252solnK1 11/26/07
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K. REGRESSION EXTENSIONS
1. Residual Analysis
Text 13.23, 13.24, 13.26, 14.18 [13.20, 13.21, 13.22, 14.9] (13.20, 13.21, 13.22, 14.9)
2. Dummy Variables
14.38-14.39, 14.41 [14.33 – 14.35] (15.6 – 15.8)
3. Nonlinear regression
15.1, 15.6, 15.7 [15.1, 15.6, 15.7] (15.1, 15.13, 15.14)
4. Runs test
K1
5. Durbin-Watson test
13.32-13.34 [13.28, 13.29, 13.30] (13.28, 13.29, 13.30)
This document includes sections 1-3
--------------------------------------------------------------------------------------------- ------------------------------------
Residual Analysis
Most answers below are from the Instructor’s Solution Manual.
Exercise 13.23 [13.20 in 9th]: You have to look at the problem in the book for 20 and 21. A residual
analysis of the data indicates no apparent pattern. The assumptions of regression appear
to be met.
Exercise 13.24 [13.21 in 9th]: A residual analysis of the data indicates a pattern, with sizeable clusters of
consecutive residuals that are either all positive or all negative. This appears to violate the
assumption of independence of errors.
Exercise 13.26 [13.22 in 9th]: (a)-(b)
The Instructor’s Solution Manual says “Based on a residual
analysis, the model appears to be adequate.” I’m not so sure. I ran the data and got the output.
————— 12/2/2003 4:01:58 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > Retrieve "C:\Documents and Settings\RBOVE.WCUPANET\My Documents\Drive
D\MINITAB\petfood.MTW".
Retrieving worksheet from file: C:\Documents and Settings\RBOVE.WCUPANET\My
Documents\Drive D\MINITAB\petfood.MTW
# Worksheet was saved on Wed Nov 12 2003
Results for: petfood.MTW
MTB > Name c7 = 'RESI1' c8 = 'SRES1'
MTB > Regress c1 1 c2;
SUBC>
Residuals 'RESI1';
SUBC>
SResiduals 'SRES1';
SUBC>
Constant;
SUBC>
Brief 3.
Regression Analysis: Sales versus Space
The regression equation is
Sales = 1.45 + 0.0740 Space
Predictor
Constant
Space
S = 0.3081
Coef
SE Coef
T
P
1.4500
0.2178
6.66
0.000
0.07400
0.01591
4.65
0.001
R-Sq = 68.4%
R-Sq(adj) = 65.2%
252solnK1 11/26/07
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Analysis of Variance
Source
Regression
Residual Error
Total
Obs
1
2
3
4
5
6
7
8
9
10
11
12
Space
5.0
5.0
5.0
10.0
10.0
10.0
15.0
15.0
15.0
20.0
20.0
20.0
DF
1
10
11
SS
2.0535
0.9490
3.0025
Sales
1.6000
2.2000
1.4000
1.9000
2.4000
2.6000
2.3000
2.7000
2.8000
2.6000
2.9000
3.1000
MS
2.0535
0.0949
Fit
1.8200
1.8200
1.8200
2.1900
2.1900
2.1900
2.5600
2.5600
2.5600
2.9300
2.9300
2.9300
F
21.64
SE Fit
0.1488
0.1488
0.1488
0.0974
0.0974
0.0974
0.0974
0.0974
0.0974
0.1488
0.1488
0.1488
P
0.001
Residual
-0.2200
0.3800
-0.4200
-0.2900
0.2100
0.4100
-0.2600
0.1400
0.2400
-0.3300
-0.0300
0.1700
MTB > %Fitline c1 c2;
SUBC>
Confidence 95.0.
Executing from file: W:\wminitab13\MACROS\Fitline.MAC
Macro is running ... please wait
Regression Analysis: Sales versus Space
The regression equation is
Sales = 1.45 + 0.074 Space
S = 0.308058
R-Sq = 68.4 %
R-Sq(adj) = 65.2 %
Analysis of Variance
Source
Regression
Error
Total
DF
1
10
11
SS
2.0535
0.9490
3.0025
MS
2.0535
0.0949
F
21.6386
P
0.001
Fitted Line Plot: Sales versus Space
MTB > %Resplots c7 c6;
SUBC>
Title "Residuals vs Fits".
Executing from file: W:\wminitab13\MACROS\Resplots.MAC
Macro is running ... please wait
Residual Plots: RESI1 vs FITS1
MTB > Plot c7*c1;
SUBC>
Symbol;
SUBC>
ScFrame;
SUBC>
ScAnnotation.
St Resid
-0.82
1.41
-1.56
-0.99
0.72
1.40
-0.89
0.48
0.82
-1.22
-0.11
0.63
252solnK1 11/26/07
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Plot RESI1 * Sales
MTB > Plot c7*c2;
SUBC>
Symbol;
SUBC>
Scram;
SUBC>
ScAnnotation.
Plot RESI1 * Space
Regression Plot
Sales = 1.45 + 0.074 Space
S = 0.308058
R-Sq = 68.4 %
R-Sq(adj) = 65.2 %
3.0
Sales
2.5
2.0
1.5
5
10
15
20
Space
The plot above looks pretty random.
Residuals vs Fits
I Chart of Residuals
Residual
Residual
Normal Plot of Residuals
0.5
0.4
0.3
0.2
0.1
0.0
-0.1
-0.2
-0.3
-0.4
1
UCL=1.081
0
Mean=1.11E-16
-1
-2
-1
0
1
Normal Score
Histogram of Residuals
0
-0.4 -0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 0.4
Residual
10
Residuals v s. Fits
Residual
Frequency
1
5
Observation Number
3
2
LCL=-1.081
0
2
0.5
0.4
0.3
0.2
0.1
0.0
-0.1
-0.2
-0.3
-0.4
2.0
2.5
3.0
Fit
These are the plots that bother me. Normal plots are described on page 227 of the text. The Normal plot
looks a lot like the rectangular distribution shown there, though rectangular distributions are not as bad as
skewed distributions. The histogram gives us a bimodal distribution.
252solnK1 11/26/07
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0.5
0.4
0.3
RESI1
0.2
0.1
0.0
-0.1
-0.2
-0.3
-0.4
5
10
15
20
Space
This plot is probably the most important and it shows very little in the way of a pattern. This is nice.
Exercise 14.18 [14.9 in 9th]: We go back to the WARECOST model. a) Perform a residual analysis to
determine adequacy of the model. b) Plot residuals against months looking for a pattern. c) Find the
Durbin-Watson statistic. d) Find evidence of autocorrelation.
The Instructor’s Solution Manual says the following.
(a)
Based upon a residual analysis the model appears adequate.
(b)
There is no evidence of a pattern in the residuals versus time.
(c)
D = 2.26
(d)
D = 2.26 > 1.55. There is no evidence of positive autocorrelation in the residuals.
For an explanation see the printout.
To run this regression I used the Statistics pull-down menu and then picked Regression twice. I had put
headings on my columns – the data is in the text and on your CD, but, since I’m lazy, I identified the
columns as C1, C2 and C3. So C1 was my response (dependent - Y) variable and C2 and C3 were my
predictor (independent – X) variables. There are just too many subcommands here to use the session
window to drive Minitab. On the Regression menu I went into Graphs and checked all the residual plots.
Under Options I picked Variance Inflation Factors and Durbin-Watson. Under Results I took the last and
most complete option, though this can also be done by using the session command ‘Brief 3’ before you
start. Under storage I picked both of the residuals, though that seems to have been unnecessary unless I
wanted to do some extra plotting. When this regression was finished and I had copied all the graphs into a
Word document, I ran Stepwise from the Regression menu using C1, C2 and C3.
MTB > Retrieve "C:\Documents and Settings\RBOVE.WCUPANET\My Documents\Drive
D\MINITAB\Warecost.MTW".
Retrieving worksheet from file: C:\Documents and Settings\RBOVE.WCUPANET\My
Documents\Drive D\MINITAB\Warecost.MTW
# Worksheet was saved on Thu Nov 20 2003
Results for: Warecost.MTW
MTB > Name c4 = 'RESI1' c5 = 'SRES1'
MTB > Regress c1 2 c2 c3;
SUBC>
Residuals 'RESI1';
SUBC>
SResiduals 'SRES1';
SUBC> GHistogram;
SUBC> GNormalplot;
SUBC> GFits;
SUBC> GOrder;
252solnK1 11/26/07
SUBC>
SUBC>
SUBC>
SUBC>
SUBC>
SUBC>
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GVars c2 c3;
RType 1;
Constant;
VIF;
DW;
Brief 3.
Regression Analysis: DistCost versus Sales, Orders
The regression equation is
DistCost = - 2.73 + 0.0471 Sales + 0.0119 Orders
Predictor
Constant
Sales
Orders
Coef
-2.728
0.04711
0.011947
S = 4.766
SE Coef
6.158
0.02033
0.002249
R-Sq = 87.6%
T
-0.44
2.32
5.31
P
0.662
0.031
0.000
VIF
2.8
2.8
R-Sq(adj) = 86.4%
Analysis of Variance
Source
Regression
Residual Error
Total
Source
Sales
Orders
DF
1
1
DF
2
21
23
SS
3368.1
477.0
3845.1
MS
1684.0
22.7
F
74.13
P
0.000
Seq SS
2726.8
641.3
Comment: The gigantic p-value tells us that the constant is insignificant, but the coefficients of
Sales and Orders were significant at the 5% level. The VIF will be discussed later (Check the end of this
document), but a value below 5 is usually fine. The low p-value for the ANOVA tells us that the 2
independent variables explained a lot of the variation in DistCost. Their sequential contribution to the
Regression sum of squares is shown below the ANOVA. This makes Order look like a fairly feeble, if
significant, contributor to the regression. Actually we will find that is not true.
Obs
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
Sales
386
446
512
401
457
458
301
484
517
503
535
353
372
328
408
491
527
444
623
596
463
DistCost
52.950
71.660
85.580
63.690
72.810
68.440
52.460
70.770
82.030
74.390
70.840
54.080
62.980
72.300
58.990
79.380
94.440
59.740
90.500
93.240
69.330
Fit
63.425
63.755
84.820
67.082
70.127
67.796
49.839
77.528
84.196
77.503
75.199
48.800
62.311
65.626
63.852
75.145
88.789
59.407
87.302
93.867
70.087
SE Fit
1.332
1.511
1.656
1.332
0.999
1.193
2.134
1.139
1.525
1.126
1.838
2.277
1.483
2.847
1.152
1.069
2.004
2.155
2.535
2.097
1.049
Residual
-10.475
7.905
0.760
-3.392
2.683
0.644
2.621
-6.758
-2.166
-3.113
-4.359
5.280
0.669
6.674
-4.862
4.235
5.651
0.333
3.198
-0.627
-0.757
St Resid
-2.29R
1.75
0.17
-0.74
0.58
0.14
0.62
-1.46
-0.48
-0.67
-0.99
1.26
0.15
1.75
-1.05
0.91
1.31
0.08
0.79
-0.15
-0.16
252solnK1 11/26/07
22
23
24
389
547
415
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53.710
89.180
66.800
59.898
87.401
66.535
1.349
1.657
1.107
-6.188
1.779
0.265
-1.35
0.40
0.06
R denotes an observation with a large standardized residual
Durbin-Watson statistic = 2.26
Comment: The Durbin-Watson is checking for
autocorrelation, which will be explained later. It’s enough to say that values close to 2
indicate that autocorrelation is not a problem. The dependent variable (Y) and the first
independent variable are printed out above, followed by Fit, which means the predicted
value of Y, SE Fit, which with the appropriate value if t will give us a confidence
interval for Y, and Residual, which is the difference between the predicted and the
actual value of Y. The standardized residual seems to be the residual after the mean
residual has been subtracted from it and the standard deviation of the residual has been
divided into it.
Residual Histogram for DistCost
Normplot of Residuals for DistCost
Residuals vs Fits for DistCost
Residuals vs Order for DistCost
Residuals from DistCost vs Sales
Residuals from DistCost vs Orders
Histogram of the Residuals
(response is DistCost)
7
6
Frequency
5
4
3
2
1
0
-10
-8
-6
-4
-2
0
2
4
6
8
Residual
Comment: The histogram seems to indicate a fairly symmetrical distribution with a
peak in the middle.
252solnK1 11/26/07
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Normal Probability Plot of the Residuals
(response is DistCost)
2
Normal Score
1
0
-1
-2
-10
0
10
Residual
Comment: The text says that a straight-line Normal Probability plot indicates a Normal
distribution.
Comment: There doesn’t seem to be much of a pattern here.
252solnK1 11/26/07
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Comment: A pattern here would indicate autocorrelation.
Comment: These two graphs show no pattern either.
252solnK1 11/26/07
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MTB > Stepwise c1 c2 c3;
SUBC>
AEnter 0.15;
SUBC>
ARemove 0.15;
SUBC>
Constant.
Stepwise Regression: DistCost versus Sales, Orders
Alpha-to-Enter: 0.15
Alpha-to-Remove: 0.15
Response is DistCost on
2 predictors, with N =
Step
Constant
1
0.4576
2
-2.7282
Orders
T-Value
P-Value
0.0161
10.92
0.000
0.0119
5.31
0.000
Sales
T-Value
P-Value
24
0.047
2.32
0.031
S
5.22
4.77
R-Sq
84.42
87.59
R-Sq(adj)
83.71
86.41
C-p
6.4
3.0
More? (Yes, No, Subcommand, or Help)
SUBC> yes
No variables entered or removed
More? (Yes, No, Subcommand, or Help)
SUBC> no
Comment: This was a stepwise regression. It was done with no options, so that all the
subcommands that you see here were generated by Minitab. It seems that the independent variable with the
most explanatory power was ‘Orders,’ and the regression was Y = 0.4576 + .0161 Orders, with an Rsquared of 84.42. Minitab then added the other independent variable and got a regression of Y = -2.7282 +
0.119 Orders + 0.047 Sales, which is the same regression we got with the ‘Regress’ command. The C-p
statistic, also explained in the text, should be near k + 1, which it is for this regression. After adding two
independent variables, Minitab paused and asked me if I wanted to try for more independent variables. I
foolishly said ‘yes,’ whereupon Minitab discovered that it didn’t have any more variables to add.
Dummy Variables
Exercise 14.38 [14.33 in 9th] (15.6 in 8th edition): The equation is Y  6  4 X 1  2 X 2 . a) Interpret the
coefficient of X1 b) Interpret the coefficient of X2
c) If the t for X2 is 3.27 and   .05 , does X2 make a significant contribution to the model?
(a)
Holding constant the effect of X2, the estimated average value of the dependent variable
will increase by 4 units for each increase of one unit of X1.
(b)
Holding constant the effects of X1, the presence of the condition represented by X2 = 1 is
estimated to increase the average value of the dependent variable by 2 units.
17 
 2.11 . You can reject H0 and say
t  3.27 . n  k  1  20  2  1. This is larger than t .05
(c)
that the presence of X2 makes a significant contribution to the model.
252solnK1 11/26/07
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Exercise 14.39 [14.34 in 9th] (15.7 in 8th edition): The chair of an accounting department wants to predict
GPA in accounting on the basis of SAT score and whether the student got a B or better in the introductory
statistics course. a) Explain the steps involved in making a model and b) assume that the variable
representing the statistics grade has a coefficient of .30. Explain its meaning.
(a)
First develop a multiple regression model using X1 as the variable for the SAT score and
X2 a dummy variable with X2 = 1 if a student had a grade of B or better in the
introductory statistics course. If the dummy variable coefficient is significantly different
than zero, you need to develop a model with the interaction term X1 X2 to make sure that
the coefficient of X1 is not significantly different if X2 = 0 or X2 = 1.
(b)
If a student received a grade of B or better in the introductory statistics course, the
student would be expected to have a grade point average in accountancy that is 0.30
higher than a student who had the same SAT score, but did not get a grade of B or better
in the introductory statistics course.
Exercise 14.41 [14.35 in 9th] (15.8 in 8th edition): Use the PETFOOD data to develop a regression
explaining sales as an effect of the amount of shelf space, and the location of the product in the aisle (back
or front). a) State the equation, b) interpret coefficients, predict sales for 8 feet of shelf space in the back of
the aisle and make it into 95% confidence and prediction intervals. Data is below.
Solution: To run this regression I used the Statistics pull-down menu and then picked Regression twice. I
had put headings on my columns – the data is in the text and on your CD, but, since I’m lazy, I identified
the columns as C1, C2 and C3. So C2 was my response (dependent - Y) variable and C1 and C3 were my
predictor (independent – X) variables. There are just too many subcommands here to use the session
window to drive Minitab. On the Regression menu I went into Graphs and checked all the residual plots
except residuals vs. order. Under Options I picked Variance Inflation Factors and set up for confidence and
prediction intervals by telling it that the independent variables for this prediction were in C5 and C6. Under
Results I took the last and most complete option, though this can also be done by using the session
command ‘Brief 3’ before you start. Under storage I picked nothing. When this regression was finished and
I had copied all the graphs into a Word document, I ran the regression again with the Interaction variable
requested in part n) of the problem. To confirm my results, I ran Stepwise from the Regression menu using
C1, C3 and C4 as candidates to explain C2. The output from the run follows with comments.
————— 12/2/2003 9:15:15 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > Retrieve "C:\Berenson\Data_Files-9th\Minitab\petfood.MTW".
Retrieving worksheet from file: C:\Berenson\Data_Files-9th\Minitab\petfood.MTW
# Worksheet was saved on Mon Apr 27 1998
Comment: I downloaded the data from the text CD, but stored it where I could get it
more easily if I needed it again.
Results for: petfood.MTW
MTB > Save "C:\Documents and Settings\RBOVE.WCUPANET\My Documents\Drive
D\MINITAB\petfood3";
SUBC>
Replace.
Saving file as: C:\Documents and Settings\RBOVE.WCUPANET\My Documents\Drive
D\MINITAB\petfood3.MTW
* NOTE * Existing file replaced.
Results for: petfood3.MTW
252solnK1 11/26/07
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MTB > regress c2
1 c1
Regression Analysis: Sales versus Space
The regression equation is
Sales = 1.45 + 0.0740 Space
Predictor
Coef
SE Coef
T
P
Constant
1.4500
0.2178
6.66
0.000
Space
0.07400
0.01591
4.65
0.001
S = 0.3081
R-Sq = 68.4%
R-Sq(adj) = 65.2%
Analysis of Variance
Source
DF
Regression
1
Residual Error
10
Total
11
SS
2.0535
0.9490
3.0025
MS
2.0535
0.0949
F
21.64
P
0.001
Comment: This is the regression referred to in Problems 13.3 and 13.14.
MTB > let c4 = c1 * c3
Comment: This command creates the interaction variable, which I have labeled ‘Inter’ in C4.
MTB > print c1-c6
Data Display
Row
Space
Sales
Locatn
Inter
C5
C6
1
2
3
4
5
6
7
8
9
10
11
12
5
5
5
10
10
10
15
15
15
20
20
20
1.6
2.2
1.4
1.9
2.4
2.6
2.3
2.7
2.8
2.6
2.9
3.1
0
1
0
0
0
1
0
0
1
0
0
1
0
5
0
0
0
10
0
0
15
0
0
20
8
0
Comment: This is the data I will use. Because part c) of this problem asks for confidence and
prediction intervals I have set up values of space and sales for these intervals in C5 and C6.
MTB > Regress c2 2 c1 c3;
SUBC> GHistogram;
SUBC> GNormalplot;
SUBC> GFits;
SUBC> GVars c1 c3;
SUBC> RType 1;
SUBC>
Constant;
SUBC>
VIF;
SUBC>
Predict c5 c6;
SUBC>
Brief 3.
Regression Analysis: Sales versus Space, Locatn
The regression equation is
Sales = 1.30 + 0.0740 Space + 0.450 Locatn
Predictor
Coef
SE Coef
T
P
Constant
1.3000
0.1569
8.29
0.000
Space
0.07400
0.01101
6.72
0.000
Locatn
0.4500
0.1305
3.45
0.007
S = 0.2132
R-Sq = 86.4%
R-Sq(adj) = 83.4%
Analysis of Variance
Source
DF
Regression
2
Residual Error
9
Total
11
SS
2.5935
0.4090
3.0025
MS
1.2967
0.0454
F
28.53
VIF
1.0
1.0
P
0.000
252solnK1 11/26/07
Source
Space
Locatn
(Open this document in 'Page Layout' view!)
DF
1
1
Seq SS
2.0535
0.5400
Comment: These results look great. Note that all my coefficients are significant, with p-values
below 1%. The VIF is way below 5, which indicates a lack of collinearity. The ANOVA gives me a p-value
of zero, indicating that the regression is quite useful.
Obs
1
2
3
4
5
6
7
8
9
10
11
12
Space
5.0
5.0
5.0
10.0
10.0
10.0
15.0
15.0
15.0
20.0
20.0
20.0
Sales
1.6000
2.2000
1.4000
1.9000
2.4000
2.6000
2.3000
2.7000
2.8000
2.6000
2.9000
3.1000
Fit
1.6700
2.1200
1.6700
2.0400
2.0400
2.4900
2.4100
2.4100
2.8600
2.7800
2.7800
3.2300
SE Fit
0.1118
0.1348
0.1118
0.0802
0.0802
0.1101
0.0802
0.0802
0.1101
0.1118
0.1118
0.1348
Residual
-0.0700
0.0800
-0.2700
-0.1400
0.3600
0.1100
-0.1100
0.2900
-0.0600
-0.1800
0.1200
-0.1300
St Resid
-0.39
0.48
-1.49
-0.71
1.82
0.60
-0.56
1.47
-0.33
-0.99
0.66
-0.79
Predicted Values for New Observations
New Obs
1
Fit
1.8920
SE Fit
0.0902
95.0% CI
1.6880, 2.0960)
(
(
95.0% PI
1.3684, 2.4156)
Values of Predictors for New Observations
New Obs
1
Space
8.00
Locatn
0.000000
Residual Histogram for Sales
Normplot of Residuals for Sales
Residuals vs Fits for Sales
Histogram of the Residuals
(response is Sales)
5
Frequency
4
3
2
1
0
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
0.4
Residual
Comment: This graph doesn’t really look bad. Given the relatively small sample size, it
could be results from a symmetrical distribution.
252solnK1 11/26/07
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Normal Probability Plot of the Residuals
(response is Sales)
2
Normal Score
1
0
-1
-2
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
0.4
Residual
Comment: This seems to be fairly close to a straight line indicating that the residuals
have a distribution that is close to Normal.
252solnK1 11/26/07
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Comment: These look fairly random, given the fact that shelves seem to only come in
lengths that are multiples of 5 and that Location is a dummy variable.
MTB > Regress c2 3 c1 c3 c4;
SUBC> RType 1;
SUBC>
Constant;
SUBC>
VIF;
SUBC>
Brief 3.
Regression Analysis: Sales versus Space, Locatn, Inter
The regression equation is
Sales = 1.20 + 0.0820 Space + 0.750 Locatn - 0.0240 Inter
Predictor
Constant
Space
Locatn
Inter
Coef
1.2000
0.08200
0.7500
-0.02400
S = 0.2124
SE Coef
0.1840
0.01344
0.3186
0.02327
R-Sq = 88.0%
T
6.52
6.10
2.35
-1.03
P
0.000
0.000
0.046
0.333
VIF
1.5
6.0
6.5
R-Sq(adj) = 83.5%
Analysis of Variance
Source
Regression
Residual Error
Total
Source
Space
Locatn
Inter
DF
1
1
1
DF
3
8
11
SS
2.64150
0.36100
3.00250
MS
0.88050
0.04513
F
19.51
P
0.000
Seq SS
2.05350
0.54000
0.04800
Comment: Things don’t look so good in this regression. The VIFs for the last two variables are
high indicating collinearity; The p-value for the coefficient of interaction is very high, indicating that it’s
not significant at the 1%, 5% or 10% level.
252solnK1 11/26/07
Obs
1
2
3
4
5
6
7
8
9
10
11
12
Space
5.0
5.0
5.0
10.0
10.0
10.0
15.0
15.0
15.0
20.0
20.0
20.0
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Sales
1.6000
2.2000
1.4000
1.9000
2.4000
2.6000
2.3000
2.7000
2.8000
2.6000
2.9000
3.1000
Fit
1.6100
2.2400
1.6100
2.0200
2.0200
2.5300
2.4300
2.4300
2.8200
2.8400
2.8400
3.1100
SE Fit
0.1257
0.1777
0.1257
0.0823
0.0823
0.1164
0.0823
0.0823
0.1164
0.1257
0.1257
0.1777
Residual
-0.0100
-0.0400
-0.2100
-0.1200
0.3800
0.0700
-0.1300
0.2700
-0.0200
-0.2400
0.0600
-0.0100
St Resid
-0.06
-0.34
-1.23
-0.61
1.94
0.39
-0.66
1.38
-0.11
-1.40
0.35
-0.09
MTB > Stepwise c2 c1 c3 c4;
SUBC>
AEnter 0.15;
SUBC>
ARemove 0.15;
SUBC>
Constant.
Stepwise Regression: Sales versus Space, Locatn, Inter
Alpha-to-Enter: 0.15
Response is
Sales
Alpha-to-Remove: 0.15
on
3 predictors, with N =
Step
Constant
1
1.450
2
1.300
Space
T-Value
P-Value
0.074
4.65
0.001
0.074
6.72
0.000
Locatn
T-Value
P-Value
12
0.45
3.45
0.007
S
0.308
0.213
R-Sq
68.39
86.38
R-Sq(adj)
65.23
83.35
C-p
13.0
3.1
More? (Yes, No, Subcommand, or Help)
SUBC> yes
No variables entered or removed
More? (Yes, No, Subcommand, or Help)
SUBC> no
MTB >
Comment: The stepwise regression confirms the results above. Minitab decides that ‘Space’ is the
best single predictor and essentially redoes our first regression. It then brings in ‘Location’ and gets our
second regression. But when it is told to go ahead and add a third independent variable, it doesn’t find
enough explanatory power in ‘Inter’ to make it worth adding. Note that C-p is close to k + 1 in the second
regression, indicating good results.
The following material is an edited version of the solution in the Instructor’s Solution Manual.
14.35
(a)
(b)
(c)
Yˆ  1.30  0.074X 1  0.45X 2 , where X1 = shelf space and X2 = aisle location.
Holding constant the effect of aisle location, for each additional foot of shelf space, sales
are expected to increase on average by 0.074 hundreds of dollars, or $7.40. For a given
amount of shelf space, a front-of-aisle location is expected to increase sales on average
by 0.45 hundreds of dollars, or $45.
Yˆ  1.30  0.074(8)  0.45(0)  1.892 or $189.20
252solnK1 11/26/07
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These intervals can only be done using Minitab or another regression program and appear
in the printout.
1.6880  Y |X  X i  2.0960
1.3684  YX  X i  2.4156
(d)
(e)
(f)
Based on a residual analysis, the model appears adequate.
2,9   4.26 . Reject H and say that there is evidence of a relationship
F  28 .53  F.05
0
between sales and the two dependent variables.
9
For X1: t  6.72  t .025
 2.262 . Reject H0 and say that Shelf space makes a significant
contribution and should be included in the model.
9
For X2: t  3.45  t .025
 2.262 . Reject H0 and say that aisle location makes a significant
contribution and should be included in the model.
Both variables should be kept in the model.
(g)
Remember that our results were
The regression equation is
Sales = 1.30 + 0.0740 Space + 0.450 Locatn
Predictor
Constant
Space
Locatn
Coef
1.3000
0.07400
0.4500
SE Coef
0.1569
0.01101
0.1305
T
8.29
6.72
3.45
P
0.000
0.000
0.007
VIF
1.0
1.0
9
Using the coefficient of space, b1  0.074, t nk 1  t.025
 2.262 , and sb1  0.01101 ,
2
you should be able to get 0.049  1  0.099 . (Is this right? I haven’t checked so tell
(h)
(i)
(j)
(k)
(l)
(m)
(n)
me.). By the same method, you should be able to get 0.155   2  0.745 for location.
The slope here takes into account the effect of the other predictor variable,
placement, while the solution for Problem 13.3 did not.
rY2.12  0.864 . So, 86.4% of the variation in sales can be explained by variation in shelf
space and variation in aisle location.
2
radj
 0.834
rY2.12  0.864 while about six pages back, for the simple regression, rY2.1  0.684 . The
inclusion of the aisle-location variable has resulted in the increase.
I’m a little bit too lazy to compute these things the way our author wants you to do it.
There’s a formula at the end of 252corr that gets you these results much faster.
 6.72 2  45 .1584
t2

rY21.2  2 2

 .8338 . Holding constant the effect of aisle
t 2  df  6.72 2  9  54 .1584
location, 83.4% of the variation in sales can be explained by variation in shelf space.
 3.45 2  11 .9025
t2

rY22.1  2 1

 .5694 rY22.1  0.569 . Holding constant the
2


20
.
9025
t1  df  3.45   9 
effect of shelf space, 56.9% of the variation in sales can be explained by variation in aisle
location.
The slope of sales with shelf space is the same regardless of whether the aisle
location is front or back.
From the last regression in the printout,
Yˆ  1.20  0.082 X 1  0.75 X 2  0.024 X 1 X 2 . Do not reject H0. There is not
(o)
evidence that the interaction term makes a contribution to the model.
The two-variable model in (a) should be used.
252solnK1 11/26/07
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Nonlinear regression
Exercise 15.1: The equation given is Yˆ  5  3X 1  1.5 X 12 . n  25 . a) Predict Y for X1 = 2. b) Assume
that the t for the x-squared term is 2.35 and   .05 , does this mean that the quadratic model is an
improvement over the linear model? c) What if the same t is 1.17? d) Assume that the coefficient of x is -3,
predict Y for X1 = 2.
Solution:
(a)
Yˆ  5  3X  1.5 X 2  5  3(2)  1.5(2 2 )  17
(b)
df  n  k  1  25  2  1  22 . t  2.35  t 22  2.0739 with 22 degrees of freedom.
Reject H0. The quadratic term is significant.
t  1.17  t 22  2.0739 with 22 degrees of freedom. Do not reject H0. The quadratic term
(c)
is not significant.
(d)
Yˆ  5  3X  1.5 X 2  5  3(2)  1.5(2 2 )  5
Exercise 15.6(15.13 in 8th edition): The given equation is ln Yˆ  3.07  0.9 X 1  1.41 ln X 2 . a) Predict Y
for x1= 8.5 and x2 = 5.2. b) Interpret the meaning of the slopes.
Solution:
(a)
ln Yˆ  3.07  0.9 ln(8.5)  1.41ln(5.2)  7.32067
Yˆ  e 7.32067  1511 .22
(b)
Holding constant the effects of X2, for each additional unit of the natural logarithm of X1
the natural logarithm of Y is expected to increase on average by 0.9. Holding constant the
effects of X1, for each additional unit of the natural logarithm of X2 the natural logarithm
of Y is expected to increase on average by 1.41.
Exercise 15.7(15.14 in 8th edition): The given equation is ln Yˆ  4.62  0.5 X 10.7 X 2 (5.2)  12.51 . a)
Predict Y for x1= 8.5 and x2 = 5.2. b) Interpret the meaning of the slopes.
Solution:
Yˆ  e12.51  271,034 .12
(a)
ln Yˆ  4.62  0.5(8.5)  0.7(5.2)  12.51
(b)
Holding constant the effects of X2, for each additional unit of X1 the natural logarithm of
Y is expected to increase on average by 0.5. Holding constant the effects of X1, for each
additional unit of X2 the natural logarithm of Y is expected to increase on average by 0.7.
Note: To deal with the VIF, the following material has been added to 252corr together with an example.
A relatively recent method to check for collinearity is to use the Variance Inflation Factor
1
. Here R 2j is the coefficient of multiple correlation gotten by regressing
VIF j 
2
1 R j
the independent variable X j against all the other independent variables  Xs  . The rule
of thumb seems to be that we should be suspicious if any VIFj  5 and positively
horrified if VIFj  10 . If you get results like this, drop a variable or change your model.
Note that, if you use a correlation matrix for your independent variables and see a large
correlation between two of them, putting the square of that correlation into the VIF
formula gives you a low estimate of the VIF, since the R-squared that you get from a
regression against all the independent variables will be higher.