252y0773 11/19/07 (Page layout view!)
ECO252 QBA2
THIRD EXAM
November 26, 2007
Version 3
Name ______Key____________
Student number_______________
Class Day and hour____________
I. (8 points) Do all the following (2points each unless noted otherwise). Make Diagrams! Show your
work! All probabilities must be between zero and (positive) 1!
x ~ N 39,21
116  39 
 69  39
z
1. P69  x  116   P 
 P1.43  z  3.67 
21
21 

 P0  z  3.67   P0  z  1.43   .4999  .4236  .0763
For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line!
Shade the area between 1.43 and 3.67. Because this is on one side of zero we must subtract the area
between zero and 1.43 from the larger area between zero and 3.67. If you wish, make a completely separate
diagram for x . Draw a Normal curve with a mean at 39. Indicate the mean by a vertical line! Shade the
area between 69 and 116. This area is on one side of the mean (39) so we subtract to get our answer.
57  39 
 30  39
z
2. P30  x  57   P 
 P0.43  z  0.86   P0.43  z  0  P0  z  0.86 
21 
 21
 .1664  .3051  .4715
For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line!
Shade the entire area between -0.43 and 0.86. Because this is on both sides of zero we must add the area
between -0.43 and zero to the area between zero and 0.86. If you wish, make a completely separate diagram
for x . Draw a Normal curve with a mean at 39. Indicate the mean by a vertical line! Shade the entire
area between 69 and 116. This area is on both sides of the mean (39) so we add to get our answer.
0  39 

3. Px  0  P  z 
 Pz  1.86   P1.86  z  0  Pz  0  .4686  .5  .9686
21 

For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line!
Shade the entire area above -1.86. Because this is on both sides of zero we must add the area between
-1.86 and zero to the area above zero. If you wish, make a completely separate diagram for x . Draw a
Normal curve with a mean at 39. Indicate the mean by a vertical line! Shade the entire area above zero,
remembering that zero is below the mean. This area is on both sides of the mean (39) so we add to get our
answer.
4.
x.085 (Do not try to use the t table to get this.) For z make a diagram. Draw a Normal curve with a
mean at 0. z .085 is the value of z with 8.5% of the distribution above it. Since 100 – 8.5 = 91.5, it is also
the 91.5th percentile. Since 50% of the standardized Normal distribution is below zero, your diagram should
show that the probability between z .085 and zero is 91.5% - 50% = 41.5% or P0  z  z.085   .4150 . The
closest we can come to this is P0  z  1.37   .4147 . (1.38 is fairly close.) So z.085  1.37 (or something
slightly larger). To get from z .085 to x.085 , use the formula x    z , which is the opposite of z 
x

. x  39  1.37 21  67.77 . If you wish, make a completely separate diagram for x . Draw a Normal curve
with a mean at 39. Show that 50% of the distribution is below the mean (39). If 8.5% of the distribution is
above x.085 , it must be above the mean and have 41.5% of the distribution between it and the mean.
Check:
67 .77  39 

Px  67.77   P  z 
  Pz  1.37   Pz  0  P0  z  1.37   .5  .4147  .0853  .085
21


1
252y0773 11/19/07 (Page layout view!)
II. (22+ points) Do all the following (2 points each unless noted otherwise). Do not answer a question
‘yes’ or ‘no’ without giving reasons. Show your work when appropriate. Use a 5% significance level except
where indicated otherwise. Note that this is extremely long and that no one will do all the problems, so look
them over!
1. Turn in your computer problems 2 and 3 marked as requested in the Take-home. (5 points, 2 point
penalty for not doing.)
2. In an ordinary 1-way ANOVA, if the computed F statistic exceeds the value from the F table at the
given significance level, we can
a. Reject the null hypothesis because the difference between the means is not significant
b. *Reject the null hypothesis because there is evidence of a significant difference between some of
the means.
c. Not reject the null hypothesis because the difference between the means is not significant.
d. Not reject the null hypothesis because the difference between the means is significant.
c. Not reject the null hypothesis because the difference between the variances is not significant.
d. Not reject the null hypothesis because the difference between the variances is significant.
e. None of the above.
[7]
3. After an analysis if variance, you would use the Tukey-Kramer procedure or similar confidence
intervals to check
a. For Normality
b. For equality of variances
c. For independence of error terms
d. *For pairwise differences in means
e. For all of the above
f. For none of the above
4. If an ordinary one-way ANOVA has 17 columns 25 rows and 17 25   425 , the degrees of freedom
for the F test has
a. 400 and 24
b. 408 and 16
c. 24 and 400
d.* 16 and 408
e. 400 and 424
f. 408 and 424
g. 424 and 400
h. 424 and 408
i. 16 and 24
j. None of the above. The correct answer is _______.
Explanation: This is a one-way ANOVA. The total number of observations is n  425 and the number
of columns is m  17 . This means there are 425-1 = 424 total degrees of freedom and that between the
columns there are 17-1 = 16 degrees of freedom. This leaves 424 – 16 = 408 degrees of freedom for the
error (within) term. Numbers are filled in below.
Source
SS
DF
MS
F
F

Between
SSB
m  1  16
Within
SSW
n  m  408
Total
SST
n  1  424
SSB
m 1
SSW
MSW 
nm
MSB 
F
MSB
MSW
16,408  1.67
F.05
2
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5. Assuming that your answer to 4 is correct and that the significance level is 5%, the correct value of F
from the table is 1.67. (This may have to be approximate. If so, what did you use?) (1)
[12]


16, 408
16, 400
Explanation: This is supposed to be F.05
. But the closest we can come is F.05
 1.67 . Since
16,1000  1.65 , F 16, 408 could only be a hair below 1.67.
the next value down on the table is F.05
.05
Note: I will check your answer against what you said in the previous question. The answer above is
16, 408 .
wrong if you did not say something close to F.05
Exhibit 1 A manager believes that the number of sales that an employee makes is related to the number of
years worked and their score on an aptitude test. He runs the data below on Minitab and gets the following
Employee
1
2
3
4
5
6
7
8
Sales
100
90
80
70
60
50
40
30
Years
10
3
8
5
5
7
1
1
Score
70
100
90
40
80
50
40
10
MTB > regress c1 2 c2 c3
Regression Analysis: Sales versus Years, Score
The regression equation is
Sales = 21.3 + 3.10 Years +
Predictor
Coef SE Coef
Constant
21.29
13.14
Years
3.103
1.984
Score
0.4698
0.2133
S = 15.1088
R-Sq = 72.8%
Analysis of Variance
Source
DF
SS
Regression
2 3058.6
Residual Error
5 1141.4
Total
7 4200.0
Source
Years
Score
DF
1
1
0.470 Score
T
P
1.62 0.166
1.56 0.179
2.20 0.079
R-Sq(adj) = 62.0%
MS
1529.3
228.3
F
6.70
P
0.038
Seq SS
1951.4
1107.3
The sum of the sales column is 520 and the sum of the squared numbers in the sales column is not needed.
The sum of the 'years' column is 40 and the sum of the squared numbers in the years column is 274.
The sum of the score column is 480 and the sum of the squared numbers in the score column is 35200
If Sales is the dependent variable and years and score are the independent variables we have found that the
sum of x1y is 2980 and the sum of x1 x2 is 2720. The sum of x2y has not been computed.
6. In the multiple regression, what coefficients are significant at the 10% significance level? (2)
Solution: Only 0.4698, the coefficient of ‘score’ has a p-value below .10.
7. In the multiple regression, what coefficients are significant at the 5% significance level? (1)
Solution: None
[15]
8. Assuming that the coefficients in the multiple regression are correct, how many sales would we predict
for someone with 5 years of experience and a score of 80? (1)
Sales = 21.29 + 3.103 Years + 0.4698 Score = 21.29 + 3.103(5) + 0.4698(80)
= 21.29 + 15.50 + 37.58 = 74.37
3
252y0773 11/19/07 (Page layout view!)
Exhibit 1
Empl
1
2
3
4
5
6
7
8
Sales
100
90
80
70
60
50
40
30
Years
10
3
8
5
5
7
1
1
Regression Analysis: Sales versus Years, Score
The regression equation is
Sales = 21.3 + 3.10 Years + 0.470 Score
Predictor
Coef SE Coef
T
P
Constant
21.29
13.14 1.62 0.166
Years
3.103
1.984 1.56 0.179
Score
0.4698
0.2133 2.20 0.079
S = 15.1088
R-Sq = 72.8%
R-Sq(adj) = 62.0%
Score
70
100
90
40
80
50
40
10
Analysis of Variance
Source
DF
SS
Regression
2 3058.6
Residual Error
5 1141.4
Total
7 4200.0
MS
1529.3
228.3
F
6.70
P
0.038
Source DF Seq SS
Years
1 1951.4
Score
1 1107.3
The sum of the sales column is 520 and the sum of the squared numbers in the sales column is not needed.
The sum of the years column is 40 and the sum of the squared numbers in the years column is 274.
The sum of the score column is 480 and the sum of the squared numbers in the score column is 35200
If Sales is the dependent variable and years and score are the independent variables we have found that the sum of x1y is 2980 and
the sum of x1 x2 is 2720. The sum of x2y has not been computed.
9. Using the information in the multiple regression printout, make your result is 8) into a rough prediction
interval. (2).
Solution: The outline says that an approximate prediction interval is Y0  Yˆ0  t s e . Remember
df  n  k  1  10  2  1  5 . In the printout s e  S = 15.1088  MSE  228.3 . So
Y0  74.37  t.5025 15.1088  74.37  2.57115.1088  74.37  38.84
10. Using the information in the printout, what is the value of R-squared for a regression of ‘sales’ against
‘years’ alone? (2)
Solution: Looking at the sequential Sum of squares the regression sum of squares is 1951.4 for ‘Years’
1951 .4
 .4646
alone. The total sum of squares is 4200.0, so we have R 2 
4200 .0
11. Do a simple regression of ‘sales’ against ‘score’ alone. Before you do something ridiculous see
252blunders!
xy that you will need for this regression. Show your work! (2)
a) Compute the sum

Don’t compute stuff that has already been done for you!
Solution: The only column that you should have computed is in bold below.
Row
1
2
3
4
5
6
7
8
Sales
100
90
80
70
60
50
40
30
520
Years
10
3
8
5
5
7
1
1
40
Score
70
100
90
40
80
50
40
10
480
Ysq
10000
8100
6400
4900
3600
2500
1600
900
38000
x1sq
100
9
64
25
25
49
1
1
274
x2sq
4900
10000
8100
1600
6400
2500
1600
100
35200
x1y
x2y
1000 7000
270 9000
640 7200
350 2800
300 4800
350 2500
40 1600
30
300
2980 35200
x1x2
700
300
720
200
400
350
40
10
2720
b) It says that you do not need to know the sum of squares in the sales column. You do
Y 2  nY 2 . Without doing any computing, tell
however need the spare part SS y 

what its value is. (1)
Solution: The ANOVA in the computer output says that the total sum of squares is 4200.0.
Of course, if you like to waste time SS y 
Y
2
2
 520 
 nY 2  38000  8
  4200 . ( Y  65 )
 8 
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252y0773 11/19/07 (Page layout view!)
Exhibit 1
Empl Sales Years Score
1
100
10
70
2
90
3
100
3
80
8
90
4
70
5
40
5
60
5
80
6
50
7
50
7
40
1
40
8
30
1
10
Regression Analysis: Sales versus Years, Score
The regression equation is
Sales = 21.3 + 3.10 Years + 0.470 Score
Predictor
Coef SE Coef
T
P
Constant
21.29
13.14 1.62 0.166
Years
3.103
1.984 1.56 0.179
Score
0.4698
0.2133 2.20 0.079
S = 15.1088
R-Sq = 72.8%
R-Sq(adj) = 62.0%
Analysis of Variance
Source
DF
SS
Regression
2 3058.6
Residual Error
5 1141.4
Total
7 4200.0
MS
1529.3
228.3
F
6.70
P
0.038
Source DF Seq SS
Years
1 1951.4
Score
1 1107.3
The sum of the sales column is 520 and the sum of the squared numbers in the sales column is not needed.
The sum of the years column is 40 and the sum of the squared numbers in the years column is 274.
The sum of the score column is 480 and the sum of the squared numbers in the score column is 35200
If Sales is the dependent variable and years and score are the independent variables we have found that the sum of x1y is 2980 and
the sum of x1 x2 is 2720. The sum of x2y has not been computed.
c) Compute the coefficients of the equation Yˆ  b0  b2 x to predict the value of ‘sales’ on the
basis of ‘score.’ (4)
[27]
X  480 ,
XY  35200 ,
Y  520 , (you computed )
Solution: First copy n  8,
X
2
 35200 and
Y
Then compute means: X 


2
is not needed. (It’s 38000.)
 X  480  60
n
8
The ‘Spare Parts’ are as follows: SS x 
You already found SS y 

Y
2
X
Y 
2
 Y  520  65 .
n
8
 nX  35200  8602  6400
2
 nY 2  4200  SST (Total Sum of Squares) .
 XY  nXY  35200  860 65   4000 .
S xy
 XY  nXY  4000  0.625 and b
So b1 

SS x  X 2  nX 2 6400
S xy 
0
 Y  b1 X  65  0.625  60   27 .5 , which
means Yˆ  27 .5  0.625 X or Y  27.5  0.625 X  e
d) Compute R 2 . (3)
Solution: SSR  b1 S xy  0.6254000  2500 . We can say R 2 
R2 
b1 S xy
SSy

SSR 2500

 .59524 or
SST 4200
S xy 2
0.625 4000 
4000 2  .59524

 .59524 or R 2 
6400 4200 
4200
SS x SS y
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252y0773 11/19/07 (Page layout view!)
MTB > regress c1 2 c2 c3
Regression Analysis: Sales versus Years, Score
The regression equation is
Sales = 21.3 + 3.10 Years + 0.470 Score
Predictor
Coef SE Coef
T
P
Constant
21.29
13.14 1.62 0.166
Years
3.103
1.984 1.56 0.179
Score
0.4698
0.2133 2.20 0.079
S = 15.1088
R-Sq = 72.8%
R-Sq(adj) = 62.0%
Exhibit 1 A manager believes that
the number of sales that an employee
makes is related to the number of
years worked and their score on an
aptitude test. He runs the data below
on Minitab and gets the following
Empl Sales Years Score
1
100
10
70
2
90
3
100
3
80
8
90
4
70
5
40
5
60
5
80
6
50
7
50
7
40
1
40
8
30
1
10
Analysis of Variance
Source
DF
SS
Regression
2 3058.6
Residual Error
5 1141.4
Total
7 4200.0
MS
1529.3
228.3
F
6.70
P
0.038
Source DF Seq SS
Years
1 1951.4
Score
1 1107.3
The sum of the sales column is 520 and the sum of the squared numbers in the sales column is not needed.
The sum of the 'years' column is 40 and the sum of the squared numbers in the years column is 274.
The sum of the score column is 480 and the sum of the squared numbers in the score column is 35200
If Sales is the dependent variable and years and score are the independent variables we have found that the sum of x1y is 2980 and
the sum of x1 x2 is 2720. The sum of x2y has not been computed.
e) Is the slope in this regression significant at the 5% level? Do not answer this question
without appropriate calculations! (4)
Solution: We can compute SSE  SST  SSR  4200  2500  1700 . Then
SS y  b1 S xy 4200  0.625 4000 
SSE 1700
s e2 

 283 .3333 or s e2 

 283 .3333
n2
6
n2
6
 1  283 .3333
s e  283 .3333  16 .83251 . So s b21  s e2 
 .04427 and

6400
 SS x 
H 0 : 1  0
b 0
use t  1
and if the null
s b  .04427  0.2104 . The outline says to test 
1
H
:


0
s b1
 1 1
hypothesis is false in that case we say that 1 is significant. So our ‘do not reject’ zone is between


0.625  0
.2104
 2.971 is outside both these zones, so that there is no doubt that the coefficient is significant.
8
8
 t .05
 1.860 if   .10 or between  t .025
 2.306 if   .05 . Our calculated t 
f) Predict the average number of sales individuals with a score of 75 will make and make
your estimate into an appropriate 95% interval. (4)
Solution: We found Yˆ  27.5  0.625X  27.5  0.62575  74.375 . The outline says that the
Confidence Interval is  Y0

1 X X
 Yˆ0  t sYˆ , where sY2ˆ  s e2   0
n
SS x

 1 75  60 2
s e2  283.3333 and SS x  6400 . So sŶ2  283 .3333  
8
6400

2  ,


X  60 ,




 283 .3333 0.12500  0.03516   283 .3333 0.16016   45.3776 . We will use
8
 2.306 . So  Y0  Yˆ0  t sYˆ  74.375  2.306 6.7363 
sYˆ  45.3776  6.7363 and t .025
 74.375  15.534 or 58.841 to 89.908.
6
252y0773 11/19/07 (Page layout view!)
Exhibit 1 A manager believes that
the number of sales that an employee
makes is related to the number of
years worked and their score on an
aptitude test. He runs the data below
on Minitab and gets the following
Empl Sales Years Score
1
100
10
70
2
90
3
100
3
80
8
90
4
70
5
40
5
60
5
80
6
50
7
50
7
40
1
40
8
30
1
10
MTB > regress c1 2 c2 c3
Regression Analysis: Sales versus Years, Score
The regression equation is
Sales = 21.3 + 3.10 Years + 0.470 Score
Predictor
Coef SE Coef
T
P
Constant
21.29
13.14 1.62 0.166
Years
3.103
1.984 1.56 0.179
Score
0.4698
0.2133 2.20 0.079
S = 15.1088
R-Sq = 72.8%
R-Sq(adj) = 62.0%
g) Do an analysis of variance using your SST, SSE and SSR for this equation or using 1,
R 2 and 1  R 2 . What have you already done that makes this table redundant? If you
don’t know what redundant means, ask! (3)
[43]
Solution: We actually have almost all this done. We have already found SS y  4200  SST ,
SSR  b1 S xy  0.6254000  2500 and SSE  SST  SSR  4200  2500  1700 . So our ANOVA
table will be as below.
Source
SS
DF
MS
F
Regression
1
2500
8.824
2500
F
1,6   5.99
F.05
Error
1700
6
283.3333
Total
4200
7
If we recall R 2  .59524 for this regression, we can rewrite the table as below.
Source
DF
‘MS’
F
F
R2
Regression
0.59524
1
0.59524
8.824
1,6   5.99
F.05
Error
0.40476
6
0.06746
Total
1.00000
7
Just for reassurance, here is the Minitab output.
MTB > regress c1 1 c3
Regression Analysis: Sales versus Score
The regression equation is
Sales = 27.5 + 0.625 Score
Predictor
Coef SE Coef
T
P
Constant
27.50
13.96 1.97 0.096
Score
0.6250
0.2104 2.97 0.025
S = 16.8325
R-Sq = 59.5%
R-Sq(adj) = 52.8%
Analysis of Variance
Source
DF
SS
Regression
1 2500.0
Residual Error
6 1700.0
Total
7 4200.0
MS
2500.0
283.3
F
8.82
P
0.025
This is redundant because we have already shown that the coefficient of ‘Score’ is significant.
Because there is only one independent variable, this shows the same thing.
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252y0773 11/19/07 (Page layout view!)
Exhibit 1 A manager believes that the number of sales that
an employee makes is related to the number of years worked
and their score on an aptitude test. He runs the data below on
Minitab and gets the following
Employee Sales Years Score
1
100
10
70
2
90
3
100
3
80
8
90
4
70
5
40
5
60
5
80
6
50
7
50
7
40
1
40
8
30
1
10
MTB > regress c1 2 c2 c3
Regression Analysis: Sales versus Years, Score
The regression equation is
Sales = 21.3 + 3.10 Years + 0.470 Score
Predictor
Coef SE Coef
T
P
Constant
21.29
13.14 1.62 0.166
Years
3.103
1.984 1.56 0.179
Score
0.4698
0.2133 2.20 0.079
S = 15.1088
R-Sq = 72.8%
R-Sq(adj) =
62.0%
Analysis of Variance
Source
DF
SS
P
Regression
2 3058.6
0.038
Residual Error
5 1141.4
Total
7 4200.0
Source
Years
Score
DF
1
1
MS
F
1529.3
6.70
228.3
Seq SS
1951.4
1107.3
h) Using the information on Regression Sums of squares or R 2 and 1  R 2 in the
ANOVA that you just did and from the multiple regression, do an F test to see if adding
‘years’ to the regression of ‘sales’ against ‘score’ is worthwhile. Do not waste our time by
repeating stuff that has already been done. (3)
[46]
Solution: In view of the original printout, we can rewrite out ANOVA tables above for the
multiple regression.
So our ANOVA table will be as below.
Source
SS
DF
MS
F
F
Regression
3058.6
2
1529.3
6.70
Error
1141.4
5
226.3
Total
4200
7
If we recall R 2  .7280 for this regression, we can rewrite the table as below.
Source
DF
‘MS’
F
F
R2
Regression
0.7280
2
0.3640
6.691
Error
0.2720
5
0.0544
Total
1.0000
7
For the regression against ‘Score’ alone, we had R 2  .59524 and SSR  2500 . So that if we
itemize the regressions above, we get the tables below.
Source
SS
DF
MS
F
F
Regression
3058.6
2
Score
2500.0
1
1,5   6.61
F.05
Years
558.6
1
558.6
2.468
Error
1141.4
5
226.3
Total
4200
7
If we use R 2 instead for this regression, we can rewrite the table as below.
Source
DF
‘MS’
F
F
R2
Regression
0.7280
2
Score
0.5952
1
1,5   6.61
F.05
Years
0.1328
1
0.1328
2.441
Error
0.2720
5
0.0544
Total
1.0000
7
In spite of the gigantic rounding error in the table using R 2 , the results are the same as in the ttest on the coefficient of ‘years’ in the second regression, the calculated F is below the table F so
that adding ‘years’ does not significantly improve the results.
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Exhibit 2 (Groebner) A product is being produced on 3 different lines using 3 different layouts for the
lines. A sample of 36 observations are taken on various days over a period of four weeks so that there are
12 observations for the daily output for each line evenly divided between the three possible layouts. Assume
  .05 .
MTB > Twoway c6 c2 c3;
SUBC>
Means c2 c3.
Two-way ANOVA: output 3 versus line, layout
Source
DF
line
2
layout
2
Interaction __
Error
__
Total
35
S = 19.86
R-Sq
line
1
2
3
Mean
132.333
127.250
127.750
layout
1
2
3
SS
MS
F
P
188.4
94.2
0.24 0.789
27832.1 13916.0 35.29 0.000
1859.1
_____
1.18 0.342
10648.0
_____
40527.6
= 73.73%
R-Sq(adj) = 65.94%
Individual 95% CIs For Mean Based on
Pooled StDev
------+---------+---------+---------+--(-------------*--------------)
(--------------*--------------)
(--------------*-------------)
------+---------+---------+---------+--120.0
128.0
136.0
144.0
Mean
116.917
167.583
102.833
Individual 95% Cis For Mean Based on
Pooled StDev
----+---------+---------+---------+----(----*---)
(----*----)
(----*----)
----+---------+---------+---------+----100
125
150
175
12. Fill in the missing degrees of freedom and the missing mean squares. (2)
[48]
Solution: We can find the degrees of freedom by multiplying the degrees of freedom for the factors that
interact. The error degrees of freedom are whatever is needed to make the column add up and the mean
squares are found by dividing the sums of squares by degrees of freedom. The corrected table reads as
below.
Two-way ANOVA: output 3 versus line, layout
Source
DF
line
2
layout
2
Interaction
4
Error
27
Total
35
S = 19.86
R-Sq
SS
MS
F
P
188.4
94.2
0.24 0.789
27832.1 13916.0 35.29 0.000
1859.1
484.8
1.18 0.342
10648.0
394.37
40527.6
= 73.73%
R-Sq(adj) = 65.94%
13. Is there significant interaction between ‘line’ and ‘layout’? Don’t answer unless you can tell me what the
evidence is. (2)
4, 27  2.73 if we like to work. It is larger than the computed F of 1.18. Or we
Solution: We can look up F.05
can simply note that since the p-value of 0.342 is well above any significance level that we are likely to use,
we cannot reject the null hypothesis that the interaction is insignificant.
14. Is the difference between lines significant? Why?(1)
2, 27  3.35 if we like to work. It is larger than the computed F of 0.24. Or we
Solution: We can look up F.05
can simply note that since the p-value of 0..789 is well above any significance level that we are likely to
use, we cannot reject the null hypothesis that the difference between the line means is insignificant.
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252y0773 11/19/07 (Page layout view!)
15. Do a confidence interval of your choice for the difference between layout 1 and layout 3. Tell what kind
of interval you are using , what its characteristics are and whether it shows a significant difference. (4)[55]
Solution: The outline gives us a choice. There are 3 rows, 3 columns and 4 observations per cell. R  3,
C  3, P  4, x1..  132 .333, x3..  127 .750 and RC P  1  333  27 . The error (within) mean square is
MSW  394 .37 . x1..  x3..  132 .333  127 .750  4.583 .
2MSW

PC
2394 .37 
 65 .7283  8.1073
12
i. A Single Confidence Interval
If we desire a single interval we use the formula for a Bonferroni Confidence Interval below with
2MSW
m  1 . For rows this would be 1   3  x1  x 3   t RC P 1
2
PC
27
8.1073   4.584  2.052 8.1073   4.58  16.64
 4.584  t .025
ii. Scheffé Confidence Interval
If we desire intervals that will simultaneously be valid for a given confidence level for all possible
intervals between means, for row means, we use
2MSW
2, 27  8.1073 
 4.584  2 F.05
1   2  x1  x 2   R  1FR 1, RC P 1
PC
 4.584  23.35 8.1073   4.58  20.99
iii. Bonferroni Confidence Interval
If we use this for row means 1   2  x1  x 2   t RC P 1
2m
2MSW
, but it is usually
PC
impractical.
iv. Tukey Confidence Interval
This is of similar meaning to the Scheffé. For row means, we use
MSW
1   2  x1  x 2   qR , RC P 1
PC

3, 27
8.1073  4.584  3.518.1073   4.58  28.46 . I’m suspicious. The Tukey
 4.584  q

interval should be smaller than the Scheffé.
Part of the Tukey table appears below.

df  24
0.05
0.01
df  30
0.05
0.01
2
3
4
5
m
6
7
8
9
10
*
2.92
3.96
*
*
3.53
4.55
*
3.90
4.91
*
*
4.17
5.17
*
4.37
5.37
*
*
4.54
5.54
*
4.68
5.69
*
*
4.81
5.81
*
4.92
5.92
*
2.89
3.89
*
*
3.49
4.45
*
3.85
4.80
*
*
4.10
5.05
*
4.30
5.24
*
*
4.46
5.40
*
4.60
5.54
*
*
4.63
5.65
*
4.73
5.76
3, 24
3,30
 3.53 and q.05
 3.49 , so
It looks to me as if q3,27 ought to be halfway between q.05
q3,27  3.51 . Note that, since all of these intervals include zero, none of these differences is
significant.
10
252y0773 11/19/07 (Page layout view!)
16. (Groebner) An industrial firm analyses the amount of breakage (in dollar cost) that occurs using 3
different shipping methods and four products. There is a strong likelihood that the data does not come from
the Normal distribution. The purpose of the test is to see if the four shipping methods differ in breakage and
the analysis is blocked by product.
Rail
Plane Truck
Product 1
7960
8053
8818
Product 2
8399
7764
9432
Product 3
9429
9196
9260
Product 4
6022
5821
5676
The most appropriate method for doing this test is:
a) *The Friedman Test
b) The Kruskal-Wallis Test
c) One-way ANOVA
d) Two-way ANOVA
e) The sign test
[57]
f) Another test (Name it!)
17. Assume that your decision is correct in 16. What is your null hypothesis or hypotheses? Be specific! Are
you talking about rows or columns or both? Are you comparing means, medians, proportions or variances?
Solution: The null hypothesis is that the medians of the columns are equal.
18. OK. Let’s see you do the test. (4)
[63]
Now compute the Friedman statistic
Rail
Plane Truck
Product 1
1
2
3
 12

2
2


SR

  3r c  1
F
i
Product 2
2
1
3
 rc c  1 i

Product 3
3
1
2
 12
Product 4
3
2
1
92  62  92   344

Sum
9
6
9
 434

rcc  1 434
1


SRi 

 24 and 9 +
Check:
  198   48  1.5 .
2
2
4

6 + 9 = 24.
If you check Table 8 (excerpt below) for c  3 and r  4 , you find that the p-value is .653, which is above
any significance level that we are likely to use, so there we cannot refute the null hypothesis of equal
medians.
c  3,
r 4
 



p  value
 F2
0.000
0.500
1.500
2.000
3.500
4.500
6.000
6.500
8.000
1.000
.931
.653
.431
.273
.125
.069
.042
.005
A Minitab verification follows.
MTB > Friedman c1 c2 c3.
Friedman Test: breakage versus mode blocked by product
S = 1.50
DF = 2
P = 0.472
Sum
of
mode
N Est Median Ranks
plane 4
8142.0
6.0
rail
4
8310.0
9.0
truck 4
8605.5
9.0
Grand median = 8352.5
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ECO252 QBA2
THIRD EXAM
Nov 26-29, 2007
TAKE HOME SECTION
Name: _________________________
Student Number: _________________________
Class days and time : _________________________
Please Note: Computer problems 2 and 3 should be turned in with the exam (2). In problem 2, the 2 way
ANOVA table should be checked. The three F tests should be done with a 1% significance level and you
should note whether there was (i) a significant difference between drivers, (ii) a significant difference
between cars and (iii) significant interaction. In problem 3, you should show on your third graph where the
regression line is. You should explain whether the coefficients are significant at the 1% level. Check what
your text says about normal probability plots and analyze the plot you did. Explain the results of the t and F
tests using a 5% significance level. (3)
III Do the following. (22+ points) Note: Look at 252thngs (252thngs) on the syllabus supplement part of
the website before you start (and before you take exams). Show your work! State H 0 and H 1 where
appropriate. You have not done a hypothesis test unless you have stated your hypotheses, run the
numbers and stated your conclusion. (Use a 95% confidence level unless another level is specified.)
Answers without reasons or accompanying calculations usually are not acceptable. Neatness and
clarity of explanation are expected. This must be turned in when you take the in-class exam. Note
that from now on neatness means paper neatly trimmed on the left side if it has been torn, multiple
pages stapled and paper written on only one side. Show your work!
1) The Lees, in their book on statistics for Finance majors, ask about the relationship of gasoline prices  y 
in cents per gallon to crude oil prices x1  in dollars per barrel and present the data for the years 1975 1988. I have obtained most of the data for the years 1980 – 2007. It is presented below.
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
Year
1980
1981
1982
1983
1984
1985
1986
1987
1988
1989
1990
1991
1992
1993
1994
1995
1996
1997
1998
1999
2000
2001
2002
2003
2004
2005
2006
2007
GasPrice
1.25
1.38
1.30
1.24
1.21
1.20
0.93
0.95
0.96
1.02
1.16
1.14
1.13
1.11
1.11
1.15
1.23
1.23
1.06
1.17
1.51
1.46
1.36
1.59
1.88
2.30
*
3.10
CrudePrice
26.07
35.24
31.87
26.99
28.63
26.25
14.55
17.90
14.67
17.97
22.22
19.06
18.43
16.41
15.59
17.23
20.71
19.04
12.52
17.51
28.26
22.95
24.10
28.53
36.98
50.23
*
90.00
Yr-1979
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
This data set also contains the year with 1979 subtracted from it x 2  . You may need to use this later.
Ignore it in Problem 1. Note that the numbers for 2006 have not yet been published in my source, Statistical
12
252y0773 11/19/07 (Page layout view!)
Abstract of the United States, and that the numbers for 2007 are my estimates for third quarter prices. These
are unleaded prices, which the Lees did not use. You are supposed to use only the numbers for 1990
through 2006 and one other observation for your data. You will thus have n  17 observations. The other
column is the value for the year 1980  a  , where a is the second to last digit of your student number. If
you are unsure of the data that you are using or if you want help with the sums that you need to do the
regression go to 3takehome072a.
Show your work – it is legitimate to check your results by running the problem on the computer. (In fact, I
will give you 2 points extra credit for checking it and annotating the output for significance tests etc.) But I
expect to see hand computations for every part of this problem.
a. Compute the regression equation Y  b0  b1 x to predict the price of gasoline on the basis of
crude oil prices. (3)
b. Compute R 2 . (2)
c. Compute s e . (2)
d. Compute s b1 and do a significance test on b1 (2)
e. Compute a confidence interval for b0 . (2)
f. You have a crude price for 2007. Using this, predict the gasoline price for 2007 and create a
prediction interval for the price of gasoline for that year. Explain why a confidence interval for the
price is inappropriate and check to see if my estimated price is in the interval. (3)
g. Do an ANOVA for this regression. (3)
f) Make a graph of the data. Show the trend line and the data points clearly. If you are not willing
to do this neatly and accurately, don’t bother. (2)
[19]
2) Now we can use the date to see if there is a trend line in addition to the effect of crude oil.
a. Do a multiple regression of the price of gasoline against crude prices and the data variable,
which has been massaged to make 1980 year 1. This involves a simultaneous equation solution.
Attempting to recycle b1 from the previous page won’t work. (7)
c. Compute the regression sum of squares and use it in an ANOVA F test to test the usefulness of
this regression. (4)
b. Compute R 2 and R 2 adjusted for degrees of freedom for both this and the previous problem.
Compare the values of R 2 adjusted between this and the previous problem. Use an F test to
compare R 2 here with the R 2 from the previous problem. The F test here is one to see if adding a
new independent variable improves the regression. This can also be done by modifying the
ANOVAs in b.(4)
d. Use your regression to predict the price of gasoline in 2007. Is this closer to the estimated
gasoline price? Do a confidence interval and a prediction interval. (3)
[37]
e. Again there is extra credit for checking your results on the computer. Use the pull-down menu or
try
Regress GasPrice on 2 CrudePrice Yr-1979 (2)
3) According to Russell Langley, three sopranos were discussing their recent performances. Fifi noted that
she got 36 curtain calls at La Scala last week, but Adalina put her down with the fact that she got 39. Could
one of the singers really say that she had more curtain calls than another or could the differences just be due
to chance?
Personalize the data below by adding the last digit of your student number to each number in the
first row. Use a 10% significance level throughout this question.
Row
1
2
3
4
Fifi
36
22
19
16
Adelina
39
14
20
18
Maria
21
32
28
22
a) State your hypothesis and use a method to compare means assuming that each column represents a
random sample of curtain calls at La Scala. (4)
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252y0773 11/19/07 (Page layout view!)
b) Still assuming that these are random samples, use a method that compares medians instead. (3)
c) Actually, these were not random samples. Though row 1 represents curtain calls at La Scala (Milan), row
2 was in Venice, row 3 in Naples and row 4 in Rome. Will this affect our results? Does this show anything
about audiences on the four cities? Use an appropriate method to compare medians. (5)
d) Do two different types of confidence intervals between Milan and the least enthusiastic opera house.
Explain the difference between the intervals. (2)
e) Assume that we want to compare medians instead. How does the fact that these data were collected at
three opera houses affect the results? (3)
f) Do you prefer the methods that compare medians or means? Don’t answer this unless you can
demonstrate an informed opinion. (1)
g) (Extra credit) Do a Levine test on these data and explain what it tests and shows.(3)
h) (Extra credit)Check your work on the computer. This is pretty easy to do. Use the same format as in
Computer Problem 2, but instead of car and driver numbers use the singers’ and cities’ names. You can use
the stat and ANOVA pull-down menus for One-way ANOVA, two-way ANOVA and comparison of
variances of the columns. You can use the stat and the non-parametrics pull-down menu for Friedman and
Kruskal-Wallis. You also probably ought to test columns for Normality. Use the Statistics pull-down menu
and basic statistics to find the normality tests. The Kolmogorov-Smirnov option is actually Lilliefors. The
ANOVA menu can check for equality of variances. In light of these tests was ANOVA appropriate? You
can get descriptions of unfamiliar tests by using the Help menu and the alphabetic command list or the Stat
guide. (Up to 7) [58]
You should note conclusions on the printout – tell what was tested and what your conclusions are using a
10% significance level.
14