CHAPTER 2 A REVIEW OF BASIC STATISTICAL CONCEPTS

ANSWERS TO ODD NUMBERED PROBLEMS 1. Descriptive Statistics Variable C1 Variable C1 a. b. c. d. e. N 28 Min 5.00 Mean Median Tr Mean 21.32 17.00 20.69 Max 54.00 Q1 Q3 11.25 28.75 StDev SE Mean 13.37 2.53

X = 21.32

S = 13.37 S2 = 178.76 If the policy is successful, smaller orders will be eliminated and the mean will increase. 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. The best forecast over the long-term is the mean of 21.32. Point estimate: X = 10.76% 1 = .95 Z = 1.96, n = 30,
X 1.96 S /

f. 3. a. b.

c.

(5.85%, 15.67%) df = 301 = 29, t = 2.045

X 2.045 S /

n =10.76 1.96 13.71 /

X =10.76,

30 =10.76 4.91

S =13.71

(5.64%, 15.88%) d.

n = 10.76 2.045 13.71 /

30 = 10.76 5.12

We see that the 95% confidence intervals in b and c are not much different. This explains why a sample of size n = 30 is often taken as the cutoff between large and small samples.

5.

H0: = 12.1 H1: > 12.1

n = 100 S = 1.7

= .05 X = 13.5 1

Reject H0 if Z > 1.645

13.5 12.1 Z = 1.7 = 8.235 100

Reject H0 since the computed Z (8.235) is greater than the critical Z (1.645). The mean has increased. 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 9. H0: = 700 H1: 700 n = 50 S = 50 = .05 X = 715

Reject H0 if Z < -1.96 or Z > 1.96

715 700 Z = 50 = 2.12 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

11.

a.

b. c.

Positive Y = 6058 X2 = 513 Y2 = 4,799,724 XY = 48,665 X = 59 r = .938

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 not reject the null hypothesis which so states this. The few students that erroneously 3

reject H0 demonstrate Type I error. 15. 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 H 1 : 44

two-sided test, = .02, critical value: |Z|= 2.33

X 44 S/ n = 45.2 44 10.3 / 175 = 1.54

Test statistic: Z =

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.

CHAPTER 3 EXPLORING DATA PATTERNS AND CHOOSING A FORECASTING TECHNIQUE

ANSWERS TO ODD NUMBERED PROBLEMS 4

1. 3.

Qualitative forecasting techniques rely on human judgment and intuition. Quantitative forecasting techniques rely more on manipulation of past historical data. 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 measures the variability of the time series after the other components have been removed. The autocorrelation coefficient measures the correlation between a variable, lagged one or more periods, and itself. a. b. c. d. nonstationary series stationary series nonstationary series stationary series

5. 7.

9.

Naive methods, simple averaging methods, moving averages, simple exponential smoothing, 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 cycle; and the number of sales resulting from a constant level of effort. Classical decomposition, census II, Winter's exponential smoothing, time series multiple regression, and Box-Jenkins methods. Examples are: electrical consumption, summer/winter activities (sports like skiing), clothing, and agricultural growing seasons, retail sales influenced by holidays, three-day weekends, and school calendars. 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 45 -77 -9 -28 28 -26 8 40 -140

11.

13.

Yes! The original series has a decreasing trend. 15. a. b. c. MPE MAPE MSE 5

17.

= 4465.7

(Yt-1 Y ) = 2396.1

(Yt Y )2 = 55,912,891 (Yt Y ) (Yt-2 Y ) = 44,081,598

(Yt Y ) (Yt-1 Y ) = 50,056,722 r1 =

50,056,722 55,912 ,891

= .895

H0: 1 = 0 H1: 1 0 Reject if t < -2.069 or t > 2.069

SE( r k ) =

1 + 2 ri 2
i =1

k 1

1 + 2 ( r1 )
i =1

11

n
t= r1 1 .895 0 = = 4.39 SE(rk) .204

24

1 = .204 24

Since the computed t (4.39) is greater than the critical t (2.069), reject the null. H0: 2 = 0H1: 2 0

Reject if t < -2.069 or t > 2.069 r2 =

44,081,598 = .788 55,912,891

SE( r k

)=

1 + 2 ri
i =1

k 1

1 + 2 ( .895)
i =1

2 1

n
t =

24

2.6 = .33 24

r1 1 .7880 = 2.39 = SE(r1) .33

Since the computed t (4.39) is greater than the critical t (2.069), reject the null. b. The data are nonstationary.

Autocorrelation Function for Loans

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

Lag

Corr

LBQ

1 2 3 4 5 6

0.90 0.79 0.67 0.56 0.43 0.31

4.39 2.39 1.68 1.25 0.92 0.63

21.74 39.37 52.85 62.57 68.73 72.04

19.

Figure 3.18 - The data are nonstationary. (Trended 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 some pattern that could be modeled.

21.

a. Time series plot follows

b. The sales time series appears to vary about a fixed level so it is stationary. c. The sample autocorrelation function for the sales series follows

The sample autocorrelations die out rapidly. This behavior is consistent with a stationary series. Note that the sales data are not random. Sales in adjacent weeks tend to be positively correlated.

Since, in this case, the residuals differ from the original observations by the constant

CHAPTER 4
8

MOVING AVERAGES AND SMOOTHING METHODS

ANSWERS TO ODD NUMBERED PROBLEMS 1. 3. 5. 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 RESI1 * * * -0.96000 0.08000 1.80667 0.74333 1.32333 -0.26000 1.27667 0.62667 1.40333 MSD: 1.17281 Exponential smoothing Moving average Winter's three-parameter linear and seasonal model

Accuracy Measures MAPE: 4.63186

MAD: 0.94222

The nave approach is better.

a.

221.2 is forecast for period 9

9.

3-month moving-average Period Yield AVER1 1 9.29 * 2 9.99 * FITS1 * * RESI1 * * 9

3 4 5 6 7 8 9 10 11 12

10.16 10.25 10.61 11.07 11.52 11.09 10.80 10.50 10.86 9.97

9.8133 10.1333 10.3400 10.6433 11.0667 11.2267 11.1367 10.7967 10.7200 10.4433

* 9.8133 10.1333 10.3400 10.6433 11.0667 11.2267 11.1367 10.7967 10.7200

* 0.436667 0.476667 0.730000 0.876667 0.023333 -0.426667 -0.636667 0.063333 -0.750000

Accuracy Measures MAPE: 4.58749

MAD: 0.49111

MSD: 0.31931

MPE: .6904

Row Period Forecast Lower Upper 1 13 10.4433 9.33579 11.5509

Moving Average
Actual

11.5
Predicted Forecast Actual Predicted Forecast

10.5

Yield
Moving Average Length: 3

9.5

MAPE: MAD: MSD:

4.58749 0.49111 0.31931

10

12

14

Time

b.

5-month moving-average Period Yield MA Predict 1 9.29 * * 2 9.99 * * 3 10.16 * * 4 10.25 * * 5 10.61 10.060 * Error * * * * * 10

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 Row Period Forecast Lower Upper 1 13 10.644 9.23041 12.0576 Accuracy Measures MAPE: 5.58295 MAD: 0.60400

MSD: 0.52015

MPE: .7100

Moving Average
Actual

12
Predicted Forecast Actual Predicted

11

Forecast

Yield
Moving Average

10

Length:

MAPE: MAD: MSD:

5.58295 0.60400 0.52015

10

12

14

Time

g. 11.

Use 3-month moving average forecast: 10.4433

Time Demand Smooth Predict 1 2 3 4 205 251 304 284 205.000 228.000 266.000 275.000 205.000 205.000 228.000 266.000 11

Error 0.0000 46.0000 76.0000 18.0000

5 6 7 8 9 10 11 12

352 300 241 284 312 289 385 256

313.500 306.750 273.875 278.938 295.469 292.234 338.617 297.309

275.000 313.500 306.750 273.875 278.938 295.469 292.234 338.617

77.0000 -13.5000 -65.7500 10.1250 33.0625 -6.4688 92.7656 -82.6172 MSD: 2943.24

Accuracy Measures MAPE: 14.67

MAD:

43.44

Row Period Forecast Lower Upper 1 13 297.309 190.879 403.738

Single Exponential Smoothing

Actual

400
Predicted Forecast Actual Predicted Forecast

300

Demand

Smoothing Constant Alpha: 0.500

MAPE:

14.67 43.44 2943.24

200

MAD: MSD:

10

12

14

Time

12

13.

a.

= .4 Accuracy Measures MAPE: 14.05 MAD: 24.02 Row Period Forecast Lower Upper 1 57 326.367 267.513 385.221 MSD: 1174.50

b.

= 0.6 Accuracy Measures MAPE: 14.68 MAD: 24.56 Row Period Forecast 1 57 334.07 Lower 273.889 MSD: 1080.21 Upper 394.251

c.

A smoothing constant of approximately .6 appears to be best.

d.

No! When the residual autocorrelations shown above are examined, some of them were found to be significant.

15.

The autocorrelation function shows that the data are seasonal with a slight trend. Therefore, the Winters model is used to forecast revenues. 13

Smoothing Constants - Alpha (level): 0.8 Beta (trend): 0.1 Gamma (seasonal): 0.1

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

14

CHAPTER 5
15

TIME SERIES AND THEIR COMPONENTS

ANSWERS TO ODD NUMBERED PROBLEMS 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. The basic forces that affect and help explain the trend-cycle of a series are population growth, price inflation, technological change, and productivity increases. Weather and the calendar year such as holidays affect the seasonal component. a.

3. 5. 7.

YEAR Y 1980 11424 1981 12811 1982 14566 1983 16542 1984 19670 1985 20770 1986 22585 1987 23904 1988 25686 1989 26891 1990 29073 1991 28189

T 11105.2 12900.6 14695.9 16491.3 18286.7 20082.1 21877.4 23672.8 25468.8 27263.6 29059.0 30854.3

C 102.871 99.306 99.116 100.307 107.565 103.426 103.234 100.977 100.855 98.633 100.048 91.362 16

1992 1993 1994 1995 1996 1997 b. c. d. 9. 11. 13.

Y Y

30450 31698 35435 37828 42484 44580

32649.7 93.263 34445.1 92.025 36240.5 97.777 38035.8 99.454 39831.2 106.660 41626.6 107.095

= 9309.81 + 1795.38X = 9309.81 + 1795.38(19) = 43422.03

If there is a cyclical affect, it is very slight.

Y = TS = 850(1.12) = $952 All of the statements are correct except d. a. The regression equation and seasonal indexes are shown below. The trend is the most important but both trend and seasonal should be used to forecast. Trend Line Equation Y = 2241.77 + 25.5111X Seasonal Indices Period 1 2 3 4 Index 0.96392 1.02559 1.00323 1.00726

Accuracy of Model MAPE: 3.2 MAD: 87.3 MSD: 11114.7 b. Forecasts Row Period Forecast 1 43 3349.53 2 44 3388.67 c. The forecast for third quarter is very accurate (3,349.5 versus 3,340) The forecast for fourth quarter is high compared to Value Line (3,388.7 versus 3,300). Following are computer generated graphs for this problem: 17

Decomposition Fit for Sales

3400 3200
Actual Predicted Forecast Actual

3000 2800

Predicted Forecast

Sales
2600

2400 2200

MAPE: MAD:

3.2 87.3 11114.7

2000 0 5 10 15 20 25 30 35 40 45

MSD:

Time

18

15.

a.

Addivitive Decomposition Ln(Sales) 19

Data LnCavSales Length 77.0000 NMissing 0 Trend Line Equation Yt = 4.61322 + 2.19E-02*t Seasonal Indices Period 1 2 3 4 5 6 7 8 9 10 11 12 Index 0.33461 -0.01814 -0.40249 -0.63699 -0.71401 -0.57058 -0.27300 -0.00120 0.46996 0.72291 0.74671 0.34220

b. c.

Pronounced trend and seasonal components. Would use both to forecast. Forecasts Date Period Forecasts(Ln(Sales)) Forecasts(Sales) 20

Jul. Aug. Sep. Oct. Nov. Dec. d. e.

79 80 81 82 83 84

Jun. 78 6.06811 6.36178 6.85482 7.12964 7.17532 6.79267

5.74866 432 579 948 1248 1307 891

314

See last column of table in part c. Forecasts of sales developed from additive decomposition are higher (for all months June 2000 through December 2000) 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.

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. Seasonal Indices (Multiplicative Decomposition for Demand) 21

c.

Period Index Period 1 0.951313 5 2 0.955088 6 3 0.966953 7 4 0.988927 8

Index 1.00495 1.00093 1.02748 1.07089

Period 9 10 11 12

Index 1.04368 0.98570 0.99741 1.00669

Demand tends to be relatively high in the summer months. d. Forecasts derived from a multiplicative decomposition of demand (see plot below). Date Oct. Nov. Dec. Period 130 131 132 Forecast 172.343 175.773 178.803

19.

a.

JAN =

600 = 500 1.2

500(1.37) = 685 people estimated for FEB b. .

T T

= 140 + 5(t); t for JAN 2003 = 72 = 140 + 5(72) = 500

Y Y Y Y Y Y Y

JAN FEB MAR APR MAY JUN JUL

= (140 + 5(72) = 500)(1.20) = 600 = (140 + 5(73) = 505)(1.37) = 692 = (140 + 5(74) = 510)(1.00) = 510 = (140 + 5(75) = 515)(0.33) = 170 = (140 + 5(76) = 520)(0.47) = 244 = (140 + 5(77) = 525)(1.25) = 656 = (140 + 5(78) = 530)(1.53) = 811 22

AUG SEP OCT NOV DEC c. 23. 5

Y Y Y Y Y

= (140 + 5(79) = 535)(1.51) = 808 = (140 + 5(80) = 540)(0.95) = 513 = (140 + 5(81) = 545)(0.60) = 327 = (140 + 5(82) = 550)(0.82) = 451 = (140 + 5(83) = 555)(0.97) = 538

1289.73(2.847) = 3,671.86

CHAPTER 6 REGRESSION ANALYSIS

ANSWERS TO ODD NUMBERED PROBLEMS 1. Option b is inconsistent because the regression coefficient and the correlation coefficient must have the same sign.

23

3.

Correlation of Sales and Expend. = 0.848

The regression equation is Sales = 828 + 10.8 Expend. Predictor Constant Expend. S = 67.19 Coef 828.1 10.787 StDev 136.1 2.384 T 6.08 4.52 P 0.000 0.000

R-Sq = 71.9%

R-Sq(adj) = 68.4%

Analysis of Variance Source DF Regression 1 Error 8 Total 9 a. b. c. d. e. SS 92432 36121 128552 MS 92432 4515 F P 20.47 0.000

Yes, because r = .848 and t = 4.52.

Y Y

= 828 + 10.8X = 828 + 10.8(50) = $1368

72% since r2 = .719 Unexplained sum of squares = 36,121 Divide by df = (n - 2) = 8 to get 4515.
4515 = 67.19 = sy.x

f.

Total sum of squares is 128,552 Divide by df = (n - 1) = 9 to get the variance, 14,283.6. The square root is Y's standard deviation, sy = 119.5.

5. .

The regression equation is Cost = 208 + 70.9 Age Predictor Constant Age S = 111.6 Coef 208.20 70.918 StDev 75.00 9.934 T P 2.78 0.027 7.14 0.000 R-Sq(adj) = 86.2%

R-Sq = 87.9%

Analysis of Variance 24

Source DF Regression 1 Error 7 Total 8

SS 634820 87197 722017

MS F P 634820 50.96 0.000 12457

a.

25

1100 1000 900 800 700 600 500 400 300 200 0 5 10

Cost

Age

b. c.

Positive Y = 6058 X2 = 513

Y

Y2 = 4,799,724 XY = 48,665

X = 59 r = .938

d. e.

= 208.2033 + 70.9181X

H0: = 0 H1: 0 Reject H0 if t < -2.365 or t > 2.365

sb1 = s y x /

( X X )

= 111.6 / 126.22 = 9.93

t=

b1 70.9181 = = 7.14 sb1 9.93

Reject the null hypothesis. Age and maintenance cost are linearly related in the population. f.
Y

= 208.2033 + 70.9181(5) = 562.7938 or $562.80

26

7.

The regression equation is Orders = 15.8 + 1.11 Catalogs T P 5.13 0.000 0.011 R-Sq(adj) = 43.8%

Predictor Coef StDev Constant 15.846 3.092 Catalogs 1.1132 0.3596 3.10 S = 5.757 a. b. c. d. R-Sq = 48.9%

r = .7 and t = 3.1 so a significant relationship exists between Y and X.

Y

= 15.846 + 1.1132X

sy.x = 5.57 Analysis of Variance Source DF SS Regression 1 317.53 Error 10 331.38 Total 11 648.92 49% since r2 = .489 H0: = 0 H1: 0 Reject H0 if t < -3.169 or t > 3.169
sb1 = s y x /

MS 317.53 33.14

F 9.58

P 0.011

e. f.

( X X )

= 5.757 / 256.25 = .36

t=

b1 1.1132 = = 3.09 sb1 .36

Fail to reject the null hypothesis. Orders received and catalogs distributed are not linearly related in the population. g. H0: = 0 H1: 0 Reject H0 if F > 10.04
F = MSR 317.53 = = 9.58 MSE 33.14

Yes and Yes h.

Y

= 13.846 + 1.1132(10) = 24.96 or 24,959

27

9.

a. b.

The two firms seem to be using very similar rationale since r = .959. If ABC bids 1.01, the prediction for Y becomes COMP = -3.257 + 1.03435(101) = 101.212, the point estimate. For an interval estimate using 95% confidence the appropriate standard error is sy.x, since only the variability of the data points around the (population) regression line need be considered: 101.205 1.96(.7431) 101.205 1.456 99.749 to 102.661 If the data constitute a sample, the point estimate for Y would be the same but the interval estimate would use the standard error of the forecast instead of Sy.x, since the variability of sample regression lines around the population regression line would have to be considered along with the scatter of sample Y values around the calculated line. The probability of winning the bid would then involve the same problem as in c. above except that sf would be used in the calculation: Z = (101 - 101.212)/.768 = -.276 where sf = .768, giving an area of .1103. (.5 .1103) = .3897.

c.

11.

The regression equation is Permits = 2217 - 145 Rate Predictor Constant Rate S = 144.3 Coef 2217.4 -144.95 StDev 316.2 27.96 T 7.01 -5.18 P 0.000 0.000

R-Sq = 79.3%

R-Sq(adj) = 76.4%

Analysis of Variance Source DF Regression 1 Error 7 Total 8 a. SS 559607 145753 705360 MS F P 559607 26.88 0.000 20822

28

b.

Y = 5375 Y2 = 3,915,429 X = 100.6 X2 = 1151.12 XY = 56,219.8 r = -0.8907

c.

Reject H0 if t < - 2.365 or t > 2.365. Computed t score = 5.1842 Reject the null hypothesis: Interest rates and building permits are linearly related in the population. 44.9474 r2 = .7934 Using knowledge of the linear relationship between interest rates and building permits (r = -0.8907) we can explain 79.3% of the building permit variable variance.

d. e. f.

13.

The regression equation is Defects = - 17.7 + 0.355 Size Predictor Coef StDev Constant -17.731 4.626 Size 0.35495 0.02332 S = 7.863 R-Sq = 95.5% Analysis of Variance Source DF Regression 1 Error 11 SS 14331 680 T -3.83 15.22 P 0.003 0.000

R-Sq(adj) = 95.1% MS 14331 62 F 231.77 29 P 0.000

Total a.

12

15011

b. c.

Defects = - 17.7 + 0.355 Size The slope coefficient, .355, is significantly different from zero. + 0.00101

d.

Examination of the residuals shows a curvilinear pattern. 30

e.

The best model transforms the predictor variable and uses X2 as its independent variable. The regression equation is Defects = 4.70 + 0.00101 Sizesqr Predictor Coef StDev Constant 4.6973 0.9997 Sizesqr 0.00100793 0.00001930 S = 2.341 R-Sq = 99.6% T P 4.70 0.000 52.22 0.000

R-Sq(adj) = 99.6%

Analysis of Variance Source DF Regression 1 Error 11 Total 12 SS 14951 60 15011 MS 14951 5 F 2727.00 P 0.000

Unusual Observations Obs Sizesqr Defects 3 5625 6.000 f. g.

Fit StDev Fit Residual 10.367 0.920 -4.367

St Resid -2.03R + 0.00101

The slope coefficient, .355, is significantly different from zero.

Examination of the residuals shows a random pattern. h. R denotes an observation with a large standardized residual 31

Fit 95.411 i.

StDev Fit 95.0% CI 95.0% PI 1.173 ( 92.828, 97.994) ( 89.645, 101.177)

Defects = 4.70 + 0.00101 Sizesqr

b. c.

The regression equation is: Market = 60.7 + 0.414 Assessed

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. 15. a. b. c. d. The regression equation is: OpExpens = 18.9 + 1.30 PlayCosts by player costs.
r 2 = .751 . About 75% of the variation in operating expenses is explained

F = 72.6 , p-value = .000 < .10. The regression is clearly significant at the = .10 level.

Coefficient on X = player costs is 1.3. Is H 0 : 1 = 2 reasonable?

t= 1.3 2.0 = 4.58 suggests 1 = 2 is not supported by the data. Appears .153

that operating expenses have a fixed cost component represented by the intercept b0 = 18.9 , and then are about 1.3 times player costs. e. f.
2 s =58.6 , Y Y f

gives 58.6 2(5.5) or (47.6, 69.6). SE Fit 17.69 Residual 3.45R St Resid

Unusual Observations Obs Play Cost OpExpens Fit 7 18.0 60.00 42.31 1.64

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

32