252x0581 12/14/05
ECO252 QBA2
Final EXAM
December 14, 2005
Name and Class hour:_________________________
I. (12+ 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)
Berenson et. al. present a file called RESTRATE. It contains Zagat ratings for 50 restaurants in New York
City and 50 more restaurants on Long Island. The data columns are described below.
‘Location’
New York or Long Island
‘Food’
Quality of food on a 0-25 scale
‘Décor’
Quality of décor on a 0-25 scale
‘Service’
Quality of service on a 0-25 scale
‘Summated Rating’
The sum of the food, décor, and service variables.
‘Locate’
A dummy variable – 1 if the restaurant is on Long Island
‘Price’
Average price for a meal ($), the dependent variable.
‘Inter’
The product of ‘Location’ and ‘Summated Rating’
The values of these data and the correlation matrix between them appear at the end of this section. We are
trying to explain price based on the Zagat rating and the dummy variable for location. The regression is run
as follows. Assume a 5% significance level.
Regression Analysis: Price versus Locate, Summated rating
The regression equation is
Price = - 13.7 - 7.54 Locate + 0.961 Summated rating
Predictor
Constant
Locate
Summated rating
S = 5.94216
Coef
-13.699
-7.537
0.96079
SE Coef
5.054
1.197
0.08960
R-Sq = 59.2%
Analysis of Variance
Source
DF
SS
Regression
2 4960.2
Residual Error 97 3425.0
Total
99 8385.2
Source
Locate
Summated rating
DF
1
1
T
-2.71
-6.30
10.72
P
0.008
0.000
0.000
VIF
1.0
1.0
R-Sq(adj) = 58.3%
MS
2480.1
35.3
F
70.24
P
0.000
Seq SS
900.0
4060.2
a) What is the difference (in dollars) in expected price of a meal at restaurants of similar quality in New
York and Long Island? (1)
b) How much would you expect to pay for a meal at a New York restaurant with a Zagat (summated) rating
of 50? (1)
c) Which of the coefficients are significant? Do not answer this without evidence from the printout. (2)
d) Compute the coefficient of partial determination (partial correlation squared) for ‘locate’ and explain its
meaning. (2)
1
252x0581 12/14/05
Now the authors suggest that we add the interaction term to the equation. The new result is.
Regression Analysis: Price versus Locate, Summated rating, Inter
The regression equation is
Price = - 26.3 + 13.1 Locate + 1.19 Summated rating - 0.368 Inter
Predictor
Constant
Locate
Summated rating
Inter
S = 5.84913
Coef
-26.291
13.13
1.1872
-0.3676
SE Coef
7.957
10.26
0.1423
0.1813
R-Sq = 60.8%
DF
1
1
1
P
0.001
0.204
0.000
0.045
R-Sq(adj) = 59.6%
Analysis of Variance
Source
DF
SS
Regression
3 5100.9
Residual Error 96 3284.4
Total
99 8385.2
Source
Locate
Summated rating
Inter
T
-3.30
1.28
8.34
-2.03
MS
1700.3
34.2
F
49.70
P
0.000
Seq SS
900.0
4060.2
140.6
d) Is this a better regression than the previous one? In order to answer this comment on R-squared, Rsquared adjusted and the sign and significance of the coefficients. (3)
[9]
At this point I was on my own. First I ran a stepwise regression and got the following.
Stepwise Regression: Price versus Food, Décor, ...
Alpha-to-Enter: 0.15
Alpha-to-Remove: 0.15
Response is Price on 6 predictors, with N = 100
Step
Constant
Summated rating
T-Value
P-Value
1
-13.66
2
-18.40
3
-15.14
4
-15.82
0.893
8.50
0.000
1.047
11.51
0.000
1.323
9.14
0.000
1.953
4.53
0.000
-0.137
-6.57
0.000
-0.137
-6.74
0.000
-0.139
-6.85
0.000
-0.93
-2.42
0.018
-1.85
-2.62
0.010
Inter
T-Value
P-Value
Food
T-Value
P-Value
Décor
T-Value
P-Value
S
R-Sq
R-Sq(adj)
-0.93
-1.55
0.125
7.02
42.46
41.87
5.87
60.16
59.34
5.73
62.44
61.27
5.69
63.37
61.83
More? (Yes, No, Subcommand, or Help)
SUBC> n
The results here don’t seem very practical. For example the fourth version of the regression this gives me is
Price = - 15.82 + 1.953 Summated rating – 0.139 Inter – 1.85 Food -0.93 Décor
Note that the number under the coefficient is a t-ratio and the number under that is a p-value for the
significance test.
e) What are the two ‘best’ variables the stepwise regression picks? Why might I be reluctant to add ‘Food’
and ‘Décor’ in spite of evidence in the significance tests? (2) [11]
2
252x0581 12/14/05
So, the next version of the regression that I did was.
MTB > regress c7 2 locate inter;
SUBC> VIF.
Regression Analysis: Price versus Locate, Inter
The regression equation is
Price = 39.7 - 52.9 Locate + 0.820 Inter
Predictor
Constant
Locate
Inter
Coef
39.740
-52.896
0.8196
S = 7.64274
SE Coef
1.081
8.541
0.1469
R-Sq = 32.4%
Analysis of Variance
Source
DF
SS
Regression
2 2719.3
Residual Error 97 5665.9
Total
99 8385.2
Source
Locate
Inter
DF
1
1
T
36.77
-6.19
5.58
P
0.000
0.000
0.000
VIF
31.2
31.2
R-Sq(adj) = 31.0%
MS
1359.7
58.4
F
23.28
P
0.000
Seq SS
900.0
1819.3
This was pretty much as I expected, so I added the components of Zagat’s rating to the equation.
MTB > regress c7 5 c6 c8 c2 c3 c4;
SUBC> VIF.
Regression Analysis: Price versus Locate, Inter, Food, Décor, Service
The regression equation is
Price = - 21.1 + 8.8 Locate - 0.294 Inter + 0.251 Food + 1.15 Décor + 1.97 Service
Predictor
Constant
Locate
Inter
Food
Décor
Service
Coef
-21.130
8.84
-0.2937
0.2505
1.1465
1.9678
S = 5.69379
SE Coef
7.979
10.22
0.1804
0.3766
0.2798
0.4322
R-Sq = 63.7%
Analysis of Variance
Source
DF
SS
Regression
5 5337.8
Residual Error 94 3047.4
Total
99 8385.2
Source
Locate
Inter
Food
Décor
Service
DF
1
1
1
1
1
T
-2.65
0.86
-1.63
0.67
4.10
4.55
P
0.009
0.390
0.107
0.508
0.000
0.000
VIF
80.6
84.9
2.7
2.3
3.2
R-Sq(adj) = 61.7%
MS
1067.6
32.4
F
32.93
P
0.000
Seq SS
900.0
1819.3
371.6
1574.8
672.1
f) Use an F test to compare
Price = 39.7 - 52.9 Locate + 0.820 Inter
with
Price = - 21.1 + 8.8 Locate - 0.294 Inter + 0.251 Food + 1.15 Décor + 1.97 Service
What does the F test show about the 3 variables that we added? (3)
[14]
3
252x0581 12/14/05
g) Now compare
Price = - 21.1 + 8.8 Locate - 0.294 Inter + 0.251 Food + 1.15 Décor + 1.97 Service
with
Price = - 26.3 + 13.1 Locate + 1.19 Summated rating - 0.368 Inter
You can’t use an F test here but you can look at R-squared and significance. Does the equation that I just
fitted look like an improvement? As always, give reasons. (2).
[16]
h) I don’t think I am finished. Should I drop some variables from
?
[18]
Price = - 21.1 + 8.8 Locate - 0.294 Inter + 0.251 Food + 1.15 Décor + 1.97 Service
Why? Which would you suggest that I drop first? Why? Check VIFs and significance. (2)
4
252x0581 12/14/05
————— 12/6/2005 8:46:21 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > print c1-c8
Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
Location
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
NYC
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
Food
19
18
19
23
23
23
20
20
19
21
20
21
24
20
17
21
21
20
17
21
23
17
22
19
21
19
19
21
24
19
17
19
22
22
14
22
20
18
18
24
21
18
20
21
19
18
20
21
19
21
21
17
17
23
23
21
21
21
22
22
23
23
21
17
23
15
Décor
21
17
16
18
20
18
17
15
18
19
17
23
20
17
18
17
19
16
11
16
20
19
14
19
19
14
17
13
21
16
15
16
19
18
15
22
15
14
20
18
17
17
19
10
14
17
16
12
17
20
18
14
17
19
22
18
19
18
18
20
20
18
14
17
23
17
Service
18
17
19
21
21
20
16
17
18
19
16
21
22
20
14
20
21
19
13
20
23
16
15
18
20
16
19
21
21
19
15
19
21
20
15
21
18
17
16
21
18
17
19
17
19
17
17
14
19
20
21
17
18
18
21
19
23
18
20
20
22
20
19
17
22
15
Summated
rating Locate
58
0
52
0
54
0
62
0
64
0
61
0
53
0
52
0
55
0
59
0
53
0
65
0
66
0
57
0
49
0
58
0
61
0
55
0
41
0
57
0
66
0
52
0
51
0
56
0
60
0
49
0
55
0
55
0
66
0
54
0
47
0
54
0
62
0
60
0
44
0
65
0
53
0
49
0
54
0
63
0
56
0
52
0
58
0
48
0
52
0
52
0
53
0
47
0
55
0
61
0
60
1
48
1
52
1
60
1
66
1
58
1
63
1
57
1
60
1
62
1
65
1
61
1
54
1
51
1
68
1
47
1
Price
50
38
43
56
51
36
25
33
41
44
34
39
49
37
40
50
50
35
22
45
44
38
14
44
51
27
44
39
50
35
31
34
48
48
30
42
26
35
32
63
36
38
53
23
39
45
37
31
39
53
37
37
29
38
37
38
39
29
36
38
44
27
24
34
44
23
Inter
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
60
48
52
60
66
58
63
57
60
62
65
61
54
51
68
47
5
252x0581 12/14/05
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
LI
19
20
20
20
23
19
15
20
21
23
27
17
22
20
20
25
17
25
19
27
21
19
20
23
24
18
15
16
18
20
21
21
23
19
14
19
15
12
19
21
13
17
17
20
16
17
11
16
12
25
17
22
18
20
11
18
21
19
27
18
16
20
16
12
24
18
15
14
17
18
17
18
20
19
15
22
18
21
19
16
17
19
16
24
18
23
19
24
17
19
20
21
23
20
14
17
17
18
21
19
20
16
50
57
52
50
62
59
43
59
56
64
62
50
50
55
48
74
52
70
56
71
49
56
61
63
74
56
45
53
51
50
66
58
58
49
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
30
32
25
29
43
31
26
34
23
41
32
30
28
33
26
51
26
48
39
55
24
38
31
30
51
30
27
38
26
28
33
38
32
25
50
57
52
50
62
59
43
59
56
64
62
50
50
55
48
74
52
70
56
71
49
56
61
63
74
56
45
53
51
50
66
58
58
49
MTB > Correlation c2 c3 c4 c5 c6 c8.
Correlations: Food, Décor, Service, Summated rating, Locate, Inter
Food
0.374
0.000
Décor
Service
0.727
0.000
0.633
0.000
Summated rat
0.800
0.000
0.824
0.000
0.914
0.000
Locate
0.089
0.381
0.084
0.406
0.137
0.175
0.120
0.235
Inter
0.203
0.043
0.201
0.045
0.253
0.011
0.257
0.010
Décor
Service
Summated rat
Locate
0.984
0.000
6
252x0581 12/14/05
II. Do at least 4 of the following 6 Problems (at least 10 each) (or do sections adding to at least 38 points –
(Anything extra you do helps, and grades wrap around). You must do parts a) and b) of problem 1. 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 151 points, but 70 is considered a perfect score.
1. a) If I want to test to see if the mean of x 2 is larger than the mean of x1 my null hypotheses are:
i) 1   2 and D  0
v) 1   2 and D  0
ii) 1   2 and D  0
iii) 1   2 and D  0
vi) 1   2 and D  0
vii) 1   2 and D  0
iv) 1   2 and D  0
viii) 1   2 and D  0 (2)
Let us revisit Problem B in the take-home. We are going to evaluate the evaluators.
Candidate
1
2
3
4
5
6
7
8
9
Moore
52
25
29
33
24
36
42
49
20
Gaston Difference
38
14
31
-6
24
5
29
4
27
-3
28
8
41
1
27
22
31
-11
Assume that we can use the Normal distribution
for these data. There are 4 columns here: first,
the number of the candidate; second Moore’s
evaluation of the candidate; third Gaston’s
evaluation of the same candidates. Finally, we
have the difference between the ratings. Don’t
forget that the data are cross-classified. Note that
x1  310
the sums for Moore’s column are
and

they are

x12
 11696 . For the difference column,
 d  34 and  d
2
 952
b) Compute the mean and standard deviation for Gaston. (2) Show your work!
c) The mean rating that has been given to hundreds of job candidates interviewed over the last year is 35.
Regard Gaston’s ratings as a random sample from a Normal population and test that his mean rating is
below 35. Use (i) Either a test ratio or a critical value for the mean (3) and (ii) an appropriate confidence
interval. (2)
d) Test the hypothesis that the population mean for Moore is higher than for Gaston. Don’t forget that the
data is cross-classified. (3)
e) To see how well they agree, compute a correlation between Moore’s and Gaston’s ratings and check it
for significance. (5)
[17]
7
252x0581 12/14/05
2. Of course it was nonsense to assume that the data was Normally distributed in Problem 1.
Candidate
1
2
3
4
5
6
7
8
9
Moore
52
25
29
33
24
36
42
49
20
Gaston Difference
38
14
31
-6
24
5
29
4
27
-3
28
8
41
1
27
22
31
-11
a. But, just in case, test Moore’s column for Normality. Remember that the sample mean and variance were
calculated from the data. (5)
b. Now test that the median for Gaston is below 35. If you can do it, use the Wilcoxon signed rank test. (4)
c. Now test to see if the median for Moore is higher than the median for Gaston. Don’t forget that the data is
cross-classified. (4)
d. Compute a rank correlation between Moore and Gaston’s ratings and test it for significance. (5) [13]
8
252x0581 12/14/05
3. (Lind et al -190) We are trying to locate a new day-care center. We want to know if the proportion of
parents who are eligible to put children in day care is larger on the south side of town than on the east side
of town.
a) If the south side is Area 1 and the north side is area 2, state the null and alternate hypotheses and do the
test. Answer the question “Should we put the center on the south side?” with a yes, no or maybe
depending on your result. (4)
b) Find a p-value for the null hypothesis in a) (2)
c) Do a 2-sided 99% confidence interval for the
South East
proportion eligible on the South side. (2)
Number eligible
88
57
Sample size
200
150
d) Do a 2-sided 89% confidence interval for the
proportion eligible on the South side. Do not use
a value of z from the t table! (1)
e) We suddenly realize that out of 100 parents
surveyed on the North side 51 are eligible. Do a
test of the hypothesis that the proportion is the
same in all 3 areas. (5)
f) Use the Marascuilo procedure to determine
whether there is a significant difference between
the areas with the largest and second largest
proportion of eligible parents. (3)
[17]
9
252x0581 12/14/05
4. From the data on the right find the following.
xy (1)
a)

b) R 2 (2)
c) The sample correlation rxy (1)
d) Test the hypothesis that the population
correlation between x and y is 0.75 (5)
e) Compute a simple regression of x against y .
Remember that y is the dependent variable. (4)
[13]
Row
1
2
3
4
5
6
7
8
9
10
X
3
8
6
9
6
9
6
4
8
5
Y
22
68
68
96
46
80
52
38
78
48
 X  64,  X
 Y  596.  Y
2
 448
2
 40000
10
252x0581 12/14/05
5. From the data in Problem 4 find the following
a) s e (3)
b) Do a significance test for the slope b1 (3)
c) Find a confidence interval for the constant
b0 (3)
d) Do an ANOVA for this regression and explain
its meaning. (3)
e) Find the value of Y when X is 9 and build a
prediction interval around it (3)
[15]
Row
1
2
3
4
5
6
7
8
9
10
X
3
8
6
9
6
9
6
4
8
5
Y
22
68
68
96
46
80
52
38
78
48
 X  64,  X
 Y  596.  Y
2
 448
2
 40000
11
252x0581 12/14/05
6. Do the following
a) Assume that n1  10, n 2  12, n3  15, s12  20, and s 22  15, s32  10,   .10 and that all come
from a Normal distribution. Test the following:
(i)
 1  15 (2)
(ii)  1   2 (2)
(iii) Explain what test would be appropriate to test  1   2   3 (1)
b) Read both parts of this question before you start.
To test the response of an 800 number, I make 20
attempts to reach the number, continuing to call
until I get through. (i) I hypothesize that the
results follow a Poisson distribution with an
unknown mean. Test this – data are at right. (6)
(ii) I hypothesize that the results follow a Poisson
distribution with a mean of 2 (5) Do not do these
two parts using the same method.
[16]
Number of Unsuccessful
Tries before success.
0
1
2
3
4
5
6
7
Observed
Frequency
4
2
8
12
10
0
2
2
40
12
252x0581 12/07/05
(Blank)
13
252x0581 12/07/05
ECO252 QBA2
Final EXAM
December 14-16, 2005
TAKE HOME SECTION
Name: _________________________
Student Number: _________________________
Class days and time: _________________________
III Take-home Exam (20+ points)
A) 4th computer problem (5+)
This is an internet project. You should do only one of the following 2 problems.
Problem 1: In his book, Statistics for Economists: An Intuitive Approach (New York, HarperCollins,
1992), Alan S. Caniglia presents data for 50 states and the District of Columbia. These data are presented as
an appendix at the end of this section.
The Data consists of six variables.
The dependent variable, MIM, the mean income of males (having income) who are 18 years of age or older.
PMHS, the percent of males 18 and older who are high school graduates.
PURBAN, the percent of total population living in an urban area.
MAGE, the median age of males.
Using his data, I got the results below.
Regression Analysis: MIM versus PMHS
The regression equation is
MIM = 2736 + 180 PMHS
Predictor
Constant
PMHS
Coef
2736
180.08
S = 1430.91
SE Coef
2174
31.31
R-Sq = 40.3%
T
1.26
5.75
P
0.214
0.000
R-Sq(adj) = 39.1%
Analysis of Variance
Source
DF
SS
Regression
1
67720854
Residual Error 49 100328329
Total
50 168049183
MS
67720854
2047517
F
33.07
P
0.000
Unusual Observations
Obs PMHS
MIM
Fit SE Fit Residual St Resid
1 69.1 12112 15180
200
-3068
-2.17R
3 71.6 12711 15630
215
-2919
-2.06R
50 81.9 21552 17485
447
4067
2.99R
R denotes an observation with a large standardized residual.
His only comment is that a 1% increase in the percent of males that are college graduates results is
associated with about a $180 increase in male income and that there is evidence here that the relationship is
significant.
He then describes three dummy variables: NE = 1 if the state is in the Northeast (Maine through
Pennsylvania in his listing); MW = 1 if the state is in the Midwest (Ohio through Kansas) and SO = 1 if the
state is in the South (Delaware through Texas). If all of the dummy variables are zero, the state is in the
West (Montana through Hawaii). I ran the regression with all six independent variables. To check these
variables, look at his data.
MTB > regress c2 6 c3-c8;
SUBC> VIF;
SUBC> brief 2.
Regression Analysis: MIM versus PMHS, PURBAN, MAGE, NE, MW, SO
The regression equation is
MIM = - 1294 + 198 PMHS + 49.4 PURBAN - 42 MAGE + 247 NE + 757 MW + 1269 SO
14
252x0581 12/07/05
Predictor
Constant
PMHS
PURBAN
MAGE
NE
MW
SO
Coef
-1294
198.13
49.36
-42.1
246.6
756.7
1268.9
S = 1271.71
SE Coef
5394
53.97
14.27
151.6
723.7
608.2
863.0
R-Sq = 57.7%
T
-0.24
3.67
3.46
-0.28
0.34
1.24
1.47
DF
1
1
1
1
1
1
VIF
3.8
1.4
1.5
2.4
2.1
5.2
R-Sq(adj) = 51.9%
Analysis of Variance
Source
DF
SS
Regression
6
96890414
Residual Error 44
71158768
Total
50 168049183
Source
PMHS
PURBAN
MAGE
NE
MW
SO
P
0.811
0.001
0.001
0.783
0.735
0.220
0.149
MS
16148402
1617245
F
9.99
P
0.000
Seq SS
67720854
23781889
281110
1416569
193443
3496549
Unusual Observations
Obs PMHS
MIM
Fit SE Fit Residual St Resid
50 81.9 21552 16999
543
4553
3.96R
R denotes an observation with a large standardized residual.
He has asked whether region affects the independent variable, on the strength of the significance tests in the
output above, he concludes that the regional variables do not have any affect on male income. (Median Age
looks pretty bad too.)
There are two ways to confirm these conclusions. Caniglia does one of these, an F test that shows whether
the regional variables as a group have any effect. He says that they do not. Another way to test this is by
using a stepwise regression.
MTB > stepwise c2 c3-c8
Stepwise Regression: MIM versus PMHS, PURBAN, MAGE, NE, MW, SO
Alpha-to-Enter: 0.15 Alpha-to-Remove: 0.15
Response is MIM on 6 predictors, with N = 51
Step
Constant
1
2736
2
2528
PMHS
T-Value
P-Value
180
5.75
0.000
134
4.46
0.000
PURBAN
T-Value
P-Value
S
R-Sq
R-Sq(adj)
Mallows C-p
50
3.86
0.000
1431
40.30
39.08
15.0
1263
54.45
52.55
2.3
More? (Yes, No, Subcommand, or Help)
SUBC> y
No variables entered or removed
More? (Yes, No, Subcommand, or Help)
SUBC> n
What happens is that the computer picks PMHS as the most valuable independent variable, and gets the
same result that appeared in the simple regression above. It then adds PURBAN and gets
MIM = 2528 + 134 PMHS + 50 PURBAN. The coefficients of the 2 independent variables are significant, the
adjusted R-Sq is higher than the adjusted R-sq with all 6 predictors and the computer refuses to add any
more independent variables. So it looks like we have found our ‘best’ regression. (See the text for
interpretation VIFs and C-p’s.)
15
252x0581 12/07/05
So here is your job. Update this work. Use any income per person variable, a mean or a median for men,
women or everybody. Find measures of urbanization or median age. Fix the categorization of states if you
don’t like it. Regress state incomes against the revised data. Remove the variables with insignificant
coefficients. If you can think of new variables add them. (Last year I suggested trying percent of output or
labor force in manufacturing.) Make sure that you pick variables that can be compared state to state.
Though you can legitimately ask whether size of a state affects per capita income, using total amount
produced in manufacturing is poor because it’s just going to be big for big states. 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. For data sources, try the sites mentioned in 252Datalinks.
Use F tests for adding the regional variables and use stepwise regression. Don’t give me anything you don’t
understand.
Problem 2: Recently the Heritage Foundation produced the graph below.
What I want to know is if you can develop an equation relating per capita income (the dependent variable)
and Economic freedom x  . Because it is pretty obvious that a straight line won’t work, you will probably
need to create a x 2 variable too. But I would like to know what parts of ‘economic freedom’ affect per
capita income. In addition to the Heritage Foundation Sources, the CIFP site mentioned in 252datalinks,
and the CIA Factbook might provide some interesting independent variables. You should probably use a
sample of no more than 50 countries and it’s up to you what variables to use. You are, of course, looking
for significant coefficients and high R-squares. For instructions on how to do a regression, try the material
in Doing a Regression.
16
252x0581 12/07/05
B. Do only Problem 1 or problem 2. (Problem Due to Donald R Byrkit). Four different job candidates are
interviewed by seven executives. These are rated for 7 traits on a scale of 1-10 and the scores are added
together to create a total score for each candidate-rater pair that is between 0 and 70. The results appear
below.
Row
1
2
3
4
5
6
7
Sum
Sum
Sum
Sum
Sum
Sum
Raters
Moore
Gaston
Heinrich
Seldon
Greasy
Waters
Pierce
of
of
of
of
of
of
Lee
52
38
54
43
58
36
52
Candidates
Jacobs
25
31
38
30
44
28
41
Wilkes
29
24
40
31
46
22
37
Delap
33
29
39
28
47
25
45
Jacobs = 237
squares (uncorrected) of Jacobs = 8331
Wilkes = 229
squares (uncorrected) of Wilkes = 7947
Delap = 246
squares (uncorrected) of Delap = 9094
Personalize the data by adding the second to last digit of your student number to Lee’s column. For example
Roland Dough’s student number is 123689, so he uses 52 + 8 = 60, 38 + 8 = 46, 62 etc. If the second to last
digit of your student number is zero, add 10.
Problem 1: a) Assume that a Normal distribution applies and use a statistical procedure to compare the
column means, treating each column as an independent random sample. If you conclude that there is a
difference between the column means, use an individual confidence interval to see if there is a significant
difference between the best and second-best candidate. If you conclude that there is no difference between
the means, use an individual confidence interval to see if there is a significant difference between the best
and worst candidate. (6)
b) Now assume that a Normal distribution does not apply but that the columns are still independent random
samples and use an appropriate procedure to compare the column medians. (4)
[16]
Problem 2: a) Assume that a Normal distribution applies and use a statistical procedure to compare the
column means, taking note of the fact that each row represents one executive. If you conclude that there is a
difference between the column means, use an individual confidence interval to see if there is a significant
difference between the best and second-best candidate. If you conclude that there is no difference between
the column means, use an individual confidence interval to see if there is a significant difference between
the kindest and least kind executive. (8)
b) Now assume that a Normal distribution does not apply but that each row represents the opinion of one
rater and use an appropriate procedure to compare the column medians. (4)
c) Use Kendall’s coefficient of concordance to show how the raters differ and do a significance test. (3)
Problem 3: (Extra Credit) Decide between the methods used in Problem 1 and Problem 2. 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, C and D above them.
MTB >
MTB >
SUBC>
SUBC>
MTB >
MTB >
AOVOneway A B C D
stack A B C D C11;
subscripts C12;
UseNames.
rank C11 C13
vartest C11 C12
MTB > Unstack (c13);
SUBC>
Subscripts c12;
SUBC>
After;
SUBC>
VarNames.
MTB > NormTest 'A';
SUBC>
KSTest.
#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.
#Unstacks the ranks in the next available
# columns. Uses IDs in c12.
#Does a test (apparently Lilliefors) for Normality
# on column A.
17
252x0581 12/07/05
C. You may do both problems. These are intended to be done by hand. A table version of the data for
problem 2 is provided in 2005data1 which can be downloaded to Minitab. I do not want Minitab results for
these data except for Problem 2e.
Problem 1: Using data from the 1970s and 1980s, Alan S. Caniglia calculated a regression of
nonresidential investment on the change in level of final sales to verify the accelerator model of investment.
This theory says that because capital stock must be approximately proportional to production, investment
will be driven by changes in output. In order to check his work I put together a data set 2005series. The last
two years of the series are in Exhibit C1 below.
Exhibit C1
Row Date
73 1988 01
74 1988 02
75 1988 03
76 1988 04
77 1989 01
78 1989 02
79 1989 03
80 1989 04
RPFI
862.406
879.330
882.704
891.502
900.401
901.643
917.375
902.298
Sales
6637.22
6716.38
6749.47
6835.07
6873.33
6933.55
7015.34
7026.76
Sales-4Q
6344.41
6431.37
6510.82
6542.55
6637.22
6716.38
6749.47
6835.07
Change
292.815
285.006
238.644
292.522
236.106
217.171
265.876
191.695
DEFL %Y
2.897
3.318
3.699
3.724
4.013
4.016
3.596
3.537
MINT %
9.88
9.67
9.96
9.51
9.62
9.79
8.93
8.92
RINT
6.983
6.352
6.261
5.786
5.607
5.774
5.334
5.383
‘Date’ consists of the year and the quarter. ‘RPFI’ consists of real fixed private investment from
2005InvestSeries1. ‘Sales’ consists of sales data (actually a version of gross domestic product) from
2005SalesSeries1. ‘Sales-4Q’ (Sales 4 Quarters earlier’ is also sales data from 2005SalesSeries1, but is the
data of one year earlier. (Note that the 1989 numbers in ‘Sales-4Q’ are identical to the 1988 numbers in
‘Sales.’ ‘Change’ is ‘Sales’ – ‘Sales-4Q. ‘DEFL %Y’ is the percent change in the gross domestic deflator
over the last year (a measure of inflation) taken from 2005deflSeries1. ‘MINT %’ is an estimate of the
percent return on Aaa bonds taken from 2005intSeries1. Only the values for January, April, July and
October are used since quarterly data was not available. ‘RINT’ (an estimate of the real interest rate) is
‘MINT %’ - ‘DEFL %Y’.
These are manipulated in the input to the regression program as in Exhibit C2 below.
Exhibit C2
Row Time
73 1988 01
74 1988 02
75 1988 03
76 1988 04
77 1989 01
78 1989 02
79 1989 03
80 1989 04
Y
86.2406
87.9330
88.2704
89.1502
90.0401
90.1643
91.7375
90.2298
X1
29.2815
28.5006
23.8644
29.2522
23.6106
21.7171
26.5876
19.1695
X2
6.98
6.35
6.26
5.79
5.61
5.77
5.33
5.38
Here Y is ‘RFPI’ divided by 10. X1 is ‘Change’ divided by 10. X2 is ‘RINT’ rounded to eliminate the last
decimal place. If you don’t understand how I got Exhibit C2 from Exhibit C1 find out before you go
any further,
Personalize the data by adding one year (four values) to the data in 2005 series. Pick the year to be added
by adding the last digit of your student number to 1990. Make sure that I know the year you are using. Then
get, for your year, ‘RPFI’ from 2005InvestSeries1, ‘Sales’ from 2005SalesSeries1, ‘Sales-4Q’ from
2005SalesSeries1 (Make sure that you use the sales of one year earlier, not 1989 unless your year is 1990.),
‘DEFL %Y’ 2005deflSeries1 and ‘MINT %’ from 2005intSeries1. Calculate ‘Change’ by subtracting
‘Sales-4Q’ from ‘Sales.’ If you are going to do Problem 2, calculate ‘RINT’ by subtracting ‘DEFL %Y’
from ‘MINT %.’ Present your four rows of new values in the format of Exhibit C1. Now manipulate your
numbers to the form in Exhibit C2 and again present your four rows of numbers. These are observations 81
through 84.
Now it’s time to compute your spare parts. The following are computed for you from the data for 1970
through 1989.
18
252x0581 12/07/05
Sum of Y = 5323.20
Sum of X1 = 1283.42
Sum of X2 = 328.33
Sum of Ysq = 371032
Sum of X1sq = 30307.57
Sum of X2sq = 2080.65
Sum of X1Y = 92676.9
Sum of X2Y = 24188.2
Sum of X1X2 = 6324.09
 Y  5323 .2
 X 1  1283.42
 X 2  328.33
 Y  371032
 X  30307 .6
 X  2080 .65
 X 1 Y  92676.9
 X 2 Y 24188.2
 X 1 X 2  6324.09
n  80
2
2
1
2
2
Add the results of your data to these sums (You only need the sums involving X1 and Y if you are not doing
Problem 2.) (Show your work!) and do the following.
a. Compute the regression equation Yˆ  b0  b1 x1 to predict investment on the basis of change in
sales only. (2)
b. Compute R 2 . (2)
c. Compute s e . (2)
d. Compute s b0 and do a significance test on b0 (1.5)
e. Compute s b1 and do a significance test on b1 (2)
f. In the first quarter of 2001 sales were 9883.167, the interest rate was 7.15% and the gdp inflation
rate was 2.176%. In the first quarter of 2000 sales were 9668.827. Get values of Y and X1 from
this and predict the level of investment for 2001. Using this, create a confidence interval or a
prediction interval for investment in 2001 as appropriate. (3)
g. Do an ANOVA table for the regression. What conclusion can you draw from the hypothesis test
in the ANOVA? (2)
[30]
Problem 2: Continue with the data in problem 1.
a. Compute the regression equation Yˆ  b0  b1 x1  b2 x 2 to predict investment on the basis of real
interest rates and change in sales. Do not attempt to use the value of b1 you got in problem 1. Is
the sign of the coefficient what you expected? Why? (5)
b. 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)
c. Using either R – squares or SST, SSR and SSE do an F tests (ANOVA). First check the
usefulness of the multiple regression and show whether the use of real interest rates gives a
significant improvement in explanatory power of the regression? (Don’t say a word without
referring to a statistical test.) (3)
d. Use the values in 1f to compute a prediction for 2001 investment. By what percent does the
predicted investment change if you add real interest rates? (2)
e. If you are prepared to explain the results of VIF and Durbin-Watson (Check the text!), run the
regression of Y on X1 and X2 using
MTB > Regress Y 2 X1 X2;
SUBC>
VIF;
SUBC>
DW;
SUBC>
Brief 2.
Explain your results. (2)
[44]
19
252x0581 12/07/05
Caniglia’s Original Data
Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
STATE
ME
NH
VT
MA
RI
CT
NY
NJ
PA
OH
IN
IL
MI
WI
MN
IA
MO
ND
SD
NE
KS
DE
MD
DC
VA
WV
NC
SC
GA
FL
KY
TN
AL
MS
AR
LA
OK
TX
MT
ID
WY
CO
NM
AZ
UT
NV
WA
OR
CA
AK
HI
MIM
12112
14505
12711
15362
13911
17938
15879
17639
15225
16164
15793
17551
17137
15417
15878
15249
14743
13835
12406
14873
15504
16081
17321
15861
15506
13998
12529
12660
13966
14651
13328
13349
13301
11968
12274
15365
14818
16135
14256
14297
17615
16672
14057
15269
15788
16820
17042
15833
17128
21552
15268
PMHS
69.1
73.0
71.6
74.0
65.1
71.8
68.9
70.0
68.0
69.0
68.8
68.9
69.3
70.9
73.5
71.9
66.2
68.0
68.3
74.2
74.5
70.4
69.2
67.9
64.3
58.6
58.2
58.2
60.4
68.0
55.8
59.0
59.9
57.2
58.3
61.3
68.7
65.3
73.8
73.5
77.9
79.1
70.6
73.4
80.4
76.0
77.5
75.1
74.3
81.9
76.9
PURBAN
47.5
52.2
33.8
83.8
87.0
78.8
84.6
89.0
69.3
73.3
64.2
83.3
70.7
64.2
66.9
58.6
68.1
48.8
46.4
62.9
66.7
70.6
80.3
100.0
66.0
36.2
48.0
54.1
62.4
84.3
50.9
60.4
60.0
47.3
51.6
68.6
67.3
79.6
52.9
54.0
62.7
80.6
72.1
83.8
84.4
85.3
73.5
67.9
91.3
64.3
86.5
MAGE
29.2
29.2
28.4
29.6
30.1
30.6
30.3
30.7
30.4
28.6
28.0
28.6
27.8
28.3
28.3
28.7
29.3
27.5
27.9
28.6
28.7
28.7
29.2
29.9
28.6
29.1
28.1
26.7
27.3
32.9
27.8
28.7
27.8
26.1
29.2
26.2
28.6
27.1
28.4
27.0
26.7
27.9
26.6
28.2
23.8
30.0
29.0
29.5
28.9
26.3
27.6
NE
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
MW
0
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
SO
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
20