252y0541 5/7/05
Introduction
What is a significant difference or a statistical test? Many of you seem to have no idea what a statistical
test is. We have been doing them every day. For example, look at question 5b) in Part II.
b. Of the campaigns that took 0 – 2 months 7 were ineffective. Of the campaigns that took more
than two months, 8 were ineffective. Is the fraction that were ineffective in the first category below
the fraction in the second category? (5)
Any answer to this question should show that you are aware of the warning in the beginning of Part II.
Show your work! State H 0 and H1 where applicable. Use a significance level of 5% unless
noted otherwise. Do not answer questions without citing appropriate statistical tests – That is,
explain your hypotheses and what values from what table were used to test them.
In other words, if all you do here is compute some proportions and tell me that they are different, you are
wasting both our time. A statistical test is required. If you don’t know what I mean by a (significant)
difference between two fractions or proportions either 1) quit right now or 2) review the course material
including the answer to 5b) until you do.
I haven’t looked at this exam since it was given and I may have missed some corrections.
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252y0541 5/7/05
ECO252 QBA2
Final EXAM
May 4, 2005
Name and Class hour:____KEY________________
I. (25+ points) Do all the following. Note that answers without reasons receive no credit. Most answers
require a statistical test, that is, stating or implying a hypothesis and showing why it is true or false by citing
a table value or a p-value. If you haven’t done it lately, take a fast look at ECO 252 - Things That You
Should Never Do on a Statistics Exam (or Anywhere Else)
The next 12 pages contain computer output. This comes from a data set on the text CD-ROM called
Auto2002. There are 121 observations. The dependent variable is MPG (miles per gallon). The columns in
the data set are:
Name
The make and model
SUV
‘Yes’ if it’s an SUV, ‘No’ if not.
Drive Type
All wheel, front wheel, rear wheel or four wheel.
Horsepower
An independent variable
Fuel Type
Premium or regular
MPG
The dependent variable
Length
In inches – an independent variable
Width
In inches – an independent variable
Weight
In pounds – an independent variable
Cargo Volume Square feet – an independent variable
Turning Circle Feet – an independent variable.
I added the following
SUV_D
A dummy variable based on ‘SUV’, 1 for an SUV, otherwise zero.
Fuel_D
A dummy variable based on ‘Fuel Type’, 1 for a Premium fuel., otherwise zero
SUVwt
An interaction variable, the product of ‘SUV_D’ and ‘Weight’
SUVtc
An interaction variable, the product of ‘SUV_D’ and ‘Turning Circle’
HPsq
Horsepower Squared.
AWD_D
A dummy variable based on ‘Drive Type’, 1 for all wheel drive, otherwise zero
FWD_D
A dummy variable based on ‘Drive Type’, 1 for front wheel drive, otherwise zero
RWD_D
A dummy variable based on ‘Drive Type’, 1 for rear wheel drive, otherwise zero
SUV_L
An interaction variable, the product of ‘SUV_D’ and ‘Length’
Questions are included with the regressions and thus cannot be in order of difficulty. It’s probably a good
idea to look over the questions and explanations before you do anything.
————— 4/28/2005 6:18:32 PM ————————————————————
Welcome to Minitab, press F1 for help.
Results for: 252x0504-4.MTW
MTB > Stepwise 'MPG' 'Horsepower' 'Length' 'Width' 'Weight' 'Cargo Volume' &
CONT>
'Turning Circle' 'SUV_D' 'Fuel_D' 'SUVwt' 'HPsq' 'AWD_D' &
CONT>
'FWD_D' 'RWD_D' 'SUV_L';
SUBC>
AEnter 0.15;
SUBC>
ARemove 0.15;
SUBC>
Best 0;
SUBC>
Constant.
Because I had relatively little idea of what to do, I ran a stepwise regression. You probably have
not seen one of these before, but they are relatively easy to read. Note that it dropped 2 observations so that
the results will not be quite the same as I got later.
The first numbered column represents the single independent variable that seems to have the most
explanatory effect on MPG, The equation reads MPG = 38.31 – 15.34 Weight The fact that Weight
2
252y0541 5/7/05
entered first with a negative coefficient should surprise no one. At the bottom appears s e , R 2 , R 2 and the
C p statistic mentioned in your text. The value of the t-ratio and its p-value appear below the coefficient.
Stepwise Regression: MPG versus Horsepower, Length, ...
Alpha-to-Enter: 0.15 Alpha-to-Remove: 0.15
Response is MPG on 14 predictors, with N = 119
N(cases with missing observations) = 2 N(all cases) = 121
Step
Constant
Weight
T-Value
P-Value
1
38.31
2
36.75
3
41.59
4
50.06
5
50.15
6
59.00
-0.00491
-15.34
0.000
-0.00436
-11.87
0.000
-0.00578
-12.82
0.000
-0.00495
-9.31
0.000
-0.00424
-6.74
0.000
-0.00339
-5.61
0.000
-1.72
-2.84
0.005
-33.71
-4.99
0.000
-35.29
-5.36
0.000
-35.12
-5.40
0.000
-18.68
-2.71
0.008
0.180
4.75
0.000
0.185
5.04
0.000
0.182
5.01
0.000
0.088
2.26
0.026
-0.285
-2.79
0.006
-0.292
-2.90
0.004
-0.255
-2.75
0.007
-0.0124
-2.01
0.046
-0.1619
-5.04
0.000
SUV_D
T-Value
P-Value
SUV_L
T-Value
P-Value
Turning Circle
T-Value
P-Value
Horsepower
T-Value
P-Value
HPsq
T-Value
P-Value
S
R-Sq
R-Sq(adj)
Mallows C-p
0.00040
4.73
0.000
2.50
66.78
66.50
71.5
2.43
68.94
68.40
61.4
2.23
74.04
73.36
34.8
2.17
75.70
74.85
27.4
2.14
76.55
75.51
24.7
1.96
80.45
79.41
4.8
More? (Yes, No, Subcommand, or Help)
SUBC> y
I’m greedy, so while I was surprised that Minitab had found six explanatory (independent) variables that
actually seemed to affect miles per gallon I wanted more. For the first time ever (for me), Minitab found
another variable
3
252y0541 5/7/05
Step
Constant
7
58.50
Weight
T-Value
P-Value
-0.00342
-5.74
0.000
SUV_D
T-Value
P-Value
-19.0
-2.79
0.006
SUV_L
T-Value
P-Value
0.090
2.36
0.020
Turning Circle
T-Value
P-Value
-0.210
-2.24
0.027
Horsepower
T-Value
P-Value
-0.175
-5.43
0.000
HPsq
T-Value
P-Value
0.00042
5.03
0.000
Fuel_D
T-Value
P-Value
0.92
2.11
0.037
S
R-Sq
R-Sq(adj)
Mallows C-p
1.93
81.21
80.02
2.5
More? (Yes, No, Subcommand, or Help)
SUBC> y
No variables entered or removed
More? (Yes, No, Subcommand, or Help)
SUBC> n
Because I was worried about Collinearity, I had the computer do a table of correlations between all the
independent variables. The table is triangular since the correlation between, say, Length and Horsepower is
going to be the same as the correlation between Horsepower and Length. So, for example, the correlation
between Horsepower and Length is .648 and the p-value of zero below it evaluates the null hypothesis that
the correlation is insignificant. The explanation of Predicted R2 that appears below the correlation table
was a new one on me, but could help you in comparing the regressions.
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252y0541 5/7/05
MTB > Correlation 'Horsepower' 'Length' 'Width' 'Weight' 'Cargo Volume' &
CONT>
'Turning Circle' 'SUV_D' 'Fuel_D' 'SUVwt' 'SUVtc' 'HPsq' 'AWD_D' &
CONT>
'FWD_D' 'RWD_D' 'SUV_L'.
Correlations: Horsepower, Length, Width, Weight, Cargo Volume, ...
Horsepower
0.648
0.000
Length
Width
0.660
0.000
0.825
0.000
Weight
0.673
0.000
0.634
0.000
0.780
0.000
Cargo Volume
0.296
0.001
0.395
0.000
0.546
0.000
0.716
0.000
Turning Circ
0.497
0.000
0.750
0.000
0.658
0.000
0.650
0.000
SUV_D
0.160
0.080
-0.102
0.265
0.180
0.049
0.535
0.000
Fuel_D
0.321
0.000
-0.013
0.886
-0.042
0.645
0.057
0.540
SUVwt
0.182
0.045
-0.077
0.403
0.206
0.023
0.562
0.000
SUVtc
0.185
0.042
-0.062
0.502
0.211
0.020
0.577
0.000
HPsq
0.989
0.000
0.632
0.000
0.645
0.000
0.668
0.000
AWD_D
0.059
0.523
-0.118
0.199
-0.037
0.691
0.065
0.483
FWD_D
-0.370
0.000
-0.001
0.994
-0.163
0.076
-0.453
0.000
RWD_D
0.334
0.000
0.070
0.445
0.151
0.101
0.351
0.000
SUV_L
0.197
0.030
-0.053
0.564
0.219
0.016
0.582
0.000
Cargo Volume
0.486
0.000
Turning Circ
SUV_D
Fuel_D
0.459
0.000
0.139
0.127
-0.245
0.007
-0.069
0.456
-0.147
0.110
SUVwt
0.473
0.000
0.161
0.078
0.999
0.000
-0.141
0.125
SUVtc
0.484
0.000
0.196
0.031
0.996
0.000
-0.142
0.121
Length
Turning Circ
SUV_D
Fuel_D
Width
Weight
5
252y0541 5/7/05
HPsq
0.289
0.001
0.480
0.000
0.173
0.058
0.296
0.001
AWD_D
0.021
0.823
-0.068
0.461
0.185
0.043
0.218
0.017
FWD_D
-0.165
0.071
-0.027
0.771
-0.517
0.000
-0.280
0.002
RWD_D
0.108
0.239
0.015
0.874
0.364
0.000
0.098
0.288
SUV_L
0.487
0.000
0.181
0.047
0.996
0.000
-0.145
0.114
SUVwt
0.998
0.000
SUVtc
HPsq
AWD_D
HPsq
0.198
0.030
0.200
0.028
AWD_D
0.184
0.044
0.174
0.057
0.040
0.667
FWD_D
-0.522
0.000
-0.526
0.000
-0.369
0.000
-0.366
0.000
RWD_D
0.367
0.000
0.374
0.000
0.347
0.000
-0.137
0.135
SUV_L
0.999
0.000
0.998
0.000
0.215
0.018
0.176
0.054
FWD_D
-0.810
0.000
RWD_D
-0.529
0.000
0.381
0.000
SUVtc
RWD_D
SUV_L
Cell Contents: Pearson correlation
P-Value
PRESS
Assesses your model's predictive ability. In general, the smaller the prediction sum of squares
(PRESS) value, the better the model's predictive ability. PRESS is used to calculate the predicted R2. PRESS, similar
to the error sum of squares (SSE), is the sum of squares of the prediction error. PRESS differs from SSE in that each
fitted value, i, for PRESS is obtained by deleting the ith observation from the data set, estimating the regression
equation from the remaining n - 1 observations, then using the fitted regression function to obtain the predicted value
for the ith observation.
Predicted R2
Similar to R2. Predicted R2 indicates how well the model predicts responses for new observations,
2
whereas R indicates how well the model fits your data. Predicted R2 can prevent overfitting the model and is more
useful than adjusted R2 for comparing models because it is calculated with observations not included in model
calculation. Predicted R2 is between 0 and 1 and is calculated from the PRESS statistic. Larger values of predicted R 2
suggest models of greater predictive ability.
6
252y0541 5/7/05
So now it’s time to get serious. My first regression was based on what I had learned from the
stepwise regression. The only one of the variables that I left out from the stepwise regression was FUEL_D.
1. Look at the results of Regression 1. But don’t forget what has gone before.
a. What does the Analysis of variance show us? Why? (1)
b. What suggests that the relation of MPG to one of the variables is nonlinear? (1)
c. What does the equation suggest that the difference is between an extra inch on an SUV
and a non_SUV? (1)
d. Why did I leave out FUEL_D (2)
e. Which coefficients are not significant? Why? (2)
f. What do the values of the VIFs tell us? (2)
MTB > Regress 'MPG' 6 'Weight' 'SUV_D' 'SUV_L' 'Turning Circle'
CONT>
'Horsepower' 'HPsq';
SUBC>
Constant;
SUBC>
Brief 2.
&
MTB > Regress 'MPG' 6 'Weight' 'SUV_D' 'SUV_L' 'Turning Circle'
CONT>
'Horsepower' 'HPsq';
SUBC>
GNormalplot;
SUBC>
NoDGraphs;
SUBC>
RType 1;
SUBC>
Constant;
SUBC>
VIF;
SUBC>
Press;
SUBC>
Brief 2.
&
Regression Analysis: MPG versus Weight, SUV_D, ... (Regression 1)
The regression equation is
MPG = 63.1 - 0.00303 Weight - 14.8 SUV_D + 0.0653 SUV_L - 0.264 Turning Circle
- 0.213 Horsepower + 0.000522 HPsq
Predictor
Constant
Weight
SUV_D
SUV_L
Turning Circle
Horsepower
HPsq
Coef
63.105
-0.0030345
-14.812
0.06527
-0.2639
-0.21251
0.00052249
SE Coef
3.978
0.0006859
7.957
0.04478
0.1050
0.03575
0.00009459
T
15.86
-4.42
-1.86
1.46
-2.51
-5.94
5.52
P
0.000
0.000
0.065
0.148
0.013
0.000
0.000
VIF
5.6
282.1
307.9
2.0
63.5
61.3
S = 2.27485
R-Sq = 77.5%
R-Sq(adj) = 76.4%
PRESS = 752.906
R-Sq(pred) = 71.34%
Analysis of Variance
Source
DF
SS
Regression
6 2037.34
Residual Error 114
589.95
Total
120 2627.29
Source
Weight
SUV_D
SUV_L
Turning Circle
Horsepower
HPsq
DF
1
1
1
1
1
1
MS
339.56
5.17
F
65.62
P
0.000
Seq SS
1605.19
47.29
132.83
52.31
41.83
157.89
Unusual Observations
Obs Weight
MPG
Fit
16
5590 13.000 15.361
34
7270 10.000
6.856
40
5590 13.000 15.361
62
4065 19.000 14.633
SE Fit
1.137
1.461
1.137
0.654
Residual
-2.361
3.144
-2.361
4.367
St Resid
-1.20 X
1.80 X
-1.20 X
2.00R
7
252y0541 5/7/05
108
111
114
115
2150
2750
2935
2940
38.000
41.000
41.000
24.000
30.489
33.473
29.806
29.791
0.632
1.133
0.777
0.778
7.511
7.527
11.194
-5.791
3.44R
3.82RX
5.24R
-2.71R
R denotes an observation with a large standardized residual.
X denotes an observation whose X value gives it large influence.
2.
Look at the results of Regression 2. But don’t forget what has gone before.
a. What variable did I drop? Why? (2)
b. Are there any coefficients that have a sign that you would not expect? Why? (1)
c. A Chevrolet Suburban is an SUV with rear wheel drive and 285 horsepower, that takes
Regular fuel, has a length of 219 inches, a width of 79 inches, a weight of 5590 pounds, a
cargo volume of 77.0 square feet and a turning circle of 46 Feet (!!! Maybe it was
inches?). What miles per gallon does the equation predict? What would it be if the vehicle
was not classified as an SUV? (3)
d. Why do I like this regression better than the previous one? (2)
[17]
MTB > Regress 'MPG' 5 'Weight' 'SUV_D'
CONT>
'HPsq';
SUBC>
GNormalplot;
SUBC>
NoDGraphs;
SUBC>
RType 1;
SUBC>
Constant;
SUBC>
VIF;
SUBC>
Press;
SUBC>
Brief 2.
'Turning Circle' 'Horsepower'
Regression Analysis: MPG versus Weight, SUV_D, ...
&
(Regression 2)
The regression equation is
MPG = 63.1 - 0.00250 Weight - 3.25 SUV_D - 0.250 Turning Circle
- 0.239 Horsepower + 0.000593 HPsq
Predictor
Constant
Weight
SUV_D
Turning Circle
Horsepower
HPsq
Coef
63.137
-0.0025020
-3.2492
-0.2501
-0.23928
0.00059313
SE Coef
3.998
0.0005834
0.6272
0.1051
0.03082
0.00008163
T
15.79
-4.29
-5.18
-2.38
-7.76
7.27
P
0.000
0.000
0.000
0.019
0.000
0.000
VIF
4.0
1.7
1.9
46.7
45.2
S = 2.28595
R-Sq = 77.1%
R-Sq(adj) = 76.1%
PRESS = 744.047
R-Sq(pred) = 71.68%
Analysis of Variance
Source
DF
SS
Regression
5 2026.35
Residual Error 115
600.94
Total
120 2627.29
Source
Weight
SUV_D
Turning Circle
Horsepower
HPsq
DF
1
1
1
1
1
MS
405.27
5.23
F
77.56
P
0.000
Seq SS
1605.19
47.29
46.32
51.65
275.90
Unusual Observations
Obs
16
34
40
108
Weight
5590
7270
5590
2150
MPG
13.000
10.000
13.000
38.000
Fit
14.381
5.945
14.381
30.081
SE Fit
0.921
1.328
0.921
0.570
Residual
-1.381
4.055
-1.381
7.919
St Resid
-0.66 X
2.18RX
-0.66 X
3.58R
8
252y0541 5/7/05
111
114
115
2750
2935
2940
41.000
41.000
24.000
33.910
30.060
30.047
1.098
0.761
0.762
7.090
10.940
-6.047
3.54RX
5.08R
-2.81R
R denotes an observation with a large standardized residual.
X denotes an observation whose X value gives it large influence.
Because I wanted to look at the effect of the three drive variables on MPG, I ran another stepwise
regression. The first part of this is identical to the last stepwise regression, but after the 6 th regression, I
forced out SUV_L and forced in AWD_D, FWD_D and RWD_D. Because I had to make the regressions
comparable, I threw an observation with an anomalous drive variable out and redid my two regressions as
Regressions 3 and 4. I then added in all the drive variables as a package in Regression 5.
MTB > Stepwise 'MPG' 'Horsepower' 'Length' 'Width' 'Weight' 'Cargo Volume' &
CONT>
'Turning Circle' 'SUV_D' 'Fuel_D' 'SUVwt' 'HPsq' 'AWD_D' &
CONT>
'FWD_D' 'RWD_D' 'SUV_L';
SUBC>
AEnter 0.15;
SUBC>
ARemove 0.15;
SUBC>
Best 0;
SUBC>
Constant.
Stepwise Regression: MPG versus Horsepower, Length, ...
Alpha-to-Enter: 0.15 Alpha-to-Remove: 0.15
Response is MPG on 14 predictors, with N = 119
N(cases with missing observations) = 2 N(all cases) = 121
Step
Constant
Weight
T-Value
P-Value
1
38.31
2
36.75
3
41.59
4
50.06
5
50.15
6
59.00
-0.00491
-15.34
0.000
-0.00436
-11.87
0.000
-0.00578
-12.82
0.000
-0.00495
-9.31
0.000
-0.00424
-6.74
0.000
-0.00339
-5.61
0.000
-1.72
-2.84
0.005
-33.71
-4.99
0.000
-35.29
-5.36
0.000
-35.12
-5.40
0.000
-18.68
-2.71
0.008
0.180
4.75
0.000
0.185
5.04
0.000
0.182
5.01
0.000
0.088
2.26
0.026
-0.285
-2.79
0.006
-0.292
-2.90
0.004
-0.255
-2.75
0.007
-0.0124
-2.01
0.046
-0.1619
-5.04
0.000
SUV_D
T-Value
P-Value
SUV_L
T-Value
P-Value
Turning Circle
T-Value
P-Value
Horsepower
T-Value
P-Value
HPsq
T-Value
P-Value
S
R-Sq
R-Sq(adj)
Mallows C-p
0.00040
4.73
0.000
2.50
66.78
66.50
71.5
2.43
68.94
68.40
61.4
2.23
74.04
73.36
34.8
2.17
75.70
74.85
27.4
2.14
76.55
75.51
24.7
1.96
80.45
79.41
4.8
More? (Yes, No, Subcommand, or Help)
SUBC> remove 'SUV_L'.
Step
7
8
9
9
252y0541 5/7/05
Constant
59.15
59.00
58.50
Weight
T-Value
P-Value
-0.00267
-5.10
0.000
-0.00339
-5.61
0.000
-0.00342
-5.74
0.000
SUV_D
T-Value
P-Value
-3.13
-5.51
0.000
-18.68
-2.71
0.008
-18.95
-2.79
0.006
0.088
2.26
0.026
0.090
2.36
0.020
SUV_L
T-Value
P-Value
Turning Circle
T-Value
P-Value
-0.236
-2.51
0.013
-0.255
-2.75
0.007
-0.210
-2.24
0.027
Horsepower
T-Value
P-Value
-0.199
-7.09
0.000
-0.162
-5.04
0.000
-0.175
-5.43
0.000
0.00050
6.75
0.000
0.00040
4.73
0.000
0.00042
5.03
0.000
HPsq
T-Value
P-Value
Fuel_D
T-Value
P-Value
S
R-Sq
R-Sq(adj)
Mallows C-p
0.92
2.11
0.037
2.00
79.56
78.66
7.8
1.96
80.45
79.41
4.8
1.93
81.21
80.02
2.5
More? (Yes, No, Subcommand, or Help)
SUBC> enter 'AWD_D' 'FWD_D' 'RWD_D'.
Step
Constant
10
60.14
11
59.11
12
58.50
13
58.50
Weight
T-Value
P-Value
-0.00355
-5.75
0.000
-0.00346
-5.72
0.000
-0.00344
-5.72
0.000
-0.00342
-5.74
0.000
SUV_D
T-Value
P-Value
-19.5
-2.82
0.006
-19.1
-2.77
0.007
-18.8
-2.74
0.007
-19.0
-2.79
0.006
SUV_L
T-Value
P-Value
0.092
2.37
0.020
0.090
2.32
0.022
0.089
2.30
0.023
0.090
2.36
0.020
Turning Circle
T-Value
P-Value
-0.207
-2.10
0.038
-0.205
-2.09
0.039
-0.202
-2.07
0.041
-0.210
-2.24
0.027
Horsepower
T-Value
P-Value
-0.175
-5.33
0.000
-0.177
-5.42
0.000
-0.176
-5.41
0.000
-0.175
-5.43
0.000
0.00042
4.98
0.000
0.00043
5.04
0.000
0.00042
5.02
0.000
0.00042
5.03
0.000
HPsq
T-Value
P-Value
10
252y0541 5/7/05
Fuel_D
T-Value
P-Value
0.73
1.49
0.139
0.80
1.66
0.099
0.87
1.92
0.057
AWD_D
T-Value
P-Value
-1.1
-0.76
0.451
FWD_D
T-Value
P-Value
-1.36
-0.98
0.331
-0.51
-0.62
0.535
-0.17
-0.32
0.752
RWD_D
T-Value
P-Value
-1.23
-0.93
0.353
-0.42
-0.55
0.586
S
R-Sq
R-Sq(adj)
Mallows C-p
1.95
81.37
79.65
7.6
1.95
81.27
79.73
6.1
0.92
2.11
0.037
1.94
81.22
79.86
4.4
1.93
81.21
80.02
2.5
More? (Yes, No, Subcommand, or Help)
SUBC> no
Results for: 252x0504-41.MTW
MTB > WSave "C:\Documents and Settings\rbove\My Documents\Minitab\252x050441.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0504-41.MTW'
MTB > erase c21
MTB > Regress 'MPG' 6 'Weight' 'SUV_D' 'SUV_L' 'Turning Circle' &
CONT>
'Horsepower' 'HPsq' ;
SUBC>
GNormalplot;
SUBC>
NoDGraphs;
SUBC>
RType 1;
SUBC>
Constant;
SUBC>
VIF;
SUBC>
Press;
SUBC>
Brief 2.
Regression Analysis: MPG versus Weight, SUV_D, ...
(Regression 3)
The regression equation is
MPG = 64.4 - 0.00284 Weight - 15.8 SUV_D + 0.0694 SUV_L - 0.305 Turning Circle
- 0.214 Horsepower + 0.000524 HPsq
Predictor
Constant
Weight
SUV_D
SUV_L
Turning Circle
Horsepower
HPsq
Coef
64.364
-0.0028431
-15.843
0.06943
-0.3045
-0.21444
0.00052386
SE Coef
3.973
0.0006832
7.867
0.04423
0.1055
0.03528
0.00009332
T
16.20
-4.16
-2.01
1.57
-2.89
-6.08
5.61
P
0.000
0.000
0.046
0.119
0.005
0.000
0.000
VIF
5.7
276.4
301.7
2.0
63.1
61.0
S = 2.24427
R-Sq = 78.3%
R-Sq(adj) = 77.2%
PRESS = 725.963
R-Sq(pred) = 72.34%
Analysis of Variance
Source
DF
SS
Regression
6 2055.21
Residual Error 113
569.15
Total
119 2624.37
Source
DF
MS
342.54
5.04
F
68.01
P
0.000
Seq SS
11
252y0541 5/7/05
Weight
SUV_D
SUV_L
Turning Circle
Horsepower
HPsq
1
1
1
1
1
1
Unusual Observations
Obs Weight
MPG
16
5590 13.000
34
7270 10.000
36
2715 24.000
40
5590 13.000
107
2150 38.000
110
2750 41.000
113
2935 41.000
114
2940 24.000
1602.61
49.58
135.39
61.04
47.88
158.71
Fit
15.259
6.907
28.432
15.259
30.543
33.747
30.000
29.985
SE Fit
1.123
1.442
0.493
1.123
0.624
1.126
0.772
0.774
Residual
-2.259
3.093
-4.432
-2.259
7.457
7.253
11.000
-5.985
St Resid
-1.16 X
1.80 X
-2.02R
-1.16 X
3.46R
3.74RX
5.22R
-2.84R
R denotes an observation with a large standardized residual.
X denotes an observation whose X value gives it large influence.
MTB > Regress 'MPG' 5 'Weight' 'SUV_D'
CONT>
'HPsq' ;
SUBC>
GNormalplot;
SUBC>
NoDGraphs;
SUBC>
RType 1;
SUBC>
Constant;
SUBC>
VIF;
SUBC>
Press;
SUBC>
Brief 2.
'Turning Circle' 'Horsepower'
Regression Analysis: MPG versus Weight, SUV_D, ...
&
(Regression 4)
The regression equation is
MPG = 64.4 - 0.00228 Weight - 3.53 SUV_D - 0.288 Turning Circle
- 0.243 Horsepower + 0.000599 HPsq
Predictor
Constant
Weight
SUV_D
Turning Circle
Horsepower
HPsq
Coef
64.352
-0.0022848
-3.5330
-0.2884
-0.24278
0.00059879
SE Coef
3.999
0.0005871
0.6366
0.1057
0.03051
0.00008071
T
16.09
-3.89
-5.55
-2.73
-7.96
7.42
P
0.000
0.000
0.000
0.007
0.000
0.000
VIF
4.2
1.8
2.0
46.6
45.0
S = 2.25865
R-Sq = 77.8%
R-Sq(adj) = 76.9%
PRESS = 720.507
R-Sq(pred) = 72.55%
Analysis of Variance
Source
DF
SS
Regression
5 2042.80
Residual Error 114
581.57
Total
119 2624.37
Source
Weight
SUV_D
Turning Circle
Horsepower
HPsq
DF
1
1
1
1
1
MS
408.56
5.10
F
80.09
P
0.000
Seq SS
1602.61
49.58
52.45
57.33
280.82
Unusual Observations
Obs Weight
MPG
Fit
16
5590 13.000 14.223
34
7270 10.000
5.938
40
5590 13.000 14.223
107
2150 38.000 30.108
SE Fit
0.914
1.312
0.914
0.563
Residual
-1.223
4.062
-1.223
7.892
St Resid
-0.59 X
2.21RX
-0.59 X
3.61R
12
252y0541 5/7/05
110
113
114
2750
2935
2940
41.000
41.000
24.000
34.201
30.262
30.251
1.095
0.759
0.760
6.799
10.738
-6.251
3.44RX
5.05R
-2.94R
R denotes an observation with a large standardized residual.
X denotes an observation whose X value gives it large influence.
Look at the results of Regression 5 and Regression 4. But don’t forget what has gone before.
a. Do an F test to see if Regression 5 is better than Regression 4. If you can include the
results from my forcing variables in the last stepwise regression. (6)
b. Should I have included another dummy variable to represent 4-wheel drive? Why? (2)
c. Are there any coefficients in Regression 5 that have a sign that you would not expect?
Why? (1)
d. A Chevrolet Suburban is an SUV with rear wheel drive and 285 horsepower, that takes
Regular fuel, has a length of 219 inches, a width of 79 inches, a weight of 5590 pounds, a
cargo volume of 77.0 square feet and a turning circle of 46 Feet (!!! Maybe it was
inches?). How do the predictions for MPG in Equations 2 and 4 differ in percentage
terms? (3)
[29]
Why do I like this regression better than the pr
3.
MTB > Regress 'MPG' 8 'Weight' 'SUV_D' 'Turning Circle' 'Horsepower'
CONT>
'HPsq' 'AWD_D' 'FWD_D' 'RWD_D';
SUBC>
GNormalplot;
SUBC>
NoDGraphs;
SUBC>
RType 1;
SUBC>
Constant;
SUBC>
VIF;
SUBC>
Press;
SUBC>
Brief 2.
Regression Analysis: MPG versus Weight, SUV_D, ...
&
(Regression 5)
The regression equation is
MPG = 66.4 - 0.00248 Weight - 3.83 SUV_D - 0.254 Turning Circle
- 0.251 Horsepower + 0.000618 HPsq - 1.21 AWD_D - 2.10 FWD_D - 1.70 RWD_D
Predictor
Constant
Weight
SUV_D
Turning Circle
Horsepower
HPsq
AWD_D
FWD_D
RWD_D
Coef
66.435
-0.0024795
-3.8302
-0.2541
-0.25082
0.00061833
-1.213
-2.103
-1.697
SE Coef
4.400
0.0006077
0.6814
0.1116
0.03122
0.00008244
1.620
1.490
1.434
T
15.10
-4.08
-5.62
-2.28
-8.03
7.50
-0.75
-1.41
-1.18
P
0.000
0.000
0.000
0.025
0.000
0.000
0.455
0.161
0.239
VIF
4.4
2.0
2.2
48.6
46.7
3.4
11.2
8.6
S = 2.26416
R-Sq = 78.3%
R-Sq(adj) = 76.8%
PRESS = 727.840
R-Sq(pred) = 72.27%
Analysis of Variance
Source
DF
SS
Regression
8 2055.33
Residual Error 111
569.03
Total
119 2624.37
Source
Weight
SUV_D
Turning Circle
Horsepower
HPsq
AWD_D
DF
1
1
1
1
1
1
MS
256.92
5.13
F
50.12
P
0.000
Seq SS
1602.61
49.58
52.45
57.33
280.82
2.00
13
252y0541 5/7/05
FWD_D
RWD_D
1
1
3.36
7.17
Unusual Observations
Obs
34
57
72
107
109
110
113
114
Weight
7270
4735
4720
2150
5435
2750
2935
2940
MPG
10.000
14.000
15.000
38.000
14.000
41.000
41.000
24.000
Fit
5.609
13.622
15.901
30.231
13.477
34.346
30.341
30.329
SE Fit
1.377
1.447
1.374
0.574
1.338
1.106
0.765
0.766
Residual
4.391
0.378
-0.901
7.769
0.523
6.654
10.659
-6.329
St Resid
2.44RX
0.22 X
-0.50 X
3.55R
0.29 X
3.37RX
5.00R
-2.97R
R denotes an observation with a large standardized residual.
X denotes an observation whose X value gives it large influence.
14
252y0541 5/7/05
II. Do at least 4 of the following 6 Problems (at least 15 each) (or do sections adding to at least 60 points –
(Anything extra you do helps, and grades wrap around) . Show your work! State H 0 and H1 where
applicable. Use a significance level of 5% unless noted otherwise. Do not answer questions without
citing appropriate statistical tests – That is, explain your hypotheses and what values from what table
were used to test them. Clearly label what section of each problem you are doing! The entire test has
175 points, but 100 is considered a perfect score.
Exhibit 1. A tear-off copy of this exhibit appears at the end of the exam.
An entrepreneur believes that her business is growing steadily and wants to compute a trend line for her
output Y against time x1  T . She also decides to repeat the regression after adding x 2  T 2 as a second
independent variable. Her data and results follow. The t statistics have been relabeled ‘t-ratio’ to prevent
confusion with T .
Regression Analysis: Y versus T
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Y
53.43
59.09
59.58
64.75
68.65
65.53
68.44
70.93
72.85
73.60
72.93
75.14
73.88
76.55
79.05
T
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
T2
1
4
9
16
25
36
49
64
81
100
121
144
169
196
225
The regression equation is
Y = 56.7 + 1.54 T
Predictor
Coef SE Coef
Constant
56.659
1.283
T
1.5377
0.1411
S = 2.36169
R-Sq = 90.1%
t-ratio
P
44.15 0.000
10.89 0.000
R-Sq(adj) = 89.4%
Regression Analysis: Y versus T, TSQ
The regression equation is
Y = 52.4 + 3.04 T - 0.0939 TSQ
Predictor
Coef SE Coef
Constant
52.401
1.545
T
3.0405
0.4444
TSQ
-0.09392 0.02701
S = 1.73483
R-Sq = 95.1%
t-ratio P
33.91 0.000
6.84 0.000
-3.48 0.005
If you need them, her means and spare parts are below.
Y  68.9600
X 22  nX 22  SSX 2  75805.3
X 1  8.00
Y 2  nY 2  SST  SSY  734.556
X 2  82.6667
X 1Y  nX 1Y  SX 1Y  430.550
X 12  nX 12  SSX1  280
X 2 Y  nX 2 Y  SX 2Y  6501.33





X
1X 2
 nX 1 X 2  SX 1X 2  4480.00
15
252x0541 4/22/05
1. Do the following using Exhibit 1.
a) Explain what numbers in the printout were used to compute the t-ratio 6.84, what table value
you would compare it with to do a 2-sided 1% significance test and whether and why the
coefficient is significant. (3)
b) The entrepreneur looked at the residual analysis of the first regression and decided that she
needs a time squared term. What is she likely to have seen to cause her to make that decision? (1)
c) Use the values of R 2 to do an F test to see if the addition of the T 2 term makes a significant
improvement in the regression. (4)
d) Get (Compute) R 2 ( R 2 adjusted for degrees of freedom) for the second regression and explain
what it seems to show. (2)
e) In the first regression, the Durbin-Watson statistic was 1.07 and for the second it was 1.94.
What do these numbers indicate? (Do a significance test.) (5)
f) For the second regression, make a prediction of the output in the 16 th year and use the suggestion
in the outline to make it into a prediction interval. Why would a confidence interval be
inappropriate here? (3)
g) (Extra Credit) Find the partial correlation, rY 1.2 . (2)
[15]
H 0 :  i   i 0
b   i0
b  0 3.0405
a) n  15 and   .01 .To test 
use t i  i
. So t1  1

 6.84 .
H
:



s
s b1i
0.4444
i0
 1 i
bi
Compare this with t nk 1  t 12  3.055 . Since the computed t-ratio is larger than the table value, reject

.005
2
H 0 : 1  0 and conclude that the coefficient is significant.
b) A curvature about the x- axis.
c) With only one independent variable R12  90.1% and after the T 2 was added R22  95.1% The
difference between the two is 5.0%. The F test show in the outline was to compute
2
2
n  k  r  1  Rk  r  Rk 
F r ,n k r 1 

 Here k is the number of independent variables in the first
2
r
 1  Rk  r 
regression and r is the number of independent variables added. This was discussed in Problem J1 and the
following fake ANOVA was suggested. The second computed F is larger than table F and so our null
hypothesis that the T 2 term has no effect is rejected. This test is identical to the t test on the T 2 term and
1,12  9.33 is
we can note that 12.42 is approximately equal to the square of the t-ratio (-3.48) and that F.01
12
 3.055 .
the square of t .005
Source
SS
1X
1 more X
Error
Total
DF
MS
F
90.1
1
90.1
220.65s
5.0
1
5.0
12.42s
4.9
100
12
14
F.05
1,12  9.33
F.01
1,12  9.33
F.01
0.40833
n  1R 2  k
15  1.951  2  .943 . For the previous regression R-Sq(adj) = 89.4% . This is a

15  2  1
n  k 1
preliminary indicator that the explanatory power of the regression rose when we added a new variable.
d) Rk2 
16
252x0541 4/22/05
e) For the first regression DW was 1.07 and for the second it was 1.94. n  15 . k  1 in the first regression
and k  2 in the second. The table in the text says that for k  1 and   .01 , d L  0.81 and d U  1.07
and that for   .05 , d L  1.08 and d U  1.36 . For k  2 and   .01 , d L  0.70 and d U  1.25
and that for   .05 , d L  0.95 and d U  1.54 . The line in the notes is shown below.
0
+
 0
dL
+
?
 0
dU
+
2
+
 0
4  dU
+
?
If we slip the 1% values in, we get the following for k  1
0   0 0.81
?
1.07   0 2   0 2.93
+
+
+
+
+
For the 5% values and k  1
0   0 1.08
?
1.36
+
+
+
 0
For the 1% values and k  2
0   0 0.70
?
1.25
+
+
+
 0
For the 5% values and k  2
0   0 0.95
?
1.54
+
+
+
 0
2
+
 0
2
+
 0
2
+
 0
?
2.64
+
?
2.75
+
?
2.46
+
?
4 dL
+
 0
4
+
3.19
+
 0
3.92
+
 0
3.30
+
 0
3.05
+
 0
4
+
4
+
4
+
4
+
For the tests for k  1 , our value of DW seems to be in the ‘?’ range, but for the second regression 1.94 is
comfortable in the ‘no autocorrelation’ range.
f) T  16 . The regression equation is Y = 52.4 + 3.04 T - 0.0939 TSQ
and s e  S =
s
The outline says “To find an approximate confidence interval  Y0  Yˆ0  t e and an
n
approximate prediction interval Y  Yˆ  t s . Remember df  n  k  1 .” The prediction interval is
1.73483.
0
0
e
appropriate here because we are making a single predication. So
Y  52 .4  3.0416   0.0939 16 2  t .12
005 `1.73486   52 .4  48 .64  24 .04  3.055 1.73483   125 .06  5.30 .
g) The t-ratio was -3.48 and the outline says that for a regression involving 3 variables, rY23.12 
t 22
 3.48 2  0.50229
 3.48 2  12
t 32
t 32  df
.
so rY 1.2   0.50229  .7079 . This is an
t 22  df
indicator of the additional explanatory power of this variable.
Similarly here rY21.2 

17
252x0541 4/22/05
2. Do the following using Exhibit 1.
a) Compute the (Pearson’s) sample correlation between output and time and test it for significance.
(5)
b) Test the hypothesis that the population correlation   0.8 . (5)
c) Do Spearman’s rank correlation between output and time and test it for significance. Why is the
rank correlation higher than Pearson’s? (6)
[16, 31]
 XY  nXY
 X  nX  Y
a) r 
2
X
2
2
 nY
So r 
Y
2
 nY 2  SST  SSY  734.556,
 X Y  nX Y  SX1Y  430.550.
 nX 12  SSX1  280 and
2
1
We were given
2
1
1
430 .550 2
 .9013  .9494 . According to the outline the test for significance is H 0 : xy  0
734 .556 280 
r
against H1 : xy  0 t n  2  

sr
r
1 r
n2
2

.9013
1  .9494
13

.9013

.003892
.9013
 14 .338 . Since this is
.0623859
13
larger than t n2  t .025
 2.179 , we reject the null hypothesis and say that the correlation is significant.
2
b) According to the outline “If we are testing H 0 : xy   0 against H 1 : xy   0 , and  0  0 , the test is
1 1 r 
z  ln 
quite different. We need to use Fisher's z-transformation. Let ~
 . This has an approximate
2  1 r 
~
n 2 
z  z
1
1  1 0 
 and a standard deviation of s z 
t

mean of  z  ln 
,
so
that
.

n3
sz
2  1 0 
(Note: To get ln , the natural log, compute the log to the base 10 and divide by .434294482. )”
1  1  r  1  1.9494  1
1
~
z  ln 
  ln 
  ln 38 .5257   1.58575   0.7929
2  1  r  2  0.0506  2
2
1 1 0
 0  0.8 so  z  ln 
2  1 0
 1  1.8  1
1
  ln 
 2  0.2   2 ln 9.00   2 0.95424   0.4771

~
n  2 
z   z 0.7929  0.4771
1
1
sz 

 0.288675 and t


 1.0940 . Since this is not larger
n3
12
sz
0.288675
than t n2  t 13  2.179 , we do not reject the null hypothesis and say that the correlation is not

.025
2
significantly different from 0.8.
c) The data appears below with ranks.
Row
Y rY T rT T 2 d d 2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
53.43
59.09
59.58
64.75
68.65
65.53
68.44
70.93
72.85
73.60
72.93
75.14
73.88
76.55
79.05
1
2
3
4
7
5
6
9
8
11
10
13
12
14
15
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
1
4
9
16
25
36
49
64
81
100
121
144
169
196
225
0 0
0 0
0 0
0 0
2 4
-1 1
-1 1
0 0
0 0
1 1
-1 1
1 1
-1 1
0 0
0 0
0 10
 d  1  610   1  60  1  0.017  .9821 .
 1
15225 `1
3360
nn  1
2
6
rs
2
According to the Table 13, the 5% critical value
is .4429, so we can reject the null hypothesis that
there is no rank correlation.
The rank correlation is larger than the Pearson
correlation because the relationship, though
basically monotonic ( x rises when y rises) it
has some curvature rather than being a straight
line.
18
252x0541 4/22/05
3. (Berenson et. al.) The operations manager of a light bulb factory wants to determine if there is any
difference between the average life expectancy of a light bulb manufactured by two different machines. A
random sample of 25 light bulbs from machine 1 has a sample mean of 375 hours. A random sample of 25
light bulbs from machine 2 has a sample mean of 362 hours.
a) Test whether the mean lives of the bulbs differ at the 5% significance level. Assume that 110 is
the population standard deviation for machine 1 and that 125 is the population standard deviation
for machine 2. Do not use a confidence interval. State your null hypothesis! (4)
b) Find a p-value for the null hypothesis in part a) and interpret it. Do not use the t-table. (3)
c) Test whether the mean lives of the bulbs differ at the 5% significance level. Assume that 110 is
the sample standard deviation for machine 1 and that 125 is the sample standard deviation for
machine 2. Do not use a confidence interval. State your null hypothesis! Make and state an
assumption about the equality of the two standard deviations. (3 or 5)
d) Test the assumption about the standard deviations that you made in c). State your null
hypothesis! (2)
e) Make the following confidence intervals.
(i) A confidence interval for the difference between the means in a). (1)
(ii) A confidence interval for the difference between the means in c) (2)
(iii) A confidence interval for the ratio of the population variances in d) (2)
[15, 46]
Solution: Part of TABLE 3 appears below.
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
Difference
H 0 : D  D0 *
d cv  D0  z d
D  d z 2  d
d  D0
z
between Two
H
:
D

D
,

1
0
d
Means (
 12  22
D







1
2
d
known)
n
n
1
2
d  x1  x 2
Difference
between Two
Means (
unknown,
variances
assumed equal)
Difference
between Two
Means(
unknown,
variances
assumed
unequal)
1
1

n1 n2
DF  n1  n2  2
sd  s p
s12 s22

n1 n2
sd 
DF 
 s12 s22 
  
n

 1 n2 
1 , DF2
F1DF


2
1
FDF1 , DF2
2
2
   
s12
2
n1
s 22
H 1 : D  D0 ,
D  1   2
H 0 : D  D0 *
D  d  t 2 s d
n1  1
Ratio of Variances
H 0 : D  D0 *
D  d  t 2 s d
2
n2
n2  1
 22 s22 DF1 , DF2

F

 12 s12 .5  .5  2 
DF1  n1  1
DF2  n 2  1
 2

.5  .5   2    or
1  
2

H 1 : D  D0 ,
t
sˆ 2p 
t
D  1   2
*Same as
H 0 : 1   2
d cv  D0  t  2 s d
d  D0
sd
n1  1s12  n2  1s22
n1  n2  2
d cv  D0  t  2 s d
d  D0
sd
H 1 : 1   2
if D0  0.
H0 : 12   22
H1 : 12   22
F DF1 , DF2 
s12
s 22
and
F DF2 , DF1 
s 22
s12
This is Exercise 10.5 in the text, which was assigned. At least some of the language in this solution is
quoted.
I am assuming that   .05 . The following hypothotheses are tested in a) – c).
19
252x0541 4/22/05
Light bulbs produced by machine 1 have the same average life expectance as
H 0 : 1   2
those produced by machine 2.
H1 : 1   2 Light bulbs produced by machine 1 have a different life expectancy from those produced
by machine 2
a)
n1  25, x1  375 , and  1  110 , n2  25, x 2  362 , and  2  125 ; d  375  362  13 .
d 
 12
n1

 22
n2

110 2 125 2

 484  625  1109  33 .307
25
25
Test Ratio: z 
d  D0
d

z 2  z.025  1.960
13  0
 0.390 . Make a diagram showing a Normal
33 .307
curve with a vertical bar at t  0 . Show one 2.5% rejection zone above z .025  1.960 and
a second 2.5% rejection zone below z .025  1.960 . Since 0.390 is not in the rejection
zone, do not reject the null hypothesis.
Critical Value: d cv  D0  z d  0  1.960 33 .307   0  65 .28 . Make a diagram
showing a Normal curve with a vertical bar at t  0 . Show one 2.5% rejection zone above
65.28 and a second 2.5% rejection zone below -65.28. Since d  13 is not in the
rejection zone, do not reject the null hypothesis.
b)
c)
`
pvalue  Pz  0.390   .5  P0  z  0.39   .5  .1517  .3484
n1  25, x1  375 , and s1  110 , n2  25, x 2  362 , and s 2  125 d  375  362  13 .
(i) Assume that population variances are equal.
n  1s12  n2  1s 22 110 2  125 2 12100  15625

s p2  1


 13862 .5 and
n1  n 2  2
2
2
1
  1
s d  s p2  
 n1 n 2

2
  13862 .5
 1109  33 .3016
25

Test Ratio: t 
df  n1  n 2  2  48
t.48
025  2.011
d  D0
13  0

 0.390 Make a diagram showing an almostsd
33 .3016
Normal curve with a vertical bar at t  0 . Show one 2.5% rejection zone above
48
t.48
025  2.011 and a second 2.5% rejection zone below  t .025  2.011. Since 0.390 is
not in the rejection zone, do not reject the null hypothesis.
Critical Value: d cv  D0  t  2 s d  0  2.01133 .3016   0  66 .97 Make a
diagram showing a Normal curve with a vertical bar at t  0 . Show one 2.5% rejection
zone above 66.97 and a second 2.5% rejection zone below 66.97. Since d  13 is not in
the rejection zone, do not reject the null hypothesis.
(ii) Do not assume that population variances are equal.
sd 
s12 s 22
110 2 125 2



 484  625  1109  33 .307
n1 n 2
25
25
20
252x0541 4/22/05


  s2 s2 2
  1  2 
  n1 n 2 
df  
2
2
  s2 
 s 22 
 
 1 
 n2 
  n1 
 


n2 1
 n1  1






1109 2
1229881


2
2
625  9760 .6667  16276 .0417
 484 


24
24



1229881
 34 .1286
36036 .7084
Test Ratio: t 
t.34
025  2.032
d  D0
13  0

 0.390
sd
33 .307
Critical Value: d cv  D0  t 2 sd  0  2.032 33 .307   0  67 .70
Writeups and conclusions are essentially the same as for equal variances.
H1 : 12   22 n1  25, x1  375 , and s1  110 , n2  25, x 2  362 , and
d) H0 : 12   22
s 2  125 d  375  362  13 . To test equality of variances, test the ratios . The larger of the two ratios is
2
s 22
 125 
25, 25

  1.29 . Compare this with F.025 , which is between 2.30 and 2.24. Since the computed F is
2
s1  110 
less than the table F do not reject the null hypothesis.
e) (i) From Table 3 D  d  z  2  d . We already know d  13 ,
d 
 12
n1

 22
n2

110 2 125 2

 33 .307 and z 2  z.025  1.960 , so
25
25
D  13  1.960 (33 .307 )  13  65.28
(ii) D  d  t  2 s d . We already know d  13 and depending on what we assumed in c), either
1
  1
s d  s p2  
 n1 n 2

  33 .3016 and t.48
025  2.011 , so D  13  2.01133 .3016   13  66 .97 or

s12 s 22

 33 .307 and t.34
025  2.032 , so D  13  2.03233.307  13  67.70 .
n1 n 2
sd 
(iii) The version of the interval on table 3 is a bit too consise. From ‘Confidence Intervals and
Hypothesis Testing for Variances’ we find the following two possibilities.
s 22
 22 s 22 ( n1 1, n2 1)
s12
 12 s12 ( n2 1, n1 1)
1
1


F


F
or
.

s12 Fn2 1,n1 1  12 s12 2
s 22 Fn1 1, n2 1  22 s 22 2
2
2
25, 25 is between 2.30 and 2.24, probably about 2.29 and that
We know F.025
2
2
s12  110 
 125 


1
.
2913

,
so


  0.7744 , so we have
s12  110 
s 22  125 
s 22
1.2913
2
2
1
1
 22  1.2913 2.29  or 0.7744
 12  0.7744 2.29  , which become
2.29  1
2.29  2
0.564 
 22
 12
 2.96 or 0.338 
 12
 22
 1.77 .
21
252x0541 4/22/05
4. (Berenson et. al.) A student team is investigating the size of the bubbles that can be blown with various
brands of bubble gum. The data below is the total diameter in inches of the bubbles and is presented in two
different ways. These Exhibits are repeated as a tear-off sheet at the end of the exam with the sums and sums of squares
computed for you.
Exhibit 2: Size of Bubbles Blocked by Blower
Row
1
2
3
4
5
Student
Loopy
Percival
Poopsy
Dizzy
Booger
Brand 1
Brand 2
Brand 3
x1
x 2
x 3
x 4
9.5
4.0
5.5
8.5
4.5
8.5
8.5
7.5
7.5
8.0
11.5
11.0
7.5
7.5
8.0
8.75
9.50
9.25
9.50
9.25
Brand 4
Exhibit 3: Size of Bubbles Blocked by Blower but ranked as four independent random samples.
Row
1
2
3
4
5
Student
Loopy
Percival
Poopsy
Dizzy
Booger
x1
r1
8.75
9.50
9.25
9.50
9.25
x 2
13.0
17.0
14.5
17.0
14.5
9.5
4.0
5.5
8.5
4.5
r2
17
1
3
11
2
x 3
8.5
8.5
7.5
7.5
8.0
r3
11.0
11.0
5.5
5.5
8.5
x 4
11.5
11.0
7.5
7.5
8.0
r4
20.0
19.0
5.5
5.5
8.5
Compare the data in 3 different ways.
a) Do only one of the following.
(i) Consider the data random samples from Normally distributed populations and compare means. (6)
(ii) Consider the data blocked data from Normally distributed populations and compare means. (8)
b) Consider the data random samples from non-Normally distributed populations and compare medians. (5)
c) Consider the data blocked data from non-Normally distributed populations and compare medians. (5)
a) Do only one of the following.
(i) Consider the data random samples from Normally distributed populations and
compare means. (6)
(ii) Consider the data blocked data from Normally distributed populations and compare
means. (8)
You were given the column sums and sums of squares.
x
x
1
 46.25,
3
 40,
x
x
2
1
 428.188,
2
3
 321,
x
x
You were also given the row sums and sums of squares.
1
3
5
x
x
x
1
 38.25
3
 29.75
5
 29.75
x
x
x
2
1
 371.313
2
3
 228.313 4
2
5
 233.813
2
x
4
2
 32,
4
 45.5,
x
2
 33.00
x
x
 33.00
x
2
4
2
2
 229,
2
4
 429.75
x
2
2
 299.500
 275.000
22
252x0541 4/22/05
Notice that the first 4 columns of my tableau would be the same whether you were doing 1-way or 2-way
ANOVA. ‘SS’ stands for sum of squares.
Brand 1
Block
x1
1
2
3
4
5
Sum
8.750
9.500
9.250
9.500
9.250
46.250
Brand 2
Brand 3
Brand 4
x 2
x 3
x 4
9.50
4.00
5.50
8.50
4.50
+32.00
8.5
8.5
7.5
7.5
8.0
+40.0
Sum
x
i..
ni
11.50
38.25 4
11.00
33.00 4
7.50
29.75 4
7.50
33.00 4
8.00
29.75 4
+45.50 = 163.75 20
9.5625
8.2500
7.4375
8.2500
7.4375
8.1875
= 20.00  n
nj
5
+5
+5
+5
x j
9.250
6.40
8.0
9.10
x
x 2j
85.563
+40.96
+64.0
Note that x is not a sum, but is
 x
SSC   n
SST 
2
ij
+82.81
 x .
371.31
299.50
228.31
275.00
233.81
1407.94
 x
91.441
68.063
55.316
68.063
55.316
338.199
2
ij
x
2
i
8.19  x This is not a sum.
 x
273.33   x
428.188 +229.00 +321.0 +429.75 = 1407.94 
SS
x i2.
SS
x i.
=
2
ij
2
j
n
 n x  1407 .938  20 8.1875 2  1407 .938  1340 .703  67 .235 .
2
 n x  5273 .3325   20 8.1875 2  1366 .665  1340 .703  25 .960 . This is SSB in a
2
2
j x j
one way ANOVA.
SSR 
 n x
2
i i.
 n x  4338 .199   20 8.1875 2  1352 .796  1340 .703  12 .093
2
( SSW  SST  SSC  SSR  29.182 )
(ii) 2-way ANOVA (Blocked by student)
Source
SS
DF
MS
F
Rows (Blocks)
12.093
4
3.02325
1.243ns
Columns(Brands)
25.960
3
8.65333
3.558s
F.05
F 4,12  3.26
F 3,12  3.49
H0
Row means equal
Column means equal
Within (Error)
29.182
12
2.43183
Total
67.235
19
So the brands (column means) are significantly different although the students are not.
(i) One way ANOVA (Not blocked by student) ( SSW  SST  SSB  .0694 )
Source
SS
DF
MS
F.01
F
Columns(Brands)
25.960
3
8.65333
3.354s
F 3,16  3.01
H0
Column means equal
Within (Error)
41.275
16
2.57969
Total
67.235
19
Once again, the brands (column means) are significantly different.
23
252x0541 4/22/05
The results in b) and c) were totally cut and pasted from the outline.
b) Consider the data random samples from non-Normally distributed populations and compare
medians. (5)
We repeat Exhibit 3. The x s are the original numbers and the r s are their ranks among the 20 data items.
The Kruskal-Wallis Test is equivalent to one-way ANOVA when the underlying distribution is non-normal.
The null hypothesis is H 0 : Columns come from same distribution or medians equal.
Row Student
x1
1 Loopy
8.75
2 Percival 9.50
3 Poopsy
9.25
4 Dizzy
9.50
5 Booger
9.25
Sum of Ranks
Size of Column
So n1  n 2  n3  n 4  4 , n 
r1
x 2
r2
x 3
r3
x 4
r4
13.0
17.0
14.5
17.0
14.5
76.0
5
9.5
4.0
5.5
8.5
4.5
17
1
3
11
2
34
5
8.5
8.5
7.5
7.5
8.0
11.0
11.0
5.5
5.5
8.5
41.5
5
11.5
11.0
7.5
7.5
8.0
20.0
19.0
5.5
5.5
8.5
58.5
5
n
i
 20 , SR1  76 , SR2  34 , SR3  41 .5 and SR4  58 .5 .
To check the ranking, note that the sum of the four rank sums is 76 + 34 + 41.5 + 58.5 = 210, and that the
nn  1 20 21

 210 .
sum of the first n  20 numbers is
2
2
 12
 SRi 2 

  3n  1
Now, compute the Kruskal-Wallis statistic H  
 nn  1 i  ni 
 12  76 2 34 2 41 .52 58 .52 

  321   12  5776  1156  1722 .25  3422 .25  63





5
5
5 
5
 420 
 20 21  5
 12 

12076 .5  63  69 .0086  63  6.0086 . The Kruskal-Wallis table (Table 9) does not have values for
 2100 

tests with 4 columns, so we must use the  2 distribution, with df  m  1 , where m is the number of
columns. Since there are m  4 columns and   .05 , test H with  .2053  7.8147 . Since H is larger than
 .205 , reject the null hypothesis of equal medians.
24
252x0541 4/22/05
c) Consider the data blocked data from non-Normally distributed populations and compare
medians. (5)
We repeat Exhibit 3 somewhat changed. The x s are the original numbers and the r s are their ranks within
the rows of 4. The Friedman Test is equivalent to two-way ANOVA with one observation per cell when the
underlying distribution is non-normal. The null hypothesis is H 0 : Columns come from same distribution or
medians equal. Note that the only difference between this and the Kruskal-Wallis test is that the data is
cross-classified in the Friedman test.
[24, 70]
Row Student
x1
r1
x 2
r2
x 3
r3
x 4
r4
1 Loopy
8.75
2 Percival 9.50
3 Poopsy
9.25
4 Dizzy
9.50
5 Booger
9.25
Sum of Ranks
Size of Column
2.0
3.0
4.0
4.0
4.0
17.0
5
9.5
4.0
5.5
8.5
4.5
3.0
1.0
1.0
3.0
1.0
9.0
5
8.5
8.5
7.5
7.5
8.0
1.0
2.0
2.5
1.5
2.5
9.5
5
11.5
11.0
7.5
7.5
8.0
4.0
4.0
2.5
1.5
2.5
14.5
5
Assume that   .05 . In the data above, the methods are represented by c  4 columns, and the groups by
r  5 rows.. In each row the numbers are ranked from 1 to c  4 . For each column, compute SRi , the rank
sum of column i . To check the ranking, note that the sum of the four rank sums is 17 + 9 + 9.5 + 14.5 = 50,
cc  1
and that the sum of the c numbers in a row is
. However, there are r rows, so we must multiply
2
rcc  1 545
SRi 

 50 .
the expression by r . So we have
2
2
 12

SRi2   3r c  1
Now compute the Friedman statistic  F2  
 rc c  1 i


 


 12
17 2  92  9.52  14 .52   355   12 289  81  90 .25  210 .25   75  80 .46  75  5.46

100

 545

Since the size of the problem is larger than those shown in Table 10, we use the  2 distribution, with
df  c  1  4  1  3 , where c is the number of columns. We compare  F2  5.46 with  .2053  7.8147 .
Since  F2 is not larger than  .205 , do not reject the null hypothesis of equal medians.
The results in b) and c) seem to indicate the greater power of ANOVA than the non-parametric
methods when the assumptions are appropriate.
[24, 70]
25
252x0541 4/22/05
5. (Berenson et. al.) The time it takes to design and launch a marketing campaign is called a cycle time.
Marketing campaigns are classified by cycle time (in months) and effectiveness. Don’t even think of
answering any part of this question without doing a statistical test!
Effectiveness
D u r a t i o n
< 1 mo.
1-2 mo.
2-4 mo.
>4 mo.
Total
Very Effective
15
28
24
6
73
Effective
9
26
33
19
87
Ineffective
5
2
3
5
15
Total
29
56
60
30
175
a. Test to see if the proportion in the various effectiveness categories is related to cycle time. (8)
b. Of the campaigns that took 0 – 2 months 7 were ineffective. Of the campaigns that took more
than two months, 8 were ineffective. Is the fraction that were ineffective in the first category below
the fraction in the second category? (5)
c. Test the hypothesis that 45% of campaigns are very effective. (4)
d. As you know a Jorcillator has two components, a Phillinx and a Flubberall. We recorded the
order in which they were replaced over the last year to see if there was a pattern or the replacement
sequence was just random. We got PPPFFFPPPFFPPFFPPFFF. Test it! (3)
e. (Anderson et. al.) The number of emergency calls our Fire department receives is believed to
have a Poisson distribution with a parameter of 3. Test this against data for a period of 120 days: 0
calls on 9 days, 1 call on 12 days, 2 calls on 30 days, 3 calls on 27 days, 4 calls on 22 days. 5 calls
on 13 days and 7 calls on 6 days. (5)
[25, 95]
5. (Berenson et. al.) The time it takes to design and launch a marketing campaign is called a cycle time.
Marketing campaigns are classified by cycle time (in months) and effectiveness. Don’t even think of
answering any part of this question without doing a statistical test!
Effectiveness
D u r a t i o n
< 1 mo.
1-2 mo.
2-4 mo.
>4 mo.
Total
Very Effective
15
28
24
6
73
Effective
9
26
33
19
87
Ineffective
5
2
3
5
15
Total
29
56
60
30
175
a. Test to see if the proportion in the various effectiveness categories is related to cycle time. (8)
b. Of the campaigns that took 0 – 2 months 7 were ineffective. Of the campaigns that took more than two months, 8 were
ineffective. Is the fraction that were ineffective in the first category below the fraction in the second category? (5)
c. Test the hypothesis that 45% of campaigns are very effective. (4)
d. As you know a Jorcillator has two components, a Phillinx and a Flubberall. We recorded the order in which they were
replaced over the last year to see if there was a pattern or the replacement sequence was just random. We got
PPPFFFPPPFFPPFFPPFFF. Test it! (3)
e. (Anderson et. al.) The number of emergency calls our Fire department receives is believed to have a Poisson distribution
with a parameter of 3. Test this against data for a period of 120 days: 0 calls on 9 days, 1 call on 12 days, 2 calls on 30
days, 3 calls on 27 days, 4 calls on 22 days. 5 calls on 13 days and 7 calls on 6 days. (5)
a) Test to see if the proportion in the various effectiveness categories is related to cycle time. (8)
. The observed data is indicated by O, the expected data by E .
H 0 : Mktg Campaigns are Homogeneous by duration.
The original numbers are rewritten to show p r , the proportion in each row and labeled O for observed.
O Columns
1
2
3
4
Total
Row 1
Row 2
Row 3
15
9
5
28
26
2
24
33
3
6
19
5
73
87
15
29
56
60
30
Total
pr
.417143
.497143
.085714
175
26
252x0541 4/22/05
p c represents the proportion in each column. n 
 O  175 . There are r  3 rows, c  4 columns and
rc  12 cells. Each cell gets p c p r n  p c (Column total) . For example, the proportion in Column 1 is
29
175  .165714 . So for the upper left corner the expected value is
29
73
175  175175  .165714 .417143 175  12 .0971 . This was rounded to 12.10. the results are E .
E Columns
Row 1
Row 2
Row 3
Total
1
2
3
4
Total
12.10 23.36 25.03
14.42 27.84 29.83
2.49 4.80 5.14
12.51
14.91
2.57
29.01 56.00 60.00
28.99
The formula for the chi-squared statistic is  2 

O  E 2
73
87
15
175
or  2 
E
formulas is shown in the table below, though only one should be used.
Row O
1
2
3
4
5
6
7
8
9
10
11
12
2
So  calc

E O
E
15 12.10
9 14.42
5
2.49
28 23.36
26 27.84
2
4.80
24 25.03
33 29.83
3
5.14
6 12.51
19 14.91
5
2.57
175 175.00

O  E 2
E
-2.90
5.42
-2.51
-4.64
1.84
2.80
1.03
-3.17
2.14
6.51
-4.09
-2.43
0.00
2
E  O2 E  O 
8.4100
29.3764
6.3001
21.5296
3.3856
7.8400
1.0609
10.0489
4.5796
42.3801
16.7281
5.9049
2
 16.01646 or  calc

E
pr
.417143
.497143
.085714

O2
 n . Both of these two
E
O2
E
0.69504 18.5950
2.03720
5.6172
2.53016 10.0402
0.92164 33.5616
0.12161 24.2816
1.63333
0.8333
0.04239 23.0124
0.33687 36.5069
0.89097
1.7510
3.38770
2.8777
1.12194 24.2119
2.29763
9.7276
16.01646 191.0165

O2
 n  191 .0165  175  16.0165 .
E
The degrees of freedom for this application are r  1c  1  3  14  1  23  6 .
2
If our significance level is 5%, compare  calc
to  .2056   12 .5916 . Since our value of our computed sum is
greater than the table chi-squared, reject the null hypothesis.
b) Of the (29 + 56 = 65) campaigns that took 0 – 2 months 7 were ineffective. Of the (60 + 30 =
90) campaigns that took more than two months, 8 were ineffective. Is the fraction that were ineffective in
the first category below the fraction in the second category? (5) (It should have been above!)
If p1 is the proportion of successes in the first sample, and p2 is the proportion of successes in the second
sample, we define p  p1  p 2 . Then our hypotheses will be
H 0 : p1  p 2 or H 0 : p  p 0  0
H 1 : p1  p 2
Let p1 
H 1 : p  p 0  0
.
x1
7
x
8
 .088889 and p  p1  p2  .0188030 where x1 is the number

 .107692 , p2  2 
n2 90
n1 65
of successes in the first sample, x 2 is the number of successes in the second sample, n1 and n 2 are the
sample sizes and q  1  p . The usual three approaches to testing the hypotheses can be used.
27
252x0541 4/22/05
 p1  p2    p1  p2   z s p , where
.107692 .892308  .088889 .911111 

 .00147838  .00089986
Confidence Interval: p  p  z s p or
2
s p 
p1 q1 p 2 q 2


n1
n2
65

2
90
The one sided interval is p  p  z s p  .0188  1.645.0487  .0989 . Compare this
interval with p 0  0 . Since p  .0989 does not contradict p  0 do not reject H 0 .
p  p 0 .0188030

 0.3907 where
Test Ratio: z 
 p
.0481245
 1
1 
1 
 1
  .0967742 .903226     .0481245 and

n
n
65
90


2 
 1
n p  n 2 p 2 x1  x 2
78
p0  1 1


 .0967742 . Because our alternate hypothesis is
n1  n 2
n1  n 2 65  90
 p  p 0 q 0 
p  0 , this is a left-sided test and the rejection zone is below z .05  1.645 . Since the
test ratio is not in that interval, do not reject H 0 .
Critical Value: pCV  p0  z  p becomes pCV  p0  z  p
2
 0  1.645 .04812   0.0792 and the rejection zone is below -0.0792. If we test this
against p  p1  p2  .0188030 , we cannot reject H 0 . For calculation of  p , see Test
Ratio above.
c) Test the hypothesis that 45% of campaigns are very effective. (4)
Out of 175 campaigns, 73 were very effective. To test H 0 : p  p 0 against H1 : p  p0 . Here p 0  .45 and
p
x 73

 .417143 .
n 175
(i) Test Ratio: z 
p  p0
p

.417143  .45
 0.874 where
.03761
p0 q0
.45.55 

 .00141  .03761 . Make a diagram with a mean at zero and
n
175
rejection zones below z .025  1.960 and z .025  1.960 . Since z  0.874 falls between
these numbers, do not reject H 0 .
p 
(ii) Critical Value: pcv  p0  z  p  .45  1.960.03761  .45  .074 . The rejection
2
regions are below .376 and above .524. Since p 
x 73

 .417 falls between these
n 175
numbers, do not reject H 0 .
(iii) Confidence Interval: p  p  z  2 s p  .4171  1.960 .03727   .4171  .073
pq
.417143 .582857

n
175
includes p 0  .45 .
sp 

.001389  .03727 . The confidence interval
28
 .00237824  .04
252x0541 4/22/05
d) As you know a Jorcillator has two components, a Phillinx and a Flubberall. We recorded the
order in which they were replaced over the last year to see if there was a pattern or the replacement
sequence was just random. We got PPPFFFPPPFFPPFFPPFFF. Test it! (3)
To test the null hypothesis of randomness for a small sample, assume that the significance level is 5% and
use the table entitled 'Critical values of r in the Runs Test.’ n  20 , n1  10 , n 2  10, and r  8.
For n1  10 and n2  10 , the table gives a critical values of 6 and 16. This means that we reject the null
hypothesis if r  6 or r  16 . In this case, since r  8 we do not reject the null hypothesis.
e) (Anderson et. al.) The number of emergency calls our Fire department receives is believed to
have a Poisson distribution with a parameter of 3. Test this against data for a period of 120 days: 0
calls on 9 days, 1 call on 12 days, 2 calls on 30 days, 3 calls on 27 days, 4 calls on 22 days. 5 calls
on 13 days and 7 calls on 6 days. (5)
No matter how we do this, we need a Poisson table with a parameter of 3. This is copied from the syllabus
supplement. The E  fn column comes from multiplying the f column by n  120 .
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
k
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
f  P(x=k) Fe  P(xk) E  fn
0.049787
0.149361
0.224042
0.224042
0.168031
0.100819
0.050409
0.021604
0.008102
0.002701
0.000810
0.000221
0.000055
0.000013
0.000003
0.000001
0.000000
0.000000
0.04979
0.19915
0.42319
0.64723
0.81526
0.91608
0.96649
0.98810
0.99620
0.99890
0.99971
0.99993
0.99998
1.00000
1.00000
1.00000
1.00000
1.00000
5.9744
17.9233
26.8850
26.8850
20.1637
12.0983
6.0491
2.5925
0.9722
0.3241
0.0972
0.0265
0.0066
0.0016
0.0004
0.0001
0.0000
0.0000
Most of the E column contains values that are below 5, so for a chi-square procedure, they need to be
added up, especially since the observed data only goes up to k  7. The chi-square procedure follows. Our
null hypothesis is H 0 : Poisson6
row
1
2
3
4
5
6
7
O
E O
E
9
5.9744
12 17.9233
30 26.8850
27 26.8850
22 20.1637
13 12.0983
7 10.0704
120 120.000
O  E 2
-3.02556
5.92332
-3.11496
-0.11496
-1.83628
-0.90172
3.07040
0.00024
2
E  O2 E  O 
9.1540
35.0857
9.7030
0.0132
3.3719
0.8131
9.4274
E
O2
E
1.53220 13.5578
1.95755
8.0342
0.36091 33.4759
0.00049 27.1155
0.16723 24.0035
0.06721 13.9689
0.93615
4.8657
5.02174 125.0215
O2
 n  125 .0215  120  5.0215 . In this case we have 6
E
E
degrees of freedom and  .2056   12 .5916 . Since our computed value is less than the table value, we cannot
2
So  calc


2
 5.0274 or  calc


reject the null hypothesis.
29
252x0541 4/22/05
This problem can also be done by a Kolmogorov-Smirnov procedure.
O
Row
O
Fo
Fe  P(xk) D  Fe  Fo
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
9
12
30
27
22
13
7
0.075000
0.100000
0.250000
0.225000
0.183333
0.108333
0.058333
0.07500
0.17500
0.42500
0.65000
0.83333
0.94167
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
0.04979
0.19915
0.42319
0.64723
0.81526
0.91608
0.96649
0.98810
0.99620
0.99890
0.99971
0.99993
0.99998
1.00000
1.00000
1.00000
1.00000
1.00000
0.0252100
0.0241500
0.0018100
0.0027700
0.0180733
0.0255867
0.0335100
0.0119000
0.0038000
0.0011000
0.0002900
0.0000700
0.0000200
0.0000000
0.0000000
0.0000000
0.0000000
0.0000000
 0.1242 . Since all the values in the D column are below the table
120
value, we cannot reject the null hypothesis.
The 5% critical value is 1.36
30
252x0541 4/22/05
6. Test to see if the price of new homes rose between 2001 and 2002. The following data represents a
random sample of typical prices in thousands in 10 zip codes in 2001 and 2002.
Row
Location
2001
2002
x1
1
2
3
4
5
6
7
8
9
10
Alexandria
Boston
Decatur
Kirkland
New York
Philadelphia
Phoenix
Raleigh
San Bruno
Tampa
245.795
391.750
205.270
326.524
545.363
185.736
170.413
210.015
385.387
194.205
293.266
408.803
227.561
333.569
531.098
197.874
175.030
196.094
391.409
199.858
Some of the following data may be of use to you.
 x  2860.458,
 d  -94.104,
1
x
d
2
1
2
change
d  x1  x 2
x2
 953941.216,
x
-47.471
-17.053
-22.291
-7.045
14.265
-12.138
-4.617
13.921
-6.022
-5.653
x
 2954.562,
2
2
2
 999628.915,
 3724.975
If you want to receive full credit, you must clearly label each section that you do!
a) Remember that the data is cross classified. Assume that the underlying distribution is not Normal and compare
medians. (5)
b) Remember that the data is cross classified. Assume that the underlying distribution is Normal and compare means. (4)
c) Forget that the data is cross classified. Assume instead that it represents two random samples from Chester County, one
for each year and that the underlying distribution is not Normal. Compare medians. (6)
[15, 110]
Solution:
a) Remember that the data is cross classified. Assume that the underlying distribution is not
Normal and compare medians. (5)
The Wilcoxon Signed Rank Test for Paired Samples is a test for equality of medians when the data is
paired. It can also be used for the median of a single sample. The Sign Test for paired data is a simpler test
to use in this situation, but it is less powerful.
As in many tests for measures of central tendency with paired data, the original numbers are discarded, and
the differences between the pairs are used. If there are n pairs, these are ranked according to absolute value
from 1 to n , either top to bottom or bottom to top. After replacing tied absolute values with their average
rank, each rank is marked with a + or – sign and two rank sums are taken, T  and T  . The smaller of
these is compared with Table 7.
Row
Location
2001
x1
1
2
3
4
5
6
7
8
9
10
Alexandria
Boston
Decatur
Kirkland
New York
Philadephia
Phoenix
Raleigh
San Bruno
Tampa
245.795
391.750
205.270
326.524
545.363
185.736
170.413
210.015
385.387
194.205
2002 difference
x2
293.266
408.803
227.561
333.569
531.098
197.874
175.030
196.094
391.409
199.858
rank signed rank
d  x 2  x1
d
r
-47.471
-17.053
-22.291
-7.045
14.265
-12.138
-4.617
13.921
-6.022
-5.653
47.471
17.053
22.291
7.045
14.265
12.138
4.617
13.921
6.022
5.653
10
8
9
4
7
5
1
6
3
2
r*
-10
-8
-9
-4
7
-5
-1
6
-3
-2
.
 H 0 : 1   2
n  10 and   .05 . If we add together the numbers in r * with a + sign we get . T   13 .

 H 1 : 1   2
If we do the same for numbers with a – sign, we get T   42 To check this, note that these two numbers
nn  1 10 11

 55 , and that
must sum to the sum of the first n numbers, and that this is
2
2
T   T   13  42  55 .
31
252x0541 4/22/05
We check 13, the smaller of the two rank sums against the numbers in table 7. For a one-sided 5% test, we
use the   .05 column. For n  10 , the critical value is 11, and we reject the null hypothesis only if our
test statistic is below this critical value. Since our test statistic is 13, we do not reject the null hypothesis.
b) Remember that the data is cross classified. Assume that the underlying distribution is Normal
and compare means. (4)
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
Difference
H 0 : D  D0 *
D  d t  2 s d
d cv  D0 t  2 s d
d  D0
t
between Two
H 1 : D  D0 ,
s
d
d  x1  x 2
Means (paired
D




1
2
s
data.)
df  n  1 where
sd  d
n1  n 2  n
n
H 0 : D  D0  0 , H1 : D  D0  0 where D  1   2 . We were given
d
s d2 
sd 
2
 3724.975. d 
d
2
 nd 2
n 1
sd

n

 d  -94.104,
94 .104
 9.4104 . Recall that
10
3724 .975  10  9.4104 2
 315 .4910 so that s d  315 .4910  17 .7621
9
315 .4910
 31 .54910  5.6169 df  n  1  9 t.905  1.833
10
Test Ratio: t 
d  D0  9.4104

 1.675 . Since this is a left-sided test, we reject the null
sd
5.6169
hypothesis if t calc is below -1.833. Since t calc is not below the critical value, we cannot reject
the null hypothesis.
Critical Value: d cv  D0 t  2 s d becomes d cv  D0 t  s d  1.833 5.6169   10 .296 . The
rejection region lies below -10.296. Since d  9.4104 is not below the critical value, we cannot
reject the null hypothesis.
Confidence Interval: D  d t  2 s d becomes D  d t  s d  9.4104  1.833 5.6169   0.8856 .
This does not contradict H 0 : D 0 , so we cannot reject the null hypothesis.
32
252x0541 4/22/05
c) Forget that the data is cross classified. Assume instead that it represents two random samples
from Chester County, one for each year and that the underlying distribution is not Normal.
Compare medians. (6)
The Wilcoxon-Mann-Whitney Test for Two Independent Samples is used if samples are independent. This
test is appropriate to test whether the two samples come from the same distribution. If the distributions are
similar, it is often called a test of equality of medians.
H :   2
If we use  for the median, our hypotheses are  0 1
and   .05 . n1  n 2  10 .
H 1 : 1   2
Row
1
2
3
4
5
6
7
8
9
10
Location
Alexandria
Boston
Decatur
Kirkland
New York
Philadephia
Phoenix
Raleigh
San Bruno
Tampa
2001
rank
x1
r1
245.795
391.750
205.270
326.524
545.363
185.736
170.413
210.015
385.387
194.205
11
17
8
13
20
3
1
9
15
4
101
2002
x2
293.266
408.803
227.561
333.569
531.098
197.874
175.030
196.094
391.409
199.858
rank
r2
12
18
10
14
19
6
2
5
16
7
109
Now compute the sums of the ranks. SR1  101, SR2  109 . As a check, note that these two rank sums must
add to the sum of the first n numbers, and that this is
nn  1 20 21

 210 , and that
2
2
SR1  SR2  101  109  210 .
This can be compared against the critical values for TL and TU ( TU is actually only needed for a 2-sided
test) in table 14b. These are 83 and 127. Since W  109 is between these values, we cannot reject the null
hypothesis.
[15, 110]
33
252x0541 4/22/05
ECO252 QBA2
Final EXAM
May 2-6, 2005
TAKE HOME SECTION
Name: _________________________
Student Number: _________________________
Class days and time : _________________________
1) Please Note: computer problems 2,3 and 4 should be turned in with the exam (2). In problem 2, the 2
way ANOVA table should be checked. The three F tests should be done with a 5% significance level and
you should note whether there was (i) a significant difference between drivers, (ii) a significant difference
between cars and (iii) significant interaction. In problem 3, you should show on your third graph where the
regression line is. Check what your text says about normal probability plots and analyze the plot you did.
Explain the results of the t and F tests using a 5% significance level. (2)
2) 4th computer problem (4+)
This is an internet project. You are trying to answer the question, ‘how well does manufacturing explain
differences in income?’ You should use some measure of income per person or family in each state as your
dependent variable and try to explain it as a function of (to start with) percent of output or labor force in
manufacturing. This should start out as a simple regression. Then you should try to see whether there are
other variables that explain the differences as well. One possibility is the per cent of the adult population
with college or high school diplomas. Possible sources of data are below, but think about what you use, and
try to find some other sources. Total income of a state, for example is a very poor choice, rather than some
per capita measure because it is simply going to be high for places with a lot of people without indicating
how well off they are. Similarly the fraction of the workforce with a certain education level is far better then
the number. For instructions on how to do a regression, try the material in Doing a Regression.
http://www.nam.org/s_nam/sec.asp?CID=5&DID=3 Manufacturing share in state economies
(http://www.nam.org/Docs/IEA/26767_20002001ManufacturingShareandChangeinStateEconomies.pdf?DocTypeID=9&TrackID=&Param=@CategoryI
D=1156@TPT=2002-2001+Manufacturing+Share+and+Change+in+State+Economics)
http://www.nemw.org/data.htm Per capita income by state.
http://www.nemw.org/data.htm State personal income per capita.
http://www.bea.doc.gov/bea/regional/data.htm Personal income per capita by state.
http://www.census.gov/statab/www/ Many state statistics, including persons with bachelor’s degrees.
http://www.epinet.org/content.cfm/datazone_index Income inequality, median income, unemployment rates.
Anyway, your job is to add whatever variable you think ought to explain your income measure. Consider all
50 states your sample. Your report should tell what numbers you used, from where and from what years.
What coefficients were significant and do you think on the basis of your results that manufacturing is an
important predictor of a state’s prosperity? Mark all significant F and t coefficients using a 5% significance
level. Explain VIFs.
Of course, if you don’t like this assignment, get approval to research something else on the internet. For
example, does the per cent of the population in prison affect the crime rate (maybe with a few years’ lag)?
Or are there better predictors? And get out the Durbin-Watson, prison vs. crime rate is a time series project.
[8]
3) Hotshot Associates is afraid of sex discrimination charges and collects the data below. The dependent
variable is income in thousands of dollars and the two independent variables are education in years and a
dummy variable indicating sex (1 means a female). The lines in the middle are missing because the totals
34
252x0541 4/22/05
are reliable and you don’t need them. The only thing that is missing is you. Add yourself to the sample as a
21st observation with 12 years of education and an income of 100.0 (thousand) plus the last two digits of
your student number as hundreds. For example Roland Dough’s student number is 123689, so he adds
$8900 to $100000 to get 108900, which he records as 108.9.
y
Row
1
2
3
4
5
INC
39.0
43.7
62.6
42.8
55.0
17 72.9
18 56.1
19 67.1
20 82.3
1168.5
x1
x2
x12
x 22
EDUC
2
4
8
8
8
SEX
0
1
0
1
0
4
16
64
64
64
0
1
0
1
0
16
16
17
21
241
0
1
0
0
7
256
256
289
441
3285
a. Compute the regression equation
y2
x1 y
1521.00
1909.69
3918.76
1831.84
3025.00
x2 y
0.0
43.7
0.0
42.8
0.0
0
4
0
8
0
0 5314.41 1166.4
0.0
1 3147.21
897.6 56.1
0 4502.41 1140.7
0.0
0 6773.29 1728.3
0.0
7 70091.67 14783.9 370.6
0
16
0
0
81
Yˆ  b0  b1 x1
78.0
174.8
500.8
342.4
440.0
x1 x 2
to predict salaries the basis of education only. (2)
2
b. Compute R . (2)
c. Compute s e . (2)
d. Compute
s b1
and do a significance test on
e. Compute
s b0
and do a confidence interval for
b1 (1.5)
b0 (1.5)
f. You are about to hire your nephew for the summer and want to know how much to pay him He has 14 years of
education. Using this create a prediction interval his salary. Explain why a confidence interval for the price is
inappropriate. (3)
g. Do an ANOVA table for the regression. What conclusion can you draw from the hypothesis test in the ANOVA? (2)
[22]
Extra credit from here on.
h. Do a multiple regression of price against education and sex. (5)
i. Compute R-squared and R-squared adjusted for degrees of freedom for this regression and
compare them with
the values for the previous problem. (2)
j. Using either R – squares or SST, SSR and SSE do F tests (ANOVA). First check the usefulness of the
simple
regression and then the value of ‘sex’ as an improvement to the regression. How should this impact Hotshot Associates’
discrimination problem? (Don’t say a word without referring to a statistical test.) (3)
k. Predict what you will pay your nephew now. How much change is there from your last prediction? (2)
Solution: a. Compute the regression equation Yˆ  b0  b1 x1 to predict salaries the basis of
education only. (2) Here is the modified data. I seem to think that Roland is a woman.
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
INC
39.0
43.7
62.6
42.8
55.0
60.6
59.4
57.1
56.5
53.5
55.7
58.8
64.1
58.8
62.5
60.0
72.9
56.1
67.1
82.3
108.9
1277.4
Solution: a) n  21,
EDUC
2
4
8
8
8
10
12
12
12
12
12
13
14
14
15
15
16
16
17
21
12
253
SEX
0
1
0
1
0
0
0
0
0
1
1
0
0
1
0
1
0
1
0
0
1
8
x1sq
4
16
64
64
64
100
144
144
144
144
144
169
196
196
225
225
256
256
289
441
144
3429
 y  1277.4,  x
1
x2sq
0
1
0
1
0
0
0
0
0
1
1
0
0
1
0
1
0
1
0
0
1
8
 253,
ysq
x1y
1521.0
78.0
1909.7
174.8
3918.8
500.8
1831.8
342.4
3025.0
440.0
3672.4
606.0
3528.4
712.8
3260.4
685.2
3192.3
678.0
2862.3
642.0
3102.5
668.4
3457.4
764.4
4108.8
897.4
3457.4
823.2
3906.3
937.5
3600.0
900.0
5314.4 1166.4
3147.2
897.6
4502.4 1140.7
6773.3 1728.3
11859.2 1306.8
81950.9 16090.7
x
2
 8,
x
2
1
x2y
0.0
43.7
0.0
42.8
0.0
0.0
0.0
0.0
0.0
53.5
55.7
0.0
0.0
58.8
0.0
60.0
0.0
56.1
0.0
0.0
108.9
479.5
 3429,
x
x1x2
0
4
0
8
0
0
0
0
0
12
12
0
0
14
0
15
0
16
0
0
12
93
2
2
 8,
35
252x0541 4/22/05
y
2
 81950.9,
 x y  116090.7,  x
1
 479.5,
2y
x x
1 2
 93.   .05 .
Note: Spare parts in this section were generated by my Minitab program 252OLS2. This program is
available from me if you wish to check your own solution.
 y  1277 .4  60.8286 , x   x
First, we compute y 
and
x2 
x
2
n

n
1
21
1
n

253
 12 .0476 (  x in simple regression),
21
8
 0.380952 . Then, we compute our spare parts:
21
SST  SSy 
y
 ny  81950 .9  2160 .8286 2  4248 .46 *
2
2
 x y  nx y  116090 .7  2112.0476 60.8286   701 .071 (  Sxy in simple regression)
Sx y   X Y  nX Y  479 .5  210.380952 60.8286   7.12857
SSx1   x12  nx12  3429  2112.04762  380.952 * (  SSx in simple regression)
SSx2   X 22  nX 22  8  210.3809522  4.95238 *
and Sx x   X X  nX X  93  2112.0476 0.380952   3.38095 .
Sx1 y 
1
1
2
2
1 2
2
1
2
1
2
Note: * These spare parts must be positive. The rest may well be negative.
We conclude that b1 
Sxy

SSx
 xy  nxy  701 .071  1.8403
 x  nx 380 .952
2
2
b0  y  b1 x  60.8286  1.8403 12.0476  38.6574
So Yˆ  b  b x becomes Yˆ  38.6574  1.8403 x .
0
1
2
b) Compute R . (2)
We already know that SST  SSy 
y
2
 ny  81950 .9  2160 .8286 2  4248 .46 ,
2
Sxy  701.071 and SSx  380 .952 .
 XY  nXY 
Sxy  701 .071 


 .3037 Hey! This stinks.

SSy 380 .952 4248 .46 
SSx
 X  nX  Y  nY 
2
R
2
2
2
2
2
2
2
c) Compute s e . (2) Recall that b1  1.8403
SSR  b1 Sxy  b1
 xy  nxy   1.8413 701 .071  1290 .9
or
SSR  b1 Sxy  R 2 SST   .30374248.46  1290.26
SSE  SST  SSR  4248 .46  1290 .26  2958 .20
or
s e2
s e2 
SSE 2528 .20

 155 .6947
n2
19
2
SST  b12 SSx 4248 .46  1.8402  380 .952  2858 .42


 155 .7067

19
19
n2
s e  155 .7  12 .4779
So
( s e2 is always positive!)
36
252x0541 4/22/05
d) Compute s b1 and do a significance test on b1 (1.5) Recall n  21 ,   .05 , SSx  380 .952 ,
s e2  155.7067 and b1  1.8403 .
 H 0 :  1   10
For most two-sided tests use t n2 2  t .19
use
025  2.093 . From the outline – “To test 
 H 1 :  1   10
b  10
. Remember  10 is most often zero – and if the null hypothesis is false in that case we
t 1
s b1
 1  155 .7067
say that 1 is significant.” s b21  s e2 
 0.4087 and s b  0.4087  0.6393 . So

1
 SS x  380 .952
b 
1.8403
t  1 10 
 2.879 . Much to my surprise, this is larger than the table value of t , so we
sb1
0.6393
can say that b1 is significant.
e) Compute s b0 and do a confidence interval for b0 (1.5) Recall n  21 , SSx  380 .952 , s e2  155.7067 ,
b0  38 .6574 and x  12.0476 .
1
s b20  s e2  
n


2


  155 .7067  1  12 .0467    155 .7067  1  0.38095   66 .7308
 21

380 .952 


X 2  nX 2 
 21

X2
s b  66 .7308  8.1689  0  b0  t  2 sb0  38.6574  2.0938.1689  38.66  17.09 .
0
Since the error part of the interval is smaller than b0  38 .6574 , we can conclude that the intercept
is significant too.
f) You are about to hire your nephew for the summer and want to know how much to pay him He has 14
years of education. Using this create a prediction interval his salary. Explain why a confidence interval for
the price is inappropriate. (3) Recall that Yˆ  38.6574  1.8403 x , n  21 , SSx  380 .952 , s e2  155.7067
and x  12.0476 .
1 X X 2

From the outline, the Prediction Interval is Y0  Yˆ0  t sY , where sY2  s e2   0
 1 . In
n

SS x


ˆ
this formula, for some specific X , Yˆ  b  b X . Here X  14 , Y  38.6574  1.8403 x

0
0
0
1
0
 38.6574  1.8403 14   64.4216 , i.e. $64,421.60. Then
1
sY2  s e2  
n

X 0  X 2
 X
2
 nX 2


0

 1 14  12 .0476 2

1

 1  155 .7067  
 1  155 .7067   .010006  1




21
380
.
952
21






 155 .7067 1.0576   164 .6790 and sY  164 .6790  12 .8327 , so that, if t n2 2  t .19
025  2.093
the prediction interval is Y0  Yˆ0  t sY  64.4216  2.093 12.8327   64.3  26.9 . The confidence
interval represents a confidence interval for the average value that Y will take when x  14 , while
the prediction interval is a confidence interval for an individual value and is thus more appropriate.
37
252x0541 4/22/05
g) Do an ANOVA table for the regression. What conclusion can you draw from the hypothesis test in the
ANOVA? (2)
Recall that SSR  1290 .26 , SSE  2958 .20 and SST  4248 .46 .
[22]
We can do a ANOVA table as follows:
Source
SS
DF
MS
F
Regression
SSR
MSR
MSR MSE
1
Error
MSE
SSE
n2
Total
SST
n 1
Source
Regression
Error
Total
1,19
Note that F.05
SS
DF
MS
F
1290.26
1
1290.26 8.287s
2958.20 19
155.69
4248.46 20
 4.38 so we reject the null hypothesis that x and y are unrelated. I have
indicated this by an ‘s’ next to the computed F. This is the same as saying that H 0 : 1  0 is false.
Extra credit from here on.
h) Do a multiple regression of price against education and sex. (5)
Here comes the fun! It’s time to repeat the whole spare parts box.
SST  SSy  4248.46 *, Sx1 y  701 .071 , Sx2 y  7.12857 , SSx1  380 .952 *,
SSx2  4.95238 *, and Sx1 x 2   3.38095 . Also y  60.8286 , x1  12.0476 and
x 2  0.380952 .
Note: * These spare parts must be positive. The rest may well be negative.
We substitute these numbers into the Simplified Normal Equations:
X 1Y  nX 1Y  b1
X 12  nX 12  b2
X 1 X 2  nX 1 X 2


 X Y  nX Y  b  X X
2
which are
2
1
1
2
 
 nX X   b  X
1
2
2
2
2

 nX  ,
2
2
701 .071  380 .952 b1  3.38095 b2

  7.12857   3.38095 b1  4.95238 b2
and solve them as two equations in two unknowns for b1 and b2 . These are a fairly tough pair of
 1.464789 we
3.38095
 1026 .97143  558 .01448 b1  4.95238 b2
get 4.95238. The equations become 
If we add these
  7.12857   3.38095 b1  4.95238 b2
1019 .84286
 1.83877 . Now
together, we get 1019 .84286  554 .63353 b1 . This means that b1 
554 .63353
remember that 701 .071  380 .952 b1  3.38095 b2 . This can be rewritten as
3.38095 b2  701 .071  380 .952 b1 . Let’s substitute
equations to solve until we notice that, if we multiply 3.38095 by 4.95238
b1  1.83877 . 3.38095 b2  701 .071  380 .952 1.83877   0.58789 . So
0.58789
 0.17385 . (It’s worth checking your work by substituting your values of
3.38095
b1 and b2 back into the normal equations.) Finally we get b0 by solving b0  Y  b1 X 1  b2 X 2
 60.8286  1.83877 12.0476   0.17385 0.380952   60.8286  22 .1528  0.0662  38 .6096 .
Thus our equation is Yˆ  b  b X  b X  38.6096  1.8388X  0.1739X .
b2  
0
1
1
2
2
1
2
38
252x0541 4/22/05
i) Compute R-squared and R-squared adjusted for degrees of freedom for this regression and compare them
with the values for the previous problem. (2) Recall SST  SSy  4248.46 *, Sx1 y  701 .071 , Sx 2 y
 7.12857 , SSx1  380 .952 *, SSx2  4.95238 *, and Sx1 x 2   3.38095 . Also y  60.8286 ,
x1  12.0476 and x 2  0.380952 and Yˆ  38.6096  1.8388X 1  0.1739X 2 . Also from the previous
regression R 2  R12  .3037 .
SSE  SST  SSR * and so
SSR  b1 Sx1 y  b2 Sx2 y  1.8388 701 .071  1.8388 7.12857   1289 .129  13.108
 1302 .23 * so SSE  4248 .46  1302 .23  2946 .23 *
SSR 1302 .23
R 2  R 22 

 .3065 *. If we use R 2 , which is R 2 adjusted for degrees of
SST 4248 .46
freedom, for the first regression, the number of independent variables was k  1 and
R2 
n  1R 2  k  20 0.3037  1  .2671
R2 
n  1R 2  k  20 0.3064  2  .2293 .
n  k 1
19
and for the second regression k  2 and
This is very bad. R-squared adjusted is supposed to
n  k 1
18
rise if our new variable has any explanatory power.
Note: * These numbers must be positive. The rest may well be negative.
j) Using either R – squares or SST, SSR and SSE do F tests (ANOVA). First check the usefulness of the
simple regression and then the value of ‘sex’ as an improvement to the regression. How should this impact
Hotshot Associates’ discrimination problem? (Don’t say a word without referring to a statistical test.) (3)
Recall SST  SSy  4248.46 , SSE  2946 .23 and SSR  1302 .23 .
The general format for a regression ANOVA table reads:
Source
SS
DF
MS
Fcalc
F
k
Regression
SSR
MSR MSR MSE F k , nk 1
n  k  1 MSE
Error
SSE
Total
SST
n 1
The ANOVA table from the first regression reads:
Source
SS
DF
MS
Regression 1290.26
1
1290.26
Error
2958.20 19
155.69
Total
4248.46 20
The ANOVA table from the second regression reads:
Source
SS
DF
MS
Fcalc
8.287s
Fcalc
F.05
1,19  4.38
F.05
F.05
2,18  3.55
F.05
Regression 1302.23
2
651.115 3.978s
Error
2946.23 18
163.68
Total
4248.46 20
Since our computed F is larger that the table F, we reject the hypothesis that the Xs and Y are
unrelated.
If we now divide the effects of the two independent variables, we get:
Source
SS
DF
MS
Fcalc
F.05
Educ
1290.26
1
1,19  4.38
Sex
11.97
1
11.97 0.0876ns F.05
Error
2946.23 18
163.68
Total
4248.46 20
Since our computed F is not larger that the table F, we do not reject the hypothesis that the second
independent variable (Sex) and Y are unrelated. But was this because I said Roland was female? If
this is correct Hotshot Associates doesn’t have much of a problem.
39
252x0541 4/22/05
k) Predict what you will pay your nephew now. How much change is there from your last prediction? (2) )
Recall SST  SSy  4248.46 , SSE  2946 .23 and SSR  1302 .23 . Also
Yˆ  38.6096  1.8388X  0.1739X
1
2
Since I forgot to include the rest of the question, which was to make the result into an approximate
prediction interval, I’ll do it anyway. We need to find s e . The best way to do this is to do an
ANOVA or remember that s e2 
SSE
2946 .23

 163 .679 .
n  k  1 21  2  1
18
So s e  163 .679  12 .7937 . For two sided tests and intervals we use t nk 1  t .025
 2.101 .
2
Our predicted earnings for a male with 14 years of education is
Yˆ  38.6096  1.838814  0.17390  64.3528
s
The outline says that an approximate confidence interval is  Y0  Yˆ0  t e
n
 64 .3528  2.10112.7937   64 .36  28.98 and an approximate prediction interval is
Y  Yˆ  t s  64.3528  2.10112.7937   64.4  29.0 .
0
0
e
Our previous prediction interval was Y0  Yˆ0  t sY  64.3  26.9 . When you consider the fact that
the larger error term wipes out the 0.1% increase in Yˆ , the prediction is essentially the same.
0
40
252x0541 4/22/05
4) An airport authority wants to compare training of air traffic controllers at three locations. Data is on the
next page. To personalize these data add the last two digits of your student number as a 9 th number to
column C.
a. Compare the performance of locations A, B, and C assuming that the underlying distribution is nonNormal. (4)
[26]
b. Use a one-way ANOVA to test the hypothesis of equal means. (5) It is legitimate to check your results by
computer, but I expect to see hand computations every step of the way.
[31]
c. (Extra Credit) Decide between the methods that you used in a) and b). To do this test for equal variances
and for Normality on the computer. What is your decision? Why?
(4)
You can do most of this with the following commands in Minitab if you put your data in 3 columns of
Minitab with A, B, and C above them.
MTB >
MTB >
SUBC>
SUBC>
MTB >
MTB >
AOVOneway A B C
stack A B C c11;
subscripts c12;
UseNames.
rank c11 c13
vartest c11 c12
#Does a 1-way ANOVA
# Stacks the data in c12, col.no. in c12.
#Puts the ranks of the stacked data in c13
#Does a bunch of tests, including Levene’s
On stacked data in c11 with IDs in c12.
MTB > Unstack (c13);
SUBC>
Subscripts c12;
SUBC>
After;
SUBC>
VarNames.
#Unstacks the ranks in the next 5 available
# columns. Uses IDs in c12.
MTB > NormTest 'A';
SUBC>
KSTest.
#Does a test (apparently Lilliefors)for Normality
# on column A.
Data for Problem 4
Row
A
B
C
1 96 65 60
2 82 74 73
3 88 72 85
4 70 66 61
5 90 79 79
6 91 82 85
7 87 73 88
8 88
79
This might help.
MTB > sum c1
Sum of A
Sum of A = 692
MTB > ssq c1
Sum of Squares of A
Sum of squares (uncorrected) of A = 60278
MTB > sum c2
Sum of B
Sum of B = 511
MTB > ssq c2
Sum of Squares of B
Sum of squares (uncorrected) of B = 37535
41
252x0541 4/22/05
a) Compare the performance of locations A, B, and C assuming that the underlying distribution is nonNormal. (4)
The modified data is shown below. The x s are the original numbers and the r s are their
ranks among the 20 data items. The Kruskal-Wallis Test is equivalent to one-way ANOVA when
the underlying distribution is non-normal. The null hypothesis is H 0 : Columns come from same
distribution or medians equal.
x1 r1
x 2
r2
x 3
r3
Row
1
2
3
4
5
6
7
8
9
A
96
82
88
70
90
91
87
88
C9_A
24.0
14.5
20.0
6.0
22.0
23.0
18.0
20.0
147.5
B
65
74
72
66
79
82
73
C9_B
4.0
10.0
7.0
5.0
12.0
14.5
8.5
61.0
So n1  8 , n 2  7 , n 3  9 , n 
C
60
73
85
61
79
85
88
79
23
n
i
C9_C
2.0
8.5
16.5
3.0
12.0
16.5
20.0
12.0
1.0
91.5
 24 , SR1  147 .5 , SR2  61 .0 and SR3  91 .5 .
To check the ranking, note that the sum of the four rank sums is 147.5 + 61.0 + 91.5 = 300, and
nn  1 24 25 

 300 .
that the sum of the first n  24 numbers is
2
2
 12
 SRi 2 

  3n  1
Now, compute the Kruskal-Wallis statistic H  
 nn  1 i  ni 
 12  147 .52 61 .02 91 .52 

  3 25    12 2719 .53125  531 .57143  930 .25000   75







24
25
8
7
9
 300 



 167 .254  75  92.254 . The Kruskal-Wallis table (Table 9) does not have values for tests with
columns this long, so we must use the  2 distribution, with df  m  1 , where m is the number of

columns. Since there are m  3 columns and   .05 , test H with  .2052   5.9915 . Since H is
larger than  .205 , reject the null hypothesis of equal medians.
[26]
b) Use a one-way ANOVA to test the hypothesis of equal means. (5) It is legitimate to check your results by
computer, but I expect to see hand computations every step of the way.
[31]
Here’s what the computer output said.
One-way ANOVA: A, B, C
Source
Factor
Error
Total
DF
2
21
23
SS
1228
3986
5214
MS
614
190
F
3.23
P
0.060
Can we get similar results by hand?
42
252x0541 4/22/05
Using the tableau suggested in the outline with a modification for unequal column lengths, we get
the following.
A
B
C
Sum
1
2
3
96
65
60
82
74
73
88
72
85
70
66
61
90
79
79
91
82
85
87
73
88
88
---79
-----23
Sum
 1836 
x ij
692 +
511 +
633

8+
nj
86.5
SS
60278 +
37535 +
47855
n j x2j
59858 +
37303 +
44521
73
70.333
 24  n
1836
 76 .5  x
24
 145668 
 x
 141682   n x
2   xij2  nx 2  145668  2476.52  5214
2
2
2
2
. j  x    n j x. j  nx  141682  24 76 .5  1228
ij
2
ij
2
j j
x
SSW  SST  SSB  3986
Source
SS
Between
9
x j
 x
SSB   x
SST 
7+
1228
DF
MS
F
F.05
2
614
3.235ns
F 2,21  3.47
H0
Column means equal
Within
3986
21
189.81
Total
5214
23
Since our computed F is smaller that the table F, we cannot reject the null hypothesis that the
locations’ performance is similar.
c. (Extra Credit) Decide between the methods that you used in a) and b). To do this test for equal variances
and for Normality on the computer. What is your decision? Why?
(4)
Here are my computer results.
MTB > vartest c7 c8
Test for Equal Variances: C7 versus C8
95%
C8
A
B
C
Bonferroni confidence intervals for standard deviations
N
Lower
StDev
Upper
8
4.7074
7.7460 18.9708
7
3.6650
6.2183 16.8724
9 12.7262 20.4145 46.2301
Bartlett's Test (normal distribution)
Test statistic = 10.63, p-value = 0.005
Levene's Test (any continuous distribution)
Test statistic = 1.61, p-value = 0.223
Test for Equal Variances: C7 versus C8
43
252x0541 4/22/05
Test for Equal Variances for C7
Bartlett's Test
Test Statistic
P-Value
A
10.63
0.005
Lev ene's Test
C8
Test Statistic
P-Value
1.61
0.223
B
C
0
10
20
30
40
95% Bonferroni Confidence Intervals for StDevs
50
Well, the intervals seem to overlap and Levene’s test has a p-value above 5%, which means we
cannot reject the hypothesis of equal variances. But the Bartlett Test is worrisome, since it says
that if the distribution is not Normal, the variances are not equal.
44
252x0541 4/22/05
MTB > normtest 'A';
SUBC> KSTest.
Probability Plot of A
Probability Plot of A
Normal
99
Mean
StDev
N
KS
P-Value
95
90
Percent
80
70
60
50
40
30
20
10
5
1
70
75
80
85
90
95
100
105
A
At this point I am inclining toward the Kruskal-Wallis test as being the right one. ANOVA
presumes a Normal parent distribution and equal variances. The p-value of the ‘K-S’ test is above
5%, so we cannot reject the Normality assumption, but then the Bartlett test says the variances are
not equal. If we use a 10% significance level instead of 5%, we would reject the Normality
assumption and use the results of the Levene test. But if the distribution is not Normal, equal
variances won’t push us toward ANOVA.
45
86.5
7.746
8
0.276
0.073