CHAPTER 2 A REVIEW OF BASIC STATISTICAL CONCEPTS ANSWERS TO ODD NUMBERED PROBLEMS 1. Descriptive Statistics Variable N Mean Median Tr Mean StDev SE Mean C1 28 21.32 17.00 20.69 13.37 2.53 Variable Min Max Q1 Q3 C1 5.00 54.00 11.25 28.75 a. X = 21.32 b. S = 13.37 c. S 2 = 178.76 d. If the policy is successful, smaller orders will be eliminated and the mean will increase. e. If the change causes all customers to consolidate a number of small orders into large orders, the standard deviation will probably decrease. Otherwise, it is very difficult to tell how the standard deviation will be affected. f. The best forecast over the long-term is the mean of 21.32. 3. a. Point estimate: 1
Transcript
CHAPTER 2
A REVIEW OF BASIC STATISTICAL CONCEPTS
ANSWERS TO ODD NUMBERED PROBLEMS
1. Descriptive Statistics
Variable N Mean Median Tr Mean StDev SE MeanC1 28 21.32 17.00 20.69 13.37 2.53
Variable Min Max Q1 Q3C1 5.00 54.00 11.25 28.75
a. X = 21.32
b. S = 13.37
c. S2 = 178.76
d. If the policy is successful, smaller orders will be eliminated and the mean will increase.
e. If the change causes all customers to consolidate a number of small orders into large orders, the standard deviation will probably decrease. Otherwise, it is very difficult to tell how the standard deviation will be affected.
f. The best forecast over the long-term is the mean of 21.32.
3. a. Point estimate:
b. 1 = .95 Z = 1.96, n = 30, (5.85%, 15.67%)
c. df = 301 = 29, t = 2.045 (5.64%, 15.88%)
d. 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 n = 100 = .05
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H1: > 12.1 S = 1.7 X = 13.5
Reject H0 if Z > 1.645
Z = = 8.235
Reject H0 since the computed Z (8.235) is greater than the critical Z (1.645). The mean has increased.
7. n = 60,
two-sided test, = .05, critical value: |Z|= 1.96
Test statistic:
Since |2.67| = 2.67 > 1.96, reject 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
9. H0: = 700 n = 50 = .05H1: 700 S = 50 X = 715
Reject H0 if Z < -1.96 or Z > 1.96
Z = = 2.12
Since the calculated Z is greater than the critical Z (2.12 > 1.96), reject the null hypothesis. The forecast does not appear to be reasonable.
p-value: P(Z < 2.12 or Z > 2.12) = 2 P(Z > 2.12) = 2(.017) = .034, strong evidence against
11. a.
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b. Positive
c. Y = 6058 Y2 = 4,799,724 X = 59X2 = 513 XY = 48,665 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
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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
reject H0 demonstrate Type I error.
15. n = 175, Point estimate: 98% confidence interval: 1 = .98 Z = 2.33
(43.4, 47.0)
Hypothesis test:
two-sided test, = .02, critical value: |Z|= 2.33
Test statistic:
Since |Z| = 1.54 < 2.33, do not reject 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 ANDCHOOSING A FORECASTING TECHNIQUE
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ANSWERS TO ODD NUMBERED PROBLEMS
1. Qualitative forecasting techniques rely on human judgment and intuition. Quantitative forecasting techniques rely more on manipulation of past historical data.
3. The secular trend of a time series is the long-term component that represents the growth or decline in the series over an extended period of time. The cyclical component is the wave- like fluctuation around the trend. The seasonal component is a pattern of change that repeats itself year after year. The irregular component measures the variability of the time series after the other components have been removed.
5. The autocorrelation coefficient measures the correlation between a variable, lagged one or more periods, and itself.
7. a. nonstationary series b. stationary series c. nonstationary series d. stationary series
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.
11. 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.
17. Y = 4465.7 (Yt-1 Y ) = 2396.1 (Yt Y )2 = 55,912,891
(Yt Y ) (Yt-1 Y ) = 50,056,722 (Yt Y ) (Yt-2 Y ) = 44,081,598
r1 = 50 056 722
55 912 891
, ,
, , = .895
H0: ρ1 = 0H1: ρ1 0
Reject if t < -2.069 or t > 2.069
SE( ) = = = = .204
= = 4.39
Since the computed t (4.39) is greater than the critical t (2.069), reject the null.
H0: ρ2 = 0H1: ρ2 0
Reject if t < -2.069 or t > 2.069
r2 = 44,081598
55 912 891
,
, , = .788
SE( ) = = =2 6
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.= .33
.
.
788 0
33= 2.39
Since the computed t (4.39) is greater than the critical t (2.069), reject the null.
b. The data are nonstationary.
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654321
1.0
0.8
0.6
0.4
0.2
0.0
-0.2
-0.4
-0.6
-0.8
-1.0
LBQTCorrLag
72.04
68.73
62.57
52.85
39.37
21.74
0.63
0.92
1.25
1.68
2.39
4.39
0.31
0.43
0.56
0.67
0.79
0.90
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3
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Autocorrelation Function for Loans
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.
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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
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CHAPTER 4
MOVING AVERAGES AND SMOOTHING METHODS
ANSWERS TO ODD NUMBERED PROBLEMS
1. Exponential smoothing
3. Moving average
5. Winter's three-parameter linear and seasonal model
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.
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CHAPTER 5
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.
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.
5. Weather and the calendar year such as holidays affect the seasonal component.
7. a.
YEAR Y T C1980 11424 11105.2 102.8711981 12811 12900.6 99.3061982 14566 14695.9 99.1161983 16542 16491.3 100.307
d. If there is a cyclical affect, it is very slight.
9. Y = TS = 850(1.12) = $952
11. All of the statements are correct except d.
13. 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 = 2241.77 + 25.5111X
Seasonal Indices
Period Index 1 0.96392 2 1.02559 3 1.00323 4 1.00726
Accuracy of Model MAPE: 3.2 MAD: 87.3 MSD: 11114.7
b. Forecasts
Row Period Forecast
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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:
d. See last column of table in part c. e. 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.
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c. Seasonal Indices (Multiplicative Decomposition for Demand)
Period Index Period Index Period Index 1 0.951313 5 1.00495 9 1.04368
Source DF SS MS F PRegression 1 634820 634820 50.96 0.000Error 7 87197 12457Total 8 722017
a.
25
1050
1100
1000
900
800
700
600
500
400
300
200
Age
Cost
b. Positive
c. Y = 6058 Y2 = 4,799,724 X = 59 X2 = 513 XY = 48,665 r = .938
d. Y = 208.2033 + 70.9181X
e. H0: = 0 H1: 0
Reject H0 if t < -2.365 or t > 2.365
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
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7. The regression equation isOrders = 15.8 + 1.11 Catalogs
Predictor Coef StDev T PConstant 15.846 3.092 5.13 0.000
Catalogs 1.1132 0.3596 3.10 0.011
S = 5.757 R-Sq = 48.9% R-Sq(adj) = 43.8%
a. r = .7 and t = 3.1 so a significant relationship exists between Y and X.
b. Y = 15.846 + 1.1132X
c. sy.x = 5.57
d. Analysis of VarianceSource DF SS MS F PRegression 1 317.53 317.53 9.58 0.011Error 10 331.38 33.14Total 11 648.92
e. 49% since r2 = .489
f. H0: = 0 H1: 0
Reject H0 if t < -3.169 or t > 3.169
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
Yes and Yes
h. Y = 13.846 + 1.1132(10) = 24.96 or 24,959
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9. a. The two firms seem to be using very similar rationale since r = .959.
b. 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
c. 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.
11. The regression equation isPermits = 2217 - 145 Rate
Source DF SS MS F PRegression 1 559607 559607 26.88 0.000Error 7 145753 20822Total 8 705360
a.
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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 = 5.1842Reject the null hypothesis: Interest rates and building permits are linearly related in the population.
d. 44.9474
e. r2 = .7934
f. 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.
Source DF SS MS F PRegression 1 14951 14951 2727.00 0.000Error 11 60 5Total 12 15011
Unusual ObservationsObs Sizesqr Defects Fit StDev Fit Residual St Resid 3 5625 6.000 10.367 0.920 -4.367 -2.03R
f. The slope coefficient, .355, is significantly different from zero. + 0.00101
g.
Examination of the residuals shows a random pattern.
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h. R denotes an observation with a large standardized residual
Fit StDev Fit 95.0% CI 95.0% PI 95.411 1.173 ( 92.828, 97.994) ( 89.645, 101.177)
i. Defects = 4.70 + 0.00101 Sizesqr
b. The regression equation is: Market = 60.7 + 0.414 Assessed
c. . 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. The regression equation is: OpExpens = 18.9 + 1.30 PlayCosts
b. . About 75% of the variation in operating expenses is explained by player costs.
c. , p-value = .000 < .10. The regression is clearly significant at the level.
d. Coefficient on = player costs is 1.3. Is reasonable?
suggests is not supported by the data. Appears that
operating expenses have a fixed cost component represented by the intercept , and then are about 1.3 times player costs.
e. , gives or (47.6, 69.6).
f. Unusual Observations
Obs Play Cost OpExpens Fit SE Fit Residual St Resid 7 18.0 60.00 42.31 1.64 17.69 3.45R
R denotes an observation with a large standardized residual Team 7 has unusually low player costs relative to operating expenses.
