252solngr4-081 4/13/08 (Open this document in 'Page Layout' view!)
Name:
Class days and time:
Please include this on what you hand in!
Graded Assignment 4
The data set GOLFBALL is in problem 11.14 of the text 9th or 10th edition or on the CD. You must answer
at least the following questions versions A, B and C of the problem.
Only neat and legible papers with written answers in complete sentences will be read!
a) At a 5% level is there evidence of a difference in the average distance traveled by the golf balls with
different designs? Why? c) What assumptions are necessary in (a)? e) What golf ball design should be
chosen?
Do this problem in Excel as follows.
Use columns A, B, C, E and F on the Excel spreadsheet for data
In the first row of Columns B, C, D and F put in Des 1, Des 2, Des 3 and Des 4. Label Column E with Des
4a and Column A with ‘golfer.’ Starting in Cell A2 Put in the letters A through J to identify the golfers –
unless, of course, you want to suggest some names.
Now put in the data in columns B, C, D and F, skipping column D
Version A
To fill column E in cell E2 write =F2 after your 'enter' this cell should read '213.9'
Use the fill handle on cell E2 to make column E identical to column F except for the heading. Do not go on
unless this is true. Save your data as gdataA.xls
Use the 'tools' pull-down menu and pick ‘data analysis' (If you cannot find this, use Tools and Add-Ins to
put in the analysis packs.)
Pick 'ANOVA: Single Factor. Set input range to $B$1:$E$11. Select 'New worksheet ply' and ‘columns’,
check 'labels in first row' hit 'OK' and save your results as gresltA.xls.
Version B
In order to check for the effect of the fact that the data is blocked by employees, repeat the analysis using
‘ANOVA: Two-Factor without replication. Set input range to $A$1:$E$11, check ‘labels,’ and save your
results as gresltB.xls
Version C
Take the last digit of your student number (if it's zero, use 10) and add 5 to it. Call this x and make sure
that I know its value. Go back to your original data or use the 'file' pull-down menu to open gresltA.xls.
To fill column E this time in cell E2 write =F2+ x .Now highlight cell E2 and use the fill handle to make
column E equal to column F plus x . Do not go on unless this is true. Save your data as gdataC.xls.
Run the one-way ANOVA again and save your results as gresltC.xls
Submit the data and results with your Student number. The most effective way to do this is to paste the
results into a Word document and then add neat hand or typed notes. Indicate what hypotheses were tested,
what the p-value was and whether, using the p-value, you would reject the null if (i) the significance level
was 5% and (ii) the significance level was 10%, explaining why. You will have two answers for each of
your two problems.
For your version C ANOVA do a Scheffe confidence interval and a Tukey-Kramer interval or procedure for
each of the C 24  6 possible differences between means and report which are different at the 5% level
according to each of the 2 methods. Answer as much as you can of the questions in Problem 11.14. The
extra credit below may be needed for a really complete answer.
Extra Credit: Take the data from your last ANOVA and perform a Levene test on it using the third
example in 252mvarex. as a pattern for your calculations. Make sure that you explain what is being tested
and what you conclude. Hand in separately – this will be treated as extra credit on your next take-home
exam. See below for all of this.
Extra Extra Credit: Do Bartlett and Levene tests using the example in 252mvar as your pattern. It turns
out that your ANOVA has just enough columns to do this test. See below for all of this.
252grass4-051 4/13/05 (Open this document in 'Page Layout' view!)
Data for 1st and 2nd ANOVA
gdataA
Golfer
Des 1
Des 2
1
206.32
203.81
2
223.85
223.85
3
207.94
206.75
4
224.79
223.97
5
206.19
205.68
6
229.75
234.3
7
204.45
204.49
8
228.51
219.5
9
209.65
210.86
10
221.44
233
Results for 1st ANOVA
gresltA
Anova: Single Factor
SUMMARY
Groups
Des 1
Des 2
Des 3
Des 4a
Des 3
217.08
230.55
221.43
227.95
218.04
213.84
224.13
224.87
211.82
229.49
Des 4a
213.9
231.1
221.28
221.53
229.43
235.45
213.54
228.35
214.51
225.09
Des 4
213.9
231.1
221.28
221.53
229.43
235.45
213.54
228.35
214.51
225.09
H 0 : 1   2   3   4
Count
10
10
10
10
ANOVA
Source of
Variation
Between Groups
Within Groups
SS
397.9091
3129.27
Total
3527.179
Sum
2162.89
2166.21
2219.2
2234.18
Average
216.289
216.621
221.92
223.418
Variance
104.6483
139.6653
43.08287
60.3002
df
MS
132.6364
86.92417
F
1.525886
3
36
P-value
0.224397
F crit
2.866266
39
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Results for 2 nd ANOVA H 01 : RowEmployeemeans equal H 02 :  1   2   3   4
gresltB
Anova: Two-Factor Without Replication
SUMMARY
Count
1
2
3
4
5
6
7
8
9
10
Des 1
Des 2
Des 3
Des 4a
4
4
4
4
4
4
4
4
4
4
Sum
841.11
909.35
857.4
898.24
859.34
913.34
846.61
901.23
846.84
909.02
Average
210.2775
227.3375
214.35
224.56
214.835
228.335
211.6525
225.3075
211.71
227.255
Variance
38.96229
16.26729
65.66647
7.024667
127.2787
99.43723
87.47603
17.81042
4.272733
25.50057
10
10
10
10
2162.89
2166.21
2219.2
2234.18
216.289
216.621
221.92
223.418
104.6483
139.6653
43.08287
60.3002
df
MS
228.6766
132.6364
39.67334
F
5.763988
3.343212
ANOVA
Source of
Variation
Rows
Columns
Error
SS
2058.09
397.9091
1071.18
Total
3527.179
9
3
27
P-value
0.00018
0.033859
F crit
2.250131
2.960351
39
Answer: In the first ANOVA we get a p-value of .224397. Since this is above any significance level we are
likely to use, we do not reject the null hypothesis that the mean distance that the golf balls go is the same for
all numbers of hours worked. . In the second ANOVA, the p-value for columns (.033859) is much lower,
so we can reject the original null hypothesis at the 5% significance level. Note that there is a very significant
difference between golfers. Too bad that this version completely differs from what it said in the problem.
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Results for 3rd ANOVA H 0 : 1   2   3   4
gdataC
Golfer
Des 1
Des 2
Des 3
Des 4c
1
206.32
203.81
217.08
223.9
2
223.85
223.85
230.55
241.1
3
207.94
206.75
221.43
231.28
4
224.79
223.97
227.95
231.53
5
206.19
205.68
218.04
239.43
6
229.75
234.3
213.84
245.45
7
204.45
204.49
224.13
223.54
8
228.51
219.5
224.87
238.35
9
209.65
210.86
211.82
224.51
10
221.44
233
229.49
235.09
Des 4
213.9
231.1
221.28
221.53
229.43
235.45
213.54
228.35
214.51
225.09
gresltC
Anova: Single Factor
SUMMARY
Groups
Des 1
Des 2
Des 3
Des 4c
ANOVA
Source of
Variation
Between Groups
Within Groups
Count
10
10
10
10
SS
1919.109
3129.27
Sum
2162.89
2166.21
2219.2
2334.18
Average
216.289
216.621
221.92
233.418
Variance
104.6483
139.6653
43.08287
60.3002
df
MS
639.703
86.92417
F
7.359323
3
36
P-value
0.000573
F crit
2.866266
Total
5048.379
39
The modified problem is giving us some very real differences in the average distance the various golf ball
designs go. The p-value is low enough to cause a rejection of the null hypothesis at any usual significance
level.
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Types of contrast between means.
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
1
1
, where s  MSW .

n1 n2
2
Scheff e  Confidence Interval
If we desire intervals that will simultaneously be valid for a given confidence level for all possible intervals
 1
1 
between column means, use 1   2  x1  x2   m  1Fm 1, n  m   s
.

 n
n2 
1

Tukey Confidence Interval
This also applies to all possible differences.
1   2  x1  x2   q m,n  m 
s
2
1
1
. This gives rise to Tukey’s HSD (Honestly Significant

n1 n 2
Difference) procedure. Two sample means x .1 and x .2 are significantly different if x.1  x.2 is greater
than q m,n  m 
s
2
1
1

n1 n 2
Contrasts
From the Excel output, x1  216 .289 , x2  216 .621, x3  221 .920 , x4  233 .418 , m  4, n  m  36,
s 110  110  86 .9242 0.2  17 .3848
4,30
4, 40
4,36
 3.85 and q.05
 3.79 . Since
 4.1695 . Assume   0.05 . We will need q.05
. The table says q.05
n1  n 2  n3  n 4  20 and s 2  MSW  86.9242 . So
36 is about halfway between 30 and 40, take a halfway point between the two table values and say
3,36  2.87 The contrasts follow.
4,36
36
q.05
 3.82 . t .025
 2.028 . F.05
1   2
Individual: 1   2  216 .289  216 .621   t 36 86 .9242
2
 0.332  8.46
ns
Scheffe: 1   2  216 .289  216 .621  
  0.332  
3 2.87
 0.332  12.233
1 1

 0.332  2.028 17 .3848
10 10
86 .9242
3F.053, 36
86 .9242
1 1

10 10
1
1

 0.332  2.934  17 .3848
10 10
ns
86 .9242
Tukey: 1   2  x1  x2   q .405,36 
2
 216 .289  216 .621   3.82
86 .9242
2
1 1

10 10
1 1

10 10
 0.332  2.701  17 .3848  0.332  11.262 ns
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1   3
From the Excel output, x1  216 .289 , x2  216 .621, x3  221 .920 , x4  233 .418 .
Individual: 1   3  216 .289  221 .920   t 36 86 .9242
2
 5.631  8.46
ns
Scheffe: 1   3  216 .289  221 .920  
  5.631  
3 2.87
 5.631  12.233
1
1

 5.631  2.028 17 .3848
10 10
86 .9242
3F.053, 36
86 .9242
1 1

10 10
1
1

 5.631  2.934  17 .3848
10 10
ns
86 .9242
Tukey: 1   3  x1  x3   q .405,36 
2
 216 .289  221 .920   3.82
1 1

10 10
86 .9242
2
1 1

10 10
 5.631  2.701  17 .3848  5.631  11.262 ns
1   4
From the Excel output, x1  216 .289 , x2  216 .621, x3  221 .920 , x4  233 .418 .
Individual: 1   4  216 .289  233 .418   t 36 86 .9242
2
 17.129  8.46
s
Scheffe: 1   4  216 .289  233 .418  
  17 .129  
3 2.87
 17.129  12.233
1 1

 17 .129  2.028 17 .3848
10 10
3F.053, 36
86 .9242
1 1

10 10
1
1

 17 .129  2.934  17 .3848
10 10
86 .9242
s
86 .9242
Tukey: 1   4  x1  x.4   q .405,36 
2
 216 .289  233 .418   3.82
1 1

10 10
86 .9242
2
1 1

10 10
 17 .129  2.701  17 .3848  17.129  11.262 s
 2  3
From the Excel output, x1  216 .289 , x2  216 .621, x3  221 .920 , x4  233 .418 .
Individual:  2   3  216 .621  221 .920   t 36 86 .9242
2
 5.299  8.46
Scheffe:  2   3  216 .621  221 .920  
  5.299  
3 2.87
 5.299  12.233
86 .9242
1
1

 5.299  2.028 17 .3848
10 10
ns
3F.053, 36
86 .9242
1 1

10 10
1 1

 5.299  2.934  17 .3848
10 10
ns
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86 .9242
Tukey:  2   3  x2  x3   q .405,36 
2
1 1

10 10
 216 .621  221 .920   3.82
1 1

10 10
86 .9242
2
 5.299  2.701  17 .3848  5.299  11.262 ns
2  4
From the Excel output, x1  216 .289 , x2  216 .621, x3  221 .920 , x4  233 .418 .
Individual:  2   4  216 .621  233 .418   t 36 86 .9242
2
 16.797  8.46
s
3F.053, 36
Scheffe:  2   4  216 .621  233 .418  
  16 .797  
3 2.87
 16.797  12.233
1
1

 16 .797  2.028 17 .3848
10 10
86 .9242
1 1

10 10
1
1

 16 .797  2.934  17 .3848
10 10
86 .9242
s
86.9242
Tukey:  2   4  x2  x.4   q .405,36 
2
1 1

10 10
 216 .621  233 .418   3.82
1 1

10 10
86 .9242
2
 16 .797  2.701 17 .3848  16.797  11.262 s
3   4
From the Excel output, x1  216 .289 , x2  216 .621, x3  221 .920 , x4  233 .418 .
Individual:  3   4  221 .920  233 .418   t 36 86 .9242
2
 11.498  8.46
s
Scheffe:  3   4  221 .920  233 .418  
  11 .498  
3 2.87
 11.498  12.233
1 1

 11 .498  2.028 17 .3848
10 10
86 .9242
3F.053, 36
86 .9242
1 1

10 10
1
1

 11 .498  2.934  17 .3848
10 10
ns
86 .9242
Tukey:  3   4  x.3  x.4   q .405,36 
2
 221 .920  233 .418   3.82
86 .9242
2
1 1

10 10
1 1

10 10
 11 .498  2.701  17 .3848  11.498  11.262 s
Conclusion: I have included individual confidence levels here for completeness. The analysis of variance
definitely tells us that the means are not the same, regardless of the significance level we might want to use,
because the p-value is small. If we compare the differences in sample means we find that there is no
difference between the means for the first 3 designs, but that most of the intervals show design 4 to be
superior. The intervals are labeled ‘ns’ for not significant and ‘s’ for significant depending on whether the
error part of the interval is larger or smaller than the difference between sample means.
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Extra Credit: Take the data from your last ANOVA and perform a Levene test on it using the third
example in 252mvarex. as a pattern for your calculations using Minitab. Make sure that you explain what is
being tested and what you conclude.
To do this copy your data into rows 1-10 of columns 1-5. Remember that your column labels should be
written in above the columns. Just to make sure that you are in the right place, print out your data and run a
one-way ANOVA using:
print c1-c5
AOVO c2-c5
The test is simply
vartest c2-c5;
unstacked.
Don’t give me results without explaining them.
————— 4/5/2005 11:23:01 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > WOpen "C:\Documents and Settings\rbove\My Documents\Minitab\2gr4051C.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\rbove\My
Documents\Minitab\2gr4-051C.MTW'
Worksheet was saved on Tue Apr 05 2005
Results for: 2gr4-051C.MTW
MTB > describe c2-c5
Descriptive Statistics: Des 1, Des 2, Des 3, Des 4
Variable
Des 1
Des 2
Des 3
Des 4
N
10
10
10
10
N*
0
0
0
0
Variable
Des 1
Des 2
Des 3
Des 4
Maximum
229.75
234.30
230.55
240.45
Mean
216.29
216.62
221.92
228.42
SE Mean
3.23
3.74
2.08
2.46
StDev
10.23
11.82
6.56
7.77
Minimum
204.45
203.81
211.82
218.54
Q1
206.29
205.38
216.27
219.36
Median
215.55
215.18
222.78
228.31
Q3
225.72
226.23
228.34
234.85
MTB > AOVOneway c2-c5.
One-way ANOVA: Des 1, Des 2, Des 3, Des 4
Source
Factor
Error
Total
DF
3
36
39
S = 9.323
SS
971.0
3129.3
4100.3
MS
323.7
86.9
R-Sq = 23.68%
F
3.72
P
0.020
This is identical to our previous ANOVA
R-Sq(adj) = 17.32%
MTB > vartest c2-c5;
SUBC> unstacked.
Test for Equal Variances: Des 1, Des 2, Des 3, Des 4
95% Bonferroni confidence intervals for standard deviations
N
Lower
StDev
Upper
Des 1 10 6.40248 10.2298 22.6233
Des 2 10 7.39650 11.8180 26.1357
Des 3 10 4.10804
6.5638 14.5159
Des 4 10 4.86006
7.7653 17.1731
Bartlett's Test (normal distribution)
Test statistic = 3.51, p-value = 0.320
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Levene's Test (any continuous distribution)
Test statistic = 3.78, p-value = 0.019
Test for Equal Variances: Des 1, Des 2, Des 3, Des 4
Test for Equal Variances for Des 1, Des 2, Des 3, Des 4
Bartlett's Test
Test Statistic
P-Value
Des 1
3.51
0.320
Lev ene's Test
Test Statistic
P-Value
Des 2
3.78
0.019
Des 3
Des 4
5
10
15
20
25
95% Bonferroni Confidence Intervals for StDevs
We have very interesting results.
If we are justified using ANOVA, then the Bartlett test should be the correct one to use and should be more
powerful than the Levene test. This is not what seems to be happening. The Levene test, with a p-value of
1.9%, which is below 5% rejects the null hypothesis of equal variances. The Bartlett test, with a much
higher p-value does not.
------------------------------------------------------------------------------
Extra Extra Credit: Do Bartlett and Levene tests by hand using the examples in 252mvar as your pattern.
It turns out that your ANOVA has just enough columns to do this test.
This is an awful lot of work unless you cheat and use the computer. If you cover your tracks, I’ll never
know. To do the Bartlett test you need logarithms of variances. Label Columns 10-12 ‘stdev,’ ‘var’ and
‘log.’ Use the data that you already have in Minitab and get the variances as follows:
stdev c2 k2
stdev c3 k3
stdev c4 k4
stdev c5 k5
print k2-k5
stack k2-k5 c10
let c11 = c10 * c10
let c12 = logten(c11)
let k11 = mean(c11)
let k12 = logten(k11)
print k11 – k12
#These are the standard deviations of the columns.
#These are the variances of the columns.
#This is the pooled variance when you have equal sized samples.
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252grass4-051 4/13/05 (Open this document in 'Page Layout' view!)
print c10 – c12.
Now you are on your own. The rest of this should be pretty easy because all your n j s are equal.
The Levene test is longer, but should be much more familiar and perhaps easier to fake.
Copy columns 1 through 5 to c21-c25. Then find their medians and subtract them from the columns and
convert the columns to absolute values.
let k22 = median(c22)
let k23 = median(c23)
let k24 = median (c24)
let k25 = median(c25)
let c22 = c22 - k22
let c23 = c23 - k23
let c24 = c24 - k24
let c25 = c25 - k25
describe c22-c25
print c21 – c25
let c22 = absolute(c22)
let c23 = absolute(c23)
let c24 = absolute(c24)
let c25 = absolute(c25)
print c21 – c25
#All the columns should have zero medians now.
You are now ready for an ANOVA using:
AOVO c2-c5
#You should get the same p-value as you got for the first Bartlett test
you did.
The Bartlett Test
This test seems to require that the underlying distribution be Normal. It should not be used to compare two
columns, since a simple F test described earlier is more appropriate. Recall that in the test for comparing 2
means with equal variances we used a pooled variance 2 n1  1s12  n2  1s22 . Assume that we have c
sˆ p 
n1  n2  2
columns representing c independent samples. Then the pooled variance would be
n  1s12  n2  1s 22  n3  1s32    nc  1s c2

. The test statistic used when there are 6 or more
s p2  1
n1  n 2  n3    nc  c
rows is  2
c 1

2.30259
d
 n  1logsˆ   n  1logs  where
j
2
p
j
2
j

1 
1
1


3c  1 
n j 1
n j  c 


For smaller examples (less than 6 rows) a special table is required and the instructions that I have found are
very confusing. Use the computer.
d  1


Bartlett Test Setup
MTB > stdev c2 k2
Standard Deviation of Des 1
Standard deviation of Des 1 = 10.2298
MTB > stdev c3 k3
Standard Deviation of Des 2
Standard deviation of Des 2 = 11.8180
MTB > stdev c4 k4
Standard Deviation of Des 3
Standard deviation of Des 3 = 6.56375
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MTB > stdev c5 k5
Standard Deviation of Des 4
Standard deviation of Des 4 = 7.76532
MTB > print k2-k5
Data Display
K2
K3
K4
K5
10.2298
11.8180
6.56375
7.76532
MTB
MTB
MTB
MTB
MTB
MTB
>
>
>
>
>
>
#These are the standard deviations of the columns
stack k2-k5 c10
let c11 = c10*c10
#These are the variances of the columns.
let c12 = logten(c11)
let k11 = mean(c11) #This is the pooled variance when you have equal sized samples.
let k12 = logten(k11)
print k11 k12
Data Display
K11
K12
86.9242
1.93914
MTB > print c10-c12
Data Display
Row
1
2
3
4
stdev
10.2298
11.8180
6.5638
7.7653
var
104.648
139.665
43.083
60.300
log
2.01973
2.14509
1.63430
1.78032
---------------------------------------------------------------------------------------------------------------2
2
2
2

s1  104 .648 s 2  139 .665 s 3  43 .083 s 4  60 .300


n 2  10
n3  10
n 4  10
 n1  10
n  1s12  n2  1s 22  n3  1s32    nc  1s c2  9104 .648  9139 .655  943 .083  960 .300

s p2  1
10  10  10  10  4
n1  n 2  n3    nc  c
 86 .9242

c 1 2.30259
1 
1
1

2

n j  1 log sˆ 2p 
n j  1 log s 2j where d  1 

d
3c  1 
n j 1
n j  c 


 
 1
2
  
   


1 4 1 
1  12 1 
1  1   1   1   1   1 
               1  15  9  36   1  15  36  36   1  .02037  1.0237
35  9   9   9   9   36 




c 1

2.30259
d
 n  1logsˆ   n  1logs 
j
2
p
j
2
j
2.30259
36 log86 .9242   9log104 .648   9log 139 .655   9log 43 .083   9 log 60 .300 
1.0237
2.30259
36 1.93914   92.01973   92.14509   91.63430   9 1.78032 

1.0237
2.30259
69 .80904  97.57944   2.30259 1.59408   3.58554 . (The computer got 3.51 – Where did I

1.0237
1.0237
go wrong?) This has c  1  4  1  3 degrees of freedom and the chi-squared table says that
3
 2 .05  7.8147. Since our computed chi-squared is less than the table chi-square, do not reject the null
hypothesis.

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252grass4-051 4/13/05 (Open this document in 'Page Layout' view!)
The Levene Test
This test is quite simple. It can be used for non-Normal data and can be used to compare two columns as
well as more than two columns.
(i) Find the median of each column. (This is the middle number or the average of the two middle numbers.)
(ii) Subtract the median of each column from the column from which it comes and take the absolute value of
the result.
(iii) Do a 1-way ANOVA on the result. If the results would lead you to reject the null hypothesis (because
the computed F is above the table F or the p-value is below your significance level), reject the null
hypothesis of equal variances.
Levene Test
MTB
MTB
MTB
MTB
MTB
MTB
MTB
MTB
MTB
MTB
MTB
MTB
MTB
>
>
>
>
>
>
>
>
>
>
>
>
>
let c22 = c2
let c23 = c3
let c24 = c4
let c25 = c5
let k22 = median(c22)
let k23 = median(c23)
let k24 = median (c24)
let k25 = median(c25)
let c22 = c22 - k22
let c23 = c23 - k23
let c24 = c24 - k24
let c25 = c25 - k25
describe c22-c25
Descriptive Statistics: C22, C23, C24, C25
Variable
C22
C23
C24
C25
N
10
10
10
10
N*
0
0
0
0
Mean
0.744
1.44
-0.860
0.108
Variable
C22
C23
C24
C25
Maximum
14.20
19.12
7.77
12.14
SE Mean
3.23
3.74
2.08
2.46
StDev
10.23
11.82
6.56
7.77
C24
-5.70
7.77
-1.35
5.17
-4.74
-8.94
1.35
2.09
-10.96
6.71
C25
-9.41
7.79
-2.03
-1.78
6.12
12.14
-9.77
5.04
-8.80
1.78
Minimum
-11.10
-11.37
-10.96
-9.77
Q1
-9.26
-9.80
-6.51
-8.95
Median
-1.42109E-14
0.000000000
0.000000000
0.000000000
Q3
10.17
11.05
5.55
6.54
MTB > print c21-c25
Data Display
Row
1
2
3
4
5
6
7
8
9
10
MTB
MTB
MTB
MTB
C21
>
>
>
>
let
let
let
let
C22
-9.225
8.305
-7.605
9.245
-9.355
14.205
-11.095
12.965
-5.895
5.895
c22
c23
c24
c25
=
=
=
=
C23
-11.37
8.67
-8.43
8.79
-9.50
19.12
-10.69
4.32
-4.32
17.82
absolute(c22)
absolute(c23)
absolute(c24)
absolute(c25)
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252grass4-051 4/13/05 (Open this document in 'Page Layout' view!)
MTB > print c22 - c25
Data Display
Row
1
2
3
4
5
6
7
8
9
10
C22
9.225
8.305
7.605
9.245
9.355
14.205
11.095
12.965
5.895
5.895
C23
11.37
8.67
8.43
8.79
9.50
19.12
10.69
4.32
4.32
17.82
C24
5.70
7.77
1.35
5.17
4.74
8.94
1.35
2.09
10.96
6.71
C25
9.41
7.79
2.03
1.78
6.12
12.14
9.77
5.04
8.80
1.78
MTB > AOVO c22-c25
One-way ANOVA: C22, C23, C24, C25
Source
Factor
Error
Total
DF
3
36
39
S = 3.741
SS
158.8
503.7
662.6
Level
C22
C23
C24
C25
MS
52.9
14.0
F
3.78
N
10
10
10
10
R-Sq = 23.97%
Mean
9.379
10.303
5.478
6.466
StDev
2.743
4.902
3.250
3.723
P
0.019
R-Sq(adj) = 17.64%
Individual 95% CIs For Mean Based on
Pooled StDev
--------+---------+---------+---------+(---------*--------)
(--------*---------)
(---------*---------)
(---------*--------)
--------+---------+---------+---------+5.0
7.5
10.0
12.5
Pooled StDev = 3.741
So anyway, here is our original data.
Row
1
2
3
4
5
6
7
8
9
10
Des 1
Des 2
Des 3
x1
x2
x3
206.32
223.85
207.94
224.79
206.19
229.75
204.45
228.51
209.65
221.44
203.81
223.85
206.75
223.97
205.68
234.30
204.49
219.50
210.86
233.00
217.08
230.55
221.43
227.95
218.04
213.84
224.13
224.87
211.82
229.49
Des 4
x4
218.90
236.10
226.28
226.53
234.43
240.45
218.54
233.35
219.51
230.09
(i) Find the median of each column. (This is the middle number or the average of the two middle numbers.)
Let’s put the numbers in order so we can see the two middle numbers.
Row
x1
x2
x3
x4
1
2
3
4
5
6
7
8
9
10
204.45
206.19
206.32
207.94
209.65
221.44
223.85
224.79
228.51
229.75
203.81
204.49
205.68
206.75
210.86
219.50
223.85
223.97
233.00
234.30
211.82
213.84
217.08
218.04
221.43
224.13
224.87
227.95
229.49
230.55
218.54
218.90
219.51
226.28
226.53
230.09
233.35
234.43
236.10
240.45
The medians for the four columns are 215.545, 215.180, 222.780 and 228.310.
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252grass4-051 4/13/05 (Open this document in 'Page Layout' view!)
(ii) Subtract the median of each column from the column from which it comes and take the absolute value of
the result.
If we take the original numbers and subtract the medians, we get the following.
Row
x1
x2
x3
x4
1
2
3
4
5
6
7
8
9
10
-9.225
8.305
-7.605
9.245
-9.355
14.205
-11.095
12.965
-5.895
5.895
-11.37
8.67
-8.43
8.79
-9.50
19.12
-10.69
4.32
-4.32
17.82
-5.70
7.77
-1.35
5.17
-4.74
-8.94
1.35
2.09
-10.96
6.71
-9.41
7.79
-2.03
-1.78
6.12
12.14
-9.77
5.04
-8.80
1.78
If we remove the signs, our result is below.
Row
x1
x2
x3
x4
1
2
3
4
5
6
7
8
9
10
9.225
8.305
7.605
9.245
9.355
14.205
11.095
12.965
5.895
5.895
11.37
8.67
8.43
8.79
9.50
19.12
10.69
4.32
4.32
17.82
5.70
7.77
1.35
5.17
4.74
8.94
1.35
2.09
10.96
6.71
9.41
7.79
2.03
1.78
6.12
12.14
9.77
5.04
8.80
1.78
(iii) Do a 1-way ANOVA on the result. If the results would lead you to reject the null hypothesis (because
the computed F is above the table F or the p-value is below your significance level), reject the null
hypothesis of equal variances.
Row
1
2
3
4
5
6
7
8
9
10
Totals
x1
x2
x3
x4
9.225
8.305
7.605
9.245
9.355
14.205
11.095
12.965
5.895
5.895
11.37
8.67
8.43
8.79
9.50
19.12
10.69
4.32
4.32
17.82
5.70
7.77
1.35
5.17
4.74
8.94
1.35
2.09
10.96
6.71
9.41
7.79
2.03
1.78
6.12
12.14
9.77
5.04
8.80
1.78
Sum
93.790
103.03
54.78
64.66
316.26 
nj
10
10
10
40  n
10
x j
9.379
10.303
5.478
6.466
SS
947.370
1277.75
395.142
542.818
x 2j
87.9656 106.0900
30.0085
41.8092
ij
316 .26
 7.9065  x
40
 x
265.9351   x
3163.08 
2
ij
2
j
2   xij2  nx 2  3163 .08  407.9065 2  662 .57
2
2
2
2
2
2
2
2
. j  x    n j x. j  nx  10 9.397   10 10 .303   10 5.478   10 6.466   40 7.9065 
 x
SSB   x
SST 
 x
ij
x
 10 265 .9351   40 7.9065 2  158 .84
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252grass4-051 4/13/05 (Open this document in 'Page Layout' view!)
Source
SS
Between
158.84
DF
3
MS
52.95
F
F.05
3.78
3,36  2.87 s
F.05
H0
Column means equal
Within
503.73
36
13.9925
Total
662.57
39
Our rejection of the null hypothesis at the 5% level means that we reject the hypothesis of equal variances.
As far as the official answer to the problem 11.14 is concerned, here is the solution in the Instructor’s
Solution Manual. It doesn’t resemble my results at all, so I can’t wait to see what you got.
11.14 (a)
To test at the 0.05 level of significance whether there is any evidence of a difference in
the average distance traveled by the golf balls differing in design, we conduct an F test:
H0: 1   2  3   4
H1: At least one mean is different.
(b)
Decision rule: df: 3, 36. If F > 2.866, reject H0.
ANOVA
Source of
SS
df
MS
F
P-value
F crit
Variation
Between
2990.99
3 996.9966 53.02982 2.73E-13 2.866265
Groups
Within
676.8244
36 18.80068
Groups
Total
3667.814
39
Since Fcalc = 53.03 is above the critical bound of F = 2.866, reject H0. There is enough
evidence to conclude that there is significant difference in the average distance traveled
by the golf balls differing in design.
To determine which of the means are significantly different from one another, we use the
Tukey-Kramer procedure to establish the critical range:
QU(c, n – c) = QU(4, 36). We use QU(4, 40) = 3.79
critical range
= QU ( c ,n c ) 
MSW
2
1
1 
18.8007  1 1  =
     3.79 
  
n n 
2
 10 10 
j
j
'


5.1967
Tukey Kramer Multiple
Comparisons
Sample
Sample
Group
Mean
Size
Comparison
10 Group 1 to Group 2
1 206.614
(c)
Absolute
Difference Results
11.902 Means are
different
10 Group 1 to Group 3
19.974 Means are
2 218.516
different
10 Group 1 to Group 4
22.008 Means are
3 226.588
different
10 Group 2 to Group 3
8.072 Means are
4 228.622
different
10.106 Means are
Group 2 to Group 4
different
2.034 Means are
MSW 18.800677
Group 3 to Group 4
not different
At 5% level of significance, there is enough evidence to conclude that average
traveling distances between all pairs of designs are different with the only exception of
the pair between design 3 and design 4.
The assumptions needed in (a) are (i) samples are randomly and independently
15
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11.14
cont.
(d)
drawn, (ii) populations are normally distributed, and (iii) populations have equal
variances.
To test at the 0.05 level of significance whether the variation within the groups is
similar for all groups, we conduct a Levene's test for homogeneity of variance:
H0:
(e)
 12   22   32   42
H1: At least one variance is different.
ANOVA
Source of
SS
df
MS
F
P-value
F crit
Variation
Between
40.63675
3 13.54558 2.093228 0.118276 2.866265
Groups
Within
232.9613
36 6.471147
Groups
Total
273.598
39
Since p-value = 0.1182 > 0.05, do not reject the null hypothesis. There is not enough
evidence to conclude that there is any difference in the variation of the distance traveled
by the golf balls differing in design.
In order to produce golf balls with the furthest traveling distance, either design 3 or 4 can
be used.
16