252y0581 1/4/06
ECO252 QBA2
Final EXAM
December 14, 2005
Name and Class hour:____KEY_________________
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) Solution: The Equation reads Y  13.7  7.54 X 1  0.961 X 2 , where X 1 is the
dummy variable ‘Locate’ and X 2 is the Zagat rating. Since X 1  1 on Long Island and X 1  0 in the city,
the expected price on Long Island is $7.54 lower than in New York City. This was, quite predictably the
easiest question on the exam. Why did so few people get it?
b) How much would you expect to pay for a meal at a New York restaurant with a Zagat (summated) rating
of 50? (1) Solution: Since Y  13.7  7.54 X 1  0.961 X 2 , and X 1  0 in New York, we have

Y  13.7  7.540  0.96150  $34.35 . Again a terribly easy question.
c) Which of the coefficients are significant? Do not answer this without evidence from the printout. (2)
Solution: As usual many people thought that this was an opinion question or had no idea what statistical
significance means. The easiest way to do this is to note that all the p-values are below 5% and 1% so that
1
252y0581 1/4/06
97
the constant and the coefficients of both X 1 and X 2 are significant. Of course, you could look up t .05
and
note that all the t-ratios in the printout are above it, but why bother?
d) Compute the coefficient of partial determination (partial correlation squared) for ‘locate’ and explain its
meaning. (2) Solution: As explained in the outline, the easy way to do this problem is to note that on the
t2
 6.30 2  .2904 . If you would
printout, t1  6.30 and DF  97 . Therefore rY22.1  2 1

t1  DF  6.30 2  97
rather work harder, note that in the regression above RY212  .592 , and in the stepwise regression on the next
page RY22  .4246 , rY22.1 
RY212  RY22
1  RY22

.592  .4246
 .2909 . This is the fraction of the remaining
1  0.4246
variation in the dependent variable explained by ‘locate’ after ‘summated rating’ is used alone.
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%
Analysis of Variance
Source
DF
SS
Regression
3 5100.9
Residual Error 96 3284.4
Total
99 8385.2
Source
Locate
Summated rating
Inter
DF
1
1
1
T
-3.30
1.28
8.34
-2.03
P
0.001
0.204
0.000
0.045
R-Sq(adj) = 59.6%
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) Solution: If we look at R-squared
alone, we go from RY212  .592 and RY212  .583 to RY2123  .608 and RY2123  .596 , so that both R-squares
have risen and there seems to be an improvement. Note, however, that the coefficient of ‘locate’ is no
longer significant and that the coefficient of ‘inter’ is only significant at the 5% level. This suggests that the
location of the restaurant changes the slope of the regression line more than the intercept and that we might
do just as well without ‘locate.’ Furthermore the coefficient of ‘locate’ is positive which seems to be telling
us that LI restaurants are more expensive than city restaurants, which sounds unlikely. Note that our first
regression Y  13.7  7.54 X 1  0.961 X 2 has been replaced by
Y  26.3  13.1X 1  1.19 X 2  0.368 X 3  26.3  13.1X 1  1.19 X 2  0.368 X 1 X 2
[9]
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252y0581 1/4/06
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
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
Summated rating
T-Value
P-Value
Inter
T-Value
P-Value
Food
T-Value
P-Value
Décor
T-Value
P-Value
-0.93
-1.55
0.125
S
R-Sq
R-Sq(adj)
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] Solution: If we look at the previous
regression, we could have predicted that ‘locate’ would be dropped, making ‘summated rating’ and ‘inter’
the ‘best’ predictors. The stepwise procedure then added ‘food’ and ‘décor,’ and the coefficient of ‘food’
was significant at the 5% level. But ‘food’ is already part of ‘summated rating,’ so it seems reasonable that
if we use ‘food’ and ‘décor’ we should drop ‘summated rating.’
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
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252y0581 1/4/06
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]
Solution: All I got on this from anybody was BS. Yes, it is true that the F tests done by the computer
showed that there was a relationship between the independent variables and the dependent variables. But as
the posted problems and the material presented in class show, the way to do a comparison for a group of
variables is to note from the sequential SS printout that the sum of squares explained by the first two
independent variables is 900.0 + 1819.3 = 2719.3. The next three variables explain 371.6 + 1574.8 + 672.1
= 2618.5. We can make the ANOVA table below. We will use X 1 and X 3 for ‘locate’ and ‘inter’ and
X 4 , X 5 and X 6 for ‘food,’ ‘décor’ and ‘service.’
Source
SS
DF
MS
F
X1, X 2
2719.3
2
X 4, X5, X6
2618.5
3
872.8
26.938
Error
Total
3047.4 94
8385.2 99
32.4
F.05
3,94  2.71
F.05
2
2
n  k  r  1 Rk  r  Rk 94 .637  .324

 28 .6
r
3 1  .637
1  Rk2 r
Since our computed F is far larger than the table F, we can reject the null hypothesis that these new
variables as a group do not help explain the variation in Y. Note, however, that the coefficients of ‘locate’
‘inter’ and ‘food are not significant. However, except for ‘locate,’ all the coefficients have a plausible sign.
We could also do this by computing
4
252y0581 1/4/06
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]
2
Solution: For the second of these equations we had RY123  .608 and RY2123  .596 and for our newest
equation we have RY213456  .637 and RY213456  .617 . This looks like an improvement. Both equations,
however, have an insignificant coefficient for ‘locate.’
h) I don’t think I am finished. Should I drop some variables from
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) [18]
Solution: Normally, the high VIFs for ‘locate’ and ‘inter’ would be an alarm. But we know that these come
from the same source. Nevertheless the poor results we have gotten generally for ‘locate’ and its high pvalue suggest that it would be dropped. The coefficient of ‘food’ is also not significant. So we should
probably drop each of them separately and then try the equation with both variables removed.
5
252y0581 1/4/06
————— 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
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31
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33
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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
6
252y0581 1/4/06
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
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24
18
15
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23
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14
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15
12
19
21
13
17
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17
11
16
12
25
17
22
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20
11
18
21
19
27
18
16
20
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12
24
18
15
14
17
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17
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15
22
18
21
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16
24
18
23
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24
17
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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
7
252y0581 1/4/06
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.
Is there some reason why most of you couldn’t be bothered to state your
hypotheses?
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:
(Note: D  1   2 )
v) 1   2 and D  0
vi) 1   2 and D  0
i) 1   2 and D  0
ii) 1   2 and D  0
iii) 1   2 and D  0
vii) 1   2 and D  0
iv) 1   2 and D  0
viii) 1   2 and D  0 (2)
Solution: I’m not going to answer this one. You all had this question in advance and there is no excuse for
the fact that most of you got it wrong.
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


x12  11696 . For the difference column,
they are
 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]
Solution:
Assume   .05 . Note that this is paired data! By now you should know why.
Spare Parts computation
Candidate Gaston
x2
1
2
3
4
5
6
7
8
9
38
31
24
29
27
28
41
27
31
276
x 22 x1 x 2
1444
961
576
841
729
784
1681
729
961
8706
1976
775
696
957
648
1008
1722
1323
620
9725
x x
1 2
 9725
n9
8
252y0581 1/4/06
 x  310 and  x  11696
 x  276 and  x  8706
 d  34 and  d  952
 x  310  34.4444
x 
1
2
1
2
2
2
x2 
2
d
x
2
n

276
 30 .6667
9
 d  34  3.7778
n
9
1
1
n
9
 x  nx  11696  934.4444   1018 .25 s  8  127 .281 s
S x1x 2   x1 x 2  nx1 x 2  9725  934.4444 30.6667   218 .3352
241 .98
SS   x  nx  8706  930 .6667   241 .98 s 
 30 .248 s 
8
SS x1 
2
1
2
1
x2
2
2
2
2
 d  nd
 x  nx

SS d 
2
2
2
2
2
2
1
2
1018 .45
1
2
2
2
 952  93.7778 2  823 .5540 s d2 
 127 .281  11 .2819
30 .248  5.4998
823 .5540
 102 .944 s d  102 .944  10 .1461
8
2
2
s 2  30 .248  5.4998
 30 .248
n 1
c) H 0 :   35 H 1 :   35 Did you state your hypotheses? From the Formula Table:
b) s 22
Interval for
Confidence
Interval
  x  t 2 s x
Mean (
unknown)
x2 
x
2
DF  n 1

Hypotheses
Test Ratio
H0 :   0
t
H1 :    0
276
 30 .6667 s 2  30 .248  5.4998
9
n

n 1
t
 t.805  1.860
n9
Critical Value
xcv   0  t  2 s x
x  0
sx
sx2 
s2
n
sx 

s
n
30 .248
 1.833
9
 0  35
x   0 30 .6667  35

 2.364 Make a diagram: The diagram should show
sx
1.833
an almost Normal curve centered at zero. Since this is a left-sided test, the ‘reject’ region is the
area below  tn1  t.805  1.860 . Shade the area below -1.860. Since t calc  2.364 falls in the
reject region, reject the null hypothesis.
Critical Value for x : This is a left-sided test, so that the critical value will be below  0  35 .
xcv   0  t s x  35  1.860 1.833   31.5906 . Make a diagram: The diagram should show an
almost Normal curve centered at  0  35 . Since this is a left-sided test, the ‘reject’ region is the
area below x cv  31 .5906 . Shade the area below 31.5906. Since x 2  30 .6667 falls in the reject
region, reject the null hypothesis.
(iii) Confidence Interval. Since the alternative hypothesis is H 1 :   35 , we want a confidence
interval of the form   x  t s x  30.6667  1.860 1.833   34.0760 . Make a diagram: The
diagram should show an almost Normal curve centered at x 2  30 .6667 . The confidence interval
is the area below 30.6667. Shade the area below 30.6667. Since  0  35 does not fall in the
confidence interval, reject the null hypothesis.
(i) Test Ratio: t 
9
252y0581 1/4/06
d) H 0 : 1   2 and D  0 H 1 : 1   2 and D  0 Did you state your hypotheses?
Interval for
Confidence
Interval
Hypotheses
Test Ratio
Difference
between Two
Means (paired
data.)
D  d t  2 s d
H 0 : D  D0 *
t
H 1 : D  D0 ,
d  x1  x 2
df  n  1 where
n1  n 2  n
D  1   2
d  3.7778 s d  102 .944  10 .1461 n  9 s d 
sd

Critical Value
d  D0
sd
sd 
d cv  D0 t  2 s d
sd
n
102 .944
 3.382 tn1  t.805  1.860 D0  0 .
9
n
Use only one of the following three methods.
d  D0 3.7778  0
Test Ratio: t 

 1.117 Make a diagram: The diagram should show an
sd
3.382
almost Normal curve centered at zero. Since this is a right-sided test, the ‘reject’ region is the area
above tn1  t.805  1.860 . Shade the area above 1.860. Since t calc  1.117 does not fall in the
reject region, do not reject the null hypothesis.
Critical Value for x : This is a right-sided test, so that the critical value will be below D0  0 .
d cv  D0  t s d  0  1.860 3.382   6.291 . Make a diagram: The diagram should show an
almost Normal curve centered at D0  0 . Since this is a left-sided test, the ‘reject’ region is the
area below d cv  6.291 . Shade the area below -6.291. Since d  3.7778 does not fall in the reject
region, do not reject the null hypothesis.
Confidence Interval: Since the alternative hypothesis is H 1 : 1   2 and D  0 , we want a
confidence interval of the form D  d  t s d  3.7778  1.860 3.382   2.513 . Make a
diagram: The diagram should show an almost Normal curve centered at d  3.7778 . The
confidence interval is the area above -2.513. Shade the area above -2.513. Since D0  0 falls in
the confidence interval, do not reject the null hypothesis.
e) r 
X X
1

2 nX 1 X 2

X 12  nX 12
X 22  nX 22
S x1x 2

SS x1 SS x 2

S x21x 2
SS x1 SS x 2
The sign is that of S x1x 2 . We
have already computed SS x1  1018 .25 S x1x 2  218 .3352 and SS x 2  241 .98 .
So r 
218 .3352 2
 .1934  .4398 . As always 1  r  1 . According to the outline the test for
1018 .25 241 .98 
significance or H 0 : xy  0 against H1 : xy  0 is to use the test ratio
t n  2  
r

sr
r
1 r
n2
2

.4398
1  .1934
7

.4398
.1152

.4398
 1.295 . If this is a two-sided test, compare this
.33945
7
ratio with t n2  t .025
 2.365 and reject the null hypothesis if t calc is below -2.365 or above 2.365 and say
2
that the correlation is significant. However, the wording of the problem is more likely to be asking for
H 0 : xy  0 against H1 : xy  0 in which case we should reject the null hypothesis if t calc is above
7
tn  2  t .05
 1.895 . Either seems to me acceptable, and in any case, since t calc is not in either ‘reject’ zone,
do not reject the null hypothesis. You can say that the amount of agreement is insignificant.
10
252y0581 1/4/06
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]
Solution: a) H 0 : Normal
We know from the previous problem that x1  34 .4444 and
s1  127 .281  11 .2819 . The best method to use here is Lilliefors because the sample is small, the data is
not stated by intervals, the distribution for which we are testing is Normal, and the parameters of the
distribution are unknown. We begin by putting the numbers in order and computing
x  x x  34 .4444
(actually t ) and proceed as in the Kolmogorov-Smirnov method.
z

s
11 .2819
From the Lilliefors table for   .05 and n  6 , the critical value is .217. Since the maximum deviation
(.1328) is below the critical value, we do not reject H 0 .
O cum O
F
F
D
x
z
o
20
24
25
29
33
36
42
49
52
-1.28
-0.93
-0.84
-0.48
-0.13
0.14
0.67
1.29
1.56
1
1
1
1
1
1
1
1
1
1
2
3
4
5
6
7
8
9
0.1111
0.2222
0.3333
0.4444
0.5556
0.6667
0.7778
0.8889
1.0000
e
.5-.3997=.1003
.5-.3238=.1762
.5-.2995=.2005
.5-.1844=.3156
.5-.0517=.4483
.5+.0557=.5557
.5+.2486=.7484
.5+.4015=.9015
.5+.4406=.9406
.0108
.0460
.1328
.1288
.1073
.1110
.0294
.0126
.0594
H 0 : 1  35
b) Wilcoxon Signed Rank Test: 
n  9 and   .05 . Did you state your hypotheses? The
H 1 : 1  35
data are below: The column d is the absolute value of d , the column r ranks absolute values, and the
column r * is the ranks corrected for ties and marked with the signs on the differences.
x 2  0  median ? d  x 2  0
d
r*
r
38
31
24
29
27
28
41
27
31
35
35
35
35
35
35
35
35
35
3
-4
-11
-6
-8
-7
6
-8
-4
3
4
11
6
8
7
6
8
4
1
2
9
4
7
6
5
8
3
1.0+
2.59.04.57.56.04.5+
7.52.5
If we add together the numbers in r * with a + sign we get . T   5.5 . If we do the same for numbers with
a – sign, we get T   39.5 To check this, note that these two numbers must sum to the sum of the first n
nn  1 910 

 45 , and that T    T   5.5  39.5  45 .
numbers, and that this is
2
2
11
252y0581 1/4/06
We check 5.5, 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  9 , the critical value is 8, and we reject the null hypothesis only if our test
statistic is below this critical value. Since our test statistic is 5.5, we reject the null hypothesis.
Sign Test: 2 out of 9 are above 35. According to the Binomial table, if n  9 and p  .5,
Px  2  .08984 . Since this is not below   .05 , do not reject the null hypothesis.
As usual, some people calculated the median of both columns. They got absolutely no credit. An
awfully large number of people did a solution that would be correct for c) for section b). There is no
excuse for failing to read a problem before you do it.
 H 0 : 1   2
c) This is a Wilcoxon Signed rank test again. 
n  9 and   .05 . Did you state your
 H 1 : 1   2
hypotheses? The data are below: The column d is the absolute value of d , the column r ranks absolute
values, and the column r * is the ranks corrected for ties and marked with the signs on the differences.
x1
52
25
29
33
24
36
42
49
20
d  x1  x 2
x2
38
31
24
29
27
28
41
27
31
d
14
-6
5
4
-3
8
1
22
-11
14
6
5
4
3
8
1
22
11
r
r*
8
5
4
3
2
6
1
9
7
8+
54+
3+
26+
1+
9+
7-
If we add together the numbers in r * with a + sign we get T   31 . If we do the same for numbers with a
– sign, we get T   14 . To check this, note that these two numbers must sum to the sum of the first n
nn  1 910 

 45 , and that T    T   31  14  45 . We check 14, the
numbers, and that this is
2
2
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  9 , the critical value is 8, and we reject the null hypothesis only if our test statistic is below
this critical value. Since our test statistic is 14, we cannot reject the null hypothesis.
H 0 :  s  0
d) In this case, we have a 1-sided test 
.
n9
H 1 :  s  0
Candidate
1
2
3
4
5
6
7
8
9
rs  1 
x1
r1
x2
r2
d
d2
52
25
29
33
24
36
42
49
20
9
3
4
5
2
6
7
8
1
38
31
24
29
27
28
41
27
31
8.0
6.5
1.0
5.0
2.5
4.0
9.0
2.5
6.5
1.0
-3.5
3.0
0.0
-0.5
2.0
-2.0
5.5
-5.5
0.0
1.00
12.25
9.00
0.00
0.25
4.00
4.00
30.25
30.25
91.00
Note that
and
d
d  0
2
(a check on ranking)
 91 .
 d  1  691  1  546  1  .7583  0.2417 . If we check the table ‘Critical Values of
720
nn  1
99  1
2
6
2
2
rs , the Spearman Rank Correlation Coefficient,’ we find that the critical value for n  9 and   .05 is
.5833 so we must not reject the null hypothesis and we conclude that we cannot say that the rankings agree.
12
252y0581 1/4/06
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)
South East
c) Do a 2-sided 99% confidence interval for the
Number eligible
88
57
proportion eligible on the South side. (2)
Sample size
200
150
d) Do a 2-sided 89% confidence interval for the
f) Use the Marascuilo procedure to determine
proportion eligible on the South side. Do not use
whether there is a significant difference between
a value of z from the t table! (1)
the areas with the largest and second largest
e) We suddenly realize that out of 100 parents
proportion of eligible parents. (3)
[17]
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)
Solution: From the formula table
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
pcv  p0  z 2  p
Difference
p  p 0
p  p  z 2 s p
H 0 : p  p 0
z

between
If p  0
 p
H 1 : p  p 0
p  p1  p 2
proportions
 1
1 
If p  0
p 0  p 01  p 02
 p  p 0 q 0   
p1 q1 p 2 q 2
q  1 p
n
n
s p 

1
2 

p 01q 01 p 02 q 02
 p 

n1
n2
or p 0  0
n1
n2
n p  n2 p 2
p0  1 1
Or use s p
n1  n 2
If p1 is the proportion of successes in the first sample (South side), and p2 is the proportion of successes
in the second sample (North side), we define p  p1  p 2 . Then our hypotheses will be
H 0 : p1  p 2
H 1 : p1  p 2
or
H 0 : p  p 0  0
H 1 : p  p 0  0
.
Let p1  x1  88  .44 , p2  x2  57  .38 and p  p1  p2  .06 where x1 is the number of successes
n1
200
n2
150
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 . Only one of the usual three approaches to testing the hypotheses should be used. As usual,
some people noted that .44 is larger than .38 and did not do a statistical test. They got no credit!
Confidence Interval: p  p  z s p or  p1  p 2    p1  p 2   z s p , where
2
s p 
2
p1 q1 p 2 q 2
.44 .56  .38.62 



 .001232  .001571  .002803  .05294 The one
n1
n2
200
150
sided interval is p  p  z s p  .06 1.645.05294  .2709 . Compare this interval with
p 0  0 . Since p  .2709 does not contradict p  0 , do not reject H 0 . Make a
diagram: The diagram should show a Normal curve centered at p  .06 . The confidence
interval is the area above -.2709. Shade the area above -.2709. Since p 0  0 falls in the
confidence interval, do not reject the null hypothesis.
p  p 0
.06

 1.1277 where
Test Ratio: z 
 p
.053207
 1
1 
1 
 1
 3 4 
  .4143 .5857 


  .24266 
  .002831  .053207
n
n
200
150


 600 
2 
 1
 p  p 0 q 0 
13
252y0581 1/4/06
and p 0 
n1 p1  n 2 p 2 x1  x 2
88  57
145



 .4143 . Because our alternate hypothesis is
n1  n 2
n1  n 2 200  150 350
p  0 , this is a right-sided test and the rejection zone is above z .05  1.645 . Since the test ratio is
not in that interval, do not reject H 0 . Make a diagram: The diagram should show a Normal curve
centered at zero. The ‘reject’ region is the area above z .05  1.645 . Shade the area above 1.645.
Since z  1.1277 does not fall in the ‘reject’ zone, do not reject the null hypothesis.
Critical Value: pCV  p0  z  p becomes pCV  p0  z  p
2
 0  1.645 .053207   0.0875 and the rejection zone is above 0.0875. If we test this against
p  .06 , we cannot reject H 0 . For calculation of  p , see Test Ratio above. Make a diagram:
The diagram should show a Normal curve centered at p 0  0 . The ‘reject’ region is the area
above p CV  0.0875 . Shade the area above 0.0875. Since p  .06 does not fall in the ‘reject’
zone, do not reject the null hypothesis.
The answer to “Should we put the center on the south side?” is a maybe since we have no proof that any
area has a greater proportion of eligibles.
b) The p-value is Pp  .06   Pz  1.127  , where 1.127 is the value of the test ratio found in a). Make a
diagram: The diagram should show a Normal curve centered at zero. The p-value is represented by the area
above 1.127. Shade the area above 1.127. Pz  1.127   Pz  0  P0  z  1.13  .5  .3708  .1292 .
c) 1    .99 , so   .01 . p  p  z  s p , where z  z.005  2.576 , p  .06 and
2
2
sp 
p1 q1
.44 .56 

 .035099 . So p  .44  2.576 .035099   .44  0.09 .
n1
200
d) 1    .89 , so   .11 and

2
 .112  .055 . So we need to find z  2  z.055 . Make a diagram: The
diagram should show a Normal curve centered at zero. Mark z .055 to the right of zero and show, by
definition, that it has 5.5% above it and 50% - 5.5% = 44.5% between it and zero. So
P0  z  z.055   .4450 . The closest we can come to this on the Normal table is P0  z  1.60   .4452 . So
z.055  1.60 and our interval is p  .44  1.60.035099   .44  0.06.
e) With the discovery of the North side, our table changes.
South East North
Number eligible
88
57
51
Sample size
200
150 100
We now have a chi-squared test of homogeneity.
H 0 : p1  p 2  p 3
H 1 : Not all proportions equal
O
Elig
S E N
total
pr
 88 57 51  196
.4356


Our observed data reads as follows. Not 112 93 49  254
.5644


total 200 150 100 450 1.0000
14
252y0581 1/4/06
If we apply the proportions in the p r column to the totals in the columns we get the following.
E
Elig
S E N
total
pr
 87 .12 65 .34 43 .56  196 .02 .4356


Not 112 .88 84 .66 56 .44  253 .98 .5644
total 200 .00 150 .00 100 .00
450
1.0000
We can now compute  2 
Row
O
E
1
2
3
4
5
6
88
112
57
93
51
49
450
87.12
112.88
65.34
84.66
43.56
56.44
450.00

O  E 2
E
O2
n.
E
O  E 2
O2
E
E
or  2 
O  E 2
OE
-0.88
0.88
8.34
-8.34
-7.44
7.44
0.00
0.7744
0.7744
69.5556
69.5556
55.3536
55.3536

0.00889
0.00686
1.06452
0.82159
1.27074
0.98075
4.15335
88.889
111.127
49.725
102.162
59.711
42.541
454.153
The degrees of freedom for this application are r  1c  1  2  13  1  2 .  .2052   5.9915 . Our
computed chi-squared is  2 

O2
 n  454 .153  450  4.153 . Since this is not larger than  .2052  , we
E
cannot reject the null hypothesis.
f) We now know p1  x1  88  .44 , p2  x2  57  .38 and p3  x3  51  .51
n1
200
n2
150
n3
100
The Marascuilo procedure says for 2 by c tests, is equivalent to using a confidence interval of
c 1  p a q a
p a  pb   p a  pb    2

 n
 a

pb qb
nb

 , where a and b represent 2 groups, the chi - squared has


2   p q
pq 
c  1 degrees of freedom. For groups 3 and 1, it gives us p 3  p1   p 3  p1    2 .05  3 3  1 1 
n1 
 n3
 .51.49  .44 .56  
 .51  .44   5.9915 

  .07   5.9915 .002499  .001232 
200 
 100
 .07   0.022354  .07  0.15 . Of course this is not significant since the error part is larger than the
difference between the sample proportions.
15
252y0581 1/4/06
Row
1
2
3
4
5
6
7
8
9
10
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]
Solution: a)
Row
1
2
3
4
5
6
7
8
9
10
X
3
8
6
9
6
9
6
4
8
5
 xy  4206.
Y
22
68
68
96
46
80
52
38
78
48
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  448
 Y  596.  Y  40000
 xy   x y  and  xy   x  y 
2
2
As was stressed in class
XY
66
544
408
864
276
720
312
152
624
240
4206
It’s time to compute spare parts.
x
 x  64  6.4
y

n
10
y
n

 x  nx
  xy  nx y
SS x 
S xy
596
 59 .6
10
2
2
 448  10 6.42  38 .4 *
 4206  106.459.6  391 .60
Note that n  10 not 20. All of you should
have learned this from the last exam.
SS y 
y
2
 ny 2
 40000  1059 .62  4478 .40 *
Note: * These spare parts must be positive.
The rest may well be negative.
 XY  nXY 
Sxy  391 .60   .8917



SSy 38.44478 .4
SSx
 X  nX  Y  nY 
2
b) R
2
2
also true that R 2 
2
2
2
2
2
As always 0  R 2  1 . It is
b1 Sxy 10 .1979 391 .60 

 .8917 , but you probably haven’t found b1 yet.
SSy
4478 .4
c) r  .8917  .9443 As always 1  r  1 .
d) H 0 : xy  .75 . H1 : xy  .75 . According to the outline “If we are testing H 0 : xy   0 against
H 1 : xy   0 , and  0  0 , the test is quite different. We need to use Fisher's z-transformation. Let
1 1 r 
1  1 0
~
z  ln 
 . This has an approximate mean of  z  ln 
2  1 r 
2  1 0
~
n 2 
z  z
1
sz 

, so that t
.
n3
sz

 and a standard deviation of


(Note: To get ln , the natural log, compute the log to the base 10 and divide by .434294482. )”
1  1  r  1  1.9443  1
1
~
z  ln 
  ln 
  ln 34 .9066   3.5527   1.7763
2  1  r  2  0.0557  2
2
16
252y0581 1/4/06
1 1 0
 0  0.75 , so  z  ln 
2  1 0
 1  1.75  1
1
  ln 
 2  0.25   2 ln 7.00   2 1.9459   0.9730

~
n  2 
z   z 0.7763  0.9730
1
1
sz 

 0.377965 and t


 0.5204 . Since this is between
n3
7
sz
0.377965
8
8
 t n2  t .025
 2.306 and t n2  t .025
 2.306 , we do not reject the null hypothesis and we can say that
2
2
the correlation is not significantly different from 0.75.
e) We conclude that b1 
Sxy

SSx
 xy  nxy  391 .6  10.1979
 x  nx 38.4
2
2
b0  y  b1 x  59.6  10.1979 6.4  5.6666 . So Yˆ  b0  b1 x becomes Yˆ  5.6666  10.1979 x . This is
the regression equation. It is not time to start substituting in values of X.
.
17
252y0581 1/4/06
Row
1
2
3
4
5
6
7
8
9
10
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]
Solution: From the last page we have n  10, SS x 
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
x
2
2
 448
2
 40000
 nx 2  38 .4 , S xy 
 xy  nxy  391 .60 ,
SST  SS y  4478 .40 , R 2  .8917 and that our equation is Yˆ  b0  b1 x  5.6666  10.1979 x .
a) Compute s e . SSR  b1 Sxy  b1
 xy  nxy   10.1979 391 .60   3993 .50 and
SSE  SST  SSR  4478 .40  3993 .50  484 .90 or SSR  R 2 SST   .89174478.40  3993.38


and SSE  1  R 2 SST  1  .8917SST  .10834478.40  485.01
s e2 
( s e2 is always positive!)
SSE 484 .90

 60 .6125 So s e  60 .6125  7.7854
n2
8
b) Compute s b1 and do a significance test on b1 (1.5) Recall n  10,   .05 , SSx  391 .60 ,
s e2  60.6125 and b1  10 .1979 .
 H 0 :  1   10
For most two-sided tests use t n2 2  t .8025  2.306 . From the outline – “To test 
use
 H 1 :  1   10
b  10
t 1
. Remember  10 is most often zero – and if the null hypothesis is false in that case we
s b1
 1  60 .6125
say that 1 is significant.” s b21  s e2 
 1.5785 and s b  1.5785  1.2564 . So

1
38 .4
 SS x 
b  10 10 .1979  0
t 1

 8.117 . Our rejection zone is below -2.306 and above 2.306. Since
s b1
1.2564
our calculated t falls in the upper reject zone, we can say that b1 is significant.
c) Compute s b0 and do a confidence interval for b0 (1.5) Recall n  10,   .05 , SSx  38.4 ,
s e2  60.6125 , x  6.4 and b0  5.6666 .
1
s b20  s e2  
n


2


  60 .6125  1  6.4   60 .6125  1  1.06667   70 .7145
10



X 2  nX 2 
10 38 .4 

X2
s b  70 .7145  8.4012  0  b0  t  2 sb0  5.6666  2.3068.4012  5.67  19.37 .
0
Since the error part of the interval is larger than b0  5.6666 , we can conclude that the intercept
is not significant.
18
252y0581 1/4/06
d) The general format for a regression ANOVA table reads:
Source
SS
DF
MS
Fcalc
Regression
SSR
k
MSR
MSR MSE
F
F k , nk 1
Error
n  k  1 MSE
SSE
Total
SST
n 1
xy  nx y  3993 .50 , SST  4478 .40 and SSE  484 .90 .
From a) SSR  b1 Sxy  b1


n  10 and the number of independent variables is k  1 .
The ANOVA table for the regression reads:
Source
SS
DF
MS
Fcalc
F.05
1,8   5.32
F.05
Regression
3993.5
1
3993.5 65.89s
Error(Within)
484.9
8
60.6125
Total
4478.4
9
Since our computed F is larger than the table F, we conclude that there is a linear relationship
between the dependent and independent variable. In a simple regression this is the same as saying
that the slope is significant.
e) Recall that our equation is Yˆ  b0  b1 x  5.6666  10.1979 x . X 0  9 , n  10,   .05 ,
SSx  38.4 , s e2  60.6125 and x  6.4 .
The Prediction Interval is Y0  Yˆ0  t sY , where Yˆ  5.6666  10.1979 X 0

1 X X
 5.6666  10.1979 9  86.1145 sY2  s e2   0
n
SS x

2  1  60.6125  1  9  6.42  1


 10

38 .4


 60.6125 1.1  0.1760   77.3441 . sY  77 .3441  8.7945 . So Y0  Yˆ0  t sY
 86.1145  2.306 8.7945   86.11  20.28 . As usual for small samples these intervals are gigantic.
19
252y0581 1/4/06
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:
 1  15 (2)
(i)
(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
40 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]
Solution: From the Formula Table.
Interval for
Confidence
Interval
Variancen  1s 2
2  2
Small Sample
.5 .5 2 
Ratio of Variances
1 , DF2
F1DF


2
1
FDF1 , DF2
2
 22 s22 DF1 , DF2

F

 12 s12 .5  .5  2 
DF1  n1  1
DF2  n 2  1
 2

.5  .5   2    or
1  
2

Number of Unsuccessful
Tries before success.
0
1
2
3
4
5
6
7
Hypotheses
Test Ratio
H 0 :  2   02
2 
H1: :  2   02
H0 : 12   22
H1 : 12   22
O
4
2
8
12
10
0
2
2
40
Critical Value
n  1s 2
2
s cv

 02
F DF1 , DF2 
Observed
Frequency
2
1
4
6
5
0
1
1
20
 .25 .5 2  02
n 1
s12
s 22
and
F DF2 , DF1 
s 22
s12
a) n1  10, n 2  12, n3  15, s12  20, and s 22  15, s32  10,   .10 . So df1  n1  1  10  1  9 and
df 2  n 2  1  12  1  11 .
(i)
H 0 :  1  15 is the same as H 0 :  12  15 , so  2 
n  1s 2
 02

920 
 12 . From the
15
9
9
chi-squared table,  2 .05  16.9190 and  2 .95  3.3251 . This is a 2-sided test, and our computed
chi-squared must be between these 2 values if we are not to reject the null hypothesis. Since the
computed F-ratio is not below the lower table chi-squared, we cannot reject the null hypothesis.
(ii) H 0 :  1   2 is the same as H0 : 12   22 . Our two possible ratios are
F 9,11 
20
15
 1.3333 and F 11,9  
 0.75 . F 11,9 is below one and thus cannot possibly
15
20
20
9
 1.3333
 3.10 since all numbers on the F-table are at least one. F 9,11 
be above F 11.,05
15
20
252y0581 1/4/06

must be compared with F 9,11
.05 2.90 . Since both our computed F’s are below the corresponding
table values, we cannot reject the null hypothesis.
(iii) Explain what test would be appropriate to test  1   2   3 . The two tests mentioned for
multiple variances are the Bartlett and Levene tests. The Bartlett test is appropriate if the
underlying distribution is Normal, as it is here.
b) To test the response of an 800 number, I make 40 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) H 0 : Poisson
The only method we have to test a Poisson distribution with an unknown mean is the chi-squared
method. First we must find a mean from the data.
Number of Unsuccessful Observed Total
Tries before success.
Frequency Tries
0
4
0
If there are 60 successes in 20 tries the
1
2
2
mean number of unsuccessful tries before a
2
8
16
success is 3.
3
12
36
4
10
40
5
0
0
6
2
12
7
2
14
40
120
The first 3 columns x, f and E  are the number of tries, the probability from the Poisson table and the
values of E that come from multiplying the probabilities by n  120 . There is a rule here that we should
avoid values of E below 5, and I expect you to have eliminated at least the last value of E . This is
2
probably not enough given the large value of D
that we get for x  5 . In any case, I left the remaining
E
cells alone and went ahead to compute the sums. To save space, I used D for O  E . As usual, I used both
O  E 2
O2
 n . Actually the regular method
E
E
should be used here to check out the effect of the undersized values of E .
x
x
O
E
E
D
f
D2
O2
D2
E
E
the regular and shortcut methods to get  2 
0
1
2
3
4
5
6
7+
Sum
.049787
.149361
.224042
.224042
.168031
.100819
.050409
.03351
1.991
5.974
8.961
8.961
6.721
4.032
2.016
1.340
0
1
2
3
4
5
6+



4
2
8
12
10
0
4
1.991
5.974
8.961
8.961
6.721
4.032
3.354
-2.009
3.974
0.961
-3.039
-3.279
4.032
-0.644
40
40.00
-0.004
4.0361
15.7927
0.9235
9.2355
10.7518
16.2570
0.4147
2.02716
2.64357
0.10306
1.03064
1.59974
4.03200
0.12358
8.0362
0.6696
7.1421
16.0696
14.8787
0.0000
4.7676
11.5637
51.5637
In any case, our value of  seems to be about 11.56. Since there are 7 rows, but we used the data to
2
estimate one parameter, we get 7 – 1 – 1 = 5 degrees of freedom.  .705  14.0671. Since this is larger than
our computed  2 , we do not reject the null hypothesis.
21
252y0581 1/4/06
(ii) I hypothesize that the results follow a Poisson distribution with a mean of 2. H 0 : Poisson2
Things suddenly got much easier. I computed the cumulative observed Fo and copied Fe from the
cumulative Poisson table. I checked the 5% critical value from the Kolmogorov-Smirnov table and found it
was .210. When I got to my second computation in the D  F0  Fe column, I had already found a value
of D above the critical value, so I decided to reject the null hypothesis and quit.
x
0
1
2
3
4
5
6
7
Above 7
Total
O
4
2
8
12
10
0
2
2
2
40
O
n
Fo
.10
.05
.20
.30
.25
.00
.05
.05
.00
1.00
.10
.15
.35
.65
.90
.90
.95
1.00
1.00
Fe
0.13534
0.49601
0.67668
0.85712
0.94735
0.98344
0.99547
0.99890
1.00000
D  F0  Fe
.03534
.34601
22
252y0581 1/4/06
ECO252 QBA2
Final EXAM
December 14-16, 2005
TAKE HOME SECTION
Name: _________________________
Student Number: _________________________
Class days and time: _________________________
III Take-home Exam (20+ points). Note that interpretation of ANOVA and OLS computer output was
presented in class and discussed in documents 252anovaex1, 252anovaex2, 252regrex1
and252regrex2.
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.
23
252y0581 1/4/06
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
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
24
252y0581 1/4/06
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.)
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.
MIM = 2528 + 134 PMHS + 50 PURBAN.
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.
25
252y0581 1/4/06
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 a and Problem 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, 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.
26
252y0581 1/4/06
Original Version
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
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.
Row Raters
Lee Jacobs Wilkes
Delap
sum cnt
mean
sq
mnsq
1 Moore
52
25
29
33
139
4 34.75 5259 1207.56
2 Gaston
38
31
24
29
122
4 30.50 3822
930.25
3 Heinrich
54
38
40
39
171
4 42.75 7481 1827.56
4 Seldon
43
30
31
28
132
4 33.00 4494 1089.00
5 Greasy
58
44
46
47
195
4 48.75 9625 2376.56
6 Waters
36
28
22
25
111
4 27.75 3189
770.06
7 Pierce
52
41
37
45
175
4 43.75 7779 1914.06
sum
333
237
229
246 1045
28
41649 10115.06
count
7
7
7
7
28
mean
47.571 33.857 32.714 35.143 Grand mean = 1045/28 = 37.321
ssq
16277
8331
7947
9094 41649
meansq
2263.00 1146.30 1070.21 1235.03 5714.54
From the above
x
 x  1045 , n  28 ,  x
 x  1045  37.321 .
n
 x
SSC   n
SST 
28
Note 2837 .321  39000 .00
2
2
ij
 41649 ,
2
ij
x
2
i.
 10115 .06
x
2
.j
 5714 .54 and
 n x  41649  2837 .321 2  2649 .00 .
2
2
j x j
 n x  75714 .54   2837 .321 2
2
 40001 .78  39000 .00  1001 .78 . This is SSB in a one way ANOVA.
SSR 
 n x
2
i i.
 n x  410115 .06   2837 .321 2  40460 .24  39000 .00  1460 .24
2
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)
Solution: SSW  SST  SSB  2649 .99  1001 .78  1648 .21
Source
SS
DF
Between
1001.78
3
Within
Total
1648.21
2649.99
24
27
MS
333.93
F
4.86S
F .05
3, 24 = 3.01
F.05
H0
Column means Equal
68.675
The ‘S’ next to the calculated F indicates that since it is larger than the table F we reject the null hypothesis.
27
252y0581 1/4/06
The outline gives a formula for an individual contrast 1   2  x1  x2   t n  m  s
2
1
1
, where

n1 n 2
s  MSW .
1   4  x1  x4   t 24 MSW
2
1 1
1
1
 47 .571  35 .143   2.064 68 .675


7 7
n1 n 4
2
 12 .398   2.064 68 .675    12.398   2.064 4.4296   12.4  9.1 . Since the error part is less than the
7
difference between the two sample means, the difference is significant.
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)
Solution: H 0 : Equal medians While doing Problem 3, I ranked the data. Since there are 28 numbers, the
ranks must extend from 1 to 28. Any ties must be given the average rank.
Data Display – This is the ranked data for Kruskal – Wallis.
Row C13_Lee
1
25.5
2
15.5
3
27.0
4
20.0
5
28.0
6
13.0
7
25.5
Sum
154.5
C13_Jacobs
3.5
10.5
15.5
9.0
21.0
5.5
19.0
84.0
C13_Wilkes
7.5
2.0
18.0
10.5
23.0
1.0
14.0
76.0
C13_Delap
12.0
7.5
17.0
5.5
24.0
3.5
22.0
91.5
To check our work, the sum of these 4 column sums is 406. The sum of the numbers from 1 to 28 is
2829 
 406 so we are unlikely to have made a mistake. As always happens, a fairly large group of
2
people lost a fair amount of credit by changing the 12 in the formula below to 28. As I pointed out in
class, this is completely unreasonable. If Kruskal and Wallis had wanted the 12 changed to n , they
 1
would have written the formula as 
 n  1

i
 SRi 2

 ni

 12
Now, compute the Kruskal-Wallis statistic H  
 nn  1
 12  154 .5 2 84 2 76 2 91 .5 2





4
4
4
 28 29   4

  3n  1 .



i
 SRi 2

 ni


  3n  1





  329    12  45074 .5   329 

4

 2829  

 12  45074 .5 


  87  166 .5314  87  79 .5314 . Our Kruskal-Wallis table has no values for tests
4

 2829  
 
with four columns, so we compare this with  2
we reject the null hypothesis.
3
.05
 7.8147 and, since our computed chi-square is larger,
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)
28
252y0581 1/4/06
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)
Solution: We already have SST 
Note 2837 .3212  39000 .00
 x  n x  41649  2837.321  2649 .00 .
SSC   n x  n x  75714 .54   2837 .321 
2
2
ij
2
2
j j
2
2
 40001 .78  3900 .00  1001 .78 . This is SSB in a one way ANOVA.
SSR 
 n x
2
i i.
 n x  410115 .06   2837 .321 2  40460 .24  39000 .00  1460 .24
2
SSW  SST  SSR  SSC  2649 .99  1460 .24  1001 .78  187 .97
Source
SS
DF
MS
F
F .05
H0
Rows
1460.24
6
243.37
23.30S
6,18  2.66
F.05
Column means Equal
Columns
1001.78
3
333.93
31.98S
3,18  3.16
F.05
Column means Equal
Within
Total
187.97
2649.99
18
27
10.443
Since both our computed Fs exceed our table Fs, it looks like there is a difference between both row and
column means. An individual confidence interval has the formula for column means
2MSW
. Where the outline says to set m  1 and that if P  1 ,
 1   2  x1  x2   t RC P 1
2m
PR
replace RC P  1 with R  1C  1 . The formula becomes
2MSW
18 210 .443 
 47 .571  35 .143   t .025
7
PR
 12.398   2.1011.727   12.4  3.6 Since the error part is less than the difference between the two sample
means, the difference is significant.
 1   4  x1  x4   t R1C 1
2
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)
Solution: This is unambiguously cross classified data, so we must use a Friedman test here. The original
data is shown with ranks in row. I ranked top down, but bottom up should give the same results.
Row
1
2
3
4
5
6
7
Sum
Raters
Moore
Gaston
Heinrich
Seldon
Greasy
Waters
Pierce
Lee r1
52 1
38 1
54 1
43 1
58 1
36 1
52 1
7
Jacobs r2
25
4
31
2
38
4
30
3
44
4
28
2
41
3
22
Wilkes r3 Delap r4
29
3 33
2
24
4 29
3
40
2 39
3
31
2 28
4
46
3 47
2
22
4 25
3
37
4 45
2
22
19
To check our work, the sum of these 4 column sums is 70. The sum of the numbers from 1 to 4 in each row
45
 10 . Since there are 7 rows, multiply 10 by 7 to get 70 as the value for the column sums, so we are
is
2
unlikely to have made a mistake. Note that the 12 in this formula is a 12, not something that you might
find more convenient.
 12

 12

 12
1378   105
7 2  22 2  22 2  19 2   37 5  
 F2  
SRi2   3r c  1  
140

 745

 rc c  1 i

 


29
252y0581 1/4/06
 118 .1143  105  13.1143 Our Friedman table has no values for tests with four columns and 7 rows, so we
 
compare this with  2
hypothesis.
3
.05
 7.8147 and, since our computed chi-square is larger, we reject the null
c) Use Kendall’s coefficient of concordance to show how the raters differ and do a significance test. (3)
The outline says to take k columns with n items in each and rank each column from 1 to n . Then compute
a sum of ranks SRi for each row. The null hypothesis is that the rankings disagree. I have ranked bottom to
top.
Row
1
2
3
4
5
6
7
Raters
Moore
Gaston
Heinrich
Seldon
Greasy
Waters
Pierce
Lee
52
38
54
43
58
36
52
r1
4.5
2.0
6.0
3.0
7.0
1.0
4.5
Jacobs
25
31
38
30
44
28
41
r2
1
4
5
3
7
2
6
Wilkes r3
29
3
24
2
40
6
31
4
46
7
22
1
37
5
Delap
33
29
39
28
47
25
45
r4
4
3
5
2
7
1
6
SRi
SRi 2
12.5
11.0
22.0
12.0
28.0
5.0
21.5
156.25
121.00
484.00
144.00
784.00
25.00
462.25
We can see that the sum of SRi = 112 and the sum of SRi 2 = 2176.5. The average of the SRi column is
SR 
S
n  1k  84  16. Then
112
 16 . This serves as a check, since the outline says SR 
7
2
2
 SR
2
 
 n SR
2
 2176 .5  716 2  384 .5 . The 5% critical value for Kendall’s S is, according to
Table 12, 217.0. Since our computed S is above the table value, reject the null hypothesis. We can also
compute the Kendall Coefficient of Concordance (which must be must be between 0 and 1)
S
384 .5
384 .5
W


 0.8583 .
2
3
2
3
1 k n n
1 4 7 7
448
12
12




Note: To find your numbers for Lee note that
Lee’s column
 x
1

 c  333  7c and
x
 x
.1
1
 333 and
x
2
1
 16277 If we add the quantity c to
 c2  16277  2c(333)  7c 2 .
30
252y0581 1/4/06
Problem 3: Computer output is annotated.
————— 12/13/2005 6:03:40 PM ————————————————————
Welcome to Minitab, press F1 for help.
Results for: 252x05081-100.MTW
MTB > WSave "C:\Documents and Settings\rbove\My Documents\Minitab\252x05081100.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x05081-100.MTW'
MTB > let c20 = c2+c3+c4+c5
MTB > let c22 = c20/c21
MTB > let c23 = c2*c2+c3*c3+c4*c4+c5*c5
MTB > let c24 = c22*c22
MTB > AOVOneway c2 c3 c4 c5
One-way ANOVA: Lee, Jacobs, Wilkes, Delap
Source
Factor
Error
Total
DF
3
24
27
S = 8.284
Level
Lee
Jacobs
Wilkes
Delap
N
7
7
7
7
SS
1001.3
1646.9
2648.1
MS
333.8
68.6
F
4.86
R-Sq = 37.81%
Mean
47.571
33.857
32.714
35.143
StDev
8.522
7.151
8.712
8.649
P
0.009
R-Sq(adj) = 30.04%
Individual 95% CIs For Mean Based on
Pooled StDev
--+---------+---------+---------+------(--------*--------)
(--------*---------)
(--------*--------)
(--------*--------)
--+---------+---------+---------+------28.0
35.0
42.0
49.0
Pooled StDev = 8.284
MTB >
MTB >
SUBC>
SUBC>
MTB >
MTB >
stack c2 c3 c4 c5 c11
stack c2 c3 c4 c5 c11;
subscripts c12;
useNames.
rank c11 c13
vartest c11 c12
Test for Equal Variances: C11 versus C12
95% Bonferroni confidence intervals for standard deviations
C12 N
Lower
StDev
Upper
Delap 7 4.99463 8.64925 24.7451
Jacobs 7 4.12969 7.15142 20.4599
Lee 7 4.92097 8.52168 24.3801
Wilkes 7 5.03106 8.71233 24.9256
Bartlett's Test (normal distribution)
Test statistic = 0.28, p-value = 0.964
Levene's Test (any continuous distribution)
Test statistic = 0.08, p-value = 0.968
31
252y0581 1/4/06
Test for Equal Variances: C11 versus C12 – The high p-value for Bartlett’s test means that we
cannot reject the null hypothesis of equal variances at any reasonable significance level.
MTB >
SUBC>
SUBC>
SUBC>
MTB >
SUBC>
unstack c13;
subscripts c12;
after;
varNames.
NormTest c2;
KSTest.
Probability Plot of Lee – The high p-value for the KS (actually Lilliefors) test means that we cannot
reject the null hypothesis of Normality. This is probably because the sample is so small since the plot hints
at deviations from Normality. Between these two, we find justification in using part a rather than part b of
Problems 1 and 2.
MTB > print c1 c2 c3 c4 c5 c20 c21 c22 c23 c24
Data Display – The material on the right is the setup for comparing rows in 2-way ANOVA
Row
1
2
3
4
5
6
7
Raters
Moore
Gaston
Heinrich
Seldon
Greasy
Waters
Pierce
Lee
52
38
54
43
58
36
52
Jacobs
25
31
38
30
44
28
41
Wilkes
29
24
40
31
46
22
37
Delap
33
29
39
28
47
25
45
sum
139
122
171
132
195
111
175
cnt
4
4
4
4
4
4
4
mean
34.75
30.50
42.75
33.00
48.75
27.75
43.75
sq
5259
3822
7481
4494
9625
3189
7779
mnsq
1207.56
930.25
1827.56
1089.00
2376.56
770.06
1914.06
MTB > print c25-c28
Data Display – This is the ranked data for Kruskal – Wallis.
Row
1
2
3
4
5
6
7
C13_Delap
12.0
7.5
17.0
5.5
24.0
3.5
22.0
C13_Jacobs
3.5
10.5
15.5
9.0
21.0
5.5
19.0
C13_Lee
25.5
15.5
27.0
20.0
28.0
13.0
25.5
C13_Wilkes
7.5
2.0
18.0
10.5
23.0
1.0
14.0
32
252y0581 1/4/06
MTB > print c11 c14 c12
Data Display – This is the input for a 2-way ANOVA.
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
C11
52
38
54
43
58
36
52
25
31
38
30
44
28
41
29
24
40
31
46
22
37
33
29
39
28
47
25
45
C14
1
2
3
4
5
6
7
1
2
3
4
5
6
7
1
2
3
4
5
6
7
1
2
3
4
5
6
7
C12
Lee
Lee
Lee
Lee
Lee
Lee
Lee
Jacobs
Jacobs
Jacobs
Jacobs
Jacobs
Jacobs
Jacobs
Wilkes
Wilkes
Wilkes
Wilkes
Wilkes
Wilkes
Wilkes
Delap
Delap
Delap
Delap
Delap
Delap
Delap
MTB > Table c14 c12
Tabulated statistics: C14, C12
Rows: C14
Columns: C12
Delap Jacobs Lee
1
1
2
1
3
1
4
1
5
1
6
1
7
1
All
7
Cell Contents:
Wilkes
All
1
1
1
1
1
1
1
7
4
4
4
4
4
4
4
28
1
1
1
1
1
1
1
7
1
1
1
1
1
1
1
7
Count
MTB > Table c14 c12;
SUBC> means c11.
Tabulated statistics: C14, C12 – Since there is only one measurement per cell, this shows all data
and all means.
Rows: C14
Columns: C12
Delap Jacobs
Lee
1
33.00
2
29.00
3
39.00
4
28.00
5
47.00
6
25.00
7
45.00
All
35.14
Cell Contents:
25.00 52.00
31.00 38.00
38.00 54.00
30.00 43.00
44.00 58.00
28.00 36.00
41.00 52.00
33.86 47.57
C11 : Mean
Wilkes
All
29.00
24.00
40.00
31.00
46.00
22.00
37.00
32.71
34.75
30.50
42.75
33.00
48.75
27.75
43.75
37.32
33
252y0581 1/4/06
MTB > Twoway c11 c14 c12;
SUBC> Means c14 c12.
Two-way ANOVA: C11 versus C14, C12
Source
C14
C12
Error
Total
DF
6
3
18
27
SS
1459.36
1001.25
187.50
2648.11
MS
243.226
333.750
10.417
F
23.35
32.04
P
0.000
0.000
S = 3.227
R-Sq = 92.92%
C14
1
2
3
4
5
6
7
Individual 95% CIs For Mean Based on
Pooled StDev
-----+---------+---------+---------+---(----*---)
(----*---)
(----*----)
(----*----)
(----*---)
(----*---)
(----*---)
-----+---------+---------+---------+---28.0
35.0
42.0
49.0
Mean
34.75
30.50
42.75
33.00
48.75
27.75
43.75
C12
Delap
Jacobs
Lee
Wilkes
Mean
35.1429
33.8571
47.5714
32.7143
R-Sq(adj) = 89.38%
Individual 95% CIs For Mean Based on Pooled StDev
+---------+---------+---------+--------(----*----)
(----*----)
(----*----)
(----*-----)
+---------+---------+---------+--------30.0
35.0
40.0
45.0
MTB > sum c2
Sum of Lee
Sum of Lee = 333
MTB > ssq c2
Sum of Squares of Lee
Sum of squares (uncorrected) of Lee = 16277
MTB > sum c20
Sum of sum
Sum of sum = 1045
MTB > sum c21
Sum of cnt
Sum of cnt = 28
MTB > sum c23
Sum of sq
Sum of sq = 41649
MTB > sum c24
Sum of mnsq
Sum of mnsq = 10115.1
34
252y0581 1/4/06
Version 10
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
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
We personalize the data by adding 10 Lee’s column.
Row Raters
Lee Jacobs Wilkes
Delap
sum cnt
mean
sq
mnsq
1 Moore
62
25
29
33
149
4 37.25 6399 1387.56
2 Gaston
48
31
24
29
132
4 33.00 4682 1089.00
3 Heinrich
64
38
40
39
181
4 45.25 8661 2047.56
4 Seldon
53
30
31
28
142
4 35.50 5454 1260.25
5 Greasy
68
44
46
47
205
4 51.25 10885 2626.56
6 Waters
46
28
22
25
121
4 30.25 4009
915.06
7 Pierce
62
41
37
45
185
4 46.25 8919 2139.06
sum
403
237
229
246 1115
28
49009 11465.11
count
7
7
7
7
28
mean
57.571 33.857 32.714 35.143 Grand mean = 1045/28 = 39.821
ssq
23637
8331
7947
9094 49009
meansq
3314.42 1146.30 1070.21 1235.03 6765.96
 x  1115 , n  28 ,  x
From the above
x
 x  1115  39.821 .
n
 x
SSC   n
SST 
28
Note 2839 .821  44399 .94
2
2
ij
2
ij
 41649 ,
x
2
i.
 11465 .11
x
2
.j
 6765 .96 and
 n x  49009  2839 .821 2  4609 .06 .
2
2
j x j
 n x  76765 .96   28 39 .821 2
2
 47361 .72  44399 .94  2961 .78 . This is SSB in a one way ANOVA.
SSR 
 n x
2
i i.
 n x  411465 .11  2839 .821 2  45860 .44  44399 .94  1460 .50
2
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)
Solution: SSW  SST  SSB  4609 .06  2961 .78  1647 .28
Source
SS
DF
Between
2961.78
3
Within
Total
1647.28
4609.06
24
27
MS
987.26
F
14.38S
F .05
3, 24
F.05
=3.01
H0
Column means Equal
68.636
The ‘S’ next to the calculated F indicates that since it is larger than the table F we reject the null hypothesis.
35
252y0581 1/4/06
The outline gives a formula for an individual contrast 1   2  x1  x2   t n  m  s
2
1
1
, where

n1 n 2
s  MSW .
1   4  x1  x4   t 24 MSW
2
1 1
1
1
 57 .571  35 .143   2.064 68 .636


7 7
n1 n 4
2
 22 .428   2.064 68 .636    22.428   2.064 4.4284   22.4  9.1 . Since the error part is less than the
7
difference between the two sample means, the difference is significant.
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)
Solution: H 0 : Equal medians While doing Problem 3, I ranked the data. Since there are 28 numbers, the
ranks must extend from 1 to 28 . Any ties must be given the average rank.
Data Display – This is the ranked data for Kruskal – Wallis.
Row C13_Lee
1
25.5
2
23.0
3
27.0
4
24.0
5
28.0
6
20.5
7
25.5
Sum
173.5
C13_Jacobs
3.5
10.5
14.0
9.0
18.0
5.5
17.0
77.5
C13_Wilkes
7.5
2.0
16.0
10.5
20.5
1.0
13.0
70.5
C13_Delap
12.0
7.5
15.0
5.5
22.0
3.5
19.0
84.5
To check our work, the sum of these 4 column sums is 406. The sum of the numbers from 1 to 28 is
2829 
 406 so we are unlikely to have made a mistake. Again, note that the 12 in this formula is, in
2
fact, 12 and not some number that you might find more convenient.
 12
 SRi 2 

  3n  1
Now, compute the Kruskal-Wallis statistic H  
 nn  1 i  ni 

 12  173 .5 2 77 .5 2 70 .5 2 84 .5 2






4
4
4
 2829   4
 12  48219


 2829   4


  329    12  48219

 2829   4


  329 


  87  178 .2599  87  91 .2599 . Our Kruskal-Wallis table has no values for tests with

 
four columns, so we compare this with  2
reject the null hypothesis.
3
.05
 7.8147 and, since our computed chi-square is larger, we
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)
Solution: We already have SST 
 x
2
ij
 n x  49009  2839 .821 2  4609 .06 .
2
36
252y0581 1/4/06
Note 2839 .8212  44399 .94
SSC 
 n
2
j x j
 n x  76765 .96   28 39 .821 2
2
 47361 .72  44399 .94  2961 .78 . This is SSB in a one way ANOVA.
SSR 
 n x
2
i i.
 n x  411465 .11  2839 .821 2  45860 .44  44399 .94  1460 .50 .
2
SSW  SST  SSR  SSC  4609 .06  2961 .78  1460 .50  186 .78
Source
SS
DF
MS
F
F .05
H0
Rows
1460.50
6
243.42
23.46S
6,18  2.66
F.05
Column means Equal
Columns
2961.78
3
987.26
95.14S
3,18  3.16
F.05
Column means Equal
Within
Total
186.78
4609.06
18
27
10.377
Since both our computed Fs exceed our table Fs, it looks like there is a difference between both row and
column means. An individual confidence interval has the formula for column means
2MSW
. Where the outline says to set m  1 and that if P  1 ,
 1   2  x1  x2   t RC P 1
2m
PR
replace RC P  1 with R  1C  1 . The formula becomes
2MSW
18 210 .377 
 57 .571  35 .143   t .025
7
PR
 22.428   2.1011.722   12.4  3.6 Since the error part is less than the difference between the two sample
means, the difference is significant.
 1   4  x1  x4   t R1C 1
2
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)
Solution: This is unambiguously cross classified data, so we must use a Friedman test here. The original
data is shown with ranks in row. I ranked top down, but bottom up should give the same results.
Row
1
2
3
4
5
6
7
Sum
Raters
Moore
Gaston
Heinrich
Seldon
Greasy
Waters
Pierce
Lee r1
62 1
48 1
64 1
53 1
68 1
46 1
62 1
7
Jacobs r2
25
4
31
2
38
4
30
3
44
4
28
2
41
3
22
Wilkes r3 Delap r4
29
3 33
2
24
4 29
3
40
2 39
3
31
2 28
4
46
3 47
2
22
4 25
3
37
4 45
2
22
19
To check our work, the sum of these 4 column sums is 70. The sum of the numbers from 1 to 4 in each row
45
 10 since there are 7 rows multiply 10 by 7 to get 70 as the value for the column sums, so we are
is
2
unlikely to have made a mistake.
 12

 12

 12
1378   105
7 2  22 2  22 2  19 2   37 5  
 F2  
SRi2   3r c  1  
140

 745

 rc c  1 i

 118 .1143  105  13.1143 Our Friedman table has no values for tests with four columns and 7 rows, so we
 
 
compare this with  2
hypothesis.
3
.05


 7.8147 and, since our computed chi-square is larger, we reject the null
c) Use Kendall’s coefficient of concordance to show how the raters differ and do a significance test. (3)
37
252y0581 1/4/06
The outline says Take k columns with n items in each and rank each column from 1 to n then compute a
sum of ranks SRi for each row. The null hypothesis is that the rankings disagree. I have ranked bottom to
top.
Row
1
2
3
4
5
6
7
Raters
Moore
Gaston
Heinrich
Seldon
Greasy
Waters
Pierce
Lee
62
48
64
53
68
46
62
r1
4.5
2.0
6.0
3.0
7.0
1.0
4.5
Jacobs
25
31
38
30
44
28
41
r2
1
4
5
3
7
2
6
Wilkes r3
29
3
24
2
40
6
31
4
46
7
22
1
37
5
Delap
33
29
39
28
47
25
45
r4
4
3
5
2
7
1
6
SRi
SRi 2
12.5
11.0
22.0
12.0
28.0
5.0
21.5
156.25
121.00
484.00
144.00
784.00
25.00
462.25
We can see that the sum of SRi = 112 and the sum of SRi 2 = 2176.5. The average of the SRi column is
SR 
S
n  1k  84  16. Then
112
 16 . This serves as a check, since the outline says SR 
7
2
2
 SR
2
 
 n SR
2
 2176 .5  716 2  384 .5 . The 5% critical value for Kendall’s S is, according to
Table 12, 217.0. Since our computed S is above the table value, reject the null hypothesis. We can also
compute the Kendall Coefficient of Concordance (which must be must be between 0 and 1)
S
384 .5
384 .5
W


 0.8583 .
1 k 2 n3  n
1 42 73  7
448
12
12




38
252y0581 1/4/06
Problem 3: Computer output is annotated.
————— 12/13/2005 10:43:10 PM ————————————————————
Welcome to Minitab, press F1 for help.
Results for: 252x05081-110.MTW
MTB > WSave "C:\Documents and Settings\rbove\My Documents\Minitab\252x05081110.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x05081-110.MTW'
Existing file replaced.
MTB > let c20 = c2 + c3+ c4 + c5
MTB > let c22 = c20/c21
MTB > let c23 = c2*c2+c3*c3+c4*c4 + c5*c5
MTB > let c24 = c22 *c22
MTB > AOVOneway c2 c3 c4 c5
One-way ANOVA: Lee, Jacobs, Wilkes, Delap
Source
Factor
Error
Total
DF
3
24
27
S = 8.284
Level
Lee
Jacobs
Wilkes
Delap
N
7
7
7
7
SS
2961.3
1646.9
4608.1
MS
987.1
68.6
F
14.38
R-Sq = 64.26%
Mean
57.571
33.857
32.714
35.143
StDev
8.522
7.151
8.712
8.649
P
0.000
R-Sq(adj) = 59.79%
Individual 95% CIs For Mean Based on
Pooled StDev
----+---------+---------+---------+----(------*-----)
(------*-----)
(------*-----)
(-----*------)
----+---------+---------+---------+----30
40
50
60
Pooled StDev = 8.284
MTB >
SUBC>
SUBC>
MTB >
MTB >
stack c2 c3 c4 c5 c11;
subscripts c12;
useNames.
rank c11 c13
vartest c11 c12
Test for Equal Variances: C11 versus C12
95% Bonferroni confidence intervals for standard deviations
C12
Delap
Jacobs
Lee
Wilkes
N
7
7
7
7
Lower
4.99463
4.12969
4.92097
5.03106
StDev
8.64925
7.15142
8.52168
8.71233
Upper
24.7451
20.4599
24.3801
24.9256
Bartlett's Test (normal distribution)
Test statistic = 0.28, p-value = 0.964
Levene's Test (any continuous distribution)
Test statistic = 0.08, p-value = 0.968
39
252y0581 1/4/06
Test for Equal Variances: C11 versus C12 – The high p-value for Bartlett’s test means that we
cannot reject the null hypothesis of equal variances at any reasonable significance level.
MTB >
SUBC>
SUBC>
SUBC>
MTB >
SUBC>
unstack c13;
subscripts c12;
after;
varNames.
NormTest c2;
KSTest.
Probability Plot of Lee– The high p-value for the KS (actually Lilliefors) test means that we cannot
reject the null hypothesis of Normality. This is probably because the sample is so small since the plot hints
at deviations from Normality. Note that, because of the results of these two tests, we can use ANOVA in the
previous two problems.
MTB > print c1 c2 c3 c4 c5 c20 c21 c22 c23 c24
Data Display – The material on the right is the setup for comparing rows in 2-way ANOVA
Row
1
2
3
4
5
6
7
Raters
Moore
Gaston
Heinrich
Seldon
Greasy
Waters
Pierce
Lee
62
48
64
53
68
46
62
Jacobs
25
31
38
30
44
28
41
Wilkes
29
24
40
31
46
22
37
Delap
33
29
39
28
47
25
45
sum
149
132
181
142
205
121
185
cnt
4
4
4
4
4
4
4
mean
37.25
33.00
45.25
35.50
51.25
30.25
46.25
sq
6399
4682
8661
5454
10885
4009
8919
mnsq
1387.56
1089.00
2047.56
1260.25
2626.56
915.06
2139.06
40
252y0581 1/4/06
MTB > print c25-c28
Data Display – This is the ranked data for Kruskal – Wallis.
Row C13_Delap C13_Jacobs
1
12.0
3.5
2
7.5
10.5
3
15.0
14.0
4
5.5
9.0
5
22.0
18.0
6
3.5
5.5
7
19.0
17.0
MTB > print c11 c14 c12
C13_Lee
25.5
23.0
27.0
24.0
28.0
20.5
25.5
C13_Wilkes
7.5
2.0
16.0
10.5
20.5
1.0
13.0
Data Display – This is the input for a 2-way ANOVA.
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
C11
62
48
64
53
68
46
62
25
31
38
30
44
28
41
29
24
40
31
46
22
37
33
29
39
28
47
25
45
C14
1
2
3
4
5
6
7
1
2
3
4
5
6
7
1
2
3
4
5
6
7
1
2
3
4
5
6
7
C12
Lee
Lee
Lee
Lee
Lee
Lee
Lee
Jacobs
Jacobs
Jacobs
Jacobs
Jacobs
Jacobs
Jacobs
Wilkes
Wilkes
Wilkes
Wilkes
Wilkes
Wilkes
Wilkes
Delap
Delap
Delap
Delap
Delap
Delap
Delap
MTB > table c14 c12
Tabulated statistics: C14, C12
Rows: C14
Columns: C12
Delap Jacobs Lee
1
1
2
1
3
1
4
1
5
1
6
1
7
1
All
7
Cell Contents:
1
1
1
1
1
1
1
7
1
1
1
1
1
1
1
7
Count
Wilkes
All
1
1
1
1
1
1
1
7
4
4
4
4
4
4
4
28
41
252y0581 1/4/06
MTB > table c14 c12;
SUBC> means c11.
Tabulated statistics: C14, C12 – Since there is only one measurement per cell, this shows all data
and all means.
Rows: C14
Columns: C12
Delap Jacobs
Lee
1
33.00
25.00 62.00
2
29.00
31.00 48.00
3
39.00
38.00 64.00
4
28.00
30.00 53.00
5
47.00
44.00 68.00
6
25.00
28.00 46.00
7
45.00
41.00 62.00
All
35.14
33.86 57.57
Cell Contents: C11 : Mean
MTB > Twoway c11 c14 c12;
SUBC> Means c14 c12.
Wilkes
All
29.00
24.00
40.00
31.00
46.00
22.00
37.00
32.71
37.25
33.00
45.25
35.50
51.25
30.25
46.25
39.82
Two-way ANOVA: C11 versus C14, C12
Source
C14
C12
Error
Total
DF
6
3
18
27
SS
1459.36
2961.25
187.50
4608.11
MS
243.226
987.083
10.417
F
23.35
94.76
P
0.000
0.000
S = 3.227
R-Sq = 95.93%
C14
1
2
3
4
5
6
7
Individual 95% CIs For Mean Based on
Pooled StDev
--+---------+---------+---------+------(----*----)
(----*----)
(----*---)
(----*----)
(----*----)
(----*----)
(----*----)
--+---------+---------+---------+------28.0
35.0
42.0
49.0
Mean
37.25
33.00
45.25
35.50
51.25
30.25
46.25
C12
Delap
Jacobs
Lee
Wilkes
Mean
35.1429
33.8571
57.5714
32.7143
R-Sq(adj) = 93.90%
Individual 95% CIs For Mean Based on
Pooled StDev
--+---------+---------+---------+------(--*--)
(--*---)
(--*--)
(--*--)
--+---------+---------+---------+------32.0
40.0
48.0
56.0
MTB > sum c2
Sum of Lee – These are column sums for one-way and 2-way ANOVA
Sum of Lee = 403
MTB > ssq c2
Sum of Squares of Lee
Sum of squares (uncorrected) of Lee = 23637
MTB > sum c20
Sum of sum
Sum of sum = 1115
MTB > sum c21
42
252y0581 1/4/06
Sum of cnt
Sum of cnt = 28
MTB > sum c23
Sum of sq
Sum of sq = 49009
MTB > sum c24
Sum of mnsq
Sum of mnsq = 11465.1
MTB > sum c25
Sum of C13_Delap
Sum of C13_Delap = 84.5
MTB > sum c26
Sum of C13_Jacobs
Sum of C13_Jacobs = 77.5
MTB > sum c27
Sum of C13_Lee
Sum of C13_Lee = 173.5
MTB > sum c28
Sum of C13_Wilkes
Sum of C13_Wilkes = 70.5
MTB
MTB
MTB
MTB
MTB
>
>
>
>
>
rank c2
rank c3
rank c4
rank c5
let c36
c32
c33
c34
c35
= c32 + c33 + c34 + c35
MTB > sum c36
Sum of C36
Sum of C36 = 112
MTB > let c37 = c36*c36
MTB > sum c37
Sum of C37
Sum of C37 = 2176.5
MTB > print c32-c37
Data Display – This is the ranked data for Kendall’s Coefficient of concordance
Row
1
2
3
4
5
6
7
C32
4.5
2.0
6.0
3.0
7.0
1.0
4.5
C33
1
4
5
3
7
2
6
C34
3
2
6
4
7
1
5
C35
4
3
5
2
7
1
6
C36
12.5
11.0
22.0
12.0
28.0
5.0
21.5
C37
156.25
121.00
484.00
144.00
784.00
25.00
462.25
43
252y0581 1/4/06
MTB > print c27 c26 c28 c25
Data Display
Row
1
2
3
4
5
6
7
C13_Lee
25.5
23.0
27.0
24.0
28.0
20.5
25.5
C13_Jacobs
3.5
10.5
14.0
9.0
18.0
5.5
17.0
C13_Wilkes
7.5
2.0
16.0
10.5
20.5
1.0
13.0
C13_Delap
12.0
7.5
15.0
5.5
22.0
3.5
19.0
MTB >
44
252y0581 1/4/06
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.
45
252y0581 1/4/06
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]
Solution: This is the solution to the original Problem. Your solution should not be very different. You have
84 observations instead of n  80.
46
252y0581 1/4/06
Spare Parts Computation:
5323 .20
Y 
 66 .5400
80
1283 .42
X1 
 16 .0428
80
328 .33
X2 
 4.1041
80
X
2
1
 X  nX  SSX 2  733 .16
 Y  nY  SST  SSY  16826 .8
 X Y  nX Y  SX1Y  7277 .87
 X Y  nX Y  SX 2Y  2341 .25
 X X  nX X  SX1X 2  1056 .79
2
2
2
2
2
2
1
1
2
2
1
 nX 12  SSX1  9717 .86
2
1
2
Problem 1:
a. Compute the regression equation Yˆ  b0  b1 x1 to predict investment on the basis of change in
sales only. (2)
b1 
Sxy

SSx
 xy  nxy  7277 .87  0.7489
 x  nx 9717 .86
2
2
b0  y  b1 x  66.54  0.7489 16.0428  54.526 , So Yˆ  b0  b1 x becomes
Yˆ  54.526  0.7489 x .
b. Compute R 2 . (2) SSR  b1 Sxy  b1
R2 
 xy  nxy   0.7489 7277 .87   5450 .4
SSR 5450 .4

 .3239 or
SST 16826 .8
 XY  nXY 
Sxy  7277 .87 


 X  nX  Y  nY  SSxSSy 9717 .86 16826 .8  .3239
2
R
2
2
2
2
2
2
2
c. Compute s e . (2) SSE  SST  SSR  16826 .8  5450 .4  11376 .4 or SSR  R 2 SST  and


SSE  1 R 2 SST should give you the same result. ( s e2 is always positive!)
SSE 11376 .4
s e2 

 145 .851 So s e  145 .851  12 .0768
n2
78
d. Compute s b0 and do a significance test on b0 (1.5) H 0 :  0  0
Recall n  80,   .05 ,
X
2
1
 nX 12  SSX1  9717 .86 , s e2  145.851, X 1  16 .0428 and
b0  y  b1 x  54.526 . t.78
025  1.991
1
s b20  s e2  
n


2


  145 .851  1  16 .0428    145 .8510.01250  0.02648   5.6853
9717 .86 
X 2  nX 2 
 80

X2
s b0  5.6853  2.3844 So t 
b0   00 54 .526  0

 22 .868 . Since this is larger than 1.991 in
s b0
2.3844
absolute value, we reject the null hypothesis.
e. Compute s b1 and do a significance test on b1 (2) H 0 : 1  0
Recall n  80,   .05 ,
X
2
1
 nX 12  SSX1  9717 .86 , s e2  145.851, X 1  16 .0428 and
145 .851
2
2 1 
 0.1501 s b  0.1501  0.1225
b1  0.7489 . t.78

025  1.991 s b1  s e 
1
 SS x  9717 .86
.
47
252y0581 1/4/06
So t 
b1  10 0.7489  0

 6.113 . Our rejection zone is below -1.991 and above 1.991. Since
s b1
0.1225
our calculated t falls in the upper reject zone, we can say that b1 is significant.
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)
Recall that our equation is Yˆ  54.526  0.7489 x . X 0  9883 .167  9668 .827  217 .34 ,
n  80,   .05 ,
X
2
1
 nX 12  SSX1  9717 .86 , s e2  145.851 and X 1  16 .0428 .
The Prediction Interval is Y0  Yˆ0  t sY , where Yˆ  54.526  0.7489 x
 54.526  0.7489 217 .34   217 .291
sY2


1
s e2 


n
2

 1 217 .34  16 .04 2


X0  X

 1  145 .851 
 1
 80



SS x


9717 .86
 145 .8511.01250  4.169816   755 .8460 . sY  755 .8460  27 .4926 . So Y0  Yˆ0  t sY
 217 .291  1.99127.4926   217  55 .
g. Do an ANOVA table for the regression. What conclusion can you draw from the hypothesis test
in the ANOVA? (2)
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
Error
SSE
n  k  1 MSE
Total
SST
n 1
SSE  SST  SSR  16826 .8  5450 .4  11376 .4 . n  80 and the number of independent variables
is k  1 .
The ANOVA table for the regression reads:
Source
SS
DF
MS
Fcalc
F.05
1,78  3.96
F.05
Regression
5450.5
1
5450.5 37.37s
Error
11376.6
78
145.85
Total
16826.8
79
Since our computed F is larger than the table F, we conclude that there is a linear relationship
between the dependent and independent variable. In a simple regression this is the same as saying
that the slope is significant.
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)
Recall the values of the Spare Parts computed above. 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
2
1
1
2
 
 nX X   b  X
1
2
2
2
2

 nX 
2
2
Actually these can be considered the 2nd and 3rd equations, the first one being
Y  b0  b1 X 1  b2 X 2 .
48
252y0581 1/4/06
The 2nd and 3rd Normal equations are
7277 .87  9717 .86 b1
2341 .25  1056 .79 b1
1056 .79
 1.44142 , multiply Equation (3) by 1.44142.
733 .16
7277 .87
3374 .72
Now, subtract Equation 3' from Equation (2)
So b1 
get
1056 .79 b2
 733 .16 b2
 9717 .86 b1
 1523 .28b1
3903 .15  8194 .58b1
2
. Since
3
1056 .79 b2
 1056 .79 b2
2
3'
0b2
3903 .15
 0.4763 . Now, substitute 0.4763 into either Equation (2) or Equation (3). We
8194 .58
7277 .87  9717 .860.4763   1056 .79b2 2" or
2341 .25  1056 .790.4763   733 .16b2
3"
Using 2" , we get 1056 .79b2  7277 .87  4628 .62 . So b2 
2649 .65
 2.5069
1056 .79
1837 .90
 2.5068 . The difference is
or, using 3" , we get 733 .16b2  2341 .25  503 .35 . So b2 
733 .16
due to rounding.
Finally, since Y  b0  b1 X 1  b2 X 2 , we can write b0  Y  b1 X 1  b2 X 2
 66.5400  0.4763 16.0428  2.5068 4.1041  66.5400  7.6412  10.2882  48.6106 . Our
equation is thus Yˆ  48.611 0.476X  2.507X . The results are questionable. We expect to get
1
2
a positive coefficient for increase in sales as the accelerator theory predicts. The positive
coefficient of interest rates looks wrong, rising interest rates should depress investment. It’s
possible that we are seeing something else. Could it be that the same forces that posh up
investment, also push up interest rates?
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)
X 1Y  nX 1Y  SX 1Y  7277 .87 and
X 2 Y  nX 2 Y  SX 2Y  2341 .25


2
2
 Y  nY  SST  SSY  16826 .8
SSE  SST  SSR * . Thus
SSR  b1 Sx1 y  b2 Sx2 y  0.4763 7277 .87   2.507 2341 .25   3466 .449  5869 .514  9336 .0 *
so SSE  16826 .8  9336 .0  7490 .8 *
In the first regression, we had R 2  RY21  .3239
SSR 9336 .0

 .5548 *. If we use R 2 , which is R 2
SST 16826 .8
adjusted for degrees of freedom, for the first regression, the number of independent variables was
n  1R 2  k  79 0.3239  1  .3152 and for the second regression k  2 and
k  1 and R 2 
n  k 1
78
In this regression, we have R 2  RY212 
R2 
n  1R 2  k  79 0.5548  2  .5432 .
n  k 1
77
49
252y0581 1/4/06
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)
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
Error
SSE
n  k  1 MSE
Total
SST
n 1
SSE  SST  SSR  16826 .8  5450 .4  11376 .4 . n  80 and the number of independent variables
is k  1 .
The ANOVA table for the first regression read:
Source
SS
DF
Regression
Error
Total
Our new table is
Source
5450.5
11376.6
16826.8
1
78
79
SS
DF
MS
Fcalc
5450.5
145.85
37.37s
MS
Fcalc
F.05
1,78  3.96
F.05
F.05
2,77  2.37
F.05
Regression
9336.0
2
4668.0 47.98s
Error
7490.8
77
97.283
Total
16826.8
79
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
X1
5450.5
1
1,77  3.96
X2
2855.5
1
2855.5 29.35s F.05
Error
27490.8 77
97.283
Total
16826.8 79
Since our computed F is larger that the table F, we reject the hypothesis that X2 and Y are
unrelated.
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)
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)
Recall that our equation was Yˆ  54.526  0.7489 x . It predicted a value of Yˆ  217 .291 .
Now Yˆ  48.611 0.476X  2.507X , X  9883 .167  9668 .827  217 .34 , and
1
2
1
X 2  7.15  2.167  4.98 , so our predicted value of Y is
Yˆ  48.611 0.476217.34  2.5074.98  48.611 103.454  12.485  164.550 , a fall of about
24%.
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.
50
252y0581 1/4/06
Explain your results. (2)
Here are my results.
MTB > regress c2 2 c3 c4;
SUBC> VIF;
SUBC> DW.
Regression Analysis: Y versus X1, X2
The regression equation is
Y = 48.6 + 0.476 X1 + 2.51 X2
Predictor
Constant
X1
X2
Coef
48.610
0.4764
2.5067
S = 9.86336
SE Coef
2.161
0.1090
0.3967
R-Sq = 55.5%
Analysis of Variance
Source
DF
SS
Regression
2
9335.8
Residual Error 77
7491.0
Total
79 16826.8
Source
X1
X2
DF
1
1
T
22.50
4.37
6.32
P
0.000
0.000
0.000
VIF
1.2
1.2
R-Sq(adj) = 54.3%
MS
4667.9
97.3
F
47.98
P
0.000
Seq SS
5451.3
3884.5
Unusual Observations
Obs
X1
Y
Fit
50 -4.2 63.70 67.00
51 -8.2 61.93 66.44
SE Fit
3.38
3.91
Residual
-3.30
-4.52
St Resid
-0.36 X
-0.50 X
X denotes an observation whose X value gives it large influence.
Durbin-Watson statistic = 0.0952666
The VIFs are quite small, so there seems to be no problem of collinearity. The Durbin-Watson statistic is
awfully low. The table in the text gives for   .05 , d L  1.59 and d U  1.69 and   .01 , d L  1.44 and
dU  1.54. The 2-sided diagram that I gave you is below.
0   0 dL
?
dU
  0 2   0 4  dU
+
+
+
+
+
?
4 dL
+
 0 4
+
Since the D-W statistic is on the right extreme of this diagram, the data seems to be showing an incredible
level of Autocorrelation.
51
252y0581 1/4/06
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
52