 1.000 1.005

Forecasts Period Q3/1996 Q4/1996

c.

Forecast 3305.39 3343.02

The forecast for third quarter is a bit low compared to Value Line’s forecast (3,305 versus 3,340). The forecast for fourth quarter is a bit high compared to Value Line’s (3,343 versus 3,300).

Additional plots associated with the decomposition follow.

60

14.

a.

Multiplicative Model Data Cavanaugh Sales Length 77 NMissing 0 Fitted Trend Equation Yt = 72.6 + 6.01*t Seasonal Indices Period 1 2 3 4 5 6 7 8 9 10 11 12

Index 1.278 0.907 0.616 0.482 0.426 0.467 0.653 0.863 1.365 1.790 1.865 1.288

61

b.

Pronounced trend and seasonal components. Would use both for forecasting.

c.

Forecasts (see plot in part a) Month Forecast Jun/2006 253 Jul/2006 358 Aug/2006 478 Sep/2006 764 Oct/2006 1012 Nov/2006 1066 Dec/2006 744

15.

a.

Additive Model Data LnSales Length 77 NMissing 0 Fitted Trend Equation Yt = 4.6462 + 0.0215*t

62

Seasonal Indices Period 1 2 3 4 5 6 7 8 9 10 11 12

b. c. & d.

e.

Index 0.335 -0.018 -0.402 -0.637 -0.714 -0.571 -0.273 -0.001 0.470 0.723 0.747 0.342

Pronounced trend and seasonal components. Would use both to forecast.

Forecasts Month Forecast of LnSales Forecast of Sales Jun/2006 5.75297 315 Jul/2006 6.07205 434 Aug/2006 6.36535 581 Sep/2006 6.85802 951 Oct/2006 7.13248 1252 Nov/2006 7.17779 1310 Dec/2006 6.79477 893 Forecasts of Cavanaugh sales developed from additive decomposition are 63

higher (for all months June 2006 through December 2006) than those developed from the multiplicative decomposition. Forecasts from multiplicative decomposition appear to be a little more consistent with recent behavior of Cavanaugh sales time series.

16.

a.

Multiplicative Model Data Disney Sales Length 63 NMissing 0 Fitted Trend Equation Yt = -302.9 + 44.9*t Seasonal Indices Period 1 2 3 4

b.

Index 0.957 1.022 1.046 0.975

There is a significant trend but it is not a linear trend. First quarter sales tend to be relatively low and third quarter sales tend to be relatively high.

However, the plot in part a indicates a multiplicative decomposition with a linear trend is not an adequate representation of Disney sales. Perhaps better to do a multiplicative decomposition with a quadratic trend. Even 64

better, in this case, is to do an additive decomposition with the logarithms of Disney sales. c.

With the right decomposition, would use both the trend and seasonal components to generate forecasts.

d.

Forecasts Quarter Forecast Q4/1995 2506 Q1/1996 2502 Q2/1996 2719 Q3/1996 2830 Q4/1996 2681 However, the plot in part a indicates that forecasts generated from a multiplicative decomposition with a linear trend are likely to be too low.

17.

a.

Variation appears to be increasing with level. Multiplicative decomposition may be appropriate or additive decomposition with the logarithms of demand. b.

Neither a multiplicative decomposition or an additive decomposition with a linear trend work well for this series. This time series is best modeled with other methods. The multiplicative decomposition is pictured below.

c.

Seasonal Indices (Multiplicative Decomposition for Demand) Period

Index

Period

Index 65

Period

Index

1 2 3 4

0.947 0.950 0.961 0.998

5 6 7 8

1.004 1.007 1.022 1.070

9 10 11 12

1.045 0.982 0.995 1.019

Demand tends to be relatively high in the late summer months. d.

Forecasts derived from a multiplicative decomposition of demand (see plot below).

Month Oct/1996 Nov/1996 Dec/1996

18.

Forecast 171.2 174.9 180.5

Multiplicative Model Data U.S. Retail Sales Length 84 NMissing 0 Fitted Trend Equation Yt = 128.814 + 0.677*t Seasonal Indices 66

Period 1 2 3 4 5 6 7 8 9 10 11 12

Index 0.880 0.859 0.991 0.986 1.031 1.021 1.007 1.035 0.973 0.991 1.015 1.210

Forecasts and Actuals Period Jan/1995 Feb/1995 Mar/1995 Apr/1995 May/1995 Jun/1995 Jul/1995 Aug/1995 Sep/1995 Oct/1995 Nov/1995 Dec/1995

Forecast 164.0 160.7 186.1 185.8 194.9 193.6 191.8 197.7 186.6 190.6 196.1 234.5

Actual 167.0 164.0 192.1 187.5 201.4 202.6 194.9 204.2 192.8 194.0 202.4 238.0

Forecasts maintain the seasonal pattern but are uniformly below the actual retail sales for 1995. However, MPE = MAPE = 2.49% is relatively small.

67

19.

.

600 = 500 1.2 500(1.37) = 685 people estimated for Feb

a.

Jan =

b.

T̂ = 140 + 5(t); t for Jan 2007 = 72 T̂ = 140 + 5(72) = 500

68

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec c.

Ŷ = 500(1.20) = 600 Ŷ = (140 + 5(73))(1.37) = 692 Ŷ = (140 + 5(74))(1.00) = 510 Ŷ = (140 + 5(75))(0.33) = 170 Ŷ = (140 + 5(76))(0.47) = 244 Ŷ = (140 + 5(77))(1.25) = 656 Ŷ = (140 + 5(78))(1.53) = 811 Ŷ = (140 + 5(79))(1.51) = 808 Ŷ = (140 + 5(80))(0.95) = 513 Ŷ = (140 + 5(81))(0.60) = 327 Ŷ = (140 + 5(82))(0.82) = 451 Ŷ = (140 + 5(83))(0.97) = 538

5

22.

Deflating a time series removes the effects of dollar inflation and permits the analyst to examine the series in constant dollars.

23.

1289.73(2.847) = 3,671.86

24.

Jan Feb Mar Apr May Jun Jul

25.

Multiplicative Model

303,589 251,254 303,556 317,872 329,551 261,362 336,417

Data Employed Men Length 130 NMissing 0 Fitted Trend Equation Yt = 65355 + 72.7*t

Seasonal Indices 69

Month Index 1 0.981 2 0.985 3 0.990 4 0.995 5 1.002 6 1.014

Month 7 8 9 10 11 12

Index 1.019 1.014 1.002 1.004 0.999 0.995

70

Forecasts Month Nov/2003 Dec/2003 Jan/2004 Feb/2004 Mar/2004 Apr/2004 May/2004 Jun/2004 Jul/2004 Aug/2004 Sep/2004 Oct/2004

Forecast 74791.4 74581.7 73607.8 73954.0 74393.4 74887.2 75454.0 76419.5 76894.1 76564.4 75757.2 76005.6

A multiplicative decomposition with a default linear trend is not quite right for these data. There is some curvature in the time series as the plot of the seasonally adjusted data indicates. Not surprisingly, there is a strong seasonal component with employment relatively high in the summer and relatively low in the winter. In spite of the not quite linear trend, the forecasts seem reasonable.

26.

A linear trend is not appropriate for the employed men data. The plot below shows a quadratic trend fit to the data of Table P-25.

71

Although better than a linear trend, the quadratic trend is not quite right. Employment for the years 2000—2003 seems to have leveled off. No simple trend curve is likely to provide an excellent fit to these data. The residual autocorrelation function below indicates a prominent seasonal component since there are large autocorrelations at the seasonal lag S = 12 and its multiples.

27.

Multiplicative Model 72

Data Wal-Mart Sales Length 56 NMissing 0 Fitted Trend Equation Yt = 1157 + 1088*t Seasonal Indices Quarter Index Q1 0.923 Q2 0.986 Q3 0.958 Q4 1.133

73

74

Forecasts and Actuals Quarter Forecast Q1/2004 58328 Q2/2004 63346 Q3/2004 62607 Q4/2004 75278

Actuals 65443 70466 69261 82819

Slight upward curvature in the Wal-Mart sales data so a linear trend is not quite right. Not surprisingly, there is a strong seasonal component with 4th quarter sales relatively high and 1st quarter sales relatively low. The forecasts for 2004 are uniformly below the actuals (primarily the result of the linear trend assumption) although the seasonal pattern is maintained. Here MPE = MAPE = 9.92%.

Multiplicative decomposition better than additive decomposition but any decomposition that assumes a linear trend will not forecast sales for 2004 well.

28.

A linear trend fit to the Wal-Mart sales data of Table P-27 is shown below. A linear trend misses the upward curvature in the data.

A quadratic trend provides a better fit to the Wal-Mart sales data (see plot below). The autocorrelation function for the residuals from a quadratic trend fit suggests a prominent seasonal component since there are large autocorrelations at the seasonal lag S = 4 and its multiples.

75

76

CASE 5-1: THE SMALL ENGINE DOCTOR 1.

2.

77

3.

SEASONAL ADJUSTMENT MONTH FACTORS

FITTED VALUES AND FORECASTS, T*S 2005 2006 2007

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

8.68 9.59 13.66 17.87 25.48 34.39 27.77 19.77 15.78 8.17 6.68 30.94

0.693 0.707 0.935 1.142 1.526 1.940 1.479 0.998 0.757 0.373 0.291 1.290

17.32 25.97 18.41 27.23 25.34 30.01 32.13 46.38 44.52 63.57 58.61 82.82 46.23 64.69 32.23 44.68 25.22 34.67 12.83 17.49 10.32 13.95 47.06 63.17

78

4.

79

5.

Trend*Seasonality (T*S): Linear Trend Model:

MAD = 1.52 MAD = 9.87

6.

If you had to limit your choices to the models in 2 and 4, the linear trend model is better (judged by MAD and MSE) than any of the Holt smoothing procedures.

However, the Trend*Seasonality (T*S) model is best. This procedure is the only one that takes account of the trend and seasonality in Small Engine Doctor sales.

CASE 5-2: MR. TUX At last, John is able to deal directly with the strong seasonal effect in his monthly data.

Students find it interesting that in addition to using these to forecast, John's banker wants them to justify variable loan payments.

To forecast using decomposition, students see that both the C and I components must be estimated. We like to emphasize that studying the C column in the computer printout is helpful, but that other study is needed to estimate the course of the economy over the next several months.

The computer is not able to make such forecasts with accuracy, as anyone who follows economic news well knows.

Thinking about John’s efforts to balance his seasonal business to achieve a more uniform sales picture can generate a good class discussion. This is usually the goal of any business; examples such as boats/skis or bikes/skis illustrate this effort in many seasonal businesses. In fact, John Mosby put a great deal of effort into expanding his Seattle business in order to balance his seasonal effect. Along with his shirt making business, he has achieved a rather uniform monthly sales volume.

80

1.

The two sentences might look something like this: A computer analysis of John Mosby's monthly sales data clearly shows the strong variation by month. I think we are justified in letting him make variable monthly loan payments based on the seasonal indices shown in the computer printout.

2.

Since John expects to do twice as much business in Seattle as Spokane, the Seattle indices he should try to achieve will be only half as far from 100 as the Spokane indices, and on the opposite side of 100: Spokane Seattle Jan 31.4 134.3 Feb 47.2 126.4 Mar 88.8 105.6 Apr 177.9 61.1 May 191.8 54.1 Jun 118.6 90.7 Jul 102.9 98.6 Aug 128.7 85.7 Sep 93.8 103.1 Oct 81.5 109.3 Nov 60.4 119.8 Dec 77.1 111.5

3.

Using the sales figures for January and February of 2005, to get “average” (100%) sales dollars, divide the actual sales by the corresponding seasonal index: Jan: 71,043/.314 = 226,252 Feb: 152,930/.472 = 324,004 Now subtract the actual sales from these target values to get the sales necessary from the shirt making machine: Jan: 226,252 - 71,043 = 155,209 Feb: 324,004 - 152,930 = 171,074

CASE 5-3: CONSUMER CREDIT COUNSELING Both the trend and seasonal components are important. The trend explains about 34% percent of the total variance.

Multiplicative Model Data Clients Length 99

81

Fitted Trend Equation Yt = 89.88 + 0.638*t Seasonal Indices Month Index 1 1.177 2 1.168 3 1.246 4 0.997 5 0.940 6 1.020 7 0.916 8 0.951 9 0.878 10 1.055 11 0.868 12 0.783

The number of new clients tends to be relatively large during the first three months of the year.

Forecasts Month Apr/2003 May/2003 Jun/2003 Jul/2003 Aug/2003 Sep/2003 Oct/2003 Nov/2003 Dec/2003

Forecast 153.207 145.121 158.062 142.440 148.560 137.749 166.161 137.261 124.277

82

There is one, possibly two, large positive residuals (irregularities) at the beginning of the series but there are no significant residual autocorrelations.

CASE 5-4: MURPHY BROTHERS FURNITURE 83

1.

An adequate model was found using Holt’s linear exponential smoothing.

Smoothing Constants Alpha (level) 0.980 Gamma (trend) 0.025 Accuracy Measures MAPE 1.1 MAD 76.2 MSD 11857.8 Forecasts Month Forecast Jan/2002 8127.8 Feb/2002 8165.1 Mar/2002 8202.4 Apr/2002 8239.7 May/2002 8277.0 Jun/2002 8314.2 Jul/2002 8351.5 Aug/2002 8388.8 Sep/2002 8426.1

2.

Forecasts and Actuals 84

Month Forecast Actual Jan/2002 7453.2 7120 Feb/2002 7462.5 7124 Mar/2002 8058.7 7817 Apr/2002 7873.1 7538 May/2002 8223.5 7921 Jun/2002 8140.9 7757 Jul/2002 8308.8 7816 Aug/2002 8611.1 8208 Sep/2002 8368.2 7828 Holt’s linear smoothing was adequate for the seasonally adjusted data, but the forecasts above are uniformly above the actual values for the first nine months of 2002.

3.

Using the same procedure as in 2, the forecast for October, 2002 is 8609.2.

4.

The pattern for the three sets of data shows a trend and monthly seasonality.

CASE 5-5: AAA WASHINGTON 1.

An additive and a multiplicative decomposition perform equally well. The multiplicative decomposition is shown below.

Multiplicative Model Data Calls Length 60 NMissing 0 Fitted Trend Equation Yt = 21851 - 17.0437*t

Seasonal Indices Month

Index 85

1 2 3 4 5 6 7 8 9 11 12

0.937 0.922 0.972 0.963 0.925 1.016 1.063 1.094 1.094 1.025 0.936

Accuracy Measures: MAPE

4

MAD 814

86

MSD 1276220

2.

Decomposition analysis works pretty well for AAA Washington data. There is a slight downward trend in emergency road service call volume with a pronounced seasonal component. Volume tends to be relatively high in the summer and early fall. There is significant residual autocorrelation at lag 1 (see plot below) so not all the association in the data has been accounted for by the decomposition.

CASE 5-6: ALOMEGA FOOD STORES 87

The sales data for the Alomega Food Stores case is subjected to a multiplicative decomposition procedure in this case. A trend line is first calculated with the actual data plotted around it (using MINITAB). Students can project this line into future months for sales forecasts, although, as the case suggests, accurate forecasts will not result: The MAPE using only the trend line is 28%.

A plot of the seasonal indices from the MINITAB output is shown below.. Students can summarize the managerial benefits to Julie from studying these values. As noted in the case, the MAPE drops to 12% when the seasonal indices along with the trend are used.

Finally, a 12-month forecast is generated using both the trend line and the seasonal indices. The forecasts seem reasonable.

Month Jan/2007 Feb/2007 Mar/2007 Apr/2007 May/2007 Jun/2007 Jul/2007 Aug/2007 Sep/2007 Oct/2007 Nov/2007 Dec/2007

Forecast 785348 326276 585307 391827 558299 453257 520615 319029 614997 394599 377580 235312

There are no significant residual autocorrelations.

Although more a management concern than a forecasting one, the attitude of Jackson Tilson in the case might generate a discussion that ties the computer assisted forecasting process into the real-life personalities of business associates. Although increasingly unlikely in the business setting, there are still those whose backgrounds do not include familiarity with computer based data analysis. Students whose careers will be spent in business might benefit from a discussion of the human element in the management process.

CASE 5-7: SURTIDO COOKIES 88

1.

Multiplicative Model Data SurtidoSales Length 41 NMissing 0 Fitted Trend Equation Yt = 907625 + 4736*t Seasonal Indices Month 1 2 3 4 5 6 7 8 9 10 11 12

Index 0.696 0.546 0.517 0.678 0.658 0.615 0.716 0.567 1.527 1.664 1.988 1.829

89

Month Jun/2003 Jul/2003 Aug/2003 Sep/2003 Oct/2003 Nov/2003

Forecast 680763 795362 633209 1710846 1872289 2246745 90

2.

3.

Dec/2003 2076183 The linear trend in sales has a slight upward slope. The seasonal indices show that cookie sales are relatively high the last four months of the year with a peak in November and relatively low the rest of the year.

The residual autocorrelation function is shown below. There are no significant residual autocorrelations.

The multiplicative decomposition adequately accounts for the trend and seasonality in the data. The forecasts are very reasonable. Jame should change his thinking about the value of decomposition analysis.

CASE 5-8: SOUTHWEST MEDICAL CENTER 1.

Decomposition of a time series involves isolating the underlying components that make up the time series. These components are the trend or trend/cycle (long term growth or decline), the seasonal (consistent within year variation typically related to the calendar) and the irregular (unexplained variation).

2.

The results and forecasts from a multiplicative decomposition and an additive decomposition are nearly the same (apart from the seasonal indices being either multiplicative or additive). For the purposes of this case, either can be considered. The results from a multiplicative decomposition follow.

91

Multiplicative Model Data Total Visits Length 114 NMissing 0 Fitted Trend Equation Yt = 955.6 + 4.02*t Seasonal Indices Month 1 2 3 4 5 6 7 8 9 10 11 12

Index 0.972 1.039 0.943 0.884 1.039 0.935 1.043 1.033 0.995 1.007 1.091 1.019

92

Forecasts Month Forecast Mar/2004 1479 Apr/2004 1469 May/2004 1419 Jun/2004 1440 Jul/2004 1564 Aug/2004 1464

Month Forecast Sep/2004 1401 Oct/2004 1502 Nov/2004 1367 Dec/2004 1284 Jan/2005 1514 Feb/2005 1367 93

3.

There is a distinct upward trend in total visits. The seasonal indices show that visits in December (4th month of fiscal year) tend to be relatively low and visits in July (11th month of fiscal year) tend to be relatively high.

4.

The residual autocorrelation function is shown below.

There are significant residual autocorrelations. The residuals are far from random The forecasts may be reasonable given the last three fiscal years of data. However, looking at the time series decomposition plot in 2, it is clear a decomposition analysis is not able to describe the middle two or three fiscal years of data. For some reason, visits for these fiscal years, in general, appear to be unusually high. A decomposition analysis does not adequately describe Mary’s data and leaves her perplexed.

94

CHAPTER 6 REGRESSION ANALYSIS ANSWERS TO PROBLEMS AND CASES 1.

Option b is inconsistent because the regression coefficient and the correlation coefficient must have the same sign.

2.

a. If GNP is increased by 1 billion dollars, we will expect earnings to increase .06 billion dollars. b. If GNP is equal to zero, we expect earnings to be .078 billion dollars.

3.

Correlation of Sales and AdvExpend = 0.848 The regression equation is Sales = 828 + 10.8 AdvExpend Predictor Coef SE Coef T P Constant 828.1 136.1 6.08 0.000 AdvExpend 10.787 2.384 4.52 0.002 S = 67.1945 R-Sq = 71.9% R-Sq(adj) = 68.4% Analysis of Variance Source DF SS MS F P Regression 1 92432 92432 20.47 0.002 Residual Error 8 36121 4515 Total 9 128552 a. Yes, the regression is significant. Reject H 0 : 1 = 0 using either the t value 2.384 and it’s p value .002, or the F ratio 20.47 and it’s p value .002. b. Y = 828 + 10.8X c. Y = 828 + 10.8(50) = $1368 d. 72% since r2 = .719 e. Unexplained variation (SSE) = 36,121

94

f. Total variation (SST) is 128,552 4.

Correlation of Time and Value = 0.967 The regression equation is Time = 0.620 + 0.109 Value Predictor Coef Constant 0.6202 Value 0.10919

SE Coef T P 0.2501 2.48 0.038 0.01016 10.75 0.000

S = 0.470952 R-Sq = 93.5% R-Sq(adj) = 92.7% Analysis of Variance Source DF SS MS F P Regression 1 25.622 25.622 115.52 0.000 Residual Error 8 1.774 0.222 Total 9 27.396 a. Yes, the regression is significant. Reject H 0 : 1 = 0 using either the t value 10.75 and it’s p value .000, or the F ratio 115.52 and it’s p value .000. b. Y = .620 + .109X e. Unexplained variation (SSE) = 1.774 f. Total variation (TSS) = 27.396 Point forecast: Y = .620 + .1092(3) = 0.948 99% Interval forecast: Y + tsf sf = sy.x

2 1 (X − X ) 1+ + n  ( X − X )2

sf = .471

1 (3 −19 .78 ) 1+ + 10 2148 .9

2

= .471 1 + .1 + .131 = .471 1.231

sf = .471(1.110) = .523

5.

.948  3.355(.523) → (–.807, 2.702) Prediction interval is wide because of small sample size and large confidence coefficient. Not useful. a, b and d.

95

.

The regression equation is Cost = 208.2 + 70.92 Age

(Positive linear relationship)

S = 111.610 R-Sq = 87.9% R-Sq(adj) = 86.2% Analysis of Variance Source DF Regression 1 Error 7 Total 8

SS MS 634820 634820 87197 12457 722017

F 50.96

P 0.000

c. Correlation between Cost and Age = .938 e. Reject H 0 : 1 = 0 at the 5% level since F = 50.96 and it’s p value = .000 < .05.

Could also use t = 7.14, the t value associated with the slope coefficient, and it’s p value = .000. The correlation coefficient is significantly different from 0 since the slope coefficient is significantly different from 0. f. Y = 208.20 + 70.92(5) = 562.80 or $562.80

6.

a, b and d.

96

The regression equation is Books = 32.46 + 36.41 Feet

(Positive linear relationship)

S = 17.9671 R-Sq = 90.3% R-Sq(adj) = 89.2% Analysis of Variance Source Regression Error Total

DF SS MS 1 27032.3 27032.3 9 2905.4 322.8 10 29937.6

F 83.74

P 0.000

c. Correlation between Books and Feet = .950 e. Reject H 0 : 1 = 0 at the 10% level since F = 83.74 and it’s p value = .000 < .10.

Could also use t = 9.15, the t value associated with the slope coefficient, and it’s p value = .000. The correlation coefficient is significantly different from 0 since the slope coefficient is significantly different from 0.

f. Based on the residuals versus the fitted values plot, there is no reason to doubt the adequacy of the simple linear regression model.

97

g. Y = 32.46 + 36.41(4) = 178 books 7.

a, b, c & d.

The regression equation is Orders = 15.8 + 1.11 Catalogs Predictor Coef SE Coef Constant 15.846 3.092 Catalogs 1.1132 0.3596

(Fitted regression line)

T P 5.13 0.000 3.10 0.011

S = 5.75660 (Standard error or estimate) R-Sq = 48.9% (Percentage of variation in Orders explained by Catelogs) R-Sq(adj) = 43.8% Analysis of Variance (ANOVA Table) Source DF SS Regression 1 317.53 Residual Error 10 331.38 Total 11 648.92

MS F 317.53 9.58 33.14

P 0.011

Predicted Values for New Observations New Obs Fit SE Fit 90% CI 90% PI 1 26.98 1.93 (23.47, 30.48) (15.97, 37.98) e. Do not reject H 0 : 1 = 0 at the 1% level since t = 3.10 and it’s p value = .011 > .01.

However, would reject H 0 : 1 = 0 at the, say, 5% level.

98

f. Do not reject H 0 : 1 = 0 at the 1% level since F = 9.58 and it’s p value = .011 > .01.

Result is consistent with the result in e as it should be. g. See Fit and 90% PI at end of computer printout above. A 90% prediction interval for mail orders when 10(000) catalogs are distributed is (16, 38)---16,000 to 38,000.

8.

The regression equation is Dollars = 3538 - 418 Rate Predictor Constant Rate

Coef SE Coef 3538.1 744.4 -418.3 150.8

T 4.75 -2.77

P 0.001 0.024

S = 356.690 R-Sq = 49.0% R-Sq(adj) = 42.7% Analysis of Variance Source DF SS MS Regression 1 978986 978986 Residual Error 8 1017824 127228 Total 9 1996810

F P 7.69 0.024

a. There is a significant (at the 5% level) negative relationship between these variables.

Reject H 0 : 1 = 0 at the 5% level since t = -2.77 and it’s p value = .024 < .05. b. The data set is small. Moreover, r2 = .49 so only 49% of the variation in investment dollars is explained by interest rate. Finally, the last observation (6.2, 1420) has a large influence on the location of the fitted straight line. If this observation is deleted, there is a considerable change in the slope (and intercept) of the fitted line. Using the original straight line equation for prediction is suspect. c. A forecast can be calculated. It is 1865. However, the 95% prediction interval is wide.

Forecast unlikely to be useful without additional information. See comments in b. d. See answer to b. e. It seems reasonable to say movements in interest rate cause changes in the level of investment.

9.

a. The firms seem to be using very similar rationale since r = .959. Also, from the fitted line plot below, notice the fitted line is not far from the 45o line through the origin (with intercept 0 and slope 1).

99

b. If ABC bids 1.01, the predicted competitor’s bid is 101.212. A 95% prediction interval (PI) is given below.

New Obs 101

Fit 101.212

SE Fit 0.164

95% CI (100.872, 101.552)

95% PI (99.637, 102.786)

c. Assume normality distributed errors about the population regression line and treat the least square line as if it were the population regression line (n is reasonably large in this case). Then at ABC bid 101, possible competitor bids are normally distributed about the fitted value 101.212 with a standard deviation estimated by sy.x = .743. Consequently, the probability that ABC will have the bid is P(Z ≥ (101-101.212)/ .743) = P(Z ≥ -.285) = .51.

10.

a. Only if the sample size is large enough. The t statistic associated with the slope coefficient or the F ratio should be consulted to determine if the population regression line slope is significantly different from a horizontal line with zero slope. b. It will typically produce significant results, not necessarily useful results.

The coefficient of determination, r2, might be small, so forecasting using the fitted line is unlikely to produce a useful result.

11.

a. Scatter diagram follows.

100

b. The regression equation is Permits = 2217 - 145 Rate Predictor Constant Rate

Coef 2217.4 -144.95

SE Coef 316.2 27.96

T P 7.01 0.000 -5.18 0.001

S = 144.298 R-Sq = 79.3% R-Sq(adj) = 76.4% Analysis of Variance Source DF Regression 1 Residual Error 7 Total 8

SS 559607 145753 705360

MS 559607 20822

F 26.88

P 0.001

c. Reject H 0 : 1 = 0 at the 5% level since t = -5.18 and it’s p value = .001 < .05. d. If interest rate increases by 1%, on average the number of building permits will decrease by 145. e. From the computer output above, r2 = .793. f. Interest rate explains about 79% of the variation in number of building permits issued.

12.

g. Memo to Ed explaining the fairly strong negative relationship between mortgage interest rates and building permits issued.

The population for this problem contains X-Y data points whose correlation coefficient is .846 ( = .846). Each student will have a different answer, however, 101

most will conclude that the Y is linearly related to X, that r is around .846, r-squared = .72, and so on. The population regression equation is Y = 0.948 + 0.00469X. Any student who fails to find a meaningful relationship between X and Y will be the victim of a Type II error.

13.

a. Scatter diagram follows.

b. The regression equation is Defectives = - 17.7 + 0.355 BatchSize Predictor Constant BatchSize

Coef SE Coef T P -17.731 4.626 -3.83 0.003 0.35495 0.02332 15.22 0.000

S = 7.86344 R-Sq = 95.5% R-Sq(adj) = 95.1% Analysis of Variance Source DF SS MS Regression 1 14331 14331 Residual Error 11 680 62 Total 12 15011

F 231.77

P 0.000

c. Reject H 0 : 1 = 0 at the 5% level since t = 15.22 and it’s p value = .000 < .05

d.

102

Residual Versus Fits plot shows curvature in scatter not captured by straight line fit. e. Model with quadratic term in Batch Size fits well. Results with Size**2 as predictor variable follow.

The regression equation is Defectives = 4.70 + 0.00101 Size**2 Predictor Constant Size**2

Coef 4.6973 0.00100793

SE Coef T 0.9997 4.70 0.00001930 52.22

P 0.001 0.000

S = 2.34147 R-Sq = 99.6% R-Sq(adj) = 99.6% Analysis of Variance Source DF SS Regression 1 14951 Residual Error 11 60 Total 12 15011

MS 14951 5

F 2727.00

P 0.000

f. Reject H 0 : 1 = 0 at the 5% level since t = 52.22 and it’s p value = .000 < .05

g. Residual plots below indicate an adequate fit.

103

h. Predicted Values for New Observations New Obs Fit SE Fit 1 95.411 1.173

95% CI 95% PI (92.829, 97.993) (89.647, 101.175)

i. Prefer second model with the quadratic predictor. j. Memo to Harry showing the value of transforming the independent (predictor) variable.

14.

a.

b. The regression equation is: Market = 60.7 + 0.414 Assessed

104

c. r 2 = .376 . About 38% of the variation in market prices is explained by assessed values (as predictor variable). There is a considerable amount of unexplained variation. d. F = 16.87 , p value = .000. Regression is highly significant. e. Yˆ = 98 .1 . Making a prediction at an assessed value, 90.5, outside of range covered by data (see scatter diagram). Linear relation may no longer hold. f. Residual plots follow.

Unusual Observations Obs Assessed 3 64.6 26 72.0

Market 87.200 97.200

Fit SE Fit 87.423 1.199 90.483 0.578

Residual -0.223 6.717

St Resid -0.10 X 2.83R

R denotes an observation with a large standardized residual.

X denotes an observation whose X value gives it large leverage.

15.

a. The regression equation is: OpExpens = 18.88 + 1.30 PlayCosts b. r 2 = .751. About 75% of the variation in operating expenses is explained by player costs. c. F = 72.6 , p value = .000 < .10. The regression is clearly significant at the  = .10 level.

105

d. Coefficient on X = player costs is 1.30. Is H 0 :  1 = 2 reasonable?

1.30 − 2.0 t= = −4.58 (p value = .000) suggests  1 = 2 is not supported by .153 the data. Appears that operating expenses have a fixed cost component represented by the intercept b0 = 18 .88 , and are then about 1.3 times player costs. e. Yˆ = 58 .6 , Yˆ  2.064s f gives 58.6  2.064(5.5) or (47.2, 70.0). f. Unusual Observations Obs PlayCosts OpExpens Fit SE Fit 7 18.0 60.00 42.31 1.64

Residual St Resid 17.69 3.45R

R denotes an observation with a large standardized residual Team 7 has unusually low player costs relative to operating expenses.

16.

a. Scatter diagram follows.

b. The regression equation is Consumption = - 811 + 0.226 Families Predictor Constant Families

Coef -811.0 0.22596

SE Coef 553.6 0.05622

T P -1.47 0.158 4.02 0.001

S = 819.812 R-Sq = 43.5% R-Sq(adj) = 40.8% Analysis of Variance 106

Source DF SS MS Regression 1 10855642 10855642 Residual Error 21 14113925 672092 Total 22 24969567

F 16.15

P 0.001

Although the regression is significant, the residual versus fit plot indicates the magnitudes of the residuals increase with the level. This behavior and the scatter diagram in a suggest that consumption is not evenly distributed about the regression line. That is, the data have a megaphone-like appearance. A straight line regression model for these data is not adequate. c & d. The response variable is converted to the natural log of newsprint consumption (LnConsum).

The regression equation is LnConsum = 5.70 + 0.000134 Families Predictor Coef Constant 5.6987 Families 0.00013413

SE Coef T P 0.3302 17.26 0.000 0.00003353 4.00 0.001

S = 0.488968 R-Sq = 43.2% R-Sq(adj) = 40.5% Analysis of Variance Source DF Regression 1 Residual Error 21 Total 22

SS 3.8252 5.0209 8.8461

MS 3.8252 0.2391 107

F 16.00

P 0.001

The regression is significant (F = 16, p value = .001) although only 43% of the variation in ln(consumption) is explained by families. The residual plots above suggest the straight line regression of ln(consumption) on families is adequate. This simple linear regression model with ln(consumption) is better than the same model with consumption as the response. e. Using the results in c, a forecast of ln(consumption) with 10,000 families is 7.040 so a forecast of consumption is 1,141. f. Other variables that will influence newsprint consumption include number of papers published and retail sales (influencing newspaper advertising).

17.

a. Can see from fitted line plot below that growth in number of steakhouses is exponential, not linear.

108

b. The slope of a regression of ln(location) versus year is related to the annual growth rate.

The regression equation is LnLocations = 0.348 + 0.820 Year Predictor Constant Year

Coef SE Coef 0.3476 0.3507 0.81990 0.09004

T 0.99 9.11

P 0.378 0.001

S = 0.376679 R-Sq = 95.4% R-Sq(adj) = 94.2% Analysis of Variance Source DF Regression 1 Residual Error 4 Total 5

SS 11.764 0.568 12.332

MS 11.764 0.142

F 82.91

P 0.001

Estimated annual growth rate is 100(e.82 – 1)% = 127% c. Forecast of ln(locations) for 2007 is .348 + .820(20) = 16.748. Hence a forecast of the number of Outback Steakhouse locations for 2007 is e16.748 or 18,774,310, an absurd number. This example illustrates the danger of extrapolating a trend (growth) curve far into the future.

18.

a, Can see from fitted line plot below that growth in number of copy centers is exponential, not linear.

109

b. The slope of a regression of ln(centers) on time (year) is related to the annual growth rate.

The regression equation is LnCenters = - 0.305 + 0.483 Time Predictor Constant Time

Coef SE Coef -0.3049 0.1070 0.48302 0.01257

T -2.85 38.42

P 0.015 0.000

S = 0.189608 R-Sq = 99.2% R-Sq(adj) = 99.1% Analysis of Variance Source DF Regression 1 Residual Error 12 Total 13

SS MS 53.078 53.078 0.431 0.036 53.509

F 1476.38

P 0.000

Estimated annual growth rate is 100(e.483 – 1)% = 62%

19.

c. Forecast of ln(centers) for 2012 is -.305 + .483(20) = 9.355. Hence a forecast of the number of On The Double copy centers for 2012 is e9.355 or 11,556, an unlikely number. This example illustrates the possible danger of extrapolating a trend (growth) curve some distance into the future. a. Intercept b0 = 17.954, Slope b1 = –.2715 b. Cannot reject H0 at the 10% level since the t value associated with the slope coefficient, –1.57, has a p value of .138 > .10. The regression is not significant.

110

There does not appear to be a relationship between profits per employee and number of employees. c. r2 = .15. Only 15% of the variation in profits per employee is explained by the number of employees. d. The regression is not significant. There is no point in using the fitted function to generate forecasts for profits per employee for a given number of employees.

20.

Deleting Dun and Bradstreet gives the following results: The regression equation is Profits = 25.0 - 0.713 Employees Predictor Constant Employees

Coef SE Coef T P 25.013 5.679 4.40 0.001 -0.7125 0.2912 -2.45 0.029

S = 9.83868 R-Sq = 31.5% R-Sq(adj) = 26.3% Analysis of Variance Source DF SS MS Regression 1 579.40 579.40 Residual Error 13 1258.40 96.80 Total 14 1837.80

F 5.99

P 0.029

The regression is now significant at the 5% level (t value = -2.45, p value = .029 < .05). r2 has increased from 15% to 31.5%. These results suggest there is a linear relationship between profits per employee and number of employees. A single observation can have a large influence on the regression analysis, particularly when the number of observations is relatively small. However, the relatively small r2 of 31.5% indicates there will be a fair amount of uncertainly associated with any forecast of profits per employee. Dun and Bradstreet should not be thrown out unless there is some good (non-numerical) reason not to include this firm with the others.

21.

a. The regression equation is Actual = 0.68 + 0.922 Estimate Predictor Constant

Coef 0.683

SE Coef 1.691

T P 0.40 0.690 111

Estimate

0.92230

0.08487

10.87

0.000

S = 5.69743 R-Sq = 83.1% R-Sq(adj) = 82.4% Analysis of Variance Source DF Regression 1 Residual Error 24 Total 25

SS 3833.4 779.1 4612.5

MS F 3833.4 118.09 32.5

P 0.000

b. The regression is significant (t value = 10.87, p value = .000 or, equivalently, F ratio = 118.09, p value = .000). c. r2 = .831 or 83.1% of the variation in actual costs is explained by estimated costs. d. If estimated costs are perfect predictor of actual costs, then  0 = 0 , 1 = 1. The estimated intercept coefficient, .683, is consistent with  0 = 0 . With the t value = .40 and its p value = .69, cannot reject the null hypothesis H 0 :  0 = 0 . To check the hypothesis H1 : 1 = 1 compute t =(.922-1)/.0849 = –.92, which is not in the rejection region for a two-sided test at any reasonable significance level. The estimated slope coefficient, .922, is consistent with 1 = 1 .

22.

e. The plot of the residuals versus the fitted values has a megaphone-like appearance.

The residuals are numerically smaller for smaller projects than for larger projects.

Estimated costs are more accurate predictors of actual costs for inexpensive (smaller) projects than for expensive (larger) projects. a. The regression is significant (t value = 14.71, p value = .000). b. r2 = .90 or 90% of the variation in ln(actual costs) is explained by ln(estimated costs).

112

c. If ln(estimated costs) are perfect predictor of ln(actual costs), then  0 = 0 , 1 = 1.

The estimated intercept coefficient, .003, is consistent with  0 = 0 . With the t value = .02 and its p value = .987, cannot reject the null hypothesis H 0 :  0 = 0 .

To check the hypothesis H1 : 1 = 1 compute t =(.968-1)/.0658 = –.49, which is not in the rejection region for a two-sided test at any reasonable significance level.

The estimated slope coefficient, .968, is consistent with 1 = 1 . d. ln(24) = 3.178, so forecast of ln(actual cost) = .0026 + .968(3.178) = 3.079. Forecast of actual cost is e3.079 = 21.737.

CASE 6-1: TIGER TRANSPORT This case asks students to summarize the analysis in a report to management. We find this a useful exercise since it requires students to put the application and results of a statistical procedure into their own words. If they are able to do this, they understand the technique.

This case illustrates the use of regression analysis in a situation where determining a good regression equation is only the first step. The results must then be priced out in order to arrive at a rational decision regarding a pricing policy. This situation can generate a discussion regarding the general nature of quantitative techniques: they aid in the decision-making process rather than replace it. Possible policies regarding the small-load charge can be discussed after the cost of such loads is determined. One approach would be to take small loads at company cost, which is low. The resultant goodwill might pay off in increased regular business. Another would be to charge a low cost for small loads but only if the customer agrees to book a certain number of large loads.

The low out-of-pocket cost involved in adding small loads can focus management attention in other directions. Since no significant costs need to be recovered by the small load charge, a policy based on other considerations is appropriate.

CASE 6-2: BUTCHER PRODUCTS, INC. 1.

The 89 degree temperature is 24 degrees off ideal (89 - 65 = 24). This value is placed into the regression equation yielding a forecast number of units per day of 338.

2.

Once again, the temperature is 24 degrees from ideal (65 - 41 = 24). For X = 24, a forecast of 338 units is calculated from the regression equation.

3.

Since there is a fairly strong relationship between output and deviation from ideal temperature (r = -.80), higher output may well result from efforts to control the temperature in the work area so that it is close to 65 degrees. Gene should consider ways to do this.

4.

Gene has made a decent start towards finding an effective forecasting tool. However, since about 36% of the variation in output is unexplained, he should look for additional important predictor variables.

CASE 6-3: ACE MANUFACTURING 113

1.

The correlation coefficient is: r = .927. The corresponding t = 8.9 for testing H 0 :  = 0 has a p value of .000. We reject H0 and conclude the correlation between days absent and employee age holds for the population.

2.

Y = –4.28 + .254X

3.

r = .859. About 86% of Y's (absent days) variability can be explained through knowledge of X (employee age).

4.

The null hypothesis H 0 : 1 = 0 is rejected using either t = 8.9, p value = .000 or the F = 79.3 with p value = .000. There is a significant relation between absent days and employee age.

5.

Placing X = 24 into the prediction equation yields a Y forecast of 1.8 absent days per year.

6.

If time and cost are not factors, it might be helpful to take a larger sample to see if these small sample results hold. If results hold, a larger sample will very likely produce more precise interval forecasts.

7.

The fitted function is likely to produce useful forecasts, although 95% prediction intervals can be fairly wide because of the small sample size.

2

CASE 6-4: MR. TUX 1.

After John uses simple regression analysis to forecast his monthly sales volume, he is not satisfied with the results. The low r-squared value (56.3%) disappoints him.

The high seasonal variation should be discussed as a cause of his poor fit when using only the month number to forecast sales. The possibility of using dummy variables to account for the monthly effect is a possibility. After this topic is covered in Chapter 7, you can have the students return to this case.

2.

Not adequate.

3.

The idea of serial correlation can be mentioned at this point. The possibility of autocorrelated residuals can be introduced based on John's Durbin-Watson statistic.

In fact, the DW is low, indicating definite autocorrelation. A class discussion about this problem and what might be done about it is useful. After this topic is covered in Chapter 8, you can have the students return to this case. We hope that by this time students appreciate the difficulties involved in real-life forecasting. Forecasting Compromises and multiple attempts are the norm, not exceptions.

114

CASE 6-5: CONSUMER CREDIT COUNSELING 1.

The correlation of Clients and Stamps = 0.431 and t = 3.24, so relationship is significant but not very useful.

The regression equation is Clients = 32.7 + 0.00349 Stamps Predictor Constant Stamps

Coef 32.68 0.003487

SE Coef 31.94 0.001076

T P 1.02 0.312 3.24 0.002

S = 23.6787 R-Sq = 18.6% R-Sq(adj) = 16.8% Analysis of Variance Source DF SS MS Regression 1 5891.9 5891.9 Residual Error 46 25791.4 560.7 Total 47 31683.2

F 10.51

P 0.002

The correlation of Clients and Index = 0.752. The relation is significant (see below).

The regression equation is Clients = - 199 + 2.94 Index Predictor Constant Index

Coef -198.65 2.9400

SE Coef T P 28.64 -6.94 0.000 0.2619 11.23 0.000

S = 19.9159 R-Sq = 56.5% R-Sq(adj) = 56.1% Analysis of Variance

2.

Source DF SS MS F P Regression 1 49993 49993 126.04 0.000 Residual Error 97 38475 397 Total 98 88468 The regression equation is Clients = - 199 + 2.94 BI Jan 1993: Clients = - 199 + 2.94 (125) = 168.5 Feb 1993: Clients = - 199 + 2.94 (125) = 168.5 Mar 1993: Clients = - 199 + 2.94 (130) = 183.2 Note: Students might develop a new equation that leaves out the first three months of data for 1993. This is a better way to determine whether the model works and the 115

results are: The regression equation is Clients = - 204 + 2.99 Index Predictor Coef SE Coef T P Constant -203.85 31.37 -6.50 0.000 Index 2.9898 0.2883 10.37 0.000 S = 20.0046 R-Sq = 53.4% R-Sq(adj) = 52.9% Analysis of Variance Source DF Regression 1 Residual Error 94 Total 95

SS 43028 37617 80645

MS 43028 400

F 107.52

P 0.000

Jan 1993: Clients= - 204 + 2.99 (125) = 169.8 Feb 1993: Clients= - 204 + 2.99 (125) = 169.8 Mar 1993: Clients = - 204 + 2.99 (130) = 184.7 Regressing Clients on the reciprocal of Index produces a little better straight line fit.

The results for this transformed predictor variable follow.

The regression equation is Clients = 470 - 37719 RecipIndex Predictor Coef SE Coef T P Constant 469.58 32.07 14.64 0.000 RecipIndex -37719 3461 -10.90 0.000 S = 19.4689 R-Sq = 55.8% R-Sq(adj) = 55.3% Analysis of Variance

3.

Source DF SS MS F P Regression 1 45015 45015 118.76 0.000 Residual Error 94 35630 379 Total 95 80645 Actual Forecast Forecast Forecast(RecipIndex predictor) Jan 1993 Feb 1993 Mar 1993

4.

152 151 199

169 169 183

170 170 185

168 168 180

Only if the business activity index could itself be forecasted accurately. Otherwise, it is not a viable predictor because the values for the business activity index are not 116

available in a timely fashion.

5.

Perhaps. This topic will be the subject of Chapter 8.

6.

If a good regression equation can be developed in which the changes in the predictor variable lead the response, it might be possible to accurately forecast the rest of 1993.

However, if the regression equation is based on coincident changes in the predictor variable and response, forecasts for the rest of 1993 could not be developed since values for the predictor variable are not known in advance.

CASE 6-6: AAA WASHINGTON 1.

The four linear regression models are shown below. Both temperature and rainfall are potential predictor variables.

The regression equation is Calls = 18366 + 467 Rate Predictor Coef SE Coef T P Constant 18366 1129 16.27 0.000 Rate 467.4 174.2 2.68 0.010 S = 1740.10 R-Sq = 11.0% R-Sq(adj) = 9.5%

The regression equation is Calls = 28582 - 137 Temp Predictor Constant Temp

Coef SE Coef 28582.2 956.0 -137.44 18.06

T P 29.90 0.000 -7.61 0.000

S = 1289.61 R-Sq = 51.3% R-Sq(adj) = 50.4%

The regression equation is Calls = 20069 + 400 Rain Predictor Constant Rain

Coef SE Coef T P 20068.9 351.7 57.07 0.000 400.30 84.20 4.75 0.000

S = 1555.56 R-Sq = 29.1% R-Sq(adj) = 27.8%

117

The regression equation is Calls = 27980 - 0.0157 Members 49 cases used, 3 cases contain missing values Predictor Constant Members

Coef 27980 -0.015670

SE Coef T 3769 7.42 0.008703 -1.80

P 0.000 0.078

S = 1628.15 R-Sq = 6.5% R-Sq(adj) = 4.5% 2. & 3. Sixty-five degrees was subtracted from the temperature variable. The variable used was the absolute value of the temperature with relative zero at 65 degrees Fahrenheit labeled NewTemp.

The correlation coefficient between Calls and NewTemp is .724, indicating a fairly strong positive linear relationship. However, examination of the fitted line plot below suggests there is a curvilinear relation between Calls and NewTemp

4.

A linear regression model with predictor variable NewTemp**2 gives a much better fit. The residual plots also indicate an adequate fit.

The regression equation is Calls = 20044 + 5.38 NewTemp**2 Predictor Constant NewTemp**2

Coef 20044.4 5.3817

SE Coef 203.1 0.5462

T 98.68 9.85 118

P 0.000 0.000

S = 1111.19 R-Sq = 63.8% R-Sq(adj) = 63.2% Analysis of Variance Source DF SS Regression 1 119870408 Residual Error 55 67910916 Total 56 187781324

MS 119870408 1234744

F 97.08

P 0.000

CHAPTER 7 MULTIPLE REGRESSION ANSWERS TO PROBLEMS AND CASES 1.

A good predictor variable is highly related to the dependent variable but not too highly related to other predictor variables.

2.

The population of Y values is normally distributed about E(Y), the plane formed by the 119

regression equation. The variance of the Y values around the regression plane is constant. The residuals are independent of each other, implying a random sample. A linear relationship exists between Y and each predictor variable.

3.

The net regression coefficient measures the average change in the dependent variable per unit change in the relevant independent variable, holding the other independent variables constant.

4.

The standard error of the estimate is an estimate of σ, the standard deviation of Y.

5.

Y = 7.52 + 3(20) - 12.2(7) = -17.88

6.

a. A correlation matrix displays the correlation coefficients between every possible pair of variables in the analysis. b. The proportion of Y's variability that can be explained by the predictor variables is given by R2. It is also referred to as the coefficient of determination. c. Collinearity results when predictor variables are highly correlated among themselves. d. A residual is the difference between an actual Y value and Ŷ , the value predicted using the sample regression plane. e. A dummy variable is used to determine the relationship between a qualitative independent variable and a dependent variable. f. Step-wise regression is a procedure for selecting the “best” regression function by adding or deleting a single independent variable at different stages of it’s development.

7.

a. Each variable is perfectly related to itself. The correlation is always 1. b. The entries in a correlation matrix reflected about the main diagonal are the same. For example, r32 = r23. c. Variables 5 and 6 with correlation coefficients of .79 and .70, respectively. d. The r14 = -.51 indicates a negative linear relationship. e. Yes. Variables 5 and 6 are to some extent collinear, r56 = .69. f. Models that include variables 4 and 6 or variables 2 and 5 are possibilities. The predictor variables in these models are related to the dependent variable and not 120

too highly related to each other. g. Variable 5.

8.

a. Correlations: Amount Items

Time Amount 0.959 0.876 0.923

The Full Model regression equation is: Time = 0.422 + 0.0871 Amount - 0.039 Items Predictor Constant Amount Items

Coef SE Coef 0.4217 0.5864 0.08715 0.01611 -0.0386 0.1131

T P VIF 0.72 0.483 5.41 0.000 6.756 -0.34 0.737 6.756

S = 0.857511 R-Sq = 92.1% R-Sq(adj) = 91.1% Analysis of Variance Source DF Regression 2 Residual Error 15 Total 17

SS 128.988 11.030 140.018

MS 64.494 0.735

F 87.71

P 0.000

Amount and Time are highly collinear (correlation = .923, VIF = 6.756). Both variables are not needed in the regression function. Deleting Items with the non-significant t value gives the best regression below.

The regression equation is Time = 0.263 + 0.0821 Amount Predictor Coef SE Coef T P Constant 0.2633 0.3488 0.75 0.461 Amount 0.082068 0.006025 13.62 0.000 S = 0.833503 R-Sq = 92.1% R-Sq(adj) = 91.6% Analysis of Variance Source DF SS Regression 1 128.90 Residual Error 16 11.12

MS 128.90 0.69 121

F 185.54

P 0.000

Total

17

140.02

b. From the Full Model, checkout time decreases by .039 which does not make sense. c. Using the best model Time = .2633 + .0821(28) = 2.5621 e = Y - Y = 2.4 - 2.5621 = -.1621 d. Using the best model, sy.x = .8335 e. The standard deviation of Y is estimated by .8335. f. Using the best model, the number of Items is not relevant so Time = .2633 + .0821(70) = 6.01 g. Using the best model, the 95% prediction interval (interval forecast) for Amount = $70 is given below.

New Obs Fit SE Fit 1 6.008 0.238

95% CI 95% PI (5.504, 6.512) (4.171, 7.845)

h. Multicollinearity is a problem. Jennifer should use the regression equation with the single predictor variable Amount.

9.

a. Correlations: Food, Income, Size Food Income Income 0.884 Size 0.737 0.867 Income is highly correlated with Food (expenditures) and, to a lesser extent, so is Size. However, the predictor variables Income and Size are themselves highly correlated indicating there is a potential multicollinearity problem. b. The regression equation is Food = 3.52 + 2.28 Income - 0.41 Size 122

Predictor Constant Income Size

Coef SE Coef 3.519 3.161 2.2776 0.8126 -0.411 1.236

T P VIF 1.11 0.302 2.80 0.026 4.016 -0.33 0.749 4.016

S = 2.89279 R-Sq = 78.5% R-Sq(adj) = 72.3% When income is increased by one thousand dollars holding family size constant, the average increase in annual food expenditures is 228 dollars. When family size is increased by one person holding income constant, the average decrease in annual food expenditures is 41 dollars. Since family size is positively related to food expenditures, r = .737, it doesn’t make sense that a decrease in expenditures would occur. c. Multicollinearity is a problem as indicated by VIF’s of about 4.0. Size should be dropped from the regression function and the analysis redone with only Income as the predictor variable.

10.

.

a. Both high temperature and traffic count are positively related to number of sixpacks sold and have potential as good predictor variables. There is some collinearity (r = .68) between the predictor variables but perhaps not enough to limit their value. b. Reject H 0 : 1 = 0 if |t| > 2.898

t=

b1 .78207 = = 3.45 sb1 .22694

Reject H0 because 3.45 > 2.898 and conclude that the regression coefficient for the high temp-variable is unequal to zero in the population.

Reject H 0 :  2 = 0 if |t| > 2.898

t=

b2 .06795 = = 3.35 sb2 .02026

Reject H0 because 3.35 > 2.898 and conclude that the regression coefficient for the traffic count variable is unequal to zero in the population. c. Y = -26.706 + .78207(60) + .06795(500) = 54 (six-packs)

123

2 2727.9  (Y − Y ) d. R = 1 =1= .81 2 14316.9  (Y − Y ) We are able to explain 81% of the number of six-packs sold variation using knowledge of daily high temperature and daily traffic count.

2

e. sy.x’s =

2  (Y − Yˆ ) = n − k −1

2727.9 = (20− 3)

160.46 = 12.67

f. If there is an increase of one degree in high temperature while the traffic count is held constant, beer sales increase on an average of .78 six-packs. g. The predictor variables explain 81% of the variation in six-packs sold. Both predictor variables are significant. It would be prudent to examine the residuals (not available in the problem) before deciding to use the fitted regression function for forecasting however.

11.

a. Scatter diagram follows. Female drivers indicated by solid circles, male divers by diamonds.

124

b. The regression equation is: Ŷ = 25.5 - 1.04 X1 + 1.21 X2 For a given age of car, female drivers expect to get about 1.2 more miles per gallon than male drivers. c. Fitted line for female drivers has equation: Yˆ = 26.21 − 1.04 X1 Fitted line for male drivers has equation: Yˆ = 25.5 − 1.04 X 1

(Parallel lines with different intercepts) d.

12.

Line falls “between” point representing female drivers and point representing male drivers. Straight line equation over-predicts mileage for male drivers and under-predicts mileage for female drivers. Important to include gender variable in this regression function. a. Correlations: Sales, Outlets, Auto Sales

Outlets 125

Outlets Auto

0.739 0.548

0.670

Number of retail outlets is positively related to annual sales, r12 = .74, and is potentially a good predictor variable. Number of automobiles registered is moderately related to annual sales, r13 = .55, and is positively correlated with number of retail outlets, r23 = .67. Given number of retail outlets in the regression function, number of automobiles registered may not be required. b. The regression equation is Sales = 10.1 + 0.0110 Outlets + 0.195 Auto Predictor Coef Constant 10.109 Outlets 0.010989 Auto 0.1947

SE Coef T P VIF 7.220 1.40 0.199 0.005200 2.11 0.068 1.813 0.6398 0.30 0.769 1.813

S = 10.3051 R-Sq = 55.1% R-Sq(adj) = 43.9% Analysis of Variance Source DF SS Regression 2 1043.7 Residual Error 8 849.6 Total 10 1893.2

MS 521.8 106.2

F 4.91

P 0.041

Predicted Values for New Observations New Obs Fit SE Fit 95% CI 95% PI 1 37.00 7.15 (20.50, 53.49) (8.07, 65.93) As can be seen from the regression output, it appears as if each predictor variable is not significant (at the 5% level), however the regression is significant at the 5% level. This is one of things that can happen when the predictor variables are collinear.

The forecast for region 1 is 37 with a prediction error of 52.3 – 37 = 15.3. However, it is not a good idea to use this fitted function for forecasting. If the regression is rerun after deleting Auto, Outlets (and the regression) is significant at the 1% level and R2 is virtually unchanged at 55%.

126

c. Y = 10.11 + .011(2500) + .195(20.2) = 41.549 (million) d. The standard error of estimate is 10.3 which is quite large. As explained in part b, the fitted function with both predictor variables should not be used to forecast.

Even if the regression is rerun with the single predictor Outlets, R2 =55% and the relatively large standard error of the estimate suggest there will be a lot of uncertainly associated with any forecast.

 (Y − Y ) = n − k −1 2

e. sy.x’s =

849.6 = 106 .2 = 10.3 (11 − 3)

f. If one retail outlet is added while the number of automobiles registered remains constant, sales will increase by an average of .011 million or $11,000 dollars. If one million more automobiles are registered while the number of retail outlets remains constant, sales will increase by an average of .195 million or $195,000 dollars. However, these regression coefficients are suspect due to collinearity between the predictor variables. g. New predictor variables should be tried.

13.

a. Correlations: Sales, Outlets, Auto, Income Outlets Auto Income

Sales Outlets 0.739 0.548 0.670 0.936 0.556

Auto

0.281

The regression equation is Sales = - 3.92 + 0.00238 Outlets + 0.457 Auto + 0.401 Income Predictor Coef SE Coef T P VIF Constant -3.918 2.290 -1.71 0.131 Outlets 0.002384 0.001572 1.52 0.173 2.473 Auto 0.4574 0.1675 2.73 0.029 1.854 Income 0.40058 0.03779 10.60 0.000 1.481 S = 2.66798 R-Sq = 97.4% R-Sq(adj) = 96.2%

Analysis of Variance Source Regression

DF SS MS 3 1843.40 614.47 127

F 86.32

P 0.000

Residual Error 7 Total 10

49.83 1893.23

7.12

Personal income by region makes a significant contribution to sales. Adding Income to the regression function results in an increase in R2 from 55% to 97%. In addition, the t value and corresponding p value for Income indicates the coefficient of this variable in the population is different from 0 given predictor variables Outlets and Sales. Notice however, the regression should be rerun after deleting the insignificant predictor variable Outlets. The correlation matrix and the VIF numbers suggest Outlets is multicollinear with Auto and Income. b. Predicted Values for New Observations New Obs Fit SE Fit 1 27.306 1.878

95% CI 95% PI (22.865, 31.746) (19.591, 35.020)

Values of Predictors for New Observations New Obs Outlets 1 2500

Auto 20.2

Income 40.0

Annual sales for region 12 is predicted to be 27.306 million.

2

c. The standard error of estimate has been reduced to 2.67 from 10.3 and R has increased to 97%. The 95% PI in part b is fairly narrow. The forecast for region 12 sales in part be should be accurate. d. The best choice is to drop Outlets from the regression function. If this is done, the regression equation is Sales = - 4.03 + 0.621 Auto + 0.430 Income Predictor Constant Auto Income

Coef SE Coef T P VIF -4.027 2.468 -1.63 0.141 0.6209 0.1382 4.49 0.002 1.086 0.43017 0.03489 12.33 0.000 1.086

S = 2.87655 R-Sq = 96.5% R-Sq(adj) = 95.6%

14.

Measures of fit are nearly the same as those for the full model and there is no longer a multicollinearity problem. a. Reject H0 : 1 = 0 if |t |> 3.1. .65 t= = 13 .05 Reject H0 and conclude that the regression coefficient for the aptitude test variable 128

is significantly different from zero in the population.

Similarly, Reject H0 : 2 = 0 if |t |> 3.1.

20.6 t= = 12.2 169 .

Reject H0 and conclude that the regression coefficient for the effort index variable is significantly different from zero in the population. b. If the effort index increases one point while aptitude test score remains constant, sales performance increases by an average of $20.600. c. Y = 16.57 + .65(75) + 20.6(.5) = 75.62 2

d. s y2 x ' s (n − 3) = (3.56) (14 - 3) = 139.4 2

e. s y2 (n − 1) = (16.57) (14 - 1) = 3569.3 f. R2 = 1 -

139.4  (Y − Y ) 2 =1= 1 - .039 = .961 2 3569.3  (Y − Y )

We can explain 96.1% of the variation in sales performance with our knowledge of the aptitude test score and the effort index.

R 2 = 1− g.

15.

SSE /(n − k − 1) 134.90 / 11 = 1− = .955 SST /(n − 1) 3569.3 / 13

a. Scatter plot for cash purchases versus number of items (rectangles) and credit card purchases versus number of items (solid circles) follows.

129

b. Minitab regression output:

Notice that for a given number of items, sales from cash purchases are estimated to be about $18.60 less than gross sales from credit card purchases. c. The regression in part b is significant. The number of items sold and whether the purchases were cash or credit card explains approximately 83% of the variation in gross sales. The predictor variable Items is clearly significant. The 130

coefficient of the dummy variable X2 is significantly different from 0 at the 10% level but not at the 5% level. From the residual plots below we see that there are a few large residuals (see, in particular, cash sales for day 25 and credit card sales for day 1); but overall, plots do not indicate any serious departures from the usual regression assumptions.

d. Y = 13.61 + 5.99(25) – 18.6(1) = $145 e. sy.x’s = 30.98

df = 47

t.025 = Z.025 = 1.96

95% (large sample) prediction interval: 145  1.96(30.98) = ($84, $206) f. Fitted function in part b is effectively two parallel straight lines given by the equations: Cash purchases: Y = 13.61 + 5.99Items – 18.6(1) = -4.98 + 5.99Items Credit card purchases: Y = 13.61 + 5.99Items If we fit separate straight lines to the two types of purchases we get: Cash purchases: Y = -.60 + 5.78Items R2 = 90.5% Y = 10.02 + 6.46Items Credit card purchases: R2 = 66.0% Predictions for cash sales and credit card sales will not be too much different for the two procedures (one prediction equation or two individual equations).

In terms of R2, the single equation model falls between the fits of the separate 131

models for cash purchases and credit card purchases but closer to the higher number for cash purchases. For convenience and overall good fit, prefer the single equation with the dummy variable.

16.

a. Correlations: WINS, ERA, SO, BA, RUNS, HR, SB WINS ERA SO BA RUNS HR ERA -0.494 SO 0.049 -0.393 BA 0.446 0.015 -0.007 RUNS 0.627 0.279 -0.209 0.645 HR 0.209 0.490 -0.215 0.154 0.664 SB 0.190 -0.404 -0.062 -0.207 -0.162 -0.305 ERA is moderately negatively correlated with WINS. SO is essentially uncorrelated with WINS. BA is moderately positively correlated with WINS and is also correlated with the predictor variable RUNS. RUNS is the predictor variable most highly correlated with WINS and will be the first variable to enter the regression function in a stepwise program. RUNS is fairly highly correlated with BA, so once RUNS is in the regression function, BA is unlikely to be needed.

HR is essentially not related to WINS. SB is essentially not related to WINS. b. The stepwise results are the same for an alpha to enter = alpha to remove = .05 or .15 (the Minitab default) or F to remove = F to enter =4.

Response is WINS on 6 predictors, with N = 26 Step Constant

1 2 20.40 71.23

RUNS T-Value P-Value

0.087 0.115 3.94 10.89 0.001 0.000

ERA T-Value P-Value

-18.0 -9.52 0.000

S 7.72 3.55 R-Sq 39.28 87.72 The fitted function from the stepwise program is: WINS = 71.23 + .115 RUNS - 18 ERA with R2 = 88% 17.

a. View will enter the stepwise regression function first since it has the largest 132

correlation with Price. After that the order of entry is difficult to determine from the correlation matrix alone. Several of the predictor variable pairs are fairly highly correlated so multicollinearity could be a problem. For example, once View is in the model, Elevation may not enter (be significant). Slope and Area are correlated so it may be only one of these predictors is required. b. As pointed out in part a, it is difficult to determine the results of a stepwise program.

However, a two predictor model will probably work as well as any in this case.

Potential two predictor models include View and Area or View and Slope.

18.

a., b., & c. The regression results follow.

The regression equation is Y = - 43.2 + 0.372 X1 + 0.352 X2 + 19.1 X3 Predictor Coef Constant -43.15 X1 0.3716 X2 0.3515 X3 19.12

SE Coef T 31.67 -1.36 0.3397 1.09 0.2917 1.21 11.04 1.73

P VIF 0.192 0.290 1.473 0.246 1.445 0.103 1.481

S = 13.9119 R-Sq = 49.8% R-Sq(adj) = 40.4% Analysis of Variance Source DF Regression 3 Residual Error 16 Total 19

SS 3071.1 3096.7 6167.8

MS 1023.7 193.5

F 5.29

P 0.010

Unusual Observations Obs X1 Y Fit SE Fit Residual 20 95 57.00 84.43 4.73 -27.43

St Resid -2.10R

R denotes an observation with a large standardized residual.

Predicted Values for New Observations New Obs Fit SE Fit 95% CI 95% PI 1 80.88 3.36 (73.77, 88.00) (50.55, 111.22) F = 5.29 with a p value = .010, so the regression is significant at the 1% level.

The predicted final exam score for within term exam scores of 86 and 77 and a GPA of 3.4 is Yˆ = 81 The variance inflation factors (VIF’s) are all small (near 1); however, the t ratios and 133

corresponding p values suggest that each of the predictor variables could be dropped from the regression equation. Since the F ratio was significant, we conclude that multicollinearity is a problem. d. Mean leverage = (3+1)/20= .20. None of the observations are high leverage points. e. From the regression output above, observation 20 has a large standardized residual.

The fitted model over-predicts the response (final exam score) for this student.

19.

Stepwise regression results, with significance level .05 to enter and leave the regression function, follow.

Alpha-to-Enter: 0.05 Alpha-to-Remove: 0.05 Response is Y on 3 predictors, with N = 20 Step Constant

1 -26.24

X3 T-Value P-Value

31.4 3.30 0.004

S R-Sq R-Sq(adj)

14.6 37.71 34.25

The “best” regression model relates final exam score to the single predictor variable grade point average.

All possible regression results are summarized in the following table.

Predictor Variables

R2

X1 .295 X2 .301 X3 .377 X1, X2 .404 X1, X3 .452 X2, X3 .460 X1, X2, X3 .498 2 The R criterion would suggest using all three predictor variables. However, the results in problem 7.18 suggest there is a multicollinearity problem with three predictors. The best two independent variable model uses predictors X2 and X3.

When this model is fit, X2 is not required. We end up with a model involving the single predictor X3, the model selected by the stepwise procedure.

20.

Best three predictor variable model selected by stepwise regression follows.

134

The regression equation is LnComp = 5.69 - 0.505 Educate + 0.255 LnSales - 0.0246 PctOwn Predictor Constant Educate LnSales PctOwn S = 0.4953

Coef 5.6865 -0.5046 0.2553 -0.0246

SE Coef 0.6103 0.1170 0.0725 0.0130

R-Sq = 42.8%

T P 9.32 0.000 -4.31 0.000 3.52 0.001 -1.90 0.064

VIF 1.0 1.0 1.0

R-Sq(adj) = 39.1%

Coefficient on education is negative. Everything else equal, as education level increases, compensation decreases. Positive coefficient on lnsales implies as sales increase, compensation increases, everything else equal. Finally, for fixed education and sales, as percent ownership increases, compensation decreases.

Unusual Observations Obs Educate LnComp 31 2.00 6.5338 33 0.00 6.3969

Fit 5.9055 7.0645

SE Fit 0.4386 0.2624

Residual St Resid 0.6283 2.73RX -0.6676 -1.59 X

R denotes an observation with a large standardized residual X denotes an observation whose X value gives it large influence.

Observation 31 has a large standardized residual and is influential. Observation 33 is also influential. The CEO’s for companies 31 and 33 own relatively large percentages of their company’s stock, 34 % and 17% respectively. They are outliers in this respect. The large residual for company 31 results from underpredicting compensation for this CEO. This CEO receives very adequate compensation in addition to owning a large percentage of the company’s stock.

All in all, this k = 3 predictor model appears to be better than the k = 2 predictor model of Example 7.12.

21.

Scatter diagram with fitted quadratic regression function:

135

a. & b. The regression equation is Assets = 7.61 - 0.0046 Accounts + 0.000034 Accounts**2 Predictor Constant Accounts Accounts**2

Coef 7.608 -0.00457 0.00003361

E Coef T P VIF 8.503 0.89 0.401 0.02378 -0.19 0.853 25.965 0.00000893 3.76 0.007 25.965

S = 12.4117 R-Sq = 97.9% R-Sq(adj) = 97.3% Analysis of Variance Source DF SS Regression 2 51130 Residual Error 7 1078 Total 9 52208

MS 25565 154

F 165.95

P 0.000

The regression is significant (F = 165.95, p value = .000). Given Accounts in the model, Accounts**2 is significant ( t value = 3.76, p value = .007). Here Accounts could be dropped from the regression function and the analysis repeated with only Accounts**2 as the predictor variable. If this is done, R2 and the coefficient of Accounts**2 remain virtually unchanged. c. Dropping Accounts**2 from the model gives:

The regression equation is Assets = - 17.1 + 0.0832 Accounts 136

Predictor Coef SE Coef T P Constant -17.121 8.778 -1.95 0.087 Accounts 0.083205 0.007592 10.96 0.000 S = 20.1877 R-Sq = 93.8% R-Sq(adj) = 93.0% The coefficient of Accounts changes from the quadratic model to the straight line model because, not surprisingly, Accounts and Accounts**2 are highly collinear (VIF = 25.965 in the quadratic model).

22.

The final model: The regression equation is Taste = - 30.7 + 4.20 H2S + 17.5 Lactic Predictor Coef SE Coef Constant -30.733 9.146 H2S 4.202 1.049 Lactic 17.526 8.412

T P VIF -3.36 0.006 4.01 0.002 2.019 2.08 0.059 2.019

S = 6.52957 R-Sq = 84.4% R-Sq(adj) = 81.8% Analysis of Variance Source DF SS MS Regression 2 2777.0 1388.5 Residual Error 12 511.6 42.6 Total 14 3288.7

F 32.57

P 0.000

The regression is significant (F = 32.57, p value = .000). Although Lactic is not a significant predictor at the 5% level, it is at the 6% level (t = 2.08, p value = .059) and we have chosen to keep it in the model. R2 indicates about 84% of the variation in Taste is explained by H2S and Lactic. The residual plots below indicate the fitted function is adequate. There is no reason to doubt the usual regression assumptions.

137

23.

Using the final model from problem 22 with H2S = 7.3 and Lactic = 1.85 Predicted Values for New Observations New Obs Fit SE Fit 1 32.36 3.02

95% CI (25.78, 38.95)

95% PI (16.69, 48.04)

Since s y x ' s = 6.53 and t.025 = 2.179 a large sample 95% prediction interval is: 32.36  2.179(6.53) → (18.13, 46.59)

Notice the large sample 95% prediction interval is not too much different than the actual 95% prediction interval (PI) above.

Although the fit in this case is relatively good, the standard error of the estimate is somewhat large, so there is a fair amount of uncertainty associated with any forecast.

It may be a good idea to collect more data and, perhaps, investigate additional predictor variables.

24.

a. Correlations: GtReceit, MediaRev, StadRev, TotRev, PlayerCt, OpExpens, ...

138

GtReceit MediaRev StadRev MediaRev 0.304 StadRev 0.587 0.348 TotRev 0.771 0.792 0.753 PlayerCt 0.423 0.450 0.269 OpExpens 0.636 0.554 0.623 OpIncome 0.562 0.672 0.547 FranValu 0.655 0.780 0.701

TotRev PlayerCt OpExpens OpIncome

0.499 0.766 0.785 0.925

0.867 -0.075 0.397

0.203 0.635

0.797

Total Revenue is likely to be a good predictor of Franchise Value. The correlation between these two variables is .925. b. Stepwise Regression: FranValu versus GtReceit, MediaRev, ...

Alpha-to-Enter: 0.05 Alpha-to-Remove: 0.05 Response is FranValu on 7 predictors, with N = 26 Step 1 Constant 2.928 TotRev T-Value P-Value

1.96 11.94 0.000

S 13.7 R-Sq 85.59 R-Sq(adj) 84.99 Results from stepwise program are not surprising given the definitions of the variables and the strong (and in some cases perfect) multicollinearity. c. The coefficient of TotRev from the stepwise program is 1.96 and the constant is relatively small and, in fact, insignificant. Consequently, Franchise Value is, on average, about twice Total Revenue. d. The regression equation is OpExpens = 18.9 + 1.30 PlayerCt Predictor Coef Constant 18.883 PlayerCt 1.3016

SE Coef 4.138 0.1528

T P 4.56 0.000 8.52 0.000

S = 5.38197 R-Sq = 75.1% R-Sq(adj) = 74.1% Analysis of Variance Source

DF

SS

MS 139

F

P

Regression 1 2101.7 Residual Error 24 695.2 Total 25 2796.9

2101.7 29.0

72.56

0.000

Unusual Observations Obs PlayerCt OpExpens 7 18.0 60.00

Fit SE Fit 42.31 1.64

Residual St Resid 17.69 3.45R

R denotes an observation with a large standardized residual.

The linear relation between Operating expenses and Player costs is fairly strong.

About 75% of the variation in Operating expenses is explained by Player costs.

Observation 7 (Chicago White Sox) have relatively low Player costs as a component of Operating expenses. e. Clearly Total revenue, Operating expenses and Operating income are multicollinear since, by definition, Operating income = Total revenue – Operating expenses. Also, Total revenue ≈ Gate receipts + Media revenue + Stadium revenue so this group of variables will be highly multicollinear.

CASE 7-1: THE BOND MARKET The actual data for this case is supplied in Appendix A. Students can either be asked to Respond to the question at the end of the case or they can be assigned to run and analyze the data.

One approach that I have used successfully is to assign one group of students the role of asking Judy Johnson's questions and another group the responsibility for Ron's answers.

1.

What questions do you think Judy will have for Ron? The students always seem to come up with questions that Ms. Johnson will ask. The key is that Ron should be able to answer them. Possible issues include: Are all the predictor variables in the final model required? Is a simpler model with fewer predictor variables feasible?

Do the estimated regression coefficients in the final model make sense and are they reliable?

Four observations have large standardized residuals. Is this a cause for concern?

Is the final model a good one and can it be confidently used to forecast the utility’s bond interest rate at the time of issuance?

Is multiple regression the appropriate statistical method to use for this situation?

140

CASE 7-2: AAA WASHINGTON 1.

The multiple regression model that includes both unemployment rate and average monthly temperature is shown below. Temperature is the only good predictor variable.

2.

Yes.

3.

Unemployment rate lagged 11 months is a good predictor of emergency road service calls. Unemployment rate lagged 3 months is not a good predictor. The Minitab output with Temp and Lagged11Rate is given below.

The regression equation is Calls = 21405 - 88.4 Temp + 756 Lag11Rate Predictor Coef SE Coef T P Constant 21405 1830 11.70 0.000 Temp -88.36 19.21 -4.60 0.000 Lag11Rate 756.3 172.0 4.40 0.000 S = 1116.80 R-Sq = 64.1% R-Sq(adj) = 62.8%

Analysis of Variance Source Regression

DF SS MS 2 120430208 60215104 141

F 48.28

P 0.000

Residual Error 54 67351116 Total 56 187781324

1247243

The regression is significant. The signs on the coefficients of the independent variables make sense. The coefficient of each independent variable is significantly different from 0 (t = –4.6, p value = .000 and t = 4.4, p value = .000, respectively).

4.

The results for a regression model with independent variables unemployment rate lagged 11 months (Lag11Rate), transformed average temperature (NewTemp) and NewTemp**2 are given below.

The regression equation is Calls = 17060 + 635 Lag11Rate - 112 NewTemp + 7.59 NewTemp**2 Predictor Coef SE Coef T P Constant 17060.2 847.0 20.14 0.000 Lag11Rate 635.4 146.5 4.34 0.000 NewTemp -112.00 47.70 -2.35 0.023 NewTemp**2 7.592 1.657 4.58 0.000 S = 941.792 R-Sq = 75.0% R-Sq(adj) = 73.5% Analysis of Variance Source DF SS MS Regression 3 140771801 46923934 Residual Error 53 47009523 886972 Total 56 187781324

F 52.90

P 0.000

Unusual Observations Obs Lag11R 11 6.10 29 5.60 32 6.19 34 5.72

Calls 24010 17424 24861 19205

Fit SE Fit Residual St Resid 22101 193 1909 2.07R 20346 191 -2922 -3.17R 24854 487 7 0.01 X 21157 201 -1952 -2.12R

R denotes an observation with a large standardized residual.

X denotes an observation whose X value gives it large leverage.

The residual plots follow. There is no significant residual autocorrelation at any lag.

142

The regression is significant. Each predictor variable is significant. R2 = 75%.

Apart from a couple of large residuals, the residual plots indicate an adequate model. There is no indication any of the usual regression assumptions have been violated. A good model has been developed.

CASE 7-3: FANTASY BASEBALL (A) 1.

The regression is significant. The R 2 of 78.1% looks good. The t statistic for each of the predictor variables is large with a very small p-value. The VIF’s are relatively small for the three predictors indicating that multicollinearity is not a problem. The residual plots shown in Figure 7-4 indicate that this model is valid. Dr. Hanke has developed a good model to forecast ERA.

2.

The matrix plot below of ERA versus each of five potential predictor variables does not show any obvious nonlinear relationships. There does not appear to be any reason to develop a new model.

143

3.

The regression results with WHIP replacing OBA as a predictor variable follow.

The residual plots are very similar to those in Figure 7-4.

The regression equation is ERA = - 2.81 + 4.43 WHIP + 0.101 CMD + 0.862 HR/9 Predictor Coef SE Coef Constant -2.8105 0.4873 WHIP 4.4333 0.3135 CMD 0.10076 0.04254 HR/9 0.8623 0.1195

T P VIF -5.77 0.000 14.14 0.000 1.959 2.37 0.019 1.793 7.22 0.000 1.135

S = 0.439289 R-Sq = 77.9% R-Sq(adj) = 77.4% Analysis of Variance Source DF SS MS Regression 3 91.167 30.389 Residual Error 134 25.859 0.193 Total 137 117.026

F 157.48

P 0.000

The fit and the adequacy of this model are virtually indistinguishable from the corresponding model with OBA instead of WHIP as a predictor. The estimated coefficients of CMD and HR/9 are nearly the same in both models. Both models are good. The original model with OBA as a predictor has a slightly higher R2 and a slightly smaller standard error of the estimate. Using these criteria, it is the preferred model.

CASE 7-4: FANTASY BASEBALL (B) 144

The project may not be doomed to failure. A lot can be learned from investigating the influence of the various independent variables on WINS. However, the best regression model does not explain a large percentage of the variation in WINS, R2 = 34%, so the experts have a point. There will be a lot of uncertainty associated with any forecast of WINS. The stepwise selection of the best predictor variables and the subsequent full regression output follow.

Stepwise Regression: WINS versus THROWS, ERA, ...

Alpha-to-Enter: 0.05 Alpha-to-Remove: 0.05 Response is WINS on 10 predictors, with N = 138 Step Constant

1 2 20.531 5.543

ERA T-Value P-Value

-2.16 -2.01 -7.00 -6.80 0.000 0.000

RUNS T-Value P-Value

0.0182 3.86 0.000

S R-Sq R-Sq(adj)

3.33 3.17 26.51 33.83 25.97 32.85

The regression equation is WINS = 5.54 - 2.01 ERA + 0.0182 RUNS Predictor Coef SE Coef T P VIF Constant 5.543 4.108 1.35 0.179 ERA -2.0110 0.2959 -6.80 0.000 1.017 RUNS 0.018170 0.004702 3.86 0.000 1.017 S = 3.17416 R-Sq = 33.8% R-Sq(adj) = 32.8% Analysis of Variance Source DF SS MS F Regression 2 695.31 347.66 34.51 Residual Error 135 1360.17 10.08 Total 137 2055.48

CHAPTER 8 145

P 0.000

REGRESSION WITH TIME SERIES DATA ANSWERS TO PROBLEMS AND CASES 1.

If not properly accounted for, serial correlation can lead to false inferences under the usual regression assumptions. Regressions can be judged significant when, in fact, they are not, coefficient standard errors can be under (or over) estimated so individual terms in the regression function may be judged significant (or insignificant) when they are not (or are) and so forth.

2.

Serial correlation often arises naturally in time series data. Series, like employment, whose magnitudes are naturally related to the seasons of the year will be autocorrelated.

Series, like sales, that arise because of a consistently applied mechanism, like advertising or effort, will be related from one period to the next (serially correlated). In the analysis of time series data, autocorrelated residuals arise because of a model specification error or incorrect functional form—the autocorrelation in the series is not properly accounted for.

3.

The independent observations (or, equivalently, independent errors) assumption is most frequently violated.

4.

Durbin-Watson statistic

5.

Reject H0 if DW < 1.10. Since 1.0 < 1.10, reject and conclude that the errors are positively autocorrelated.

6.

Reject H0 if DW < 1.55, Do not reject H0 if DW > 1.62. Since 1.6 falls between 1.55 and 1.62, the test is inconclusive.

7.

Serial correlation can be eliminated by specification of the regression function (using the best predictor variables) consistent with the usual regression assumptions. This can often be accomplished by using variables defined in terms of percentage changes rather than magnitudes, or autoregressive models, or regression models involving first differenced or generalized differenced variables.

8.

A predictor variable is generated by using the Y variable lagged one or more periods.

9.

The regression equation is Fuel = 113 - 8.63 Price - 0.137 Pop 146

Predictor Coef SE Coef T P Constant 113.01 16.67 6.78 0.000 Price -8.630 2.798 -3.08 0.009 Pop -0.13684 0.08054 -1.70 0.113 S = 2.29032 R-Sq = 76.6% R-Sq(adj) = 73.0% Analysis of Variance Source DF Regression 2 Residual Error 13 Total 15

SS MS 223.39 111.69 68.19 5.25 291.58

F 21.29

P 0.000

Durbin-Watson statistic = 0.612590 The null and alternative hypotheses are: H0:  = 0

H1:  > 0

Using the .05 significance level for a sample size of 16 with 2 predictor variables, dL = .98. Since DW = .61 < .98, reject H0 and conclude the observations are positively serially correlated.

10.

The regression equation is Visitors = 309899 + 24431 Time - 193331 Price + 217138 Celeb.

Predictor Coef SE Coef T P Constant 309899 59496 5.21 0.000 Time 24431 7240 3.37 0.007 Price -193331 97706 -1.98 0.076 Celeb.

217138 47412 4.58 0.001 S = 70006.1 R-Sq = 81.8% R-Sq(adj) = 76.4% Analysis of Variance Source DF Regression 3 Residual Error 10 Total 13

SS MS 2.20854E+11 73617995859 49008480079 4900848008 2.69862E+11

F 15.02

P 0.000

Durbin-Watson statistic = 1.14430

11.

With n = 14, k =3 and α = .05, DW = 1.14 gives an indeterminate test for serial correlation.

Serial correlation is not a problem. However, it is interesting to see whether the students realize that collinearity is a likely problem since Customer and Charge are highly correlated.

Correlation matrix: 147

Use Charge Customer

Revenue 0.187 0.989 0.918

Use

Charge

0.109 0.426

0.891

The regression equation is Revenue = - 65.6 + 0.00173 Use + 29.5 Charge + 0.000197 Customer Predictor Coef Constant -65.63 Use 0.001730 Charge 29.496 Customer 0.0001968

SE Coef T P VIF 14.83 -4.43 0.000 0.001483 1.17 0.255 2.151 2.406 12.26 0.000 8.515 0.0001367 1.44 0.163 10.280

S = 6.90038 R-Sq = 98.5% R-Sq(adj) = 98.4% Analysis of Variance Source DF SS MS Regression 3 77037 25679 Residual Error 24 1143 48 Total 27 78180

F 539.30

P 0.000

Durbin-Watson statistic = 2.20656 (Cannot reject H 0 :  = 0 at any reasonable significance level) Deleting Customer from the regression function gives: The regression equation is Revenue = - 57.6 + 0.00328 Use + 32.7 Charge Predictor Coef SE Coef T P VIF Constant -57.60 14.03 -4.11 0.000 Use 0.003284 0.001039 3.16 0.004 1.012 Charge 32.7488 0.8472 38.66 0.000 1.012 S = 7.04695 R-Sq = 98.4% R-Sq(adj) = 98.3%

Analysis of Variance Source DF SS MS Regression 2 76938 38469 Residual Error 25 1241 50

F 774.66 148

P 0.000

Total

27

78180

Durbin-Watson statistic = 1.82064 (Cannot reject H 0 :  = 0 at any reasonable significance level) 12.

a. Correlations: Share, Earnings, Dividend, Payout Earnings Dividend Payout

Share Earnings Dividend 0.565 0.719 0.712 0.435 -0.049 0.662

The best model, after taking account of the initial multicollinearity, uses the predictor variables Earnings and Payout (ratio).

The regression equation is Share = 4749 + 6651 Earnings + 171 Payout Predictor Coef SE Coef T P VIF Constant 4749 5844 0.81 0.424 Earnings 6651 1546 4.30 0.000 1.002 Payout 171.40 50.49 3.39 0.002 1.002 S = 3922.16 R-Sq = 53.4% R-Sq(adj) = 49.7% Analysis of Variance Source DF Regression 2 Residual Error 25 Total 27

SS MS 440912859 220456429 384584454 15383378 825497313

F 14.33

P 0.000

Durbin-Watson statistic = 0.293387 b. With n = 28, k = 2 and α = .01, DW = .29 < dL = 1.04 so there is strong evidence of positive serial correlation. c. An autoregressive model with lagged Shareholders as a predictor might be a viable option. A regression using generalized differences with ˆ = .8 is another possibility.

13.

a.

149

b. No. The residual autocorrelation function for the residuals from the straight line fit indicates significant positive autocorrelation. The independent errors assumption is not viable.

c. The fitted line plot with the natural logarithms of Passengers as the dependent variable and the residual autocorrelation function follow.

150

The residual autocorrelation function looks a little better than that in part b, but there is still significant positive autocorrelation at lag 1. d. Exponential trend plot for Passengers follows along with residual autocorrelation function.

151

Still some residual autocorrelation. Errors are not independent. e. Models in parts c and d are equivalent. If you take the natural logarithms of fitted exponential growth model you get the fitted model in part c. f. As we have pointed out, the errors for either of the models in parts c and d are not independent. Using a model that assumes the errors are independent can lead to inaccurate forecasts and, in this case, unwarranted precision.

14.

g. Using the exponential growth model with t = 26, gives Yˆ2007 = 195. a. The best model lags permits by 2 quarters (Lg2Permits): 152

Sales = 20.2 + 9.23 Lg2Permits Predictor Coef SE Coef T P Constant 20.24 27.06 0.75 0.467 Lg2Permits 9.2328 0.8111 11.38 0.000 S = 66.2883 R-Sq = 90.2% R-Sq(adj) = 89.6% b. DW = 1.47. No evidence of autocorrelation. c. The regression equation is Sales = 16.6 + 8.80 Lg2Permits + 30.0 Season Predictor Coef SE Coef T P Constant 16.61 27.99 0.59 0.563 Lg2Permits 8.801 1.020 8.63 0.000 Season 30.02 41.67 0.72 0.484 S = 67.4576 R-Sq = 90.6% R-Sq(adj) = 89.2%

d. No. For Season: t = .72, p value = .484. e. No. DW = 1.44. No evidence of autocorrelation. f. 2007

1st quarter forecast 177 2nd quarter forecast 113

Forecasts for the 3rd and 4th quarters can be done using several different approaches. This is best left to the student with a discussion of why they used a particular method. One method that is to average the past values of Permits for the 1st and 2nd quarters and use these averages in the model.

This will result in forecasts: 3rd quarter 514; 4th quarter 235.

15.

Quarter 1 2 3 4

Sales S2 S3 S4 16.3 17.7 28.1 34.3

0 1 0 0

0 0 1 0

0 0 0 1

 The regression equation is Sales = 19.3 - 1.43 S2 + 11.2 S3 + 33.3 S4 153

Predictor Coef SE Coef T P Constant 19.292 2.074 9.30 0.000 S2 -1.425 2.933 -0.49 0.630 S3 11.163 2.999 3.72 0.001 S4 33.254 2.999 11.09 0.000 S = 7.18396 R-Sq = 80.1% R-Sq(adj) = 78.7% Analysis of Variance Source DF SS MS F P Regression 3 8726.5 2908.8 56.36 0.000 Residual Error 42 2167.6 51.6 Total 45 10894.1 Durbin-Watson statistic = 1.544 1996(3rd Qt) Ŷ = 19.3 - 1.43(0) + 11.2(1) + 33.3(0) = 30.5 1996(4th Qt) Ŷ = 19.3 - 1.43(0) + 11.2(0) + 33.3(1) = 52.6 The regression is significant. The model explains 80.1% of the variation in Sales.

There is no lag 1 autocorrelation but a significant residual autocorrelation at lag 4.

16.

a. & b. The regression equation is Dickson = - 6.40 + 2.84 Industry Predictor Coef SE Coef T P Constant -6.4011 0.8435 -7.59 0.000 Industry 2.83585 0.02284 124.14 0.000 S = 0.319059 R-Sq = 99.9% R-Sq(adj) = 99.9% Durbin-Watson statistic = 0.8237 → Consistent with positive autocorrelation.

See also plot of residuals versus time and residual autocorrelation function.

154

c. ˆ = .585 . Calculate the generalized differences Yt ' = Yt − .585Yt −1 and

X t' = X t − .585X t −1 , and fit the model given in equation (8.5). The result is Yˆ ' = −2.31 + 2.81X ' with Durbin-Watson statistic = 1.74. In this case, the t

t

estimate of  1 , ˆ = 2.81 , is nearly the same as the estimate of  1 in part a.

1

Here the autocorrelation in the data is not strong enough to have much effect on the least squares estimate of the slope coefficient. d. The standard error of ̂1 is smaller in the initial regression than it is in the regression involving generalized differences. The standard error in the initial regression is under estimated because of the positive serial correlation. The standard error in the regression with generalized differences, although larger, is the one to be trusted.

17.

The regression equation is DiffSales = 149 + 9.16 DiffIncome 20 cases used, 1 cases contain missing values Predictor Coef SE Coef T P Constant 148.92 97.70 1.52 0.145 DiffIncome 9.155 2.034 4.50 0.000 S = 239.721 R-Sq = 53.0% R-Sq(adj) = 50.3% 155

Analysis of Variance Source DF SS MS Regression 1 1164598 1164598 Residual Error 18 1034389 57466 Total 19 2198987

F 20.27

P 0.000

Durbin-Watson statistic = 1.1237 Here DiffSales = Yt ' = Yt − Yt −1 and DiffIncome = X t' = X t − X t −1 . The results involving simple differences are close to the results obtained by the method of generalized differences in Example 8.5. The estimated slope coefficient is 9.16 versus an estimated slope coefficient of 9.26 obtained with generalized. The intercept coefficient 149 is also somewhat consistent with the intercept coefficient 54483(1−.997) = 163 for the generalized differences procedure. We would expect the two methods to produce similar results since ˆ = .997 is nearly 1.

18.

a. The regression equation is Savings = 4.98 + 0.0577 Income Predictor Coef SE Coef T P Constant 4.978 5.149 0.97 0.346 Income 0.05767 0.02804 2.06 0.054 S = 10.0803 R-Sq = 19.0% ← (2) 19% of variation in Savings explained by Income Analysis of Variance Source Regression

DF 1

SS MS 430.0 430.0

F 4.23

P 0.054 ← (1) Regression is not significant at .01 level

Residual Error 18 1829.0 101.6 Total 19 2259.0 Durbin-Watson statistic = 0.4135 ← With α = .05, dL = 1.20 so positive autocorrelation is indicated.

Can improve model by allowing for autocorrelated observations (errors). b. The regression equation is Savings = - 3.14 + 0.0763 Income + 20.2 War Year Predictor Coef SE Coef T P Constant -3.141 2.504 -1.25 0.227 Income 0.07632 0.01279 5.97 0.000 War Year 20.165 2.375 8.49 0.000 ← (1) Given Income, War Year makes 156

a significant contribution at the .01 level.

S = 4.53134 R-Sq = 84.5% R-Sq(adj) = 82.7% Analysis of Variance Source DF SS MS Regression 2 1909.94 954.97 Residual Error 17 349.06 20.53 Total 19 2259.00

F P 46.51 0.000

Durbin-Watson statistic = 2.010 ← (2) No significant autocorrelation of any kind is indicated.

Using all the usual criteria for judging the adequacy of a regression model, this model is much better than the simple linear regression model in part a.

19.

a.

157

The data are clearly seasonal with fourth quarter sales large and sales for the remaining quarters relatively small. Seasonality is confirmed by the autocorrelation function with significant autocorrelation at the seasonal lag 4. b. From the autocorrelation function observations 4 periods apart are highly positively correlated. Therefore an autoregressive model with sales lagged 4 time periods as the predictor variable might be appropriate. c. The regression equation is Sales = 421 + 0.853 Lg4Sales 24 cases used, 4 cases contain missing values Predictor Coef Constant 421.4 Lg4Sales 0.85273

SE Coef T P 230.0 1.83 0.081 0.09286 9.18 0.000

S = 237.782 R-Sq = 79.3% R-Sq(adj) = 78.4% Analysis of Variance Source DF SS MS Regression 1 4767638 4767638 Residual Error 22 1243890 56540 Total 23 6011528

F 84.32

P 0.000

Significant lag 1 residual autocorrelation. d. May 31 (2003) Y = 421.4 + .85273(2118) = 2227.5 compared to 2150 Aug 31 (2003) Y = 421.4 + .85273(2221) = 2315.3 compared to 2350 158

Nov 30 (2003) Y = 421.4 + .85273(2422) = 2486.7 compared to 2600 Feb 28 (2004) Y = 421.4 + .85273(3239) = 3183.4 compared to 3400 Forecasts are not bad but they are below the Value Line estimates for the last 3 quarters and the difference becomes increasingly larger. e. Value line estimates for the last 3 quarters of 2003-04 seem increasingly optimistic. f. Model in part c can be improved by allowing for significant lag 1 residual autocorrelation. One approach is to include sales lagged 1 quarter as an additional predictor variable.

20.

a. Correlations: ChickConsum, Income, ChickPrice, PorkPrice, BeefPrice ChickConsum Income 0.922 ChickPrice 0.794 PorkPrice 0.871 BeefPrice 0.913

Income ChickPrice PorkPrice 0.932 0.957 0.986

0.970 0.928

0.941

Correlations: LnChickC, LnIncome, LnChickP, LnPorkP, LnBeefP LnChickC LnIncome LnChickP LnPorkP LnIncome 0.952 LnChickP 0.761 0.907 LnPorkP 0.890 0.972 0.947 LnBeefP 0.912 0.979 0.933 0.954 The correlations are similar for both the original and natural log transformed data.

Correlations among the potential predictor variables are large implying a multicollinearity problem. Chicken consumption is most highly correlated with Income and BeefPrice for both the original and log transformed data. Must be careful interpreting correlations with time series data since autocorrelation in the individual series can result in apparent linear association. b. Stepwise Regression: ChickConsum versus Income, ChickPrice, ...

Alpha-to-Enter: 0.05 Alpha-to-Remove: 0.05 Response is ChickConsum on 4 predictors, with N = 23 Step Constant Income T-Value P-Value ChickPrice

1 28.86

2 37.72

0.00970 0.01454 10.90 6.54 0.000 0.000 -0.29 159

T-Value P-Value S R-Sq R-Sq(adj)

-2.34 0.030 2.58 84.98 84.27

2.34 88.21 87.03

c. There is high multicollinearity among the predictor variables so the final model depends on which non-significant predictor variable is deleted first. If BeefPrice is deleted, the final model is the one selected by stepwise regression (using a .05 level for determining significance of individual terms) with significant lag 1 residual autocorrelation. If Income is deleted first, then the final model involves the three Price predictor variables as shown below. There is no significant residual autocorrelation but large VIFs, although the coefficients of the predictor variables have the right signs. In this data set, Income is essentially a proxy for the three price variables.

The regression equation is ChickConsum = 37.9 - 0.665 ChickPrice + 0.195 PorkPrice + 0.123 BeefPrice Predictor Coef SE Coef T P VIF Constant 37.859 3.672 10.31 0.000 ChickPrice -0.6646 0.1702 -3.90 0.001 17.649 PorkPrice 0.19516 0.05874 3.32 0.004 21.109 BeefPrice 0.12291 0.02625 4.68 0.000 9.011 S = 2.11241 R-Sq = 90.9% R-Sq(adj) = 89.4% Analysis of Variance Source DF SS MS F Regression 3 844.44 281.48 63.08 Residual Error 19 84.78 4.46 Total 22 929.22

P 0.000

Durbin-Watson statistic = 1.2392 21. Stepwise Regression: LnChickC versus LnIncome, LnChickP, ...

Alpha-to-Enter: 0.05 Alpha-to-Remove: 0.05 Response is LnChickC on 4 predictors, with N = 23 Step Constant

1 1.729

2 2.375

LnIncome T-Value

0.283 0.440 14.32 15.40 160

P-Value

0.000

LnChickP T-Value P-Value

0.000 -0.445 -6.06 0.000

S 0.0528 0.0321 R-Sq 90.71 96.72 R-Sq(adj) 90.27 96.40 Final model is the one selected by stepwise regression. There is no significant residual autocorrelation.

The regression equation is LnChickC = 2.37 + 0.440 LnIncome - 0.445 LnChickP Predictor Coef SE Coef T P VIF Constant 2.3748 0.1344 17.67 0.000 LnIncome 0.43992 0.02857 15.40 0.000 5.649 LnChickP -0.44491 0.07342 -6.06 0.000 5.649 S = 0.0321380 R-Sq = 96.7% R-Sq(adj) = 96.4%

Analysis of Variance Source DF Regression 2 Residual Error 20 Total 22

SS MS 0.61001 0.30500 0.02066 0.00103 0.63067

F 295.30

P 0.000

Durbin-Watson statistic = 1.7766 The coefficient of .44 on LnIncome implies as Income increases 1% chicken consumption increases by .44%, chicken price held constant. Similarly, the coefficient of –.44 on LnChickP implies as chicken price increases by 1% chicken consumption decreases by .44%, income held constant.

To obtain a forecast of chicken consumption for the following year, forecasts of income and chicken price for the following year would be required. After taking logarithms, these values would be used in the final regression equation to get a forecast of LnChickC. A forecast of chicken consumption is then generated by taking the antilog.

22. The regression equation is DiffChickC = 1.10 + 0.00075 DiffIncome - 0.145 DiffChickP 161

22 cases used, 1 cases contain missing values Predictor Coef SE Coef T P VIF Constant 1.0967 0.4158 2.64 0.016 DiffIncome 0.000746 0.003477 0.21 0.832 1.029 DiffChickP -0.14473 0.06218 -2.33 0.031 1.029 S = 1.21468 R-Sq = 22.3% R-Sq(adj) = 14.1% Analysis of Variance Source DF SS MS F Regression 2 8.039 4.020 2.72 Residual Error 19 28.033 1.475 Total 21 36.073

P 0.091

Durbin-Watson statistic = 1.642 Very little explanatory power in the predictor variables. If the non-significant DiffIncome is dropped from the model, the resulting regression is significant at the .05 level, R 2 is virtually unchanged and the standard error of the estimate decreases slightly. The residual plots look good and there is no evidence of autocorrelation. With the very low R 2, the fitted function is not useful for forecasting the change (difference) in chicken consumption.

23. The regression equation is ChickConsum = 1.94 + 0.975 LagChickC 22 cases used, 1 cases contain missing values Predictor Coef SE Coef T P Constant 1.945 1.823 1.07 0.299 LagChickC 0.97493 0.04687 20.80 0.000 S = 1.33349 R-Sq = 95.6% R-Sq(adj) = 95.4% Analysis of Variance Source DF SS Regression 1 769.45 Residual Error 20 35.56 Total 21 805.01

MS F 769.45 432.71 1.78

162

P 0.000

Fitted regression function implies this year’s chicken consumption is likely to be a very good predictor of next year’s chicken consumption. The coefficient on lagged chicken consumption (LagChickC) is almost 1. The intercept in not significant.

Chicken consumption is essentially a “random walk”—next year’s chicken consumption is this year’s chicken consumption plus a random amount with mean 0. The residual plots look good and there is no residual autocorrelation.

We cannot infer the effect of a change in chicken price on chicken consumption with this model since chicken price does not appear as a predictor variable.

24.

Yt − Yt −1 = X t − X t −1 +  t −  t −1 =  t +  t −  t −1 = t say X t − X t −1 =  t Here the independent error  t has mean 0 and variance 3σ2. So the first differences for both Yt and X t are stationary and X and Y are cointegrated of order 1. The cointegrating 163

linear combination is: Yt − X t =  t .

CASE 8-2: BUSINESS ACTIVITY INDEX FOR SPOKANE COUNTY 1.

Why did Young choose to solve the autocorrelation problem first?

Answer: Autocorrelation must be solved for first to create data (or model) consistent with the usual regression assumptions.

2.

Would it have been better to eliminate multicollinearity first and then tackle autocorrelation?

Answer: No. In order to solve the autocorrelation problem, the nature of the data was changed (first differenced). If multicollinearity were solved first, one or more important variables may have been eliminated. Autocorrelation must be accounted for first so the usual regression assumptions apply; then multicollinearity can be tackled.

3.

How does the small sample size affect the analysis?

Answer: A sample size of 15 is small for a model that uses three independent variables (ideally, n should be in the neighborhood of 30 or more). A larger sample size would almost certainly be helpful.

4.

Should the regression done on the first differences have been through the origin?

Answer: Perhaps. An intercept can be included in the regression model and then checked for significance. Ordinarily, regressions with first differenced data does not require an intercept term.

5.

Is there any potential for the use of lagged data?

Answer: Perhaps. Although using lagged dependent and independent variables would constructing an index more difficult. Since the first differenced data work well in this case, there is no real need to consider lagged variables.

6.

What conclusions can be drawn from a comparison of the Spokane County business activity index and the GNP? Answer: The Spokane business activity seems to be extremely stable. It was not affected by the national recessions of 1970 and 1974. The large peak in 1974 was caused by Expo 74 (a world fair). It would be inappropriate in this case to expect the Spokane economy to follow national patterns.

CASE 8-3: RESTAURANT SALES 1.

Was Jim’s use of a dummy variable correct?

Answer: Jims’s use of a dummy variable to represent periods when Marquette was in session or out of session seems very reasonable. A good use of a dummy variable.

2.

Was it correct to use lagged sales as a predictor variable?

164

Answer: Jim's use of lagged sales as a predictor variable was eminently sensible.

This independent variable likely to have good predictor variable and can account for autocorrelation. This is a good time to I 3.

Do you agree with Jim’s conclusions?

Answer: Yes. Model 6 is the best. However, there may be other predictor variables that would improve this model; the number of students enrolled at Marquette during a particular quarter or semester is an example.

4.

Would another type of forecasting model be more effective for forecasting weekly sales?

Answer: Possibly! Jim will investigate Box-Jenkins ARIMA models in Chapter 9.

CASE 8-4: MR. TUX John is correct to be disappointed with the model run with seasonal dummy variables since the residual autocorrelations have a spike at lag 12. From a forecasting perspective, the autoregressive model is better. The intercept term allows for a time trend, seasonality is accounted for by sales lagged 12 months as the predictor variable, R2 is large (91%) and there is no residual autocorrelation. However, this model does not include predictor variables directly under John’s control, like price, so he would not be able to determine how a change in price (or changes in other operational variables) might affect future sales.

CASE 8-5: CONSUMER CREDIT COUNSELING Nonseasonal model: The regression equation is Clients = - 292 + 3.38 Index + 0.370 Bankrupt - 0.0656 Permits Predictor Coef SE Coef T P Constant -292.27 41.23 -7.09 0.000 Index 3.3783 0.3404 9.93 0.000 Bankrupt 0.37001 0.09740 3.80 0.000 Permits -0.06559 0.02882 -2.28 0.026 S = 16.6533 R-Sq = 61.0% R-Sq(adj) = 59.5% Analysis of Variance Source DF SS MS F Regression 3 34630 11543 41.62 Residual Error 80 22187 277

P 0.000 165

Total

83 56816

Durbin-Watson statistic = 1.605 The best nonseasonal regression model used the business activity index, number of bankruptcies filed, and number of building permits to forecast number of clients seen. The Durbin-Watson test for serial correlation is inconclusive at the .05 level. The residual autocorrelation function shows some significant autocorrelation around lag 4.

Best seasonal model: The regression equation is Clients = - 135 + 2.51 Index - 3.79 S2 + 5.69 S3 - 15.9 S4 - 21.1 S5 - 13.6 S6 - 20.6 S7 - 19.6 S8 - 25.9 S9 - 6.87 S10 - 19.0 S11 - 33.1 S12 Predictor Coef Constant -135.08 Index 2.5099 S2 -3.793 S3 5.686 S4 -15.869 S5 -21.146 S6 -13.580 S7 -20.641 S8 -19.650 S9 -25.857 S10 -6.869 S11 -19.014 S12 -33.143

SE Coef T P 26.96 -5.01 0.000 0.2421 10.37 0.000 8.443 -0.45 0.655 8.469 0.67 0.504 8.445 -1.88 0.064 8.441 -2.51 0.015 8.443 -1.61 0.112 8.441 -2.45 0.017 8.443 -2.33 0.023 8.441 -3.06 0.003 8.445 -0.81 0.419 8.448 -2.25 0.027 8.441 -3.93 0.000

S = 15.7912 R-Sq = 68.8% R-Sq(adj) = 63.6% Analysis of Variance Source DF SS Regression 12 39111.7 Residual Error 71 17704.7 Total 83 56816.3

MS F 3259.3 13.07 249.4

P 0.000

Durbin-Watson statistic = 1.757 The best seasonal model uses Index and 11 seasonal dummy variables to represent the months Feb through Dec. We retain all the seasonal dummy variables for forecasting purposes even though some are non-significant. The Durbin-Watson test is inconclusive at the .05 level. The residual autocorrelations have a just significant spike at lag 6 but are otherwise non-significant. Forecasts for the first three months of 1993 follow.

166

Jan 1993 Feb 1993 Mar 1993

Forecast 179 175 197

Actual 151 152 199

Forecasts for Jan and Feb 1993 are high compared to actual numbers of clients but forecast for Mar 1993 is very close to the actual number of new clients Autoregressive model: Autoregressive models with number of new clients lagged 1, 4 and 12 months were tried. None of these models proved to be useful for forecasting. The best model had number of new clients lagged 1 month. The results are displayed below.

The regression equation is Client = 61.4 + 0.487 LagClients 95 cases used, 1 cases contain missing values Predictor Coef Constant 61.41 LagClients 0.48678

SE Coef T P 10.91 5.63 0.000 0.08796 5.53 0.000

S = 24.9311 R-Sq = 24.8% R-Sq(adj) = 24.0%

Analysis of Variance Source DF SS Regression 1 19035 Residual Error 93 57805 Total 94 76840

MS F 19035 30.62 622

P 0.000

There is just significant residual autocorrelation at lag 12 but the remaining residual autocorrelations are small. The best model of the ones attempted is the final seasonal model with predictor variables Index and the seasonal dummies.

CASE 8-6: AAA WASHINGTON 1.

The results for the best model are shown below (see also solution to Case 7-2). Each of the independent variables is significantly different from 0 at the .05 level. The signs of the coefficients are what we would expect them to be.

The regression equation is Calls = 17060 + 635 Lg11Rate - 112 NewTemp + 7.59 NewTemp**2 Predictor

Coef

SE Coef

T 167

P

Constant 17060.2 Lg11Rate 635.4 NewTemp -112.00 NewTemp**2 7.592

847.0 20.14 0.000 146.5 4.34 0.000 47.70 -2.35 0.023 1.657 4.58 0.000

S = 941.792 R-Sq = 75.0% R-Sq(adj) = 73.5% Analysis of Variance Source DF SS MS Regression 3 140771801 46923934 Residual Error 53 47009523 886972 Total 56 187781324

F 52.90

P 0.000

Durbin-Watson statistic = 1.6217 2.

Serial correlation is not a problem. The value of the Durbin-Watson statistic (1.62) would not reject the null hypothesis of no serial correlation. There are no significant residual autocorrelations. Restricting attention to integer powers, 2 is the best choice for the exponential transformation. Allowing other choices for powers, e.g. 2.4, may improve the fit a bit but is not as “nice” as an integer power.

3.

The memo to Mr. DeCoria should use all the usual inferential and descriptive summaries to defend the model in part 1. A residual analysis should also be included.

CASE 8-7 ALOMEGA FOOD STORES 1.

Julie appears to have a good regression equation with an R-squared of 91%.

Additional significant explanatory variables may be available but there is not much variation left to explain. However, it good to have students search for a good equation using the full range of available variables. Along with the R-squared value, they should check the t values for the variables in their final equation, and the F value and the residual autocorrelations. Their results can be used effectively as individual or team presentations to the class, or as a hand-in writeup or even a small term paper.

2.

“Selling” the final regression model to management, including the irascible Jackson Tilson, ties the statistical exercise in the Alomega case to the real world of business management. The idea of selling the statistical results to management can be the focus of team presentations to the class with the instructor playing the role of Tilson. Working through the presentation of results to the class adds an important “real world” element to the statistical analysis.

3.

As noted in the case, the advertising predictor variables are under the control of Alomega management. Students can demonstrate the usefulness of this result by choosing reasonable future values for these advertising variables and generating forecasts.

However, students must recognize the regression equation does not necessarily imply a cause and effect relationship between advertising expenditures and sales. In 168

addition, conditions under which the model was developed may change in the future.

4.

All forecasts, including the ones using Julie’s regression equation, assume a future that is identical to the past except for the identified predictor variables. If her model is used to generate forecasts for Alomega, she should check the model accuracy on a regular basis. The errors encountered as the future unfolds should be compared to those in the data used to generate the model. If significant changes or trends are observed, the model should be updated to include the most recent data, along with possibly discarding some of the oldest data. Alternatively, a different approach to the forecasting problem can be sought if the forecasting errors suggest that the current regression model is inadequate.

CASE 8-8 SURTIDO COOKIES 1.

The positive coefficient on November makes sense because cookie sales are seasonal sales relatively high each year in November, the month before the Christmas holidays.

2.

Jame’s model looks good. Almost 94% of the variation in cookie sales is explained by the model. The residual analysis indicates the usual regression assumptions are tenable, including the independence assumption.

3.

Forecasts: June 2003 July 2003 August 2003 September 2003 October 2003 November 2003 December 2003

4.

733,122 799,823 737,002 1,562,070 1,744,477 2,152,463 1,932,194

The regression equation is SurtidoSales = 115672 + 0.950 Lg12Sales 29 cases used, 12 cases contain missing values Predictor Coef SE Coef T P Constant 115672 91884 1.26 0.219 Lg12Sales 0.94990 0.08732 10.88 0.000 S = 243748 R-Sq = 81.4% R-Sq(adj) = 80.7% Analysis of Variance Source

DF

SS

MS 169

F

P

Regression 1 Residual Error 27 Total 28

7.03141E+12 1.60415E+12 8.63556E+12

7.03141E+12 59412957997

118.35

0.000

Durbin-Watson statistic = 1.3524 This regression model is very reasonable. About 81% of the variation in cookie sales is explained with the single predictor variable, sales lagged 12 months (Lg12Sales). The usual residual plots look good and there is no significant residual autocorrelation.

Forecasts: June 2003 July 2003 August 2003 September 2003 October 2003 November 2003 December 2003

5.

717,956 632,126 681,996 1,642,130 1,801,762 2,113,392 1,844,434

Both models fit the data well. Apart from July 2003, the forecasts generated by the models are very close to one another. Dummy variable regression explains more of the variation in cookie sales but the autoregression is simpler. Could make a case for either model.

CASE 8-9 SOUTHWEST MEDICAL CENTER 1.

The regression results along with residual plots and the residual autocorrelation function follow.

The regression equation is Total Visits = 997 + 3.98 Time - 81.4 Sep + 5.3 Oct - 118 Nov - 149 Dec - 24.2 Jan - 116 Feb + 23.8 Mar + 18.2 Apr - 30.5 May - 39.4 Jun + 35.2 Jul Predictor Coef SE Coef Constant 996.97 58.42 Time 3.9820 0.4444 Sep -81.38 71.69 Oct 5.34 71.67 Nov -118.34 71.66 Dec -148.62 71.66 Jan -24.21 71.65

T P 17.06 0.000 8.96 0.000 -1.14 0.259 0.07 0.941 -1.65 0.102 -2.07 0.041 -0.34 0.736 170

Feb Mar Apr May Jun Jul

-116.39 23.80 18.15 -30.50 -39.37 35.20

71.65 -1.62 73.55 0.32 73.53 0.25 73.53 -0.41 73.52 -0.54 73.51 0.48

0.107 0.747 0.806 0.679 0.593 0.633

S = 155.945 R-Sq = 48.9% R-Sq(adj) = 42.9% Analysis of Variance Source DF SS MS Regression 12 2353707 196142 Residual Error 101 2456198 24319 Total 113 4809905

F 8.07

Durbin-Watson statistic = 0.4339

171

P 0.000

Mary has a right to be disappointed. This regression model does not fit well. Even allowing for seasonality, only the Dec seasonal dummy variable is significant at the .05 level. The residual plots clearly show a poor fit in the middle of the series and there is a considerable amount of significant residual autocorrelation.

2.

Mary might try an autoregression with different choices of lags of total visits as predictor variable(s). She might try to fit a Box-Jenkins ARIMA model to be discussed in Chapter 9. Regardless, finding an adequate model for this time series will be challenging.

172

CHAPTER 9 BOX-JENKINS (ARIMA) METHODOLOGY ANSWERS TO PROBLEMS AND CASES 1.

a. 0  .196 b. Series is random c. Series could be a stationary autoregressive process or series could be non-stationary.

Interpretation depends on how fast the autocorrelations decay to 0. d. Seasonal series with period of 4

2.

t 1 2 3 4

Yt 32.5 36.6 33.3 31.9

et Ŷt 35.000 -2.500 34.375 2.225 36.306 -3.006 33.581 -1.681

Ŷ 5 = 35 + .25(-1.681) - .3(-3.006) = 35.482 Ŷ 6 = 35 + .25(0) - .3(-1.681) = 35.504 Ŷ 7 = 35

3.

a. Ŷ61 = 75.65

Ŷ62 = 84.04

b. Ŷ62 = 76.55

Ŷ63 = 84.45

Ŷ63 = 87.82

c. 75.65  2√3.2 4.

a. Model Autocorrelations AR die out MA cut off ARIMA die out

5.

a. MA(2)

Partial Autocorrelations cut off die out die out

b. AR(1) c. ARIMA(1,0,1)

173

6.

a. Model is not adequate. b. Q = 44.3 df = 11 α = .05 Reject H0 if  2 > 19.675 Since Q = 44.3 > 19.675, reject H0 and conclude model is not adequate. Also, there is a significant residual autocorrelation at lag 2. Add a MA term to the model at lag 2 and fit an ARIMA(1,1,2) model.

7.

a. Autocorrelations of original series fail to die out, suggesting that demand is non-stationary. Autocorrelations for first differences of demand, do die out (cut off relative to standard error limits) suggesting series of first differences is stationary. Low lag autocorrelations of series of second differences increase in magnitude, suggesting second differencing is too much. A plot of the demand series shows the series is increasing linearly in time with almost a perfect (deterministic) straight line pattern. In fact, a straight line time trend fit to the demand data represents the data well as shown in the plot below.

If an ARIMA model is fit to the demand data, the autocorrelations and plots of the original series and the series of first differences, suggest an ARIMA(0,1,1) model with a constant term might be good starting point. The first order moving average term is suggested by the significant autocorrelation at lag 1 for the first differenced series. b. The Minitab output from fitting an ARIMA(0,1,1) model with a constant is shown below.

174

The least squares estimate of the constant term, .7127, is virtually the same as The least squares slope coefficient in the straight line fit shown in part a. Also, The first order moving average coefficient is essentially 1. These two results are consistent with a straight line time trend regression model for the original data.

Suppose Y t is demand in time period t. The straight line time trend regression model is: Yt =  0 +  1t +  t . Thus Yt −1 =  0 +  1 (t − 1) +  t −1 and Yt − Yt −1 =  1 +  t −  t −1 . The latter is an ARIMA(0,1,1) model with a constant term (the slope coefficient in the straight line model) and a first order moving average coefficient of 1.

There is some residual autocorrelation (particularly at lag 2) for both the straight line fit and the ARIMA(0,1,1) fit, but the usual residual plots indicate no other problems. c. Prediction equations for period 53.

Straight line model: Yˆ53 = 19.97 + .71(53) ARIMA model: Yˆ = Y + .71 − 1.00ˆ 53

52

52

d. The forecasts for the next four periods from forecast origin t = 52 for the ARIMA model follow.

These forecasts are essentially the same as the forecasts obtained by extrapolating the fitted straight line in part a.

8.

Since the autocorrelation coefficients drop off after one time lag and the partial 175

autocorrelation coefficients trail off, an MA(1) model should be adequate. The best model is Ŷ t = 56.1853 - (-0.7064)t-1

The forecast for period 127 is Ŷ 127 = 56.1853 + 0.7064)125

Ŷ 127 = 56.1853 + 0.7064)(-5.4) = 52.37

The critical 5% chi-square value for 10 df is 18.31. Since the calculated chi-square Q for the residual autocorrelations equals 7.4, the model is deemed adequate.

The autocorrelation and partial autocorrelation plots for the original series follow.

Autocorrelation Function for Yt 1.0 0.8 0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6 -0.8 -1.0

10

20

30

Lag

Corr

T

LBQ

Lag

T

LBQ

Lag

T

LBQ

Lag

Corr

T

LBQ

1

0.39

Corr

Corr

4.33

19.20

10 -0.12 -1.13

25.16

19 -0.06 -0.51

36.49

28

0.19 1.60

62.90

2 -0.08 -0.80

20.06

11 -0.08 -0.75

26.02

20 -0.13 -1.22

39.26

29

0.03 0.21

63.01

3

0.06

0.62

20.59

12 0.10

0.95

27.43

21 -0.04 -0.36

39.51

30 -0.05 -0.45

63.51

4

0.02

0.22

20.65

13 0.14

1.36

30.41

22 -0.11 -1.02

41.53

31

64.52

5 -0.07 -0.65

21.24

14 0.02

0.20

30.47

23 -0.25 -2.22

51.35

6

0.18

21.29

15 0.13

1.21

32.95

24 -0.03 -0.26

51.49

7 -0.01 -0.12

21.31

16 0.13

1.21

35.49

25

0.50

52.05

8 -0.08 -0.82

22.27

17 0.05

0.47

35.89

26 -0.16 -1.36

56.19

9 -0.08 -0.81

23.23

18 0.03

0.27

36.02

27 -0.06 -0.54

56.87

0.02

0.06

176

0.08 0.63

Partial Autocorrelation Function for Yt 1.0 0.8 0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6 -0.8 -1.0

10

Lag PAC

20

30

T

Lag PAC

T

Lag PAC

T

0.39

4.33

10 -0.08

-0.90

19 -0.04

-0.40

28

0.05

0.53

2 -0.27

-3.03

11 0.02

0.24

20 -0.10

-1.10

29 -0.11

-1.26

1

3

Lag PAC

T

0.26

2.95

12 0.13

1.41

21 0.05

0.61

30 -0.00

-0.05

4 -0.20

-2.25

13 0.04

0.50

22 -0.20

-2.30

31

0.45

5

0.08

0.92

14 -0.02

-0.27

23 -0.05

-0.59

6 -0.01

-0.10

15 0.21

2.36

24 0.14

1.53

7 -0.06

-0.65

16 -0.11

-1.18

25 -0.10

-1.14

8 -0.02

-0.22

17 0.17

1.94

26 -0.08

-0.92

9 -0.08

-0.95

18 -0.15

-1.64

27 0.10

1.17

0.04

ARIMA model for Yt Final Estimates of Parameters Type Coef StDev T MA 1 -0.7064 0.0638 -11.07 Constant 56.1853 0.5951 94.42 Mean 56.1853 0.5951 Number of observations: 126 Residuals: SS = 1910.10 (backforecasts excluded) MS = 15.40 DF = 124 Modified Box-Pierce (Ljung-Box) Chi-Square statistic Lag 12 24 36 48 Chi-Square 7.4(DF=10) 36.4(DF=22) 64.8(DF=34) 80.5(DF=46) Period 127 128 129 9.

Forecast 52.3696 56.1853 56.1853

95 Percent Limits Lower Upper 44.6754 60.0637 46.7651 65.6054 46.7651 65.6054

Since the autocorrelation coefficients trail off and the partial autocorrelation coefficients cut off after one time lag, an AR(1) model should be adequate.

The best model is

177

Ŷ t = 109.628 - 0.9377Yt-1

The forecast for period 81 is Ŷ 81 = 109.628 - 0.9377Y80 Ŷ 81 = 109.628 - 0.9377(85) = 29.92 Autocorrelation Function for Yt 1.0 0.8 0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6 -0.8 -1.0

5

Lag

Corr

10

T

LBQ

Lag

Corr

1 -0.88 -7.86

64.18

8

0.17

2

0.80

4.50 118.31

T

15

20

LBQ

Lag

Corr

T

LBQ

0.60 229.81

15

0.21

0.74

238.24

9 -0.09 -0.31 230.51

16 -0.24 -0.84 244.21

3 -0.66 -3.03 155.90

10

17

4

0.59

0.05

0.19 230.78

0.26

0.90

251.34

2.45 186.36

11 -0.01 -0.04 230.79

18 -0.33 -1.13 262.96

5 -0.48 -1.83 206.31

12 -0.02 -0.08 230.84

19

6

13

20 -0.39 -1.28 293.47

0.40

1.47 220.35

7 -0.28 -1.00 227.24

0.08

0.28 231.46

14 -0.15 -0.53 233.73

178

0.36

1.22

277.15

Partial Autocorrelation Function for Yt 1.0 0.8 0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6 -0.8 -1.0

5

Lag

10

15

PAC

T

Lag PAC

T

1 -0.88

-7.86

8 -0.16

2

0.13

1.16

9 -0.16

3

0.28

2.54

10

0.13

4

0.16

1.43

11

5

0.15

1.30

6 -0.04

-0.40

13

7

1.10

0.12

Lag

20

PAC

T

-1.44

15 -0.00

-0.04

-1.44

16

0.03

0.28

1.19

17

0.01

0.08

0.11

1.01

18 -0.26

-2.32

12 -0.02

-0.19

19 -0.10

-0.87

0.11

0.98

20

1.42

14 -0.22

-1.95

ARIMA model for Yt Final Estimates of Parameters Type Coef StDev AR 1 -0.9377 0.0489 Constant 109.628 0.611 Mean 56.5763 0.3151

0.16

T -19.17 179.57

Number of observations: 80 Residuals: SS = 2325.19 (backforecasts excluded) MS = 29.81 DF = 78 Modified Box-Pierce (Ljung-Box) Chi-Square statistic Lag 12 24 36 48 Chi-Square 24.8(DF=10) 39.4(DF=22) 74.0(DF=34) 83.9(DF=46) Period 81 82 83

Forecast 29.9234 81.5688 33.1408

95 Percent Limits Lower Upper 19.2199 40.6269 66.8957 96.2419 15.7088 50.5728

The critical 5% chi-square value for 10 df's is 18.31. Since the calculated chi-square Q for the residual autocorrelations equals 24.8, the model is deemed inadequate. An examination of the individual residual autocorrelations suggests it might be possible to 179

improve the model by adding a MA term at lag 2.

10.

As can be seen below, the autocorrelations for the original series are slow to die out. This behavior indicates the series may be non-stationary. The autocorrelations for the differenced data cut off after lag 1 and the partial autocorrelations die out. This suggests an ARIMA(0,1,1) model. When this model is fit (see the computer output below), there are no significant residual autocorrelations and the residual plots look good. The forecasting equation from the fitted model is Ŷ t = Yt-1 - (-0.3714)t-1

The forecast for period 81 is Ŷ 81 = Y80 - (-0.3714)80 Ŷ 81 = 266.9 - (-0.3714)(3.4647) = 268.19 Autocorrelation Function for Yt 1.0 0.8 0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6 -0.8 -1.0

5

10

Lag

Corr

T

LBQ

Lag

Corr

1

0.92

8.24

70.42

8

2

0.83

4.50 127.89

9

3

0.74

3.28 174.43

4

0.65

2.54 210.36

5

0.55

6

0.45

7

0.35

T

15

20

LBQ

Lag

Corr

0.28

0.92 271.99

15

0.01

0.23

0.75 276.74

16 -0.04 -0.11 283.92

10

0.18

0.58 279.67

17 -0.09 -0.28 284.72

11

0.14

0.45 281.49

18 -0.14 -0.45 286.81

2.00 236.52

12

0.12

0.38 282.83

19 -0.19 -0.62 290.72

1.55 254.16

13

0.09

0.28 283.55

20 -0.22 -0.71 296.13

1.18 265.00

14

0.05

0.16 283.78

180

T

LBQ

0.03 283.80

ARIMA model for Yt Final Estimates of Parameters Type Coef StDev T MA 1 -0.3714 0.1052 -3.53 Differencing: 1 regular difference Number of observations: Original series 80, after differencing 79 Residuals: SS = 10637.3 (backforecasts excluded) MS = 136.4 DF = 78 Modified Box-Pierce (Ljung-Box) Chi-Square statistic Lag 12 24 36 48 Chi-Square 9.2(DF=11) 14.1(DF=23) 28.6(DF=35) 39.2(DF=47) Period 81 82 83

95 Percent Limits Lower Upper 245.848 291.635 229.885 307.597 218.787 318.695

Forecast 268.741 268.741 268.741

The critical 5% chi-square value for 11 df's is 19.68. Since the calculated chi-square Q for the residual autocorrelations equals 9.2, the model is deemed adequate.

11.

The slow decline in the early, non-seasonal lags indicates the need for regular differencing.

Autocorrelation Function for Yt 1.0 0.8 0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6 -0.8 -1.0

2

12

Lag

Corr

T

LBQ

1

0.71

6.92

49.43

2

0.63

4.34

88.66

3

0.63

4

Lag

Corr

T

22

LBQ

Lag

Corr

8 0.54

2.07 309.89

15

9 0.50

1.85 337.18

16

3.69 128.66

10 0.45

1.61 359.38

0.62

3.23 168.41

11 0.50

5

0.63

2.98 210.04

6

0.56

2.41 242.61

7

0.59

2.39 279.06

LBQ

Lag

Corr

0.40

1.22 506.38

22

0.23 0.64 602.29

0.40

1.20 525.09

23

0.26 0.73 610.88

17

0.42

1.26 546.38

24

0.42 1.18 634.13

1.74 387.14

18

0.33

0.97 559.60

12 0.70

2.34 441.38

19

0.35

1.03 574.97

13 0.49

1.56 468.47

20

0.30

0.87 586.38

14 0.41

1.28 487.93

21

0.27

0.78 595.80

181

T

T

LBQ

Autocorrelation Function for Regular 1.0 0.8 0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6 -0.8 -1.0

5

Lag

Corr

T

LBQ

1 -0.35 -3.44 2 -0.17 -1.49 3

15

Lag

Corr

25

T

LBQ

Lag

T

LBQ

Lag

12.19

8 -0.01 -0.11

25.78

15 -0.03 -0.19

90.84

22 -0.13 -0.72 109.87

15.08

9 0.05

0.40

26.06

16 -0.05 -0.28

91.10

23 -0.25 -1.43 118.13

0.07

15.09

10 -0.17 -1.35

29.20

17 0.25

98.33

24

4 -0.03 -0.23

15.16

11 -0.29 -2.22

38.14

18 -0.24 -1.38 105.13

5

1.57

18.65

12 0.65

4.80

85.04

19 0.14

6 -0.21 -1.78

23.41

13 -0.19 -1.14

89.05

20 -0.02 -0.14 107.52

7

25.76

14 -0.12 -0.73

90.72

21 0.05

0.01

0.18

0.15

1.21

Corr

1.47

Corr

T

LBQ

0.54 2.98 156.06

25 -0.14 -0.71 158.67

0.79 107.44

0.26 107.79

The peaks at lags 12 and 24 are apparent. The seasonal autocorrelation coefficients seem to be decaying slowly. Seasonal differencing is necessary.

The autocorrelation coefficient and partial autocorrelation coefficient plots for the regular and seasonal differenced data are shown on the next page.

182

Autocorrelation Function for Seasonal 1.0 0.8 0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6 -0.8 -1.0

5

Lag

Corr

T

LBQ

1 -0.49 -4.44

20.42

2 -0.03 -0.19 3 0.04 0.30

15

Lag

Corr

25

T

LBQ

Lag

T

LBQ

Lag

T

LBQ

8 0.07 0.48

23.42

15 -0.03 -0.15

61.85

22 0.02 0.12

70.43

20.47

9 0.01 0.08

23.43

16 -0.11 -0.66

63.13

23 -0.02 -0.12

70.48

20.61

10 -0.07 -0.50

23.89

17 0.21 1.22

67.63

24 0.02 0.10

70.51

4 0.03 0.23

20.70

11 0.27 2.00

31.19

18 -0.13 -0.78

69.58

25 0.03 0.20

70.65

5 -0.10 -0.76

21.64

12 -0.50 -3.48

55.77

19 0.07 0.40

70.11

6 0.09 0.67

22.38

13 0.24 1.45

61.38

20 -0.05 -0.28

70.36

7 -0.08 -0.62

23.02

14 0.06 0.37

61.78

21 -0.01 -0.07

70.38

Corr

Corr

Partial Autocorrelation Function for Seasonal 1.0 0.8 0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6 -0.8 -1.0

5

Lag

15

25

PAC

T

Lag PAC

T

Lag PAC

T

1 -0.49

-4.44

8 -0.07

-0.66

15 -0.04

2 -0.34

-3.14

9 -0.01

-0.07

16 -0.09

3 -0.21

-1.95

10 -0.09

-0.79

17

4 -0.09

-0.81

11

0.34

3.10

5 -0.17

-1.55

12 -0.32

-2.92

6 -0.07

-0.64

13 -0.20

7 -0.15

-1.36

14 -0.12

PAC

T

-0.37

22 -0.04

-0.38

-0.81

23

0.15

1.36

0.01

0.09

24 -0.20

-1.79

18

0.04

0.36

25 -0.05

-0.47

19

0.01

0.09

-1.86

20

0.02

0.17

-1.06

21

0.02

0.19

183

Lag

Concentrating on the non-seasonal lags, the autocorrelation coefficients drop off after one time lag and the partial autocorrelation coefficients trail off, so a regular moving average term of order 1 is indicated. Concentrating on the seasonal lags (12 and 24), the autocorrelation coefficients cut off after lag 12 and the partial autocorrelation coefficients trail off, so a seasonal moving average term of order 12 is suggested. An ARIMA(0,1,1)(0,1,1) model for Yt is identified.

Final Estimates of Parameters Type Coef StDev MA 1 0.7486 0.0742 SMA 12 0.8800 0.0893

T 10.09 9.85

Differencing: 1 regular, 1 seasonal of order 12 Number of observations: Original series 96, after differencing 83 Residuals: SS = 5744406210 (backforecasts excluded) MS = 70918595 DF = 81 Modified Box-Pierce (Ljung-Box) Chi-Square statistic Lag 12 24 36 48 Chi-Square 3.0(DF=10) 19.3(DF=22) 23.0(DF=34) 25.1(DF=46) Period 97 98 99 100 101 102 103 104 105 106 107 108

Forecast 163500 158300 177084 178792 188706 184846 191921 188746 185194 187669 188084 221521

95 Percent Limits Lower Upper 146991 180009 141277 175322 159562 194606 160785 196798 170227 207185 165907 203785 172532 211310 168918 208574 164936 205451 166991 208348 166993 209175 200025 243016

The critical 5% chi-square value for 10 df's is 18.31. Since the calculated chi-square Q for the residual autocorrelations equals 3, the model is deemed adequate.

12.

a. See part b. b. The autocorrelation coefficient plot below indicates that the data are non-stationary. Therefore, the data should be first differenced. The autocorrelation coefficient and partial autocorrelation coefficient plots for the first differenced data are also shown.

184

A

Autocorrelation Function for IBM 1.0 0.8 0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6 -0.8 -1.0

2

7

12

Lag

Corr

T

LBQ

Lag

Corr

T

LBQ

1

0.87

6.30

42.05

8

0.22

0.66

143.50

2

0.76

3.44

74.38

9

0.16

0.46

145.12

3

0.66

2.48

99.23

10

0.13

0.37

146.21

4

0.54

1.84

116.50

11

0.12

0.35

147.18

5

0.44

1.40

128.06

12

0.08

0.23

147.60

6

0.34

1.06

135.27

13

0.07

0.19

147.91

7

0.28

0.85

140.30

Autocorrelation Function for Diff.

1.0 0.8 0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6 -0.8 -1.0

2

7

12

Lag

Corr

T

LBQ

Lag

Corr

T

1

0.30

2.12

4.77

8

-0.08

-0.52

6.86

2

0.08

0.56

5.17

9

-0.31

-2.01

13.24

3

0.06

0.37

5.34

10

-0.22

-1.31

16.44

4

0.03

0.20

5.40

11

-0.01

-0.05

16.45

5

-0.02

-0.14

5.42

12

-0.06

-0.37

16.74

6

-0.11

-0.74

6.19

7

-0.06

-0.41

6.44

185

LBQ

ion

Partial Autocorrelation Function for Diff.

1.0 0.8 0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6 -0.8 -1.0

2

7

Lag

PAC

1

T

Lag

12

PAC

T

0.30

2.12

8 -0.06

-0.43

2 -0.00

-0.03

9 -0.30

-2.11

3

0.04

0.25

10 -0.05

-0.39

4

0.01

0.04

11

0.10

0.72

5 -0.04

-0.27

12 -0.09

-0.67

6 -0.11

-0.77

7 -0.00

-0.00

c. There is not much going on in either the autocorrelations or partial autocorrelations for the differenced series. Could make a case for a first order AR term in a model for the differenced data. The results from fitting an ARIMA(0,1,1) model are shown below. Successive changes are not random if an ARIMA(0,1,1) model is appropriate.

ARIMA model for IBM Final Estimates of Parameters Type Coef StDev T AR 1 0.3780 0.1496 2.53

P .015

Differencing: 1 regular difference Number of observations: Original series 52, after differencing 51 Residuals: SS = 1710.50 (backforecasts excluded) MS = 34.21 DF = 50 Modified Box-Pierce (Ljung-Box) Chi-Square statistic Lag 12 24 36 48 Chi-Square 7.3(DF=11) 15.8(DF=23) 28.5(DF=35) 38.1(DF=47)

Period 53 54

Forecast 311.560 314.418

95 Percent Limits Lower Upper 300.094 323.026 294.895 333.941 186

d. The residual plots look good and there are no significant residual autocorrelations.

There is no reason to doubt the adequacy of the model. e. Ŷ t = Yt-1 + .378(Yt - Yt-1) Ŷ 53 = Y52 + .378(Y53 - Y52) Ŷ 53 = 304 + .378(304 - 284) = 311.56 The naïve forecast is Yˆ = Y = 304 .

53

13.

52

One question that might arise is should the student use the first 145 observations or all 150 observations. With this many observations, it will not make much difference.

The autocorrelation function using all the data below is slow to die out and suggests the DEF time series is non-stationary. Therefore, the differenced data should be investigated.

The autocorrelation coefficient and partial autocorrelation coefficient plots for the first differenced data follow.

187

It appears that the autocorrelations for the differenced data cut off after lag one and that the partial autocorrelations die out. This suggests a regular MA term in a model for the differenced data so an ARIMA(0,1,1) model is identified. If 145 observations are used, the forecasting equation from the fitted model is Ŷ t = Yt-1 - 0.7179t-1

The computer output for the fitted model is given below.

188

Final Estimates of Parameters Type Coef SE Coef T P MA 1 0.7179 0.0582 12.34 0.000 Constant -0.00049 0.06024 -0.01 0.994 Differencing: 1 regular difference Number of observations: Original series 145, after differencing 144 Residuals: SS = 917.134 (backforecasts excluded) MS = 6.459 DF = 142 Modified Box-Pierce (Ljung-Box) Chi-Square statistic Lag 12 24 36 48 Chi-Square 12.3 29.5 57.2 66.1 DF 10 22 34 46 P-Value 0.266 0.131 0.008 0.028 Forecasts from period 145 95% Limits Period Forecast Lower Upper Actual 146 133.815 128.832 138.797 135.2 147 133.814 128.637 138.991 139.2 148 133.814 128.450 139.178 136.8 149 133.813 128.268 139.358 136.0 150 133.813 128.092 139.533 134.4 This model fits well. The usual residual analysis indicates no model inadequacies.

Comparing the forecasts with the actuals for the five days from forecast origin t = 145 using MAPE gives MAPE = 1.82%.

14.

The time series plot follows.

189

The sample autocorrelation and partial autocorrelation functions below suggest and AR(2) or, equivalently, an ARIMA(2,0,0) model. The computer output follows along with the residual autocorrelation function.

Final Estimates of Parameters Type Coef SE Coef T P AR 1 1.4837 0.0732 20.26 0.000 AR 2 -0.7619 0.0729 -10.45 0.000 Constant 17.181 1.381 12.44 0.000 Mean 61.757 4.965 Number of observations: 90 Residuals: SS = 14914.5 (backforecasts excluded) MS = 171.4 DF = 87 Modified Box-Pierce (Ljung-Box) Chi-Square statistic Lag 12 24 36 48 Chi-Square 19.9 25.9 41.7 55.9 DF 9 21 33 45 P-Value 0.018 0.209 0.142 0.128 Forecasts from period 90 Period Forecast 91 110.333

95% Limits Lower Upper Actual 84.665 136.001

190

The forecast of 110 accidents for the 91st week seems reasonable given the history of the series near that point.

There is no evidence of annual seasonality in these data but since there is less than two years of weekly observations, seasonality, if it exists, would be virtually impossible to detect.

15.

The time series plot that follows suggests the Price series is non-stationary. This is corroborated by the autocorrelations which are slow to die out. The differenced series should be investigated.

191

The autocorrelation function for the differenced data below suggests the differenced series is random. The partial autocorrelation function for the differenced data has a similar appearance.

An ARIMA(0,1,0) model is identified for the price of corn. For this model a forecast of the next observation at forecast origin t is given by Yˆt +1 = Yt . Forecasts two steps ahead are the same, similarly for three steps ahead and so forth. In other words, this model produces “flat line” forecasts whose intercept is given by Yt .

So, forecasts of the price of corn for the next 12 months are all given by the last observation or 251 cents per bushel.

16.

The variation in the Cavanaugh sales series increases with the level, so a log 192

transformation seems appropriate. Let Y t be the natural log of sales and Wt = Yt − Yt −12 be the seasonally differenced series. Two ARIMA models that represent the data reasonably well are given by the expressions ARIMA(0,0,2)(0,1,0)12 and ARIMA(1,0,0)(0,1,1)12. Both models contain a constant term. Another possibility is the ARIMA(0,1,1)(0,1,1)12 model (without a constant), but the latter doesn’t fit quite as well as the former models. The results for the ARIMA(1,0,0)(0,1,1)12 process are displayed below.

Fitted model: Wt = .54Wt −1 + .119 +  t − .81 t −12 Final Estimates of Parameters Type Coef SE Coef T P AR 1 0.5400 0.1080 5.00 0.000 SMA 12 0.8076 0.1162 6.95 0.000 Constant 0.1187 0.0060 19.70 0.000 Differencing: 0 regular, 1 seasonal of order 12 Number of observations: Original series 77, after differencing 65 Forecasts: Date Jun. 2000 Jul. 2000 Aug. 2000 Sep. 2000 Oct. 2000 Nov. 2000 Dec. 2000

ForecastLnSales 5.76675 6.11484 6.40039 6.80928 7.09153 7.14969 6.85211

193

ForecastSales 320 453 602 906 1202 1274 946

The residual autocorrelation a lag 2 can be ignored or, alternatively, can fit the ARIMA(0,0,2)(0,1,1)12 model.

17.

The variation in Disney sales increases with the level, so a log transformation seems appropriate. Let Y t be the natural log of sales and Wt = Yt − Yt − 4 be the seasonally differenced series. Two ARIMA models that represent the data reasonably well are given by the representations ARIMA(1,0,0)(0,1,1)4 and ARIMA(0,1,1)(0,1,1)4. The former model contains a constant. The results for the ARIMA(1,0,0)(0,1,1)4 process are displayed below.

Fitted model: Wt = .50Wt −1 + .089 +  t − .49 t − 4 Final Estimates of Parameters Type Coef SE Coef T P AR 1 0.4991 0.1164 4.29 0.000 SMA 4 0.4863 0.1196 4.07 0.000 Constant 0.0886 0.0063 14.07 0.000 Differencing: 0 regular, 1 seasonal of order 4 Number of observations: Original series 63, after differencing 59 Forecasts: Date ForecastLnSales Q4 1995 8.25008 Q1 1996 8.12423 Q2 1996 8.11642 Q3 1996 8.24372 Q4 1996 8.43698

194

ForecastSales 3828 3375 3349 3804 4615

18.

The data were transformed by taking natural logs; however, an ARIMA model may be fit to the original observations. Let Y t be the natural log of demand and let Wt =  12Yt = Yt − Yt −1 − Yt −12 + Yt −13 be the series after taking one seasonal difference followed by a regular difference. An ARIMA(0,1,1)(0,1,1)12 model represents the log demand series well. The results follow.

Fitted model: Wt =  t − .63 t −1 − .57  t −12 + (. 63)(. 57 ) t −13 Final Estimates of Parameters Type Coef SE Coef T P MA 1 0.6309 0.0724 8.71 0.000 SMA 12 0.5735 0.0849 6.75 0.000 Differencing: 1 regular, 1 seasonal of order 12 Number of observations: Original series 129, after differencing 116 Forecasts: Date ForecastLnDemand Oct. 1996 5.23761 Nov. 1996 5.29666 Dec. 1996 5.33704

195

ForecastDemand 188 200 208

19.

Let Wt =  12Yt = Yt − Yt −1 − Yt −12 + Yt −13 be the series after taking one seasonal difference followed by a regular difference. Examination of the autocorrelation function for Wt leads to the identification of an ARIMA(0,1,0)(0,1,1)12 model.

The results follow.

Fitted model: Wt =  t − .84 t −12 Final Estimates of Parameters Type Coef SE Coef T P SMA 12 0.8438 0.0733 11.51 0.000 Differencing: 1 regular, 1 seasonal of order 12 Number of observations: Original series 130, after differencing 117 Residuals:

SS = 7165296 (backforecasts excluded) MS = 61770 DF = 116

Modified Box-Pierce (Ljung-Box) Chi-Square statistic 196

Lag 12 24 36 Chi-Square 13.2 19.3 26.2 DF 11 23 35 P-Value 0.280 0.681 0.858

48 52.6 47 0.266

Forecasts from period 130 Period Forecast 131 73653.4 132 73448.7 133 72571.8 134 72904.3 135 73200.8 136 73711.5 137 74218.7 138 75021.6 139 75459.7 140 75114.5 141 74519.0 142 74681.4

20.

95% Limits Lower Upper 73166.1 74140.6 72759.7 74137.8 71727.9 73415.7 71929.8 73878.7 72111.4 74290.3 72518.1 74905.0 72929.6 75507.7 73643.5 76399.7 73998.0 76921.4 73573.8 76655.3 72903.0 76134.9 72993.6 76369.2

The variation in Wal-Mart sales increases with the level, so a log transformation seems appropriate. Let Y t be the natural log of sales and Wt = Yt − Yt − 4 be the seasonally differenced series. Examination of the autocorrelation function for Wt leads to the identification of an ARIMA(0,1,0)(0,1,1)4 model.

The results follow.

Fitted model: Wt =  t − .52  t − 4 197

Final Estimates of Parameters Type Coef SE Coef T P SMA 4 0.5249 0.1185 4.43 0.000 Differencing: 1 regular, 1 seasonal of order 4 Number of observations: Original series 60, after differencing 55 Residuals:

SS = 0.0323112 (backforecasts excluded) MS = 0.0005984 DF = 54

Modified Box-Pierce (Ljung-Box) Chi-Square statistic Lag 12 24 36 48 Chi-Square 12.5 20.3 30.3 47.7 DF 11 23 35 47 P-Value 0.327 0.626 0.697 0.445 Forecasts from period 60 Forecasts Period LnSales Sales Q1/05 11.1671 70,764 Q2/05 11.2514 76,988 Q3/05 11.2408 76,176 Q4/05 11.4233 91,427 Q1/06 11.2660 78,120 Q2/06 11.3503 84,991 Q3/06 11.3397 84,095 Q4/06 11.5223 100,942

21.

Autocorrelations and partial autocorrelations for number of severe earthquakes suggest an AR(1) model.

198

Summary of model fit and forecasts for the next 5 years follow.

Final Estimates of Parameters Type Coef SE Coef T P AR 1 0.5486 0.0845 6.49 0.000 Constant 9.0295 0.6082 14.85 0.000 Mean 20.001 1.347 Number of observations: 100 Residuals:

SS = 3624.87 (backforecasts excluded) MS = 36.99 DF = 98

Modified Box-Pierce (Ljung-Box) Chi-Square statistic Lag

12

24

36

48 199

Chi-Square 11.0 23.2 30.7 45.9 DF 10 22 34 46 P-Value 0.358 0.388 0.631 0.477 Forecasts from period 100 Period 101 102 103 104 105 106

22.

Forecast 21.6463 20.9037 20.4964 20.2729 20.1504 20.0831

95% Limits Lower Upper 9.7236 33.5691 7.3049 34.5026 6.4323 34.5605 6.0718 34.4741 5.9082 34.3925 5.8287 34.3376

Since the variation in the series increases with the level, a log transformation is indicated.

An examination of the autocorrelations and partial autocorrelations for LnGapSales leads to the identification of an ARIMA(0,1,0)(0,1,1)4 model. Summary of model fit and forecasts for the next 8 quarters follow.

Final Estimates of Parameters Type Coef SE Coef T P SMA 4 0.2780 0.1004 2.77 0.007 Differencing: 1 regular, 1 seasonal of order 4 Number of observations: Original series 100, after differencing 95 Residuals:

SS = 0.311695 (backforecasts excluded) MS = 0.003316 DF = 94

Modified Box-Pierce (Ljung-Box) Chi-Square statistic 200

Lag 12 24 36 48 Chi-Square 14.8 19.7 24.2 27.3 DF 11 23 35 47 P-Value 0.194 0.659 0.916 0.990 Forecasts from period 100 Period LnGapSales GapSales 101 8.19679 3,629 102 8.23373 3,766 103 8.30268 4,035 104 8.51475 4,988 105 8.21496 3,696 106 8.25189 3,835 107 8.32085 4,109 108 8.53291 5,079

23.

The long strings of 0’s (no Influenza A positive cases) of uneven lengths might create identification and fitting problems for ARIMA modeling. On the other hand, a simple AR(1) model with an AR coefficient of about .8 and no constant term might provide reasonable one week ahead forecasts for the number of positive cases. These forecasts can be generated with the understanding that any non-integer forecast less than 1 is set to 0 and any non-integer forecast greater than 1 is rounded to the closest integer.

CASE 9-1: RESTAURANT SALES 1. & 2. & 3.

AR(1) model is appropriate. See summary, forecasts and actuals below.

201

Final Estimates of Parameters Type Coef AR 1 0.5997 Constant 1921.7 Mean 4800.8

SE Coef T P 0.0817 7.34 0.000 100.2 19.18 0.000 250.3

Number of observations: 104 Residuals:

SS = 105964742 (backforecasts excluded) MS = 1038870 DF = 102

Modified Box-Pierce (Ljung-Box) Chi-Square statistic Lag 12 24 36 48 Chi-Square 8.9 24.5 36.8 48.5 DF 10 22 34 46 P-Value 0.545 0.322 0.342 0.372 Forecasts and actuals for first four weeks in January 1983: Period Forecast 105 3249.49 106 3870.48 107 4242.89 108 4466.23

95% Limits Lower Upper Actual 1251.36 5247.62 2431 1540.58 6200.38 2796 1804.68 6681.10 4432 1990.23 6942.23 5714

Forecasts are too high for first two weeks of January 1983 and too low for next Two weeks. Note however, that actual sales fall within the 95% prediction interal limits for each of the four weeks.

4.

5.

The best model in Chapter 8 for the original Restaurant Sales data is an autoregressive model with an added dummy variable to represent the period during the year when Marquette University is in session. So, because of the additional dummy variable, this model fits the data better than the AR(1) model in part 1. If the dummy variable were not present, the two models would be the same. Consequently, we would expect better forecasts with the AR + dummy variable model than with the simple AR model.

Regardless, however, if forecasts are compared to actuals from forecast origin 104 (last week in 1982), the usual measures of forecast accuracy (RMSE, MAPE, etc.) are likely to be relatively large since a large portion of the variation in sales is not accounted for by the AR + dummy variable model.

At the very least the parameters in the AR(1) model should be re-estimated if the new data are combined with the old data. A better approach is to combine the data and the go through the usual ARIMA model building process again. It may be the combined data suggest the form of the ARIMA has changed. In this case, an AR(1) is still appropriate when the new data are combined with the old data.

202

CASE 9-2: MR. TUX 1.

Box-Jenkins ARIMA models account for the autocorrelation in the observed series using possibly differenced data, lagged dependent variables and current and previous errors.

There are no potential causal (exogenous) independent variables in these models so they are often difficult to explain to management. Best to demonstrate the results.

2.

Autocorrelation and partial autocorrelation plots for the regular and seasonally differenced data suggest a non-seasonal AR(2) term (the partial autocorrelations cut off after lag 2 and the autocorrelations die out). No seasonal MA or AR terms should be included. However, here is a case where, say, the ARIMA(2, 1, 0)(0, 1, 0)12 model is more complex than necessary and a much simpler model works well. A time series plot of the seasonally differenced Mr. Tux data is shown below along with the sample autocorrelation function for these differences.

203

It is evident an ARIMA(0, 0, 0)(0, 1, 0)12 model of the form Yt − Yt −12 =  0 +  t might provide a good fit to the Mr. Tux data.

3.

To fit the model Wt = Yt − Yt −12 =  0 +  t to the Mr. Tux data, simply set ̂ 0 = W , the mean of the seasonal differences. Here ˆ 0 = 32 ,174 . Since the residuals from this model differ from the seasonal differences by the constant ˆ 0 = 32 ,174 , the residual autocorrelation function will be identical to the autocorrelation function for the seasonal differences shown in part 2. The forecasting equation is simply Yˆt = 32,174 + Yt −12 .

Setting t = 97 through t = 108, we have the forecasts for the 12 months of 2006: Yˆ97 = 32,174 + 71,043 = 103,217 Yˆ = 185,104 98

Yˆ99 = 282,733 Yˆ = 441,741 100

Yˆ101 = 426,921  Y102 = 305,048 Yˆ = 262,477 103

Yˆ104 = 407,576 Yˆ = 227,583 105

Yˆ106 = 205,692 Yˆ = 213,876 107

Yˆ108 = 290,887

The sales forecasts for 2006 are obtained by adding 32,174 to the sales for each of the 12 months of 2005.

CASE 9-3: CONSUMER CREDIT COUNSELING 2.

The autocorrelation function plot below indicates that the data are non-stationary.

The autocorrelations are slow to die out. In addition, there is a spike at lag 12 and a smaller spike at lag 24 indicating some seasonality.

204

The autocorrelation functions for the differenced series (DiffClients), the seasonally differenced series (Diff12Clients) and the series with one regular and one seasonal difference (DiffDiff12Clients) follow.

205

Relative to the autocorrelations for DiffClients and Diff12Clients, the autocorrelations for DiffDiff12Clients are much more pronounced, indicating one regular difference and one seasonal difference is too much. The autocorrelations for Diff12Clients are the cleanest with a significant spike at lag 12 and a slightly smaller spike at lag 24. This autocorrelation pattern suggests an ARIMA(0,0,0)(0,1,1)12 or an ARIMA(0,0,0)(1,1,0) model. The former model is the better choice. Summary results and forecasts follow.

Final Estimates of Parameters Type Coef SMA 12 0.4614

SE Coef T P 0.1055 4.37 0.000

Differencing: 0 regular, 1 seasonal of order 12 Number of observations: Original series 99, after differencing 87 Residuals:

SS = 61091.9 (backforecasts excluded) MS = 710.4 DF = 86 Modified Box-Pierce (Ljung-Box) Chi-Square statistic 206

Lag 12 24 36 48 Chi-Square 10.9 20.3 30.8 37.4 DF 11 23 35 47 P-Value 0.452 0.623 0.669 0.842 Forecasts from period March 1993 Period Forecast Apr 1993 123.181 May 1993 122.960 Jun 1993 140.803 Jul 1993 150.944 Aug 1993 140.056 Sep 1993 134.285 Oct 1993 146.517 Nov 1993 146.953 Dec 1993 126.243

95% Limits Lower Upper 70.931 175.431 70.710 175.210 88.553 193.053 98.694 203.194 87.806 192.306 82.035 186.535 94.267 198.767 94.703 199.203 73.993 178.493

CASE 9-4: THE LYDIA E. PINKHAM MEDICINE COMPANY 1.

The forecast for 1961 using the AR(2) model is 1290. The revised error measures are: MAD = 114

2.

MAPE = 7.1%

The results from fitting an ARIMA(1,1,0) model, one step ahead forecasts and actuals follow.

Final Estimates of Parameters Type Coef SE Coef T P AR 1 0.4551 0.1408 3.23 0.002 Differencing: 1 regular difference Number of observations: Original series 42, after differencing 41 Residuals: SS = 2139107 (backforecasts excluded) MS = 53478 DF = 40 Modified Box-Pierce (Ljung-Box) Chi-Square statistic Lag 12 24 Chi-Square 5.5 22.1 DF 11 23 P-Value 0.905 0.514

36 25.3 35 0.885

48 * * *

One step ahead forecasts from 1948 207

Period 1949 1950 1951 1952 1953 1954 1955 1956 1957 1958 1959 1960

Actual 1984 1787 1689 1866 1896 1684 1633 1657 1569 1390 1397 1289

Forecast 1905 2018 1697 1644 1947 1910 1588 1610 1668 1529 1309 1400

Error 79 -231 -8 222 -51 -226 45 47 -99 -139 88 -111

The forecasting equation is Yˆt = 1.455Yt −1 − .455Yt − 2 .

Comparing the forecasting equation for the ARIMA(1,1,0) model with the forecasting equation for the AR(2) model given in the case, we see the two equations are very similar and, consequently, would expect the one step ahead forecasts and forecast errors to be similar. The error measures for ARIMA(1,1,0) forecasts are MAD = 112

MAPE = 6.9%

the same as those for the AR(2) model. The choice of one model over the other depends upon whether one believes the sales series in non-stationary or “nearly” non-stationary.

3.

This question is intended to stimulate thinking about technological advances in products (such as the automobile) which could affect sales versus fairly standard products (such as copper) whose demand may be impacted by technological advances which require them (such as wiring). There are no right or wrong answers here--just some that are better than others.

CASE 9-5: CITY OF COLLEGE STATION 1. & 2.

208

Fitted model: Yˆt = Yt −12 + 50.479 +  t − .792 t −12

209

Model fits well and forecasts seem very reasonable.

CASE 9-6: UPS AIR FINANCE DIVISION 1.

ARIMA(0,1,0)(0,1,1)12 model for Funding

Fitted model: Yˆt = Yt −1 + Yt −12 − Yt −13 +  t − .874 t −12 A constant term is not required with a regular and a seasonal difference.

2.

The model in part 1 is adequate. The Ljung-Box chi-square statistics show no significant autocorrelation. The residual autocorrelations and residual plots below confirm the model is adequate. (There is one large residual in period 49.)

210

3.

The forecasts follow.

211

CASE 9-7: AAA WASHINGTON The results from fitting an ARIMA(0,0,1)(0,1,1)12 model and forecasts follow.

Fitted model: Yˆt = Yt −1 + Yt −12 − Yt −13 +  t + .56 t −1 − .8515 t −12 The Ljung-Box chi-square statistics show no significant autocorrelation.

The residual autocorrelations are shown below. The residual plots look good.

The model is adequate.

212

CASE 9-8: WEB RETAILER 1.

Results from fitting an ARIMA(0,1,0)(0,0,1)12 model to the Contacts data follows.

Final Estimates of Parameters Type Coef SE Coef T P SMA 12 -0.6376 0.3022 -2.11 0.046 Differencing: 1 regular difference Number of observations: Original series 25, after differencing 24 Residuals: SS = 202359762350 (backforecasts excluded) MS = 8798250537 DF = 23 Modified Box-Pierce (Ljung-Box) Chi-Square statistic Lag 12 Chi-Square 10.2 DF 11 P-Value 0.513

24 * * *

36 48 * * * * * *

This model was suggested by an examination of the plots of the autocorrelation and partial autocorrelation functions for the original series and the first differenced series.

Another potential model is an ARIMA(1,0,0)(0,0,1)12 model. But if this model is fit to the data, the estimate of the autoregressive parameter turns out to be very nearly 1, confirming the choice of the initial ARIMA(0,1,0)(0,0,1)12.

2.

The model in part 1 is adequate. The is no residual autocorrelation and the residual plots that follow look good.

213

3.

Forecasts from period 25 Period Forecast 26 426280 27 492809 28 527275 29 535656 30 545614 31 692161 32 554640 33 494570 34 484265 35 471355 36 462995

95% Limits Lower Upper 242397 610163 232759 752859 208780 845770 167890 903422 134439 956789 241741 1142580 68131 1041149 -25530 1014669 -67384 1035914 -110135 1052844 -146876 1072867 214

37

491232

-145757 1128222

The pattern of the forecasts is reasonable but the forecast of the seasonal peak in December (recall this series starts in June) is very likely to be much too low. The actual December peak may be captured by the 95% prediction limits but, because of the small sample size, these limits are wide. The lower prediction limit is even negative for some lead times.

4.

The sample size in this case is small. With only two years of monthly data, it is difficult to estimate the seasonality precisely. Although an ARIMA model does provide some insights into the nature of this series, another modeling approach may produce more readily acceptable forecasts.

CASE 9-9: SURTIDO COOKIES 1.

Results from fitting an ARIMA(0,0,0,)(0,1,1)12 model to Surtido cookie sales follow.

Final Estimates of Parameters Type Coef SMA 12 0.7150

SE Coef T P 0.1910 3.74 0.001

Differencing: 0 regular, 1 seasonal of order 12 Number of observations: Original series 41, after differencing 29 Residuals: SS = 650837704391 (backforecasts excluded) MS = 23244203728 DF = 28 Modified Box-Pierce (Ljung-Box) Chi-Square statistic 215

Lag 12 24 Chi-Square 7.4 14.2 DF 11 23 P-Value 0.770 0.921

36 * * *

48 * * *

Cookie sales have a strong and quite consistent seasonal component but with little or no growth. Following the usual pattern of looking at autocorrelations and partial autocorrelations for the original series and its various differences, the best patterns for model identification appear to be those for the original series and the seasonally differenced series. In either case, a seasonal moving average term of order 12 is included in the model to accommodate seasonality and can be deleted if non-significant. Fitting an ARIMA(1,0,0)(0,0,1)12 model gives an estimated autoregressive coefficient of about .9, suggesting perhaps a model with a regular difference, residual autocorrelations and unattractive forecasts. This line of inquiry is not useful. The ARIMA model above involving the seasonally differenced data fits well and, as we shall see, produces reasonable forecasts.

2.

As demonstrated by the residual autocorrelation function and the residual plots below, the ARIMA(0,0,0,)(0,1,1)12 model is adequate.

216

3.

The forecasts for the next 12 months follow. Judging from the time series plot, they seem very reasonable.

Forecasts from period 41

Period 42 43 44 45 46 47 48 49 50 51 52 53

95% Limits Forecast Lower Upper 627865 328983 926748 721336 422453 1020219 658579 359696 957461 1533503 1234620 1832386 1628889 1330007 1927772 2070440 1771557 2369323 1805503 1506620 2104385 778148 479265 1077031 534265 235382 833148 525169 226286 824052 697168 398285 996051 624876 325994 923759

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CASE 9-10: SOUTHWEST MEDICAL CENTER 1.

Various plots follow. Given these plots, Mary’s initial model seems reasonable.

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219

2.

Results from fitting an ARIMA(0,1,1)(0,1,1)12 model follow along with a residual analysis and forecasts for the next 12 months.

Final Estimates of Parameters Type Coef SE Coef T P MA 1 0.3568 0.0931 3.83 0.000 SMA 12 0.8646 0.0800 10.80 0.000 Differencing: 1 regular, 1 seasonal of order 12 Number of observations: Original series 114, after differencing 101 Residuals: SS = 988551 (backforecasts excluded) MS = 9985 DF = 99 Modified Box-Pierce (Ljung-Box) Chi-Square statistic 220

Lag 12 24 Chi-Square 21.2 53.0 DF 10 22 P-Value 0.020 0.000

36 72.9 34 0.000

48 88.1 46 0.000

Forecasts from period 114 Period Forecast 115 1419.59 116 1438.07 117 1386.09 118 1376.53 119 1459.48 120 1431.27 121 1365.43 122 1456.48 123 1324.46 124 1303.44 125 1442.69 126 1350.23

95% Limits Lower Upper 1223.70 1615.49 1205.16 1670.99 1121.28 1650.90 1083.27 1669.79 1140.30 1778.66 1088.12 1774.41 999.88 1730.98 1069.83 1843.14 917.79 1731.12 877.70 1729.17 998.71 1886.68 888.71 1811.74

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Collectively, the residual autocorrelations are larger than they would be for random errors; however, they suggest no obvious additional terms to add to the ARIMA model.

Apart from the large residual at month 68, the residual plots look good. The forecasts seem reasonable but the 95% prediction limits are fairly wide.

3.

Total visits for fiscal years 4, 5 and 6 seem somewhat removed from the rest of the data.

Total visit for these fiscal years are, as a group, somewhat larger than the remaining observations. Did something unusual happen during these years? Was total visits defined differently? This particular feature makes modeling difficult.

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CHAPTER 10 JUDGMENTAL FORECASTING AND FORECAST ADJUSTMENTS ANSWERS TO PROBLEMS AND CASES 1.

The Delphi method can be used in any forecasting situation where there is little or no historical data and there is expert opinion (experience) available. Two examples might be: • First year sales for a new product • Full capacity employment at a new plant Potential difficulties associated with the Delphi method include: • Assembling the “right” group of experts.

• Overcoming individual biases or agendas • Not being able to arrange timely feedback

2.

a. Month 1 2 3 4 5 6 7 8 9 10

Averaged Forecast 4924.5 5976.0 6769.0 4708.0 4964.0 6102.0 8212.5 6178.5 4806.5 4228.5

b. Month 1 2 3 4 5 6 7 8 9 10

WtAvg Forecast 4721.4 5956.8 6731.2 4601.2 4991.6 6385.2 8362.2 6320.4 4596.8 4474.2

c. Winters’ forecasts MAPE = 9.8%, Regression forecasts MAPE = 7.7% d. Averaged forecasts MAPE = 7.4%, Weighted averaged forecasts MAPE = 8.8%.

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So based on MAPE, forecasts created by taking simple average of Winters’ forecasts and Regression forecasts are preferred.

CASE 10-1: GOLDEN GARDENS RESTAURANT 1. & 2. Sue and Bill have tackled a very tough business project: designing a restaurant that will succeed. Restaurants seem to come and go on a regular basis so their planning efforts prior to opening are important.

They have already tried focus groups and have some ideas to add to their own.

Since they have a number of "expert" friends, some way must be found to use this expertise. The Delphi method suggests itself as a way to utilize their friends' knowledge. A written description of the project along with the question of proper motif could be supplied to each of their friends, along with a request to design the restaurant. These descriptions would then be mailed back to each participant with a request to re-design the business based on all the written replies. This process could be continued until changes are no longer generated.

An optional step would then be to bring the participants together for a discussion.

This expert focus group could argue their cases and respond to Sue and Bill's objections or insights. At the end of this process Sue and Bill would probably have a better idea of how a successful restaurant would look and could begin their project with more confidence. Also, financial backers would probably be more enthusiastic after reviewing the extensive planning that Sue and Bill have undertaken prior to opening their business.

CASE 10-2: ALOMEGA FOOD STORES 1.

The naïve forecasting model is not very accurate. The MSE equals 8,648,047,253.

2.

The MSE for the multiple regression model (from the regression output) equals 2,097,765,646 which is quite a bit less than the naïve model.

3.

If the naïve approach had been more accurate, combining methods would have been worth a try.

4.

If Julie did combine forecasts, she should use a weighted average that definitely favored the multiple regression model.

CASE 10-3: THE LYDIA E. PINKHAM MEDICINE COMPANY REVISITED 224

1.

These articles are more abundant than many realize. More "popular” journals, particularly financial markets titles such as Technical Analysis of Stocks & Commodities, Financial Analysts Journal, and Futures present several articles.

In addition, the proceedings from the neural network conferences (published by IEEE) will usually have some business applications. Finally, this approach is beginning to appear in more scholarly journals such as Management Science and Decision Sciences.

2.

The interested student with access to a neural network simulator should enjoy this assignment. In addition to the "backpropagation" approach, students might try radial basis functions and least mean squares if they are available.

3.

Model specification is as much an art as it is a science. for example, look at Case 9-4 where the choice between ARIMA(1,1,0) and AR(2) models is not clearcut. Neural networks, however, do not require the analyst to specify the form of the model -- they have been called "model free" function approximators (see Bart Kosko, Neural Networks and Fuzzy Systems: A Dynamical Systems Approach to Machine Intelligence, Prentice-Hall, 1992, for example).

CHAPTER 11 MANAGING THE FORECASTING PROCESS 225

ANSWERS TO PROBLEMS AND CASES

1.

a. One response: Forecasts may not be right, but they improve the odds of being close to right. More importantly, if there are no agreed upon set of forecasts to drive planning, then different groups may develop own procedures to guide planning with potential chaos as the result. b. One response: Analogy—If you think education is expensive, try ignorance.

Having a good set of forecasts is like walking while looking ahead instead of at your shoes. Planning without forecasts will lead to inefficient operations, sub optimal returns on investment, poor customer service, and so forth. c. One response: Good forecasts require not only good quantitative skills, they also require an in-depth understanding of the business or, more generally, the forecasting environment and, ultimately, good communication skills to sell forecasts to management.

CASE 11-1: BOUNDARY ELECTRONICS 1.

This case invites students to think about how to use some of the forecasting techniques discussed in Chapter 11. Guy Preston is trying to get his managers to think about the long-range position of the company, as opposed to the short range thinking that most managers are involved in on a daily basis. The case might generate a class discussion about the tendency of managers to shorten their planning horizons too much in the daily press of business.

Guy has asked his managers to write scenarios for the future: a worst case, a status quo, and a most likely scenario. His next task might be to discuss each of these three possibilities, and to discuss any differences of opinion that might emerge. A second round of written scenarios by each participant could then follow this.

2.

The instructor should point out that the purpose of Guy's retreat is to expand the planning horizon of his managers. He should be prepared to continue this effort after the first round of written scenarios: it is quite possible that his team is still caught up in the affairs of the day and is not really engaged in long range thinking. He should encourage expanded thinking after the discussion phase and try during the day to continue such thinking.

3.

There are two possible benefits from Guy's retreat. First, he may gain valuable insights into the company's future to use in his own long range thinking. Second, and probably more important, his managers may come away with an increased awareness of the importance of expanding their planning horizons. If this is true, the company will probably be in a better position to face the future.

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CASE 11-2: BUSBY ASSOCIATES 1.

Since Holt’s linear smoothing incorporates simple exponential smoothing as a special case (β = 0), would expect Holt’s procedure to fit and forecast better here.

Therefore, there is no reason to consider a combination of forecasts. Combining forecasts is best considered when the sets of forecasts are produced by different procedures.

2.

Jill should definitely update her historical data as new data points arrive. Since she is using a computer program to do the forecasting, there would be very little effort involved in this process. Why not update and re-run every quarter for a while?

3.

After the results for a few additional quarters (say 4) become available, the analysis can be re-done to see if the current model is still viable. Model parameters can be re-estimated after each new observation if appropriate computer software is available.

4.

Box-Jenkins ARIMA methodology is not well suited for small sample sizes and can be difficult to explain to a non-statistician.

This case illustrates the practical problems that are typically encountered when attempting to forecast a time series in a business setting. Among the problems Jill encounters are: • She chooses to forecast a national variable for which data values are available in the Survey of Current Business. Will this variable correlate well with the actual Y value of interest (her firm's export sales)?

• Her initial sample size is only 13.

• When she attempts to gather more data, she finds that the series underwent a definition change during the recent past, resulting in inconsistent data. She must shift her focus to another surrogate variable.

• Her data plot indicates a bump in the data and she decides a more consistent series would result if she dropped the first few data points.

A real life-forecasting project could very likely involve difficulties such as those Jill encountered in this case, or perhaps even more. For this reason this case is a "good read" for forecasting students as they finish their studies since it shows that judgment and skill must be involved in the forecasting effort: forecasting problems are not usually as clean and straightforward as textbook problems.

CASE 11-3: CONSUMER CREDIT COUNSELING Students should summarize the results of the analyses of these data in the cases at the ends of chapters 4 (smoothing), 5 (decomposition), 6 (simple linear regression), 8 (regression with time series data) and 9 (Box-Jenkins methods). Fits, residual analyses, and forecasts can be compared. Regardless of the method, there is a fair amount of unexplained variation in the number of new clients. This may be a situation where combining forecasts makes sense.

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CASE 11-4: MR. TUX We collected the data from the Mr. Tux rental shop so that real data could be used at the end of each chapter instead of contrived data. We didn't know what would happen when we tried to forecast this variable, but we think it turned out well because no one method was superior.

The case in Chapter 11 summarizes the different ways John used to forecast his monthly sales, and asks students to comment on his efforts. We think a key point is that a lot of real data sets do not lend themselves to accurate forecasting, and that continually trying different methods is required. For the Mr. Tux data, there are fairly simple seasonal models (see the cases in Chapters 8 and 9) that represent the data well and provide reasonable forecasts.

What advice should we give to John Mosby for the future? Some suggestions to offer might include: 1. Update the data set as future monthly values become available and re-run the most promising analyses to see if the current forecasting model is still viable.

2. Consider combining forecasts from two different methods.

3. Try to develop a useful relationship between monthly sales and regional economic variables. Perhaps the area unemployment rate or an economic activity index would correlate well with John's sales. Perhaps some demographic variables would correlate well. If several variables were collected over the months of John's sales data, a good regression equation might result.

This would allow John to understand how is sales are tied to the local environment.

CASE 11-5: ALOMEGA FOOD STORES 1.

Julie has to choose between two different methods of forecasting her company’s monthly sales. Students should review the results of these two efforts and decide which offers the better choice. We find that class presentations by student teams are valuable as they move the analysis beyond the computer results to simulate implementing these results in a “real” situation.

2.

Having students, either individually or in teams, prepare a memo to Julie outlining their analysis and choice of forecasting method is an alternative to class presentations. Again, the results of this case do not point to a “right” answer, but rather to the necessity of choosing a forecasting method and justifying its use. Nonquantitative considerations should come into play: the fact that Julie is the first female president of Alomega, that she jumped over several qualified candidates for the job, and that one of her subordinates (Jackson Tilson) seems to be unimpressed with both her and any computer analysis.

3.

Other forecasting methods are certainly possible in this case. An assignment 228

beyond a consideration of choosing between decomposition and multiple regression would be to find a superior forecasting method using any available software. Again, the qualitative considerations should be considered, including the necessity of balancing the complexity and accuracy of a forecasting method with its acceptance and use by the management team.

4.

Only if she finds two good methods.

CASE 11-6: SOUTHWEST MEDICAL CENTER Students should summarize the results of Mary’s forecasting efforts describing the fits, residual analyses and forecasts. Moreover, they should point out the apparent difficulty in finding an adequate model for Mary’s total visits series. If Mary’s data is accurate—there is no reason for the apparent inconsistency in her time series—then it would probably be wise to collect another year or so of data and attempt to model the entire data set or, perhaps, just the data following fiscal year 6. In the interim, she may have to settle for the forecasts from the best ARIMA model developed in Case 9-10.

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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