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ECO252 QBA2
THIRD HOUR EXAM
April 16, 2007
Name KEY
Student number_____________
I. (8 points) Do all the following (2points each unless noted otherwise). Make Diagrams! Show your
work!
x ~ N 13, 7.2
Material in italics below is a description of the diagrams you were asked to make or a general explanation
and will not be part of your written solution. The x and z diagrams should look similar. If you know what
you are doing, you only need one diagram for each problem. General comment - I can't give you much
credit for an answer with a negative probability or a probability above one, because there is no such
 x 
thing!!! In all these problems we must find values of z  
 corresponding to our values of x before
  
we do anything else. A diagram for z will help us because, if areas on both sides of zero are shaded, we
will add probabilities, while, if an area on one side of zero is shaded and it does not begin at zero, we will
subtract probabilities. Note: All the graphs shown here are missing a vertical line. They are also to
scale. A hand drawn graph should exaggerate the distances of the points from the mean. Note also
that, because of the rounding error necessary to use a conventional normal table, the results for x will
disagree with results for z, especially for small values of z.
0  13 

 Pz  1.81  Pz  0  P1.81  z  0  .5  .4649  .0351
1. Px  0  P  z 
7.2 

This is how we usually find a p-value for a left-sided test when the distribution of the test ratio is Normal.
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 below -1.81. Because this is entirely on the left side of zero, we must subtract the area
between -1.81 and zero from the area below zero. If you wish, make a completely separate diagram for x .
Draw a Normal curve with a mean at 13. Indicate the mean by a vertical line! Shade the area below 0.
This area is completely to the left of the mean (13), so we subtract the smaller area between zero and the
mean from the larger area below the mean.
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20  13 
 6  13
z
 P 0.97  z  0.97   2P0  z  0.97   2.3340   .6680
2. P6  x  20   P 
7.2 
 7.2
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 -0.97 and 0.97. Because this is entirely on both sides of zero, we must add the
area between -0.97 and zero to the area from zero to 0.97. If you wish, make a completely separate diagram
for x . Draw a Normal curve with a mean at 13. Indicate the mean by a vertical line! Shade the area from
6 to 20. This area is on both sides of the mean (13), so we add the area between 6 and the mean to the area
between the mean and 20.
12  13 
  1  13
z
 P 1.94  z  0.14   P1.94  z  0  P0.14  z  0
3. P1  x  12   P 
7.2 
 7.2
 .4738  .0557  .4181
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 -1.94 and -0.14. Because this is entirely on the left side of zero, we must subtract
the area between -0.14 and zero from the larger area between -0.14 and zero. If you wish, make a
completely separate diagram for x . Draw a Normal curve with a mean at 13. Indicate the mean by a
vertical line! Shade the area between -1 and 12. This area is completely to the left of the mean (13), so we
subtact the smaller area between 12 and the mean from the larger area between -1 and the mean.
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4. x.145 (Find z .145 first) Solution: x.145 is, by definition, the value of x with a probability of 14.5% above
it. Make a diagram. The diagram for z will show an area with a probability of 85.5% below z .145 . It is
split by a vertical line at zero into two areas. The lower one has a probability of 50% and the upper one a
probability of 85.5% - 50% = 35.5%. The upper tail of the distribution above z .145 has a probability of
14.5%, so that the entire area above 0 adds to 35.5% + 14.5% = 50%. From the diagram, we want one
point z .145 so that Pz  z .145   .855 or P0  z  z.145   .3550 . If we try to find this point on the
Normal table, the closest we can come is P0  z  1.06   .3555 . So we will use z.145  1.06 , though
1.05 would be acceptable.
Since x ~ N 13, 7.2 , the diagram for x would show 85.5% probability split in two regions on
either side of 13 with probabilities of 50% below 13 and 35.5% above 13 and below x.145 , and with 14.5%
above x.145 . z.145  1.06 , so the value of x can then be written x.145    z .145
 13  1.067.2  13  7.63  20.63.
20 .63  13 

To check this: Px  20 .63   P  z 
  Pz  1.06   Pz  0  P0  z  1.06   .5 + .3554
7.2


20 .63  13 

 .8554  85.5% . Or Px  20 .63   P  z 
  Pz  1.06   Pz  0  P0  z  1.06 
7.2


 .5  .3554  .1446  14.5%
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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.
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. (Dummeldinger) As part of a study to investigate the effect of helmet design on football injuries,
head width measurements were taken for 30 subjects randomly selected from each of 3 groups (High
school football players, college football players and college students who do not play football – so that
there are a total of 90 observations) with the object of comparing the typical head widths of the three
groups. Before the data was compared, researchers ran two tests on it. They first ran a Lilliefors test
and found that the null hypothesis was not rejected. They then ran a Bartlett test and found that the null
hypothesis was not rejected. On the basis of these results they should use the following method.
a) The Kruskal-Wallis test.
b) *One-way ANOVA
c) The Friedman test
d) Two-Way ANOVA
e) Chi-square test
[7]
Explanation: The null hypothesis of the Lilliefors test is that the distribution is Normal. The null
hypothesis of the Bartlett test is equal variances. These two results are the requirements of a one-way
ANOVA to compare the means of the three groups.
3. The coefficient of determination
a) Is the square of the sample correlation
b) Cannot be negative
c) Gives the fraction of the variation in the dependent variable explained by the
independent variable.
d) *All of the above.
e) None of the above.
S xy 2
SSR
Explanation: The coefficient of determination R 
is the ratio of two sums or

SST SS x SS y
2
squares, which must be positive quantities. The correlation is the positive or negative square root of
S xy 2
SS x SS y
.
4. The Kruskal-Wallis test is an extension of which of the following for two independent samples?
a) Pooled-variance t test
b) Paired-sample t test
c) * (Mann – Whitney -) Wilcoxon rank sum test
d) Wilcoxon signed ranks test
e) A test for comparison of means for samples with unequal variances
f) None of the above.
Explanation: The Wilcoxon rank sum test ranks all the number in the samples, just like the KruskalWallis test. Both are tests for comparison of medians.
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5. One-way ANOVA is an extension of which of the following for two independent samples?
a) *Pooled-variance t test
b) Paired-sample t test
c) (Mann – Whitney -) Wilcoxon rank sum test
d) Wilcoxon signed ranks test
e) A test for comparison of means for samples with unequal variances
f) None of the above.
[13]
Explanation: The variances are pooled in the pooled-variance t test because equal variances is an
assumption of ANOVA. Both are tests for comparison of means.
Exhibit 1: As part of an evaluation program, a sporting goods retailer wanted to compare the downhill
coasting speeds of 4 brands of bicycles. She took 3 of each brand and determined their maximum
downhill speeds. The results are presented in miles per hour in the table below. She assumes that the
Normal distribution is not applicable and uses a rank test. She treats the columns as independent
random samples.
Barth Tornado Reiser Shaw
43
37
41
43
46
38
45
45
43
39
42
46
6. What are the null and alternative hypotheses?
Solution: H 0 : All medians equal H 1 : Not all medians equal
7. Construct a table of ranks for the data in Exhibit 1.
Barth
Tornado
Reiser
Shaw
x1 r1
x 2 r2
x3 r3
x 4 r4
43 7.0 37 1.0
41 4.0 43 7.0
46 11.5 38 2.0
45 9.5 45 9.5
43 7.0 39 3.0
42 5.0 46 11.5
25.5
6.0
18.5
28.0
8. Complete the test using a 5% significance level (3)
[20]
2
 12
 SRi  

  3n  1
Now, compute the Kruskal-Wallis statistic H  
 nn  1 i  ni 
 12  25 .52 6.02 18 .52 28 .02 

  313   1 604 .1667   39  7.4744 . This is too large for the








12
13
3
3
3
3
13



3
K-W table, so we compare this to  2 .05  7.8145 . Since the computed statistic is below the table statistic,

do not reject H 0 .
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Exhibit 2: (Webster) A placement director is trying to find out if a student’s GPA can influence how
many job offers the student receives. Data is assembled and run on Minitab, with the results below.
The regression equation is
Jobs = - 0.248 + 1.27 GPA
Predictor
Coef SE Coef
Constant
-0.2481
0.7591
GPA
1.2716
0.2859
S = 1.03225
R-Sq = 71.2%
Analysis of Variance
Source
DF
SS
Regression
1 21.076
Residual Error
8
8.524
Total
9 29.600
T
P
-0.33 0.752
4.45 0.002
R-Sq(adj) = 67.6%
MS
21.076
1.066
F
19.78
P
0.002
9. In Exhibit 2, how many job offers would you predict for someone with a GPA of 3.
Solution: Jobs = - 0.248 + 1.27(3)= 3.562
10. In Exhibit 2, what percent of the variation in job offers is explained by variation of GPA? (1)
Solution: 71.2%
11. For the regression in Exhibit 2 to give us ‘good’ estimates of the coefficients.
a) The variances of ‘Jobs’ and ‘GPA’ must be equal.
b)* The amount of variation of ‘Jobs’ around the regression line must be independent of ‘GPA.’
c) ‘Jobs’ must be independent of ‘GPA.’
d) The p-value for ‘Constant’ must be higher than the p-value for ‘GPA.’
e) All of the above must be true.
f) None of the above need to be true.
[25]
12. Assuming that the Total of 29.600 in the ANOVA in Exhibit 2 is correct, but that you cannot read any
other part of the printout, what are the largest and smallest values that the Regression Sum of Squares could
take?
[27]
Solution: 0 to 29.600.
13. A corporation wishes to test the effectiveness of 3 different designs of pollution-control devices. It takes
its 6 factories and puts each type of device for one week during each month of the year, for a total of 216
measurements. Pollution in the smoke is measured and an ANOVA is begun. Let factor A be months. Factor
B will be factories and Factor C will be the devices. Fill in the following table. Note that we have too little
data to compute interaction ABC. The Total sum of squares is 106. SSA is 44, SSB is 11, SSC is 21, SSAB
is 9, SSAC is 7 and SSBC is 3. Use a significance level of 1%. Do we have enough information to decide
which of the devices is best? (Do not answer this without showing your work!) (6) [33]
Row
1
2
3
4
5
6
7
8
Factor
A
B
C
AB
AC
BC
Within
Total
SS
DF
MS
F
F01
Significant?
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Solution:
Row
1
2
3
4
5
6
7
8
Factor
A
B
C
AB
AC
BC
Within
Total
(Page layout view!)
SS
44
11
21
9
7
3
11
106
DF
11
5
2
55
22
10
110
215
MS
4.0000
2.2000
10.5000
0.1636
0.3182
0.3000
0.1000
F
40.000
22.000
105.000
1.636
3.182
3.000
F05
1.88
2.30
3.08
1.47
1.64
1.96
C7
s
s
s
s
s
s
F01
2.41
3.19
4.80
1.76
2.00
2.47
pval
s
ns
ns
ns
s
s
The SS Column must add so the missing number is 11. As is implied by the diagram and your experience
with 2-way ANOVA, there is no ABC interaction. 12 months means 11 A df, 6 plants means 5 B df, 3
devices means 2 C DF. AB df are 11(5) = 55. AC df are 11(2) = 22, BC df is 5(2) = 10. Total df is
12(6)3 – 1 = 216 – 1 = 215. Within df can be gotten by making sure that the DF column adds or by noting
11(5)2 = 110. MS is SS divided by DF. F is MS divided by MSW = 0.1000. Though all are significant at
the 5% level, at the 1% level we cannot say that the Factor C means are different since the F is not
significant at 1%.
Exhibit 3: A plant manager is concerned for the rising blood pressure of employees and believes that it is
related to the level of satisfaction that they get out of their jobs. A random sample of 10 employees rates
their satisfaction on a 1to 60 scale and their blood pressure is tested.
Row
1
2
3
4
5
6
7
8
9
10
Satisfaction BloodPr
34
124
23
128
19
157
43
133
56
116
47
125
32
147
16
167
55
110
25
156
We also know that Sum of Satisfaction = 350, Sum of BloodPr = 1363, Sum of squares (uncorrected) of
Satisfaction = 14170 and Sum of squares (uncorrected) of BloodPr = 189113. It is also true that the sum of
the first eight numbers of the product of Satisfaction and Blood is 35609.
14.
a) Finish the computation of
 xy (1).
The sum of the first eight numbers of the product of Satisfaction and Blood is 35609.
So 35609 + 110(55) + 156(25) = 35609 + 6050 + 3900 = 45559
b) Try to do a regression explaining blood pressure as a result of job satisfaction (4). Please do not
recompute things that I have already done for you.
xy  45559 and
y  1363,
x  350,
x 2  14170,
We now have n  10,





350
1363
y 2  189113. We can begin by computing x 
 35 and y 
 136.3. Now it’s
10
10
time for spare parts calculations SSx 
SST  SS y 
y
2
x
2
 nx 2  14170  10352  1920,
 ny 2  18911310136.32  3336.10,
S xy  45559  1035136.3  -2146.
 2146
 1.1177 means b0  Y  b1 X  136 .3  1.1177 35   175 .4195 .
SS x
1920
So our regression equation is Yˆ  175 .4195  1.1177 X or Y  175 .4195  1.1177 X  e .
b1 
S xy

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c) Find the slope and test its significance(4).
SSR  b1 S xy  1.11772146  2398.5842 , df  n  2  8
s e2 
SSE SST  SSR SS y  b1 S xy 3336 .10  2398 .5842



 117 .1895 ,
n2
n2
n2
8
s e  117 .1895  10 .8254
 1  117 .1895
s b21  s e2 
 0.06104 . So s b  0.06104  0.24706

1
1920
 SS x 
H :   0
b  0 1.117
8
To test  0 1
use t  1
 1.860 .

 4.5211 . We compare this with t .05
sb1
0.24706
H 1 : 1  0
Since our computed t is larger that the table t, we reject the null hypothesis.
d) Do the prediction or confidence interval as appropriate for the blood pressure of the highly
satisfied employee (satisfaction level is 57) that you hired last month. (3)
[45]
2
Recall that x  35, SS x  1920.00, s e  117.1895 and
Yˆ  175.4195 1.1177X  175.4195 1.117757  111.71 .


1 X X 2

The prediction Interval is Y0  Yˆ0  t sY , where sY2  s e2   0
 1
n

SS x


2
 1 57  35 

484


 117 .1895  
 1  117 .1895  0.1 
 1  117 .1895 1.35208   158.500. This
 10

1920
1920




8
 1.860 , and we put the whole mess together.
implies that sY  158 .500  12 .5876 Now t .05
Y0  Yˆ0  t sY  111.71 1.860158.5  11.71 294.21 . A prediction interval is appropriate for a
one-shot situatuin like this one. A confidence interval is an interval for the average response.
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ECO252 QBA2
THIRD EXAM
April 16, 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 5% significance level and you
should note whether there was (i) a significant difference between drivers, (ii) a significant difference
between cars and (iii) significant interaction. In problem 3, you should show on your third graph where the
regression line is. Check what your text says about normal probability plots and analyze the plot you did.
Explain the results of the t and F tests using a 5% significance level. (2)
III Do the following. (20+ 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.
To personalize these data, let the last digit of your student number be a . 1) (Ken Black) A national travel
organization wanted to determine whether there is a significant difference between the cost of premium gas
in various regions of the country. The following data was gathered in July 2006. Because the brand of gas
may confuse the results, data is blocked by brand.
Data Display
Row
1
2
3
4
5
6
Brand
A
B
C
D
E
F
City 1
3.47
3.43
3.44
3.46
3.46
3.44
City 2
3.40
3.41
3.41
3.45
3.40
3.43
City 3
3.38
3.42
3.43
3.40
3.39
3.42
City 4
3.32
3.35
3.36
3.30
3.39
3.39
City 5
3.50
3.44
3.45
3.45
3.48
3.49
Mark your problem ‘VERSION a .’ If a is zero, use the original numbers. If a is 1 through 4, add a
to the city 4 column. If a is 5 through 9, add 0.04  a
number is 090909, so he adds 0.04  9
100
100
100
to the city 5 column. Example: Seymour Butz’
 .05 to column 5 (subtracts .05) and gets the table below.
Version 9
Row
1
2
3
4
5
6
Brand
A
B
C
D
E
F
City 1
3.47
3.43
3.44
3.46
3.46
3.44
City 2
3.40
3.41
3.41
3.45
3.40
3.43
City 3
3.38
3.42
3.43
3.40
3.39
3.42
City 4
3.32
3.35
3.36
3.30
3.39
3.39
City 5
3.45
3.39
3.40
3.40
3.43
3.44
a) Do a 2-way ANOVA on these data and explain what hypotheses you test and what the conclusions are.
Show your work. (6)
b) Using your results from a) present two different confidence intervals for the difference between numbers
of defects for the best and worst city and for the defects from the best and second best brands. Explain
which of the intervals show a significant difference and why. (3)
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c) What other method could we use on these data to see if city makes a difference while allowing for crossclassification? Under what circumstances would we use it? Try it and tell what it tests and what it shows. (3)
d) (Extra credit) Check you results on the computer. The Minitab setup for both 2-way ANOVA and the
corresponding non-parametric method is the same as the extra credit in the third graded assignment or the
third computer problem. You may want to use and compare the Statistics pull-down menu and ANOVA or
Nonparametrics to start with. Explain your results. (4) [12]
2) A local official wants to know how land value affects home price. The official collects the following
data. 'Home price' is the dependent variable and 'Land Value' is the independent variable. Both are in
thousands of dollars. If you don’t know what dependent and independent variables are, stop work
until you find out.
Row
1
2
3
4
5
6
7
8
9
10
HomePr
67
63
60
54
58
36
76
87
89
92
LandVal
7.0
6.9
5.5
3.7
5.9
3.8
8.9
9.6
9.9
10.0
To personalize the data let g be the second to last digit in your student number. Mark your problem
‘VERSION g . ’Subtract g from 92, which is the last home price.
a) Compute the regression equation Y  b  b x to predict the home price on the basis of land value.
0
1
(3).You may check your results on the computer, but let me see real or simulated hand calculations.
b) Compute R 2 . (2)
c) Compute s e . (2)
d) Compute s b0 and do a significance test on b0 (2)
e) Use your equation to predict the price of the average home that is on land worth $9000. Make this into a
95% interval. (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)
[26]
3) A sales manager is trying to find out what method of payment works best. Salespersons are randomly
assigned to payment by commission, salary or bonus. Data for units sold in a week is below.
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
Commission
25
35
20
30
25
30
26
17
31
30
Salary
25
25
22
20
25
23
22
25
20
26
24
Bonus
28
41
21
31
26
26
46
32
27
35
23
22
26
To personalize the data let g be the second to last digit in your student number. Subtract 2 g from 46 in
the last column. Mark your problem ‘VERSION g .’Example: Payme Well’s student number is 000090, so
she subtracts 18 and gets the following numbers.
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Version 9
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
Commission
25
35
20
30
25
30
26
17
31
30
Salary
25
25
22
20
25
23
22
25
20
26
24
Bonus
28
41
21
31
26
26
28
32
27
35
23
22
26
a) State your null hypothesis and test it by doing a 1-way ANOVA on these data and explain whether the
test tells us that payment method matters or not. (4)
b) Using your results from a) present individual and Tukey intervals for the difference between earnings for
all three possible contrasts. Explain (i) under what circumstances you would use each interval and (ii)
whether the intervals show a significant difference. (2)
c) What other method could we use on these data to see if payment method makes a difference? Under what
circumstances would we use it? Try it and tell what it tests and what it shows. (3) [37]
d) (Extra Credit) Do a Levene test on these data and explain what it tests and shows. (4)
f) (Extra credit) Check you results on the computer. The Minitab setup for 1-way ANOVA can either be
unstacked (in columns) or stacked (one column for data and one for column name). There is a Tukey
option. The corresponding non-parametric methods require the stacked presentation. You may want to use
and compare the Statistics pull-down menu and ANOVA or Nonparametrics to start with. You may also
want to try the Mood median test and the test for equal variances which are on the menu categories above.
Even better, do a normality test on your columns. Use the Statistics pull-down menu 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 the 1-way 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) [12]
A 5% significance level is assumed throughout this section.
1) (Ken Black) A national travel organization wanted to determine whether there is a significant difference
between the cost of premium gas in various regions of the country. The following data was gathered in July
2006. Because the brand of gas may confuse the results, data is blocked by brand.
a) Do a 2-way ANOVA on these data and explain what hypotheses you test and what the conclusions are.
Show your work. (6) An example of 2-way ANOVA with one measurement per cell is in 252anovaex3.
Note that all three solutions below are essentially identical. To save space the row and column giving
17 .07
 3.414 . The sum of
column and row counts is omitted. In row 1 the mean price for brand A was
5
squares for the first row is 3.47 2  3.40 2  3.38 2  3.32 2  3.50 2  58.2977 . The mean price squared was
3.414 2  11 .6554 . 3.41867, the number at the bottom of the mean column and on the right of the mean row
is the overall mean, which can be found by averaging over the row or column or over the 30 original
numbers. In every case we seem to reject the similarity of city means but not the similarity of brand means.
11
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Solution: Version 0
Row Brand
1 A
2 B
3 C
4 D
5 E
6 F
Sum
SS
Mean
Meansq
City 1
3.47
3.43
3.44
3.46
3.46
3.44
20.70
71.4162
3.45000
11.9025
City 2
3.40
3.41
3.41
3.45
3.40
3.43
20.50
70.0436
3.41667
11.6736
City 3
3.38
3.42
3.43
3.40
3.39
3.42
20.44
69.6342
3.40667
11.6054
City 4
3.32
3.35
3.36
3.30
3.39
3.39
20.11
67.4087
3.35167
11.2337
City 5 Sum
SS
mean
3.50 17.07 58.2977 3.414
3.44 17.05 58.1455 3.410
3.45 17.09 58.4187 3.418
3.45 17.06 58.2266 3.412
3.48 17.12 58.6262 3.424
3.49 17.17 58.9671 3.434
20.81 102.56 350.682 3.41867
72.1791 350.682
3.46833 3.41867
12.0293 58.4445
meansq
11.6554
11.6281
11.6827
11.6417
11.7238
11.7924
70.1241
 x  nx  350.682  303.41867  350.682  350.6191 0.0629
SSR  C x  nx  570.1241  303.41867  350.6205  350.6191  0.0014
SSC  R x  nx  658.4445  303.41867  350.6667  350.6191  0.0479
SST 
2
2
2
2
i.
2
2
2
.j
2
2
Source
SS
Rows
0.0014
Columns
Within
Total
DF
MS
F
F.05
F 5,20  2.71
F 4, 20  2.87
5
0.00028
0.41 ns
0.0479
4
0.011975
17.61 s
0.0136
0.0629
20
29
0.00068
City 3
3.38
3.42
3.43
3.40
3.39
3.42
20.44
69.6342
3.40667
11.6054
City 4
3.36
3.39
3.40
3.34
3.43
3.43
20.35
69.0271
3.39167
11.5034
H0
Row means equal
Column means equal
s  MSW  .0261
Solution: Version 4
Row Brand City 1
1 A
3.47
2 B
3.43
3 C
3.44
4 D
3.46
5 E
3.46
6 F
3.44
Sum
20.70
SS
71.4162
Mean
3.45000
Mean sq 11.9025
City 2
3.40
3.41
3.41
3.45
3.40
3.43
20.50
70.0436
3.41667
11.6736
City 5
3.50
3.44
3.45
3.45
3.48
3.49
20.81
72.1791
3.46833
12.0293
sum
SS
mean
17.11 58.5649 3.422
17.09 58.4151 3.418
17.13 58.6891 3.426
17.10 58.4922 3.420
17.16 58.8990 3.432
17.21 59.2399 3.442
102.8 352.300 3.42667
352.300
3.42667
58.7142
mean sq
11.7101
11.6827
11.7375
11.6964
11.7786
11.8474
70.4527
 x  nx  352.300  303.42667  352.300  352.2620  0.0380
SSR  C x  nx  570.4527  303.42667  352.2627  352.2620  0.0007
SSC  R x  nx  658.7142  303.42667  352.2852  352.2620  0.0232
SST 
2
2
2
2
i.
2
2
2
.j
2
2
Source
SS
DF
MS
Rows
0.0007
5
0.00014
0.197 ns
Columns
0.0232
4
0.0058
8.169 s
Within
Total
0.0141
0.0380
20
29
0.00071
F
F.05
F 5,20  2.71
F 4, 20  2.87
H0
Row means equal
Column means equal
s  MSW  .0266
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Solution: Version 9
Row Brand City 1 City 2 City 3 City 4
1 A
3.47
3.40
3.38
3.32
2 B
3.43
3.41
3.42
3.35
3 C
3.44
3.41
3.43
3.36
4 D
3.46
3.45
3.40
3.30
5 E
3.46
3.40
3.39
3.39
6 F
3.44
3.43
3.42
3.39
Sum
20.70
20.50
20.44
20.11
SS
71.4162 70.0436 69.6342 67.4087
Mean
3.45000 3.41667 3.40667 3.35167
MeanSq 11.9025 11.6736 11.6054 11.2337
City 5
sum
SS
mean
3.45 17.02 57.9502 3.404
3.39 17.00 57.8040 3.400
3.40 17.04 58.0762 3.408
3.40 17.01 57.8841 3.402
3.43 17.07 58.2807 3.414
3.44 17.12 58.6206 3.424
20.51 102.26 348.616 3.40867
70.1131 348.616
3.41833 3.40867
11.6850 58.1002
mean sq
11.5872
11.5600
11.6145
11.5736
11.6554
11.7238
69.7145
 x  nx  348.616  303.40867  348.616  348.5709  0.04506
SSR  C x  nx  569.7145  303.40867  348.5725  348.5709  0.00156
SSC  R x  nx  658.1002  303.40867  348.6012  348.5709  0.03030
SST 
2
2
2
2
i.
2
2
2
.j
2
2
Source
SS
DF
MS
Rows
0.00156
5
0.00031
0.473 ns
Columns
0.03030
4
0.00758
11.48 s
Within
Total
0.01320
0.04506
20
29
0.00066
F.05
F
F 5,20  2.71
F 4, 20  2.87
H0
Row means equal
Column means equal
s  MSW  .0257 Note that s is essentially the same for all three versions so there is no point in sticking
to separate solutions for b).
b) Using your results from a) present two different confidence intervals for the difference between numbers
of defects for the best and worst city and for the defects from the best and second best brands. Explain
which of the intervals show a significant difference and why. (3)
Solution: Since I blew the statement of this section, but no one complained, foll credit will be given for any
sets of meaningful comparison of rows and columns.
The intervals presented in the outline were
Scheffé Confidence Interval
If we desire intervals that will simultaneously be valid for a given confidence level for all possible intervals
between means, use the following formulas.
For row means, use 1   2  x1  x 2  
R  1FR 1, RC P 1 2MSW
PC
.
For column means, use
C  1FC 1, RCP 1 2MSW
. Note that if P  1 we replace RC P  1 with
PR
R  1C  1 . We have 6 rows and 5 columns with 1 measurement per cell. So R 1  5 , C  1  4 and
 1   2  x1  x2  
RC P  1 is replaced by 54  20 . We already know that s  MSW  .026
The row mean contrast is 1   2  x1  x 2  
 1   2  x1  x2  
4F4,20 2MSW
6
5F5,20 2MSW
5
. The column mean contrast is
We already know that the 5% values are F 4, 20  2.87 and
F 5,20  2.71 .
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252y0731 4/25/07
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Bonferroni Confidence Interval with m  1 is an individual interval, not simultaneously valid at 95%.
Note that if P  1 we replace RC P  1 with R  1C  1 . We have 6 rows and 5 coloumns with 1
measurement per cell. So R  1  5 , C  1  4 and RC P  1 is replaced by 54  20
The row mean contrast is 1   2  x1  x 2   t 20
2MSW
and the column mean contrast is
5
2
 1   2  x1  x2   t 20
2MSW
20
. t .025
 2.086 .
6
2
Tukey Confidence Interval These are sort of a loose version of the Scheffe intervals.
Note that if P  1 we replace RC P  1 with R  1C  1 . We have 6 rows and 5 coloumns with 1
measurement per cell. So R  1  5 , C  1  4 and RC P  1 is replaced by 54  20
The row mean contrast is 1   2  x1  x 2   q6, 20
MSW
and the column mean contrast is
5
 1   2  x1  x2   q5, 20
MSW
. The Tukey table is shown below and it seem that the values we
6
5, 20
4, 20
 4.23 and q.05
 3.96
need are q.05

df  20
0.05
0.01
*
2
3
*
*
2.95
4.02
*
*
*
3.58
4.64
*
4
*
*
3.96
5.02
*
*
m
6
5
*
4.23
5.29
*
7
*
*
4.45
5.51
*
*
*
4.62
5.69
*
8
9
*
*
4.77
5.84
*
*
*
4.90
5.97
*
10
*
5.01
6.09
c) What other method could we use on these data to see if city makes a difference while allowing for crossclassification? Under what circumstances would we use it? Try it and tell what it tests and what it shows. (3)
Solution: The Friedman method is similar in intent to 2-way ANOVA with one measurement per cell. It is
used to compare medians when the underlying distribution cannot be considered Normal. The data is ranked
within rows.
Solution: Version 0
If the cities are x1 through x5 and the ranks are r1 through r5 we can make the table below.
Brand
x1 r1
x2
r2
x3 r3
x4
r4
x5
r5
A
B
C
D
E
F
Sum
3.47
3.43
3.44
3.46
3.46
3.44
4
4
4
5
4
4
25
3.40 3.0
3.41 2.0
3.41 2.0
3.45 3.5
3.40 2.0
3.43 2.5
15.0
3.38 2
3.42 3
3.43 3
3.40 2
3.39 1
3.42 1
12
The sums of ranks sum to 90, which checks our as
 12
is  F2  
 rc c  1

 SR

 12
 SR   3r c  1   656 25
2
i
i
i
2
3.36
3.39
3.40
3.34
3.43
3.43

1.0
1.0
1.0
1.0
3.0
2.5
9.5
3.50
3.44
3.45
3.45
3.48
3.49
5.0
5.0
5.0
3.5
5.0
5.0
28.5
rcc  1 656

 90 . The Friedman statistic
2
2


 15 2  12 2  9.5 2  28 .5 2   366

2
1896 .5  108  126 .43333  108  18 .4333 . This has the chi-squared distribution with 4 degrees of
30
freedom. The upper right corner of the chi-squared table appears below. Since the computed value of  2
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252y0731 4/25/07
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exceeds any value on the table, we can say that the p-value is below .005 and reject the null hypothesis of
equal medians for cities.
Degrees of Freedom
1
2
3
4

0.005 7.87946 10.5966 12.8382 14.8603
0.010 6.63491 9.2103 11.3449 13.2767
0.025 5.02389 7.3778 9.3484 11.1433
0.050 3.84146 5.9915 7.8147 9.4877
0.100 2.70554 4.6052 6.2514 7.7794
Solution: Version 4
Brand
x1 r1
x2
r2
x3 r3
x4
r4
x5
r5
A
B
C
D
E
F
Sum
3.47
3.43
3.44
3.46
3.46
3.44
4
4
4
5
4
4
25
3.40
3.41
3.41
3.45
3.40
3.43
3.0
2.0
2.0
3.5
3.0
3.0
16.5
3.38
3.42
3.43
3.40
3.39
3.42
The sums of ranks sum to 90, which checks our as
 12
is  F2  
 rc c  1

 SR
i

 12
 SR   3r c  1   656 25
2
i
i
2.0
3.0
3.0
2.0
1.5
2.0
13.5
2

3.32
3.35
3.36
3.30
3.39
3.39
1.0
1.0
1.0
1.0
1.5
1.0
6.5
3.50
3.44
3.45
3.45
3.48
3.49
5.0
5.0
5.0
3.5
5.0
5.0
28.5
rcc  1 656

 90 . The Friedman statistic
2
2


 16 .5 2  13 .5 2  6.5 2  28 .5 2   36 6

2
1934   108  128 .9333  108  20 .93 . This has the chi-squared distribution with 4 degrees of
30
freedom. The upper right corner of the chi-squared table appears above. Since the computed value of  2
exceeds any value on the table, we can say that the p-value is below .005 and reject the null hypothesis of
equal medians for cities.
Solution: Version 9
Brand
x1 r1
x2
r2
x3 r3
x4
r4
x5
r5
A
B
C
D
E
F
Sum
3.47
3.43
3.44
3.46
3.46
3.44
5.0
5.0
5.0
5.0
5.0
4.5
29.5
3.40
3.41
3.41
3.45
3.40
3.43
3
3
3
4
3
3
19
3.38
3.42
3.43
3.40
3.39
3.42
The sums of ranks sum to 90, which checks our as
2.0
4.0
4.0
2.5
1.5
2.0
16
 SR
i

3.32
3.35
3.36
3.30
3.39
3.39
1.0
1.0
1.0
1.0
1.5
1.0
6.5
3.45
3.39
3.40
3.40
3.43
3.44
4.0
2.0
2.0
2.5
4.0
4.5
19
rcc  1 656

 90 . The Friedman statistic
2
2
 12

 12

29 .5 2  19 2  16 2  6.5 2  19 2   36 6 
SRi2   3r c  1  
is  F2  
 656 

 rc c  1 i

2
1890 .50   108  126 .0333  108  18 .033 . This has the chi-squared distribution with 4 degrees of

30
 


freedom. The upper right corner of the chi-squared table appears above. Since the computed value of  2
exceeds any value on the table, we can say that the p-value is below .005 and reject the null hypothesis of
equal medians for cities.
d) (Extra credit) Check your results on the computer. The Minitab setup for both 2-way ANOVA and the
corresponding non-parametric method is the same as the extra credit in the third graded assignment or the
third computer problem. You may want to use and compare the Statistics pull-down menu and ANOVA or
Nonparametrics to start with. Explain your results. (4) [12]
15
252y0731 4/25/07
(Page layout view!)
Solution: Note that because of rounding and small numbers there is quite a bit of difference between the
numbers in the ANOVAs here and above. There is no difference in conclusions and little in the value of the
standard errors. The Friedman test results are essentially identical.
Version 0
MTB > WSave "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x0703101.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\251x07031-01.MTW'
MTB > print c1-c6
Data Display
Row
1
2
3
4
5
6
Brand
A
B
C
D
E
F
City 1
3.47
3.43
3.44
3.46
3.46
3.44
City 2
3.40
3.41
3.41
3.45
3.40
3.43
City 3
3.38
3.42
3.43
3.40
3.39
3.42
MTB > Stack c2-c6 c50;
SUBC>
Subscripts c11;
SUBC>
UseNames.
MTB > STACK c1 c1 c1 c1 c1 c51
MTB > print c50 - c52
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
C50
3.47
3.40
3.38
3.32
3.50
3.43
3.41
3.42
3.35
3.44
3.44
3.41
3.43
3.36
3.45
3.46
3.45
3.40
3.30
3.45
3.46
3.40
3.39
3.39
3.48
3.44
3.43
3.42
3.39
3.49
Rowno
A
A
A
A
A
B
B
B
B
B
C
C
C
C
C
D
D
D
D
D
E
E
E
E
E
F
F
F
F
F
City
City
City
City
City
City
City
City
City
City
City
City
City
City
City
City
City
City
City
City
City
City
City
City
City
City
City
City
City
City
City
City 4
3.32
3.35
3.36
3.30
3.39
3.39
City 5
3.50
3.44
3.45
3.45
3.48
3.49
#Puts all columns into C50 and column labels into C52.
#This could be done by hand, but pull-down
#menus are easy to use.
#Stacks row labels in 51
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
16
252y0731 4/25/07
(Page layout view!)
MTB > Twoway C50 C51 C52;
SUBC>
Means C51 C52.
Two-way ANOVA: C50 versus Rowno, City
Source
Rowno
DF
5
SS
0.0020267
MS
0.0004053
F
0.63
P
0.677 #High p-value. Do not reject equal firm
#means.
City
4 0.0485133 0.0121283 18.94 0.000 #Low p-value. Reject equal city means.
Error
20 0.0128067 0.0006403
Total
29 0.0633467
S = 0.02530
R-Sq = 79.78%
R-Sq(adj) = 70.69%
Rowno
1
2
3
4
5
6
City
City
City
City
City
City
Mean
3.414
3.410
3.418
3.412
3.424
3.434
1
2
3
4
5
Individual 95% CIs For Mean Based on
Pooled StDev
-------+---------+---------+---------+-(-----------*-----------)
(-----------*-----------)
(-----------*-----------)
(-----------*-----------)
(-----------*-----------)
(-----------*-----------)
-------+---------+---------+---------+-3.400
3.420
3.440
3.460
Mean
3.45000
3.41667
3.40667
3.35167
3.46833
Individual 95% CIs For Mean Based on
Pooled StDev
-------+---------+---------+---------+-(-----*----)
(----*-----)
(-----*----)
(----*----)
(----*----)
-------+---------+---------+---------+-3.360
3.400
3.440
3.480
MTB > Friedman C50 C52 C51.
Friedman Test: C50 versus City blocked by Rowno
S = 20.93
S = 21.29
City
City
City
City
City
City
1
2
3
4
5
N
6
6
6
6
6
DF = 4
DF = 4
P = 0.000
#Low p-value. Reject equal city medians.
P = 0.000 (adjusted for ties)
Sum
of
Est Median Ranks
3.4375
25.0
3.4095
16.5
3.4025
13.5
3.3525
6.5
3.4555
28.5
Grand median = 3.4115
Version 4
Results for: 252x07031-01a4.MTW
MTB > Stack c2-c6 c10;
SUBC>
Subscripts c11;
SUBC>
UseNames.
MTB > STACK c1 c1 c1 c1 c1 c12
17
252y0731 4/25/07
(Page layout view!)
MTB > Twoway c10 c12 c11;
SUBC>
Means c12 c11.
Two-way ANOVA: C10 versus C12, C11
Source
C12
DF
5
SS
0.0020267
MS
0.0004053
F
0.63
P
0.677
#High p-value. Do not reject equal firm
#means.
0.000 #Low p-value. Reject equal city means.
C11
4 0.0240333 0.0060083 9.38
Error
20 0.0128067 0.0006403
Total
29 0.0388667
S = 0.02530
R-Sq = 67.05%
R-Sq(adj) = 52.22%
C12
A
B
C
D
E
F
C11
City
City
City
City
City
Individual 95% CIs For Mean Based on
Pooled StDev
---+---------+---------+---------+-----(-----------*-----------)
(-----------*-----------)
(-----------*-----------)
(-----------*-----------)
(-----------*-----------)
(-----------*-----------)
---+---------+---------+---------+-----3.400
3.420
3.440
3.460
Mean
3.422
3.418
3.426
3.420
3.432
3.442
1
2
3
4
5
Mean
3.45000
3.41667
3.40667
3.39167
3.46833
Individual 95% CIs For Mean Based on
Pooled StDev
-------+---------+---------+---------+-(------*------)
(------*------)
(-------*------)
(-------*------)
(------*------)
-------+---------+---------+---------+-3.390
3.420
3.450
3.480
MTB > Friedman c50 c52 c51.
Friedman Test: C50 versus C52 blocked by C51
S = 18.43
S = 18.75
DF = 4
DF = 4
P = 0.001
#Low p-value. Reject equal city medians.
P = 0.001 (adjusted for ties)
Sum
of
C52
N Est Median Ranks
City 1 6
3.4375
25.0
City 2 6
3.4095
15.0
City 3 6
3.4025
12.0
City 4 6
3.3925
9.5
City 5 6
3.4555
28.5
Grand median = 3.4195
Version 9
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x0703101a9.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x07031-01a9.MTW'
Worksheet was saved on Wed Apr 04 2007
Results for: 252x07031-01a9.MTW
MTB > Stack c2-c6 c10;
SUBC>
Subscripts c12;
SUBC>
UseNames.
MTB > stack c1 c1 c1 c1 c1 c11
18
252y0731 4/25/07
(Page layout view!)
MTB > Twoway c10 c11 c12;
SUBC>
Means c11 c12.
Two-way ANOVA: C10 versus C11, C12
Source
C11
DF
5
SS
0.0020267
MS
0.0004053
F
0.63
P
0.677 #High p-value. Do not reject equal firm
#means.
C12
4 0.0307133 0.0076783 11.99 0.000 #Low p-value. Reject equal city means
Error
20 0.0128067 0.0006403
Total
29 0.0455467
S = 0.02530
R-Sq = 71.88%
R-Sq(adj) = 59.23%
C11
A
B
C
D
E
F
C12
City
City
City
City
City
Mean
3.404
3.400
3.408
3.402
3.414
3.424
1
2
3
4
5
Individual 95% CIs For Mean Based on
Pooled StDev
--+---------+---------+---------+------(-----------*-----------)
(-----------*-----------)
(-----------*-----------)
(-----------*-----------)
(-----------*-----------)
(-----------*-----------)
--+---------+---------+---------+------3.380
3.400
3.420
3.440
Mean
3.45000
3.41667
3.40667
3.35167
3.41833
Individual 95% CIs For Mean Based on
Pooled StDev
-------+---------+---------+---------+-(-----*----)
(----*-----)
(-----*----)
(----*----)
(-----*----)
-------+---------+---------+---------+-3.360
3.400
3.440
3.480
MTB > Friedman c50 c52 c51.
Friedman Test: C50 versus C52 blocked by C51
S = 18.03
S = 18.50
DF = 4
DF = 4
P = 0.001
#Low p-value. Reject equal city medians.
P = 0.001 (adjusted for ties)
Sum
of
C52
N Est Median Ranks
City 1 6
3.4375
29.5
City 2 6
3.4095
19.0
City 3 6
3.4025
16.0
City 4 6
3.3525
6.5
City 5 6
3.4055
19.0
Grand median = 3.4015
19
252y0731 4/25/07
(Page layout view!)
2) A local official wants to know how land value affects home price. The official collects the following
data. 'Home price' is the dependent variable and 'Land Value' is the independent variable. Both are in
thousands of dollars. If you don’t know what dependent and independent variables are, stop work
until you find out.
Row
1
2
3
4
5
6
7
8
9
10
HomePr
67
63
60
54
58
36
76
87
89
92
LandVal
7.0
6.9
5.5
3.7
5.9
3.8
8.9
9.6
9.9
10.0
To personalize the data let g be the second to last digit in your student number. Mark your problem
‘VERSION g . ’Subtract g from 92, which is the last home price.
a) Compute the regression equation Y  b  b x to predict the home price on the basis of land value.
0
1
(3).You may check your results on the computer, but let me see real or simulated hand calculations.
Version 0
Y
i
1
2
3
4
5
6
7
8
9
10
67
63
60
54
58
36
76
87
89
92
682
7.0
6.9
5.5
3.7
5.9
3.8
8.9
9.6
9.9
10.0
71.2
To summarize n  10,
Y
2
49.00 469.0 4489
47.61 434.7 3969
30.25 330.0 3600
13.69 199.8 2916
34.81 342.2 3364
14.44 136.8 1296
79.21 676.4 5776
92.16 835.2 7569
98.01 881.1 7921
100.00 920.0 8464
559.18 5225.2 49364
 X  71.2,  Y  682 ,  XY  5225 .2,  X
 X  71.2  7.12
n
10
Spare Parts: SS x 
S xy 
Y2
XY
2
 559 .18 and
 4364 . df  n  2  10  2  8.
Means: X 
SS y 
X2
X
Y
2
X
2
Y 
 Y  682  68.2
n
10
 nX  559.18  107.122  52.236
2
 nY 2  49364  1068.22  2851.6  SST (Total Sum of Squares)
 XY  nXY  5225 .2  107.12 68.2  369 .36
Coefficients:
b1 
S xy
SS x

 XY  nXY
 X  nX
2
2

369 .36
 7.0710
52 .236
b0  Y  b1 X  68 .2  70 .0710  7.12   17 .85458
So our equation is Yˆ  17.85458  7.0710 X
Check: Regression Analysis: HomePr versus LandVal
The regression equation is
HomePr = 17.9 + 7.07 LandVal
Predictor
Coef SE Coef
T
P
Constant
17.855
5.665 3.15 0.014
LandVal
7.0710
0.7576 9.33 0.000
S = 5.47564
R-Sq = 91.6%
R-Sq(adj) = 90.5%
Analysis of Variance
20
252y0731 4/25/07
(Page layout view!)
Source
Regression
Residual Error
Total
DF
1
8
9
SS
2611.7
239.9
2851.6
Unusual Observations
Obs LandVal HomePr
4
3.7
54.00
MS
2611.7
30.0
Fit
44.02
SE Fit
3.12
F
87.11
P
0.000
Residual
9.98
St Resid
2.22R
b) Compute R 2 . (2)
Please do not repeat any calculations from a), but recall the following.
Spare Parts: SS x 
SS y 
S xy 
Y
2
X
2
 nX 2  559.18  107.122  52.236
 nY 2  49364  1068.22  2851.6  SST (Total Sum of Squares)
 XY  nXY  5225 .2  107.12 68.2  369 .36
S xy
369 .36
b0  Y  b1 X  68 .2  70 .0710  7.12   17 .85458
 7.0710
SS x 52 .236
So SSR  b1 Sxy  7.0710 369 .36   2611 .745 , and
b1 
R2 

S xy 2
369 .36 2  .9159 .
SSR b1 S xy 2611 .745


 .9159 or R 2 

SST
SSy
2851 .6
SS x SS y 52 .236 2851 .6
c) Compute s e . (2) SSE  SST  SSR  2851 .6  2611 .745  239 .855
s e2 
SSE 239 .855

 29 .9819 s e  29 .9819  5.47557 This is more accurate than the next calculation.
n2
8
2
Note also that if R is available
s e2


SS y 1  R 2
n2
  2851 .61  .9159   29.97745
8
and
s e  29 .97745  5.47517 .
d) Compute s b0 and do a significance test on b0 (2). Recall X  7.12 and SS x  52.236 .
H 0 :  0   00
b   00
8
We are now testing 
with t  0
. We are using t n2  t.025
 2.306.
2
H
:



s
1
0
00

b0
 1 7.12 2 
1 X 2 
s b20  s e2  
  29 .9819 1.07049   32 .0953 . So b0  17 .85458 and
  29 .9819  
 n SS x 
10 52 .236 
H 0 :  0  0
b  0 17 .8548
sb0  39.0953  5.6653 . If we are testing 
t 0

 3.152 . Since the
s b0
5.6653
H 1 :  0  0
rejection region is above 2.306 and below -2,306, we reject the null hypothesis and say that  0 is
significant. A confidence interval would be  0  b0  t  sb0  17.855  2.3065.6653  17.855  13.064 .
2
Since this confidence interval does not include zero, we reject the null hypothesis of insignificance. Note
that at the 1% significance level we could not reject the null hypothesis.
e) Use your equation to predict the price of the average home that is on land worth $9000. Make this into a
95% interval. (3)
Solution: Recall that n  10, X  7.12 , SS x  52.236 , s e2  29.9819 (or s e  29 .9819  5.47557 ),
that both the dependent and independent variables are measured in thousands and that our equation is
Yˆ  17.85458  7.0710 X .
21
252y0731 4/25/07
(Page layout view!)
So if X  9 , Yˆ  17.85458 7.07109  81.49 . Since we are concerned with an average value, we do not
use the prediction interval The formula for the Confidence Interval is   Yˆ  t s ˆ , where
Y0

1 X X
sY2ˆ  s e2   0
n
SS x


0
Y
2

  29 .9819  1  9  7.12 
 10

52 .236



  29.9819 0.1  0.06766   29.9819 0.16766 


8
 5.02683 or s ˆ  5.02683  2.2421 . We already know that t n2  t.025
 2.306. and that Yˆ  81.49 .
2
Y
2
So that our confidence interval is Y0  Yˆ0  t sY  81.49  2.3062.2421  81.49  5.17 , so that we can say
that we are 95% sure that the average house standing on $9(thousand) worth of land will sell for between
$76.32 and $86.66 (thousand).
f) Make a graph of the data. Show the regression line and the data points clearly. If you are not willing to
do this neatly and accurately, don’t bother. (2)
[26]
Fitted Line: HomePrice versus LandVal
Version 9
Because time was short, I was only able to run one other version of the the regression and that only on the
computer, but because it is the most extreme version, all other results should fall between the results below
and the results above. The run follows with interpretation. This is a run of my ‘exec’ 252sols. If you wish to
check your results, you are welcome to use it. As you can see, it is easy to run and internally documented.
To prepare your data, put y in c1 and x in c2. Then store it using the files pull-down menu and ‘save
worksheet.’ This is to establish a path for Minitab to find 252sols. Give Minitab the command echo in order
to be able to track what it is doing. Then download 252sols.txt from the courses server to your storage
location, and start it with the Minitab command, exec '252sols.txt'.
Here is the annotated output.
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x0703201a9.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x07032-01a9.MTW'
Worksheet was saved on Thu Apr 05 2007
A worksheet of the same name is already open... current worksheet
overwritten.
MTB > Save "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x0703201a9.MTW";
22
252y0731 4/25/07
(Page layout view!)
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x07032-01a9.MTW'
Existing file replaced.
MTB > #252sols
Fakes simple ols regression by hand
MTB > regress c1 1 c2
Regression Analysis: HomePr versus LandVal
The regression equation is
HomePr = 20.5 + 6.57 LandVal
Predictor
Coef SE Coef
Constant
20.488
5.644
LandVal
6.5748
0.7548
S = 5.45503
R-Sq = 90.5%
T
P
3.63 0.007
8.71 0.000
R-Sq(adj) = 89.3%
Analysis of Variance
Source
DF
SS
Regression
1 2258.0
Residual Error
8
238.1
Total
9 2496.1
MS
2258.0
29.8
F
75.88
P
0.000
Unusual Observations
Obs LandVal HomePr
Fit SE Fit Residual St Resid
4
3.7
54.00 44.81
3.10
9.19
2.05R
6
3.8
36.00 45.47
3.04
-9.47
-2.09R
R denotes an observation with a large standardized residual.
MTB
MTB
MTB
MTB
MTB
MTB
MTB
MTB
MTB
>
>
>
>
>
>
>
>
>
let c3=c2*c2
let c4=c1*c2
let c5=c1*c1
let k1=sum(c1)
let k2=sum(c2)
let k3=sum(c3)
let k4=sum(c4)
let k5=sum(c5)
print c1-c5
Data Display
Row
1
2
3
4
5
6
7
8
9
10
MTB
MTB
MTB
MTB
MTB
MTB
MTB
MTB
MTB
MTB
HomePr
67
63
60
54
58
36
76
87
89
83
>
>
>
>
>
>
>
>
>
>
C3
49.00
47.61
30.25
13.69
34.81
14.44
79.21
92.16
98.01
100.00
in c1, x in c2
x squared, #c4 is xy
y squared
sum of y
sum of x
sum of x squared
sum of xy
sum of y squared
C4
469.0
434.7
330.0
199.8
342.2
136.8
676.4
835.2
881.1
920.0
C5
4489
3969
3600
2916
3364
1296
5776
7569
7921
8464
let k17=count(c1)
#k17 is n
let k21=k2/k17
#k21 is mean of x
let k22=k1/k17
#k22 is mean of y
let k18=k21*k21*k17
let k18=k3-k18
#k18 is SSx
let k19=k22*k22*k17
let k19=k5-k19
#k19 is SSy
let k20=k21*k22*k17
let k20=k4-k20
#k20 is Sxy
print k1-k5
Data Display
K1
K2
K3
K4
K5
LandVal
7.0
6.9
5.5
3.7
5.9
3.8
8.9
9.6
9.9
10.0
#Put y
#c3 is
#c5 is
#k1 is
#k2 is
#k3 is
#k4 is
#k5 is
673.000
71.2000
559.180
5225.20
49364.0
#k1
#k2
#k3
#k4
#k5
is
is
is
is
is
sum
sum
sum
sum
sum
of
of
of
of
of
y
x
x squared
xy
y squared
23
252y0731 4/25/07
(Page layout view!)
MTB > print k17-k22
Data Display
K17
K18
K19
K20
K21
K22
10.0000
52.2360
4071.10
433.440
7.12000
67.3000
#k17
#k18
#k19
#k20
#k21
#k22
is
is
is
is
is
is
n
SSx
SSy
Sxy
mean of x
mean of y
MTB > end
24
252y0731 4/25/07
(Page layout view!)
3) A sales manager is trying to find out what method of payment works best. Salespersons are randomly
assigned to payment by commission, salary or bonus. Data for units sold in a week is below.
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
Commission
25
35
20
30
25
30
26
17
31
30
Salary
25
25
22
20
25
23
22
25
20
26
24
Bonus
28
41
21
31
26
26
46
32
27
35
23
22
26
To personalize the data let g be the second to last digit in your student number. Subtract 2 g from 46 in the
last column. Mark your problem ‘VERSION g .’Example: Payme Well’s student number is 000090, so she
subtracts 18 and gets the following numbers.
Version 9
Row Commission
1
25
2
35
3
20
4
30
5
25
6
30
7
26
8
17
9
31
10
30
11
12
13
Salary
25
25
22
20
25
23
22
25
20
26
24
Bonus
28
41
21
31
26
26
28
32
27
35
23
22
26
a) State your null hypothesis and test it by doing a 1-way ANOVA on these data and explain whether the
test tells us that payment method matters or not. (4)
Version 0: You should be able to do the calculations below. Only the three columns of numbers
were given to us.
Payment Method
Sum
Line
Commission
Salary
Bonus
1
2
3
4
5
6
7
8
9
10
11
12
13
25
35
20
30
25
30
26
17
31
30
25
25
22
20
25
23
22
25
20
26
24
269 +
257 +
384
 910 
nj
10 +
11 +
13
 34  n
x j
26.9
23.3636
29.5385
26 .7647  x
SS
7501 +
6049 +
12002
 25552 
Sum
x 2j
723.61
545.8595
28
41
21
31
26
26
46
32
27
35
23
22
26
 x
ij
 x
2
ij
872.5207
25
252y0731 4/25/07
(Page layout view!)
Note that all the squared means were gotten by going back to the original ratios used to calculate the means.
 x  nx  25552 3426.7647  2552  2435.8824  1196.1177
SSB   n x  nx  1026.9  1123.3636  1329.5385  3426.7647
SST 
2
ij
2
j .j
2
2
2
2
2
2
2
 10723 .61  11545 .8595   13872 .5207   24355 .8824
 7236 .1  6004 .4545  11342 .7691  24355 .8824  24583 .3236  24355 .8824  227 .4413
Source
Between
SS
DF
MS
F
F.05
2
113.7207
3.6393 s
F 2,31  3.30
227.4413
H0
Column means equal
Within
968.6765
31
31.2476
Total
1196.1177
33
Because our computed F statistic exceeds the 5% table value, we reject the null hypothesis of equal means
for each payment method and conclude that method of payment matters.
This was verified by the Minitab printout .
MTB > AOVOneway c1-c3;
SUBC>
Tukey 5;
SUBC>
Fisher 5.
One-way ANOVA: Commission, Salary, Bonus
Source DF
Factor
2
Error
31
Total
33
S = 5.590
Level
Commission
Salary
Bonus
SS
MS
227.4 113.7
968.7
31.2
1196.1
R-Sq = 19.01%
N
10
11
13
Mean
26.900
23.364
29.538
F
3.64
P
0.038
R-Sq(adj) = 13.79%
Individual 95% CIs For Mean Based on
Pooled StDev
StDev ---+---------+---------+---------+-----5.425
(---------*---------)
2.111 (---------*---------)
7.412
(--------*--------)
---+---------+---------+---------+-----21.0
24.5
28.0
31.5
Pooled StDev = 5.590
b) Using your results from a) present individual and Tukey intervals for the difference between earnings for
all three possible contrasts. Explain (i) under what circumstances you would use each interval and (ii)
whether the intervals show a significant difference. (2)
Let’s start with s  MSW  31.2476  5.5899, x1  26 .9, x2  23.3636 , x3  29.5385 ,
n1  10, n2  11 and n3  13 . There are m  3 columns and the ANOVA printout gives us n  m  31 for
3,31
31
 3.49
 2.040 and q.05
the within degrees of freedom. From our tables we find t .025
26
252y0731 4/25/07
(Page layout view!)
The outline gives the following for these two intervals.
Contrast 1-2
Individual Confidence Interval
If we desire a single interval, we use the formula for the difference between two means when the variance is
known. For example, if we want the difference between means of column 1 and column 2.
1   2  x1  x2   t n  m  s
2
1
1

n1 n 2
1 1
 26 .9  23 .3636   2.040 31 .2476   
 10 11 
 3.54  2.040 5.96545  3.54  2.040 2.442   3.54  4.98 ns
Tukey Confidence Interval
This applies to all possible differences.
1   2  x1  x2   q m,n  m 
 3.54  3.49
s
2
1
1

n1 n 2
 26 .9  23 .3636   3.49
31 .2476  1 1 
  
2
 10 11 
1
5.96545  3.54  3.49 1.727   3.54  6.03 ns
2
Contrast 1-3
Individual Confidence Interval
If we desire a single interval, we use the formula for the difference between two means when the variance is
known. For example, if we want the difference between means of column 1 and column 2.
1   3  x1  x3   t n  m  s
2
1
1
1 1
 26 .9  29 .5285   2.040 31 .2476   

n1 n3
 10 13 
 2.63  2.040 5.52842  2.63  2.040 2.351   2.63  4.80 ns
Tukey Confidence Interval
This applies to all possible differences.
1   3  x1  x3   q m,n  m 
 2.63  3.49
s
2
31 .2476
1
1
 26 .9  29 .5285   3.49

2
n1 n3
1 1
  
 10 13 
1
5.52842  2.63  3.49 1.662   2.63  5.80 ns
2
Contrast 2-3
Individual Confidence Interval
If we desire a single interval, we use the formula for the difference between two means when the variance is
known. For example, if we want the difference between means of column 1 and column 2.
 2   3  x2  x3   t n  m  s
2
1
1

n 2 n3
1 1
 23 .3636  29 .5285   2.040 31 .2476   
 11 13 
 6.16  2.040 5.24435  6.16  2.040 2.290   6.16  4.67
Tukey Confidence Interval
This applies to all possible differences.
 2   3  x2  x3   q m,n  m 
s
2
31 .2476  1 1 
1
1
 23 .3636  29 .5285   3.49

  
2
n 2 n3
 11 13 
1
5.24435  6.16  3.49 1.619   6.16  5.65
2
The remainder of the Minitab printout for this problem follows.
 6.16  3.49
27
252y0731 4/25/07
(Page layout view!)
Tukey 95% Simultaneous Confidence Intervals
All Pairwise Comparisons
Individual confidence level = 98.04%
Commission subtracted from:
Salary
Bonus
Lower
-9.547
-3.147
Center
-3.536
2.638
Upper
2.474
8.424
+---------+---------+---------+--------(---------*---------)
(--------*---------)
+---------+---------+---------+---------12.0
-6.0
0.0
6.0
Salary subtracted from:
Bonus
Lower
0.540
Center
6.175
Upper
11.810
+---------+---------+---------+--------(--------*---------)
+---------+---------+---------+---------12.0
-6.0
0.0
6.0
Fisher 95% Individual Confidence Intervals
All Pairwise Comparisons
Simultaneous confidence level = 88.04%
Commission subtracted from:
Salary
Bonus
Lower
-8.518
-2.157
Center
-3.536
2.638
Upper
1.445
7.434
--------+---------+---------+---------+(-------*-------)
(-------*-------)
--------+---------+---------+---------+-6.0
0.0
6.0
12.0
Salary subtracted from:
Bonus
Lower
1.504
Center
6.175
Upper
10.845
--------+---------+---------+---------+(------*-------)
--------+---------+---------+---------+-6.0
0.0
6.0
12.0
c) What other method could we use on these data to see if payment method makes a difference? Under what
circumstances would we use it? Try it and tell what it tests and what it shows. (3) [37]
The alternative to one-way ANOVA for situations in which the parent distribution is not Normal is the
Kruskal-Wallis test. The original data is replaced by ranks.
Data Display
Row
Commission
1
2
3
4
5
6
7
8
9
10
11
12
13
Rank Sum
ni
Rank Commission
Salary
Rank Salary
Bonus
x1
r1
x2
r2
x3
25
35
20
30
25
30
26
17
31
30
14.5
31.5
3.0
26.0
14.5
26.0
20.0
1.0
28.5
26.0
25
25
22
20
25
23
22
25
20
26
24
14.5
14.5
7.0
3.0
14.5
9.5
7.0
14.5
3.0
20.0
11.0
28
41
21
31
26
26
46
32
27
35
23
22
26
191
10
Note that the sum of the first thirty-four numbers is
118.5
11
Rank Bonus
r3
24.0
33.0
5.0
28.5
20.0
20.0
34.0
30.0
23.0
31.5
9.5
7.0
20.0
285.5
13
34 35 
 595 , so that we can check our ranking by
2
noting that 191 + 118.5 + 285.5 = 595.
28
252y0731 4/25/07
 12
H 
 nn  1

i
(Page layout view!)
 SRi 2

 ni

2
2
2 


  3n  1   12  191   118 .5  285 .5   335 

11
13 
 34 35   10

1211194 .6701 
12
 105  112 .8874  105  7.8874 . The
3648 .1  1276 .5682  6270 .0019   105 
34 35 
3435 
2
design of the data is too large for the Kruskal-Wallis table, so we compare this with  2 .05  5.9915 . Since
the computed chi_squared exceeds the table value we can reject the null hypothesis of equal medians.
The computer agrees. The Minitab printout follows.

MTB > Kruskal-Wallis c4 c5.
Kruskal-Wallis Test: CSB0 versus Pay0
Kruskal-Wallis Test on CSB0
Pay0
N Median Ave
Bonus
13
27.00
Commission 10
28.00
Salary
11
24.00
Overall
34
H = 7.89 DF = 2 P = 0.019
H = 7.97 DF = 2 P = 0.019
Rank
22.0
19.1
10.8
17.5
Z
2.06
0.60
-2.72
(adjusted for ties)
d) (Extra Credit) Do a Levene test on these data and explain what it tests and shows. (4)
The Levene test is a test for equal variances. The original data is below.
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
Commission
25
35
20
30
25
30
26
17
31
30
Salary
25
25
22
20
25
23
22
25
20
26
24
Bonus
28
41
21
31
26
26
46
32
27
35
23
22
26
To find the median, put the data in order and pick the center number or an average of the two center
numbers. We find that the median of Commission is 28.00, the median of Salary is 24.00 and the median
of Bonus is 27.00.We subtract the column median from each column.
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
Commission
-3
7
-8
2
-3
2
-2
-11
3
2
Salary
1
1
-2
-4
1
-1
-2
1
-4
2
0
Bonus
1
14
-6
4
-1
1
19
5
0
8
-4
-5
-1
29
252y0731 4/25/07
(Page layout view!)
We now take absolute values.
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
Commission
3
7
8
2
3
2
2
11
3
2
Salary
1
1
2
4
1
1
2
1
4
2
0
Bonus
1
14
6
4
1
1
19
5
0
8
4
5
1
An ANOVA is done on these 3 columns.
Source
SS
DF
Between
79.2
2
MS
39.6
F.05
F
2.538 ns
F 2,31  3.30
H0
Column means equal
Within
485.1
31
15.6
Total
564.3
33
Because we cannot reject the null hypothesis, we cannot say that the variances are not equal. As long as we
use the Levine test, Minitab agrees. However, the more powerful Bartlett test causes us to reject the null
hypothesis of equal variances if the data is Normally distributed.
Test for Equal Variances: Commission, Salary, Bonus
95% Bonferroni confidence intervals for standard deviations
Commission
Salary
Bonus
N
10
11
13
Lower
3.45609
1.36993
4.96227
StDev
5.42525
2.11058
7.41188
Upper
11.5454
4.2692
13.8626
Bartlett's Test (normal distribution)
Test statistic = 12.69, p-value = 0.002
Levene's Test (any continuous distribution)
Test statistic = 2.53, p-value = 0.096
f) (Extra Credit) Check you results on the computer. The Minitab setup for 1-way ANOVA can either be
unstacked (in columns) or stacked (one column for data and one for column name). There is a Tukey
option. The corresponding non-parametric methods require the stacked presentation. You may want to use
and compare the Statistics pull-down menu and ANOVA or Nonparametrics to start with. You may also
want to try the Mood median test and the test for equal variances which are on the menu categories above.
Even better, do a normality test on your columns. Use the Statistics pull-down menu 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 the 1-way 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) [12]. Most of this
was done above. The printout below is for the Normality tests and Mood’s median test.
————— 4/18/2007 1:42:53 PM ————————————————————
Welcome to Minitab, press F1 for help.
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x07031-03-v0.MTW'
Worksheet was saved on Tue Apr 17 2007
Results for: 252x07031-03-v0.MTW
MTB > Save
v0.MTW";
"C:\Documents and Settings\RBOVE\My Documents\Minitab\252x07031-03-
30
252y0731 4/25/07
(Page layout view!)
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x07031-03-v0.MTW'
Existing file replaced.
MTB > NormTest c1;
SUBC>
KSTest.
Probability Plot of Commission
MTB > NormTest c2;
SUBC>
KSTest.
Probability Plot of Salary
MTB > NormTest c3;
SUBC>
KSTest.
Probability Plot of Bonus
Unfortunately, this is hard to read in this format. In the little boxes the last item is the p-value for the test
and in each case the p-value is reported as ‘> .150.’ Because this p-value is above any significance level we
might use, we cannot reject the null hypothesis of Normality.
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x0703103.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x07031-03.MTW'
Worksheet was saved on Wed Apr 04 2007
31
252y0731 4/25/07
(Page layout view!)
Results for: 252x07031-03.MTW
MTB > print c5 c6
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
Pay0
Commission
Commission
Commission
Commission
Commission
Commission
Commission
Commission
Commission
Commission
Salary
Salary
Salary
Salary
Salary
Salary
Salary
Salary
Salary
Salary
Salary
Bonus
Bonus
Bonus
Bonus
Bonus
Bonus
Bonus
Bonus
Bonus
Bonus
Bonus
Bonus
Bonus
C6
14.5
31.5
3.0
26.0
14.5
26.0
20.0
1.0
28.5
26.0
14.5
14.5
7.0
3.0
14.5
9.5
7.0
14.5
3.0
20.0
11.0
24.0
33.0
5.0
28.5
20.0
20.0
34.0
30.0
23.0
31.5
9.5
7.0
20.0
MTB > Mood C6 c5.
Mood Median Test: C6 versus Pay0
Mood median test for C6
Chi-Square = 11.53
DF = 2
Pay0
Bonus
Commission
Salary
N<=
3
4
10
N>
10
6
1
Median
23.0
23.0
11.0
P = 0.003
Individual 95.0% CIs
Q3-Q1
+---------+---------+---------+-----16.0
(--------*----------)
15.0
(-----------------*----)
7.5
(-----*----)
+---------+---------+---------+-----7.0
14.0
21.0
28.0
Overall median = 17.3
The Mood test gives us a low p-value which means that the medians are not similar. Because we have
concluded that the data is Normally distributed, we have to go with the Bartlett test, which says the
variances are unequal. If we had only the Levine result and the Normality tests, we could use ANOVA, but
the Bartlett test says that we have unequal variances. Under the circumstances, we have a quandary and
would probably go with the result of the Mood or Kruskal-Wallis test. In any case, we can conclude that the
means are different and that we are probably better off using commissions.
32
252y0731 4/25/07
(Page layout view!)
Version 9
MTB > print c11 - c13
Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
Co0
25
35
20
30
25
30
26
17
31
30
Sa0
25
25
22
20
25
23
22
25
20
26
24
Bo9
28
41
21
31
26
26
28
32
27
35
23
22
26
MTB > AOVOneway c11-c13;
SUBC>
Tukey 5;
SUBC>
Fisher 5.
One-way ANOVA: Co0, Sa0, Bo9
Source DF
Factor
2
Error
31
Total
33
S = 4.667
Level
Co0
Sa0
Bo9
N
10
11
13
#Note that we do not reject equal means.
SS
MS
F
P
143.0 71.5 3.28 0.051
675.1 21.8
818.1
R-Sq = 17.48%
R-Sq(adj) = 12.15%
Mean
26.900
23.364
28.154
StDev
5.425
2.111
5.520
Individual 95% CIs For Mean Based on
Pooled StDev
--+---------+---------+---------+------(---------*---------)
(---------*--------)
(--------*--------)
--+---------+---------+---------+------21.0
24.0
27.0
30.0
Pooled StDev = 4.667
Tukey 95% Simultaneous Confidence Intervals
All Pairwise Comparisons
Individual confidence level = 98.04%
Co0 subtracted from:
Sa0
Bo9
Lower
-8.554
-3.576
Center
-3.536
1.254
Upper
1.481
6.084
---------+---------+---------+---------+
(---------*---------)
(---------*--------)
---------+---------+---------+---------+
-5.0
0.0
5.0
10.0
Sa0 subtracted from:
---------+---------+---------+---------+
(---------*--------)
---------+---------+---------+---------+
-5.0
0.0
5.0
10.0
Fisher 95% Individual Confidence Intervals
All Pairwise Comparisons
Simultaneous confidence level = 88.04%
Bo9
Lower
0.086
Center
4.790
Upper
9.495
Co0 subtracted from:
Sa0
Bo9
Lower
-7.695
-2.750
Center
-3.536
1.254
Upper
0.622
5.257
-------+---------+---------+---------+-(-------*-------)
(-------*-------)
-------+---------+---------+---------+--
33
252y0731 4/25/07
(Page layout view!)
-5.0
0.0
5.0
10.0
Sa0 subtracted from:
Bo9
Lower
0.891
Center
4.790
Upper
8.689
-------+---------+---------+---------+-(-------*------)
-------+---------+---------+---------+--5.0
0.0
5.0
10.0
MTB > vartest c11-c13;
SUBC> unstacked.
Test for Equal Variances: Co0, Sa0, Bo9
95% Bonferroni confidence intervals for standard deviations
N
Lower
StDev
Upper
Co0 10 3.45609 5.42525 11.5454
Sa0 11 1.36993 2.11058
4.2692
Bo9 13 3.69589 5.52036 10.3248
Bartlett's Test (normal distribution)
Test statistic = 8.75, p-value = 0.013
Levene's Test (any continuous distribution)
Test statistic = 2.25, p-value = 0.123
MTB > Stack c11 c12 c13 c14;
SUBC>
Subscripts c15;
SUBC>
UseNames.
MTB >
MTB >
SUBC>
SUBC>
MTB >
rank c14 c16
unstack c16 c17 c18 c19;
subscripts c15;
varnames.
Kruskal-Wallis c14 c15
Kruskal-Wallis Test: CSB9 versus Pay 9
Kruskal-Wallis Test on CSB9
Pay 9
N Median Ave Rank
Z
Bo9
13
27.00
21.6
1.88
Co0
10
28.00
19.6
0.79
Sa0
11
24.00
10.8 -2.72
Overall 34
17.5
H = 7.64 DF = 2 P = 0.022
H = 7.73 DF = 2 P = 0.021 (adjusted for ties)
MTB > print c11 c17 c12 c18 c13 c19
Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
Co0
25
35
20
30
25
30
26
17
31
30
C16_Bo9
24.5
34.0
5.0
29.5
20.0
20.0
24.5
31.0
23.0
32.5
9.5
7.0
20.0
Sa0
25
25
22
20
25
23
22
25
20
26
24
C16_Co0
14.5
32.5
3.0
27.0
14.5
27.0
20.0
1.0
29.5
27.0
Bo9
28
41
21
31
26
26
28
32
27
35
23
22
26
MTB > copy c11-c13 c31-c33
* NOTE * Column lengths not equal.
C16_Sa0
14.5
14.5
7.0
3.0
14.5
9.5
7.0
14.5
3.0
20.0
11.0
We are setting up for a Levine test.
MTB > name k31 'medco0a'
MTB > name k32 'medsal0a'
MTB > name k33 'medbon9a'
34
252y0731 4/25/07
MTB
MTB
MTB
MTB
>
>
>
>
(Page layout view!)
let k31 = median(c31)
let k32 = median(c32)
let k33 = median(c33)
print k31-k33
Data Display
medco0a
medsal0a
medbon9a
MTB
MTB
MTB
MTB
>
>
>
>
28.0000
24.0000
27.0000
let c31 = c31 - k31
let c32 = c32 - k32
let c33 = c33 - k33
describe c31 - c33
Descriptive Statistics: Co0a, Sal0a, Bon9a
Variable
Co0a
Sal0a
Bon9a
N
10
11
13
N*
0
0
0
Variable
Co0a
Sal0a
Bon9a
Maximum
7.00
2.000
14.00
Mean
-1.10
-0.636
1.15
SE Mean
1.72
0.636
1.53
StDev
5.43
2.111
5.52
Minimum
-11.00
-4.000
-6.00
Q1
-4.25
-2.000
-2.50
Median
0.00
0.000
0.00
Q3
2.25
1.000
4.50
MTB > print c31 - c33
Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
MTB
MTB
MTB
MTB
Co0a
-3
7
-8
2
-3
2
-2
-11
3
2
>
>
>
>
Sal0a
1
1
-2
-4
1
-1
-2
1
-4
2
0
let c31 = abs(c31)
let c32 = abs(c32)
let c33 = abs(c33)
print c31-c33
Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
Bon9a
1
14
-6
4
-1
-1
1
5
0
8
-4
-5
-1
Co0a
3
7
8
2
3
2
2
11
3
2
Sal0a
1
1
2
4
1
1
2
1
4
2
0
Bon9a
1
14
6
4
1
1
1
5
0
8
4
5
1
35
252y0731 4/25/07
(Page layout view!)
MTB > AOVO c31 - c33
One-way ANOVA: Co0a, Sal0a, Bon9a
Source
Factor
DF
2
Error
31
Total
33
S = 3.065
Level
Co0a
Sal0a
Bon9a
N
10
11
13
SS
42.24
MS
21.12
291.20
9.39
333.44
R-Sq = 12.67%
Mean
4.300
1.727
3.923
StDev
3.199
1.272
3.904
F
2.25
P
0.123
#This is the last act of the Levene Test.
#The high p-value means we cannot reject
#the null hypothesis of equal means
R-Sq(adj) = 7.03%
Individual 95% CIs For Mean Based on
Pooled StDev
-+---------+---------+---------+-------(--------*---------)
(---------*--------)
(--------*-------)
-+---------+---------+---------+-------0.0
2.0
4.0
6.0
Pooled StDev = 3.065
36