Statistics for Business and Economics chapter 16 Regression Analysis: Model Building

The Minitab output is shown below:The regression equation is 3.. The scatter diagram is shown below:050100150200250 A simple linear regression model does not appear to be appropriate.. T

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Regression Analysis: Model Building

Learning Objectives

1 Learn how the general linear model can be used to model problems involving curvilinear

relationships

2 Understand the concept of interaction and how it can be accounted for in the general linear model

3 Understand how an F test can be used to determine when to add or delete one or more variables.

4 Develop an appreciation for the complexities involved in solving larger regression analysis problems

5 Understand how variable selection procedures can be used to choose a set of independent variables for an estimated regression equation

6 Learn how analysis of variance and experimental design problems can be analyzed using a regression model

7 Know how the Durbin-Watson test can be used to test for autocorrelation

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1 a The Minitab output is shown below:

The regression equation is

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10 15 20 25 30 35 40 45

x

The scatter diagram suggests that a curvilinear relationship may be appropriate

d The Minitab output is shown below:

Y = 9.32 + 0.424 X

Predictor Coef SE Coef T p

Constant 9.315 4.196 2.22 0.113

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b The Minitab output is shown below:

3 a The scatter diagram shows some evidence of a possible linear relationship

Y = 2.32 + 0.637 X

Constant 2.322 1.887 1.23 0.258

X 0.6366 0.3044 2.09 0.075

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6 5

4 3

transformation and the corresponding standardized residual pot are shown below

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0.200 0.175

Total 5 37395

b p-value = 005 <  = 01; reject H0

5 The Minitab output is shown below:

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b Since the linear relationship was significant (Exercise 4), this relationship must be significant

Note also that since the p-value of 003 <  = 05, we can reject H0.

c The fitted value is 1302.01, with a standard deviation of 9.93 The 95% confidence interval is 1270.41 to 1333.61; the 95% prediction interval is 1242.55 to 1361.47

6 a The scatter diagram is shown below:

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8

b No; the relationship appears to be curvilinear

c Several possible models can be fitted to these data, as shown below:

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050100150200250

A simple linear regression model does not appear to be appropriate There appears to be a

curvilinear relationship between the two variables

R denotes an observation with a large standardized residual

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The corresponding standardized residual plot is shown below:

120 100

80 60

40 20

0

4 3 2 1 0 -1 -2

There is an unusual trend in the points There is also some indication that the variance may not be constant

c The Minitab output is shown below:

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4.5 4.0

3.5 3.0

d The Minitab output is shown below:

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0.04 0.03

0.02 0.01

0.00 -0.01

050010001500200025003000350040004500

Rating

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A simple linear regression model does not appear to be appropriate There appears to be a curvilinear relationship between the two variables.

Price = 33829 - 4571 Rating + 154 RatingSq

Predictor Coef SE Coef T P

c The Minitab output is shown below:

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A simple linear regression model appears to be appropriate.

b

Note the line drawn through the data This line indicates a possible curvilinar relationship between these two variables

c In the Minitab output that follows IndexSq denotes the square of the Cost-of-Living Index

Creative Class (%) = 49.2 - 0.673 Cost-of-Living Index + 0.00282 IndexSq + 0.404 Income

Constant 49.24 17.25 2.85 0.006

Cost-of-Living Index -0.6725 0.2888 -2.33 0.024

IndexSq 0.002821 0.001223 2.31 0.026

Income 0.40418 0.06772 5.97 0.000

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10 a SSR = SST - SSE = 1030

Using Excel or Minitab, the p-value corresponding to F = 49.52 is 000.

Because p-value ≤ α, x1 is significant

100 / 23

Because p-value ≤ α, the addition of variables x2 and x3 is significant

11 a SSE = SST - SSR = 1805 - 1760 = 45

F = 440/1.8 = 244.44

Because p-value ≤ α, the overall relationship is significant.

b SSE(x1, x2, x3, x4) = 45

c SSE(x2, x3) = 1805 - 1705 = 100

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d (100 45) / 2 15.28

1.8

Because p-value ≤ α, x1 and x4 contribute significantly to the model

12 a A portion of the Minitab output follows:

Scoring Avg = 46.3 + 14.1 Putting Avg.

b A portion of the Minitab output follows:

Scoring Avg = 59.0 - 10.3 Greens in Reg + 11.4 Putting Avg - 1.81 Sand Saves

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SSE(reduced) - SSE(full) 7.2998 - 4.3240

The p-value associated with F = 8.95 (2 degrees of freedom numerator and 26 denominator)

is 001 With a p-value < α =.05, the addition of the two independent variables is statistically

significant

Earnings ($1000) = 14528 - 7640 Putting Avg.

Earnings ($1000) = 5214 + 6873 Greens in Reg - 5623 Putting Avg + 2217 Sand Saves

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significant

d A portion of the Minitab output follows:

Earnings ($1000) = 36697 - 501 Scoring Avg.

Risk = - 111 + 1.32 Age + 0.296 Pressure

Total 19 4190.9

Source DF Seq SS

Age 1 1772.0

Pressure 1 1607.7

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Unusual Observations

Obs Age Risk Fit SE Fit Residual St Resid

17 66.0 8.00 25.05 1.67 -17.05 -2.54R

Risk = - 123 + 1.51 Age + 0.448 Pressure + 8.87 Smoker -

The p-value associated with F = 4.23 (2 numerator and 15 denominator DF) is 000

Because p-value ≤ α = 05, the addition of the two terms is significant.

ERA = - 0.253 + 0.453 H/9

Constant -0.2535 0.7351 -0.34 0.732

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significant

16 a The sample correlation coefficients are as follows:

Weeks Age Educ Married Head Tenure Manager Age 0.577

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Cell Contents: Pearson correlation

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Weeks = - 0.07 + 1.73 Age - 28.7 Manager - 15.1 Head - 17.4 Sales

d The results using Minitab’s Backward Elimination procedure are shown below:

Backward elimination Alpha-to-Remove: 0.05

Response is Weeks on 7 predictors, with N = 50

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e The results using Mintab’s Best-Subset procedure are shown below:

Weeks = 13.1 + 1.64 Age - 9.76 Married - 19.4 Head - 29.0 Manager - 19.0 Sales

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17 The output obtained using Minitab’s Best Subset Regression is shown below:

Response is Scoring Avg.

Scoring Avg = - 88.1 + 0.591 Drive Average + 209 Greens in Reg.

+ 9.74 Putting Avg - 0.868 DriveGreens

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18 a Because the independent variable most highly correlated with RPG is OBP, it

will provide the best one-variable estimated regression equation The Minitab

output using OBP to predict RPG is shown below:

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RPG = - 0.909 + 32.2 OBP + 0.109 HR - 21.5 AVG + 0.244 3B - 0.0223 BB

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8 7 6 5 4 3 2 1

BB appears to be a good choice

19 See the solution to Exercise 14 in this chapter The Minitab output using the best subsets

regression procedure is shown below:

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Risk = - 91.8 + 1.08 Age + 0.252 Pressure + 8.74 Smoker

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x3 = 0 if block 1 and 1 if block 2

d The p-value of 004 is less than  = 05; therefore, we can reject H0 and conclude that the

mean time to mix a batch of material is most the same for each manufacturer

24 a The dummy variables are defined as follows:

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The Minitab output is shown below:

b Note: Estimating the mean drying for paint 2 using the estimated regression equations developed

in part (a) may not be the best approach because at the 5% level of significance, we cannot reject

H0 But, if we want to use the output, we would proceed as follows

D1 = 1 D2 = 0 D3 = 0

TIME = 133 + 6(1) + 3(0) +11(0) = 139

25 X1 = 0 if computerized analyzer, 1 if electronic analyzer

X2 and X3 are defined as follows:

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To test for any significant difference between the two analyzers we must test H0: 1 Since

the p-value corresponding to t = -4.54 is 045 <  = 05, we reject H0: 0 the time to do a tuneup is not the same for the two analyzers

26 Size = 0 if a small advertisement and 1 if a large advertisement

DesignB and DesignC are defined as follows:

DesignB DesignC AdvertisementDesign

LargeDesignB denotes the interaction between Large and DesignB

LargeDesignC denotes the interaction between Large and DesignC

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The complete data set and the Minitab output are shown below:

Number = 10.0 + 0.00 Size + 8.00 DesignB + 4.00 DesignC + 10.0 LargeDesignB

The Minitab output using only Design B follows:

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Residual Error 10 250.00 25.00

Total 11 544.00

Thus, DesignB is significant using α = 05 However, the model involving just the interaction

between Large and DesignB also provides some interesting results:

conclusions about the relationships among the variables

b The Durbin-Watson statistic is 798118 At the 05 level of significance, dL = 1.20 and dU =1.41

Because d < dL, there is significant positive autocorrelation

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28 From Minitab, d = 1.60 At the 05 level of significance, dL = 1.04 and dU = 1.77 Since dL  d

dU, the test is inconclusive

50 55 60 65 70 75 80 85

The curvature in the scatter diagram indicates that a simple linear regression model may not be appropriate

Rating = 49.9 + 14.9 Speed - 1.83 SpeedSq

c The Minitab output for the transformed nonlinear model is shown below:

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The estimated regression equation developed in part (b) provides a much better fit.

30 a

There appears to be a curvilinear relationship between weight and price

Price = 11376 - 728 Weight + 12.0 WeightSq

c A portion of the Minitab output follows:

Price = 1284 - 572 Type_Fitness - 907 Type_Comfort

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Price = 5924 - 215 Weight - 6343 Type_Fitness - 7232 Type_Comfort + 261 WxF

Delay = 80.4 + 11.9 Industry - 4.82 Public - 2.62 Quality - 4.07 Finished Predictor Coef SE Coef T P

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Regression 4 2587.7 646.9 5.42 0.002

Residual Error 35 4176.3 119.3

Total 39 6764.0

b The low value of the adjusted coefficient of determination (31.2%) does not indicate a good fit

c The scatter diagram is shown below:

The scatter diagram suggests a curvilinear relationship between these two variables

d The output from Minitab’s best subsets procedure is shown below, where FinishedSq is the square

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The estimated regression equation using Industry, Quality, Finished, and FinishedSq has an adjusted coefficient of determination of 54.4%

32 The computer output is shown below:

Total 39 6764.0

Durbin-Watson statistic = 1.55

At the 05 level of significance, dL = 1.44 and dU = 1.54 Since d = 1.55 > dU, there is no significant positive autocorrelation

Delay = 70.6 + 12.7 Industry - 2.92 Quality

Total 39 6764.0

Durbin-Watson statistic = 1.43

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b The residual plot as a function of the order in which the data are presented is shown below:

Order In Which Data Are Presented

30 20 10 0 -10 -20 -30

There is no obvious pattern in the data indicative of positive autocorrelation

c At the 05 level of significance, dL = 1.39 and dU = 1.60 Since dL ≤ d ≤ dU, the test is inconclusive

34 The dummy variables are defined as follows:

The Minitab output is shown below:

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Since the p-value = 034 is less than  = 05, there are significant differences between comfort

levels for the three types of browsers

35 Let Mid-size = 1 if a mid-size car, 0 otherwise; Luxury = 1 if a luxury car, 0 otherwise; and Sports

= 1 if a sports car, 0 otherwise

The Minitab output is shown below

Resale = 32.9 - 1.70 Mid-size + 4.30 Luxury + 7.30 Sports

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