252solnF1 11/7/05
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F. ANALYSIS OF VARIANCE
1. 1-Way Analysis of Variance
Text 11.1-11.6, 11.7**, 11.8 [11.1- 11.7, 11.8*] (11.1- 11.7, 11.8* (Same problem, different numbers – both answers will be posted)
2. 2 -Way Analysis of Variance
Text 11.15-11.18, 11.23, 11.29-11.32, 11.36 [11.15-11.18, 11.23, 11.28-11.30, 11.34] (11.15-11.18, 11.23, 11.28-11.30, 11.34), F1,
F2, F4
3. More than 2-Way analysis of Variance
F3
4. Kruskal-Wallis Test
Text 12.86-12.87, 12.89 [11.39-11.40, 11.42] (11.39-11.40, 11.42), Downing and Clark 18-12, 18-13 (in chapter 17 in D&C 3rd
edition),
5. Friedman Test
Text 12.93-12.95 [11.46-11.48] (11.65-11.67 on CD) Downing and Clark 18-4, 18-6 (in chapter 17 in D&C 3rd edition),
Graded Assignment 4 (Will be posted)
This document includes exercises 11.1 - 11.8
------------------------------------------------------------------------------------------------------------------------------------------------------------------
1-Way ANOVA Problems.
Exercise 11.1 (11.1 in 8th edition):
Solution: The table in the outline reads:
Source
SS
Between
SSB
DF
MS
dfB
SSB
MSB 
 m 1
m 1
Within
SSW
dfW
SSW
MSW 
 nm
nm
Total
SST
F
F
MSB
MSW
dfT
 n 1
We use B(Between) in dfB for the text’s less-standard A(among) in dfA . m  5 levels(columns), and
n1  n 2  n3  n 4 n 5  7 . So n 
n
i
 57   35 . dfB  m  1  5  1  4. dfT  n  1  35  1  34.
dfW  n  m is best found as a residual.
If we fill in the table with the numbers above we get:
Source
SS
DF
MS
Between
Within
Total
.
F
F.05
H0
4
30
34
1
252solnF1 11/7/05
Exercise 11.2(11.2 in 8th edition): The table now includes the given values. SSB  60 and SST  210 .
Source
SS
DF
MS
F.05
H0
F
Between
Within
Total
60
.
210
4
30
34
SSW = SST – SSB = 210 – 60 = 150 is the missing number.
SSB
60
(b)
MSB 

 15
m 1 5 1
SSW 150
(c)
MSW 

 5.00
n  m 30
MSB 15
F
3
(d)
=
MSW 5
The table now reads as below.
Source
SS
DF
MS
F.05
F
(a)
Between
Within
Total
60
150
210
4
30
34
15
5
3.00
H0
Column means equal
Exercise 11.3(11.3 in 8th edition): (a)Take the table above and add the appropriate table value of F and the
null hypothesis.
Source
SS
DF
MS
F.05
H0
F
Between
Within
Total
(b)
60
150
210

4,30
F .05 2.69
4
15
30
34
5
3.00 s

F 4,30
.05 2.69
Column means equal
(c)
Decision rule: If F > 2.69, reject H0.
(d)
Decision: Since Fcalc = 3.00 is above the critical bound F = 2.69, reject H0. The s for
‘significant’ shows regression.
Exercise 11.4(11.4 in 8th edition): We use B(Between) in dfB for the text’s less-standard A(among) in
dfA . m  3 levels(columns), and n1  n2  n3  7 . So n 
(b)
dfB  m  1  3  1  2.
dfW  n  m  21  3.
(c)
dfT  n  1  35  1  34.
(a)
n
i
 37   21 .
Exercise 11.5(11.5 in 8th edition): The solution is repeated, edited, from the Instructor’s Solution Manual .
11.5
Source
df
SS
MS
F
Between groups
4 – 1 =3
80 x (3) = 240
80
80  20 = 4
Within groups
32 – 4 = 28
560
560  28 = 20
Total
32 – 1 = 31
240 + 560 = 800
m  3. n1  n2  n3  n4  8 . n  48  32.
2
252solnF1 11/7/05
Exercise 11.6(11.6 in 8th edition): The solution is repeated, edited, from the Instructor’s Solution Manual .

11.6
(a)-(b) F 3,28
.05 2.95 Decision rule: If F > 2.95, reject H0.
(c)
(d)
(e)
Decision: Since Fcalc = 4.00 is above the critical bound of F = 2.95, reject H0. There is
enough evidence to conclude that the four group means are not all the same.
To perform the Tukey-Kramer procedure, we use m  4 in the numerator and n – m =
32 – 4 = 28 degrees of freedom in the denominator.
Since Table E.7 or Table 21 in the syllabus supplement does not contain a value for 4 and
28 degrees of freedom, we will use the next larger (and more conservative) value for 4
4, 24
and 24 degrees of freedom as an upper bound. q.05
 3.90 (On the other hand, a guess
4, 24
4,30
that the value of the number we want is between q.05
 3.90 and q.05
 3.85 and
(f)
thus about 3.875 would work here.).
From the outline we get the following confidence intervals..
i. A Single 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
, where s  MSW and n  m  dfW .

n1 n 2
ii. Scheffé Confidence Interval. If we desire intervals that will simultaneously be valid
For a given confidence level for all possible intervals between column means,

1
1 
.
1   2  x1  x2   m  1Fm1,nm   s

 n
n 2 
1

iii. Bonferroni Confidence Interval
If we only need k different intervals, use
1   2  x1  x2   t n  m  s
1
1

n1 n 2
2k
iv. Tukey Confidence Interval
This also applies to all possible differences.
1   2  x1  x2   q m,n  m 
s
1
1
This gives rise to Tukey’s HSD

n1 n 2
2
(Honestly Significant Difference) procedure. Two sample means x .1 and x .2 are
significantly different if x.1  x.2 is greater than
1  2  qm, n  m 
s
2
1
1

n1 n2
All of these could be used, but the Tukey is most popular. dfW  n  m  28. s  MSW  20 ,
3,28  2.95 and   .05 . s
28
t .025
 2.048 , F.05
1
1
1 1

 20     5  2.236
n1 n 2
8 8
i. A Single Confidence Interval
1   2  x1  x2   t n  m  s
2
1
1

 x1  x2   2.048 2.236 
n1 n 2
 x1  x2   4.579 . So the difference between two means is significant if it is
above 4.579.
3
252solnF1 11/7/05
ii. Scheffé Confidence Interval.
1   2  x1  x2  

m  1Fm1,nm  s

1
1 

n1 n 2 
 x1  x2   32.95 2.236   x1  x2   6.652 . So the difference between
two means is significant if it is above 6.652.
iii. Bonferroni Confidence Interval
1   2  x1  x2   t n  m  s
2k
1
1
- Similar to the single confidence

n1 n 2
interval with a larger and hard-to find t.
iv. Tukey Confidence Interval
1   2  x1  x2   q m,n  m 
1
1
1
2.236 
 x1  x2   3.90 

2
2 n1 n 2
 x1  x2   6.166 So the difference between
two means is significant if it is above 6.166. This is Tukey’s HSD test and the
text calls 6.166 a critical range.
s
Exercise 11.7 in 10th edition [for 9th and 8th editions, see next problem. The solution is repeated, edited,
from the Instructor’s Solution Manual .
11.7 (a) The problem, which involves the Computer Anxiety Rating Scale gives the DF and SS
columns below. You will compute MS by dividing SS by DF. You will then get a computed F statistic
by dividing MSB by MSW.
One-Way ANOVA
Source
DF
SS
MS
F Statistic
Between
5
3172
634.4
4.9567
?
Majors
Within
166
21246 127.9880
Majors
Total
171
24418
Note that F has 5 and 166 degrees of freedom. The problem says   .05 . It also gives means and
values of n for different majors as shown on the Tukey-Kramer table below. The solution given has a
5,166  2.2686 . This cannot be found exactly on a table. If you are working
reference value of F of F.05
from my table go to page 11 (or 13). Look in the 5 column . The closest denominator df’s that you can
5,125  2.29 and F 5,200  2.26. You can guess that the
find are 125 and 200. You should find F.05
.05
number you want is 2.27 or 2.28. Since your computed F is larger than either of these, reject your null
hypothesis.
(b)
H0:
 A  B  C  D  E
H1: At least one mean is different.
F5,166 = 2.2686. Since F = 4.9567 > 2.2686, reject H0. There is enough evidence to
conclude there is a difference in the mean computer anxiety experienced by different
majors.
To determine which of the means are significantly different from one another, we use the
5, 
 4.03
Tukey-Kramer procedure to establish the critical range: We use q.m,nm  q.05
4
252solnF1 11/7/05
The ‘critical range’ for Marketing and Management is
q m, n  m 
s
2
1
1
127 .9880

 4.03
n1 n 2
2
1 1
    4.03 9.1857  4.933.0308   12 .214
 19 11 
The absolute difference between the means is 44.37  43.18  1.19 . Since 1.19 is less
than the critical range, we cannot say that there is a significant difference between the
means for marketing and management. Since there are six majors, the number of possible
6! 6  5

 15 . Partial PHStat output for Tukey-Kramer multiple
comparisons is C 26 
4!2! 2.1
comparison appears below. Note that the absolute difference of 1.19, the standard error of
3.0308 and the critical value of 12.21 appear in the first row on the right side of the table.
(c)
Tukey Kramer Multiple
Comparisons
Sample Sample
Group
Mean
Size
Comparison
Marketing 1 44.37
19 Group 1 to
Group 2
Management
43.18
11 Group 1 to
2
Group 3
Other 3
42.21
14 Group 1 to
Group 4
Finance 4
41.8
45 Group 1 to
Group 5
Accounting 5 37.56
36 Group 1 to
Group 6
MIS 6 42.21
47 Group 2 to
Group 3
Group 2 to
Group 4
Group 2 to
Other Data
Group 5
Group 2 to
Level of
0.05
Group 6
significance
Group 3 to
Numerator d.f.
6
Group 4
Group 3 to
Denominator
166
Group 5
d.f.
Group 3 to
MSW
127.988
Group 6
Group 4 to
Q Statistic
4.03
Group 5
Group 4 to
Group 6
Group 5 to
Group 6
Absolute
Std. Error
Critical
Difference of Difference Range Results
1.19
3.03079827
12.21 Means are
not different
2.16
2.81764126
11.36 Means are
not different
2.57
2.18865081
8.82 Means are
not different
6.81
2.26841672
9.142 Means are
not different
2.16
2.17478228
8.764 Means are
not different
0.97
3.22314015
12.99 Means are
not different
1.38
2.69067325
10.84 Means are
not different
5.62
2.75594714
11.11 Means are
not different
0.97
2.67940444
10.8 Means are
not different
0.41
2.44807815
9.866 Means are
not different
4.65
2.51964456
10.15 Means are
not different
0
2.43568722
9.816 Means are
not different
4.24
1.78877019
7.209 Means are
not different
0.41
1.66843109
6.724 Means are
not different
4.65
1.77177436
7.14 Means are
not different
The Tukey-Kramer procedure does not show any pair-wise significant difference. The
inconsistency of the results obtained in (a) and (c) could possibly be due to the violations of any of the
assumptions needed.
Exercise [11.7 in 9th edition] (11.7 in 8th edition): The solution is repeated, heavily edited, from the
Instructor’s Solution Manual .
ni  520   100 . dfB  m  1  5  1  4.
n1  n 2  n3  n 4 n 5  20 . So n 
(a)
dfT  n  1  100  1  99.

5
252solnF1 11/7/05
H 0 : 1  2  3  4  5
where 1 = Wendy, 2 = McDonald,
3 = Checkers, 4 = Burger King, 5 = Long John Silver
H1 : Not all  j are equal
where j = 1, 2, 3, 4, 5
Decision Rule: If p-value < 0.05, reject H0.
One-Way ANOVA
Source
DF
SS
MS
F Statistic
p Value
Between
4
6536
1634 12.51148545 3.24067E-08
Within
95
12407
130.6
Total
99
18943
4,95  2.70 . There is
Since p-value is virtually 0, reject H0. Or note that Fcalc  F.05
(b)
sufficient evidence to conclude that there is a difference in the average drive-through
time of the 5 chains.
To determine which of the means are significantly different from one another, we use the
5,95
 3.92
Tukey-Kramer procedure to establish the critical range: We use q.m,nm  q.05
1
1
130 .6  1
1 

 3.92
    10 .017
n
n
2
20
20


2
1
2
Note that the means in this table are given by the problem statement. Partial PHStat
output for Tukey-Kramer multiple comparison follows.
The ‘critical range’ is q m,n  m 
s
Sample Sample
Absolute
Mean
Size
Comparison Difference Results
1
150
20 Group 1 to
17 Means are
Group 2
different
2
167
20 Group 1 to
19 Means are
Group 3
different
3
169
20 Group 1 to
21 Means are
Group 4
different
4
171
20 Group 1 to
22 Means are
Group 5
different
5
172
20 Group 2 to
2 Means are not
Group 3
different
Group 2 to
4 Means are not
Group 4
different
Group 2 to
5 Means are not
Group 5
different
Group 3 to
2 Means are not
Group 4
different
Group 3 to
3 Means are not
Group 5
different
Group 4 to
1 Means are not
Group 5
different
From the above PHStat output, the drive-through times are different in the pairs between
Wendy and each of the other 4 chains.
From the sample averages, Wendy has the shortest drive-through time at 150 seconds.
Group
(c)
6
252solnF1 11/7/05
Exercise 11.8(11.8 in 8th edition): The solution is repeated, edited almost beyond recognition, from the
Instructor’s Solution Manual . We start with the problem as stated in the 9th edition. It is much more
valuable to study that the answer in the 8th edition.
H 0 : 1  2  3
H1 : Not all  j are equal
(a)
where 1 = Experts, 2 = Readers, 3 = Darts
where j = 1, 2, 3
You should be able to do the calculations below. Only the three columns of numbers were given to us.
Sum
1
39.5
-1.1
-4.5
-8.0
Sample
2
-31.0
-20.7
-45.0
-73.3
3
39.0
31.9
14.1
5.4
25.9 +
-170.0 +
90.4
nj
4+
x j
Sum
4+
4
6.475
-42.500
22.600
SS
1645.7 +
8787.4 +
2766.6
x 2j
41.925625 
1806.250000 +
 53 .7 
2
ij
2
j .j
53 .7
 4.475  x
12
 13199 .7 
x ij2

 2358 .935625   x
510.760000
2
j
2
2
2
2
ij
 12  n
 x  nx  13199.7 12 4.475  12959.39
SSB   n x  nx  46.475  4 42.5  422.60  12 4.475
SST 
 x
2
2
2
 42358.935625  124.4752
= 9435.74253 – 240.3075 = 9195.44
Source
Between
Within
Total
SS
DF
9195.44
2
3763.95
12959.39
9
11
MS
4597.72
F
F.05
10.99 s
F 2,9  4.26
H0
Column means equal
418.2167
Minitab output follows. Decision Rule: If p-value < 0.05, reject H0.
Results for: 252CONTEST2001.mtw
MTB > Retrieve "C:\Berenson\Data_Files-9th\Minitab\CONTEST2001.mtw".
Retrieving worksheet from file: C:\Berenson\Data_Files9th\Minitab\CONTEST2001.mtw
# Worksheet was saved on Fri Jan 18 2002
Results for: 252CONTEST2001.mtw
MTB > AOVOneway c1 c2 c3.
One-way ANOVA: Experts, Readers, Darts
Analysis of Variance
Source
DF
SS
Factor
2
9195
Error
9
3764
Total
11
12959
MS
4598
418
F
10.99
P
0.004
7
252solnF1 11/7/05
Level
Experts
Readers
Darts
N
4
4
4
Pooled StDev =
Mean
6.48
-42.50
22.60
StDev
22.20
22.82
15.53
20.45
Individual 95% CIs For Mean
Based on Pooled StDev
---------+---------+---------+------(------*-----)
(------*-----)
(-----*------)
---------+---------+---------+-------35
0
35
Since p-value = 0.004 < 0.05, reject the null hypothesis. There is enough evidence to
conclude that there is a significant difference in the average returns for the three
categories.
(b)
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(3, 9) = 3.95
Data was rerun using Minitab. To use comparison procedures, the data had to be stacked
with all data in column 5 and identifiers in column 6. The data is shown in the first data display. The
command to start ANOVA is now ‘Oneway c5 c6;’ with the subcommand ‘Tukey 5.’ to do a Tukey
procedure with a 5% significance level.
————— 11/5/2003 8:17:19 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > Retrieve "C:\Documents and Settings\RBOVE.WCUPANET\My Documents\Drive
D\MINITAB\252CONTEST2001.mtw".
Retrieving worksheet from file: C:\Documents and Settings\RBOVE.WCUPANET\My
Documents\Drive D\MINITAB\252CONTEST2001.mtw
# Worksheet was saved on Wed Nov 05 2003
Results for: 252CONTEST2001.mtw
MTB > Stack c1 c2 c3 c5;
SUBC>
Subscripts c6;
SUBC>
UseNames.
MTB > print c5 c6
#This stacks the data in column 5 with column names
#in C6.
Data Display
#These are the original numbers given above.
Row
1
2
3
4
5
6
7
8
9
10
11
12
C5
39.5
-1.1
-4.5
-8.0
-31.0
-20.7
-45.0
-73.3
39.0
31.9
14.1
5.4
C6
Experts
Experts
Experts
Experts
Readers
Readers
Readers
Readers
Darts
Darts
Darts
Darts
MTB > Oneway c5 c6;
SUBC>
Tukey 5.
One-way ANOVA: C5 versus C6 #This is the same ANOVA as above.
Analysis of Variance for C5
Source
DF
SS
C6
2
9195
Error
9
3764
Total
11
12959
Level
Darts
Experts
Readers
N
4
4
4
Mean
22.60
6.48
-42.50
MS
4598
418
StDev
15.53
22.20
22.82
F
10.99
P
0.004
Individual 95% CIs For Mean
Based on Pooled StDev
---------+---------+---------+------(-----*------)
(------*-----)
(------*-----)
8
252solnF1 11/7/05
Pooled StDev =
---------+---------+---------+-------35
0
35
20.45
Tukey's pairwise comparisons
Family error rate = 0.0500
Individual error rate = 0.0209
Critical value = 3.95
Intervals for (column level mean) - (row level mean)
Darts
Experts
Experts
-24.26
56.51
Readers
24.71
105.49
(c)
(d)
8.59
89.36
We now have 5% Tukey confidence intervals for the three means note that the interval
for mean of Experts – mean of Darts is -24.26 to 56.51, which includes zero. Since the
interval includes zero, the difference is not significant. At 5% level of significance, the
Tukey Kramer multiple comparison test shows that there is enough evidence to conclude
that Experts and Readers, and Readers and Darts differ in average return.
The data collected are the returns of the selected stocks by the 3 categories not the
amount of drops compared to the previous returns of the stocks. Even if average returns
are concerned, the experts have the sample average return of 6.475% while the readers
have a sample average return of –42.5%, and the stocks chosen using the darts have a
sample average return of 22.6%. However, these differences in sample averages do not
lead to the conclusion of significant differences between the Expert and Darts in
population averages according to the result of the Tukey-Kramer multiple comparison
procedure in part (b).
H0:
 12   22   32
H1: At least one variance is different.
The procedure with the stacked data was continued by running Levene’s and Bartlett’s
tests. Levene’s test is described in the text. The important thing to note here is the pvalues.
MTB > %Vartest c5 c6;
SUBC>
Confidence 95.0.
Executing from file: W:\wminitab13\MACROS\Vartest.MAC
Macro is running ... please wait
Test for Equal Variances
Response
Factors
ConfLvl
C5
C6
95.0000
Bonferroni confidence intervals for standard deviations
Lower
Sigma
Upper
N
Factor Levels
7.8509
11.2208
11.5367
15.5300
22.1962
22.8209
84.464
120.720
124.118
4
4
4
Darts
Experts
Readers
9
252solnF1 11/7/05
Bartlett's Test (normal distribution)
Test Statistic: 0.437
P-Value
: 0.804
Levene's Test (any continuous distribution)
Test Statistic: 0.101
P-Value
: 0.905
Test for Equal Variances: C5 vs C6
Test for Equal Variances for C5
95% Confidence Intervals for Sigmas
Factor Levels
Darts
Bartlett's Test
Test Statistic: 0.437
P-Value
: 0.804
Experts
Levene's Test
Test Statistic: 0.101
P-Value
: 0.905
Readers
0
50
100
Since, in the Levene test, p-value = 0.905 > 0.05, do not reject H0. There is not enough
evidence of a significant difference in the variation in the return for the three categories.
This validates the use of ANOVA.
Now let’s try the version of the problem in the 8 th edition:
H 0 : 1  2  3
11.8
(a)
H1 : Not all  j are equal
where 1 = Experts, 2 = Readers, 3 = Darts
where j = 1, 2, 3
b) You should be able to do the calculations below. Only the three columns of numbers were given to
us.
Sample
Sum
1
2
3
23.4
54.0
3.9
21.6
2.4
-21.8
-25.3
-45.9
-46.8
-45.9
-79.8
-96.2
 256 .4 
x ij
Sum
-26.2 +
-69.3 +
-160.9

nj
4+
4+
4
x j
-6.55
-17.325
-40.225
SS
3761.0 +
11397.0 +
11935.0
x 2j
42.9025 
300.1556 +
1618.0506
 12  n
256 .4
 21 .367  x
12
 27093 .0 
x ij2

 1961 .1087   x
2
j
10
252solnF1 11/7/05
 x  nx  2709312 21.367  21614.59
SSB   n x  nx  4 6.55  4 17.325  440.225  12 21.350
SST 
2
ij
2
2
2
j .j
2
2
2
2
2
 4191.1087  1221.3672
= 7844.43 –5478.41 = 2366.02
Source
SS
Between
DF
2366.02
2
MS
1183.01
F
F.05
0.55 ns
F 2,9  4.26
H0
Column means equal
Within
19248.57
9
2138.73
Total
21614.59
11
Note that the differences between the means for this version are not significant.
The data were run on Minitab.
Decision Rule: If p-value < 0.05, reject H0.
Results for: 252CONTEST.MTW
MTB > print c1-c3
Data Display
Row
Experts
Readers
Darts
1
2
3
4
23.4
21.6
-25.3
-45.9
54.0
2.4
-45.9
-79.8
3.9
-21.8
-46.8
-96.2
One-way ANOVA: Experts, Readers, Darts
Analysis of Variance
Source
DF
SS
Factor
2
2366
Error
9
19248
Total
11
21614
Level
Experts
Readers
Darts
N
4
4
4
Mean
-6.55
-17.33
-40.23
Pooled StDev =
MTB >
Variable
Experts
Readers
Darts
Variable
Experts
Readers
Darts
(b)
MS
1183
2139
StDev
34.59
58.30
42.67
46.25
F
0.55
P
0.594
Individual 95% CIs For Mean
Based on Pooled StDev
----+---------+---------+---------+-(------------*------------)
(------------*------------)
(------------*------------)
----+---------+---------+---------+--80
-40
0
40
N
4
4
4
Mean
-6.6
-17.3
-40.2
Median
-1.8
-21.8
-34.3
TrMean
-6.6
-17.3
-40.2
Minimum
-45.9
-79.8
-96.2
Maximum
23.4
54.0
3.9
Q1
-40.8
-71.3
-83.9
Q3
23.0
41.1
-2.5
StDev
34.6
58.3
42.7
SE Mean
17.3
29.1
21.3
Since p-value = 0.594 > 0.05, do not reject the null. There is not enough evidence to
conclude that there is a significant difference in the average returns for the three
categories.
From (a), none of the 3 categories differ in average return. Nevertheless, the whole thing
was rerun with stacked data below.
11
252solnF1 11/7/05
(c)
(d)
The data collected are the returns of the selected stocks by the 3 categories not the
amount of drops compared to the previous returns of the stocks. Even if average returns
are concerned, the experts have the smallest sample average return of -6.55% while the
readers have a sample average return of -17.325%, and the stocks chosen using the darts
have a sample average return of -40.225%. However, these differences in sample
averages do not lead to the conclusion of significant differences in population averages
according to the results of the one-way ANOVA F test and the Tukey-Kramer multiple
comparison procedure.
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 variances:
H 0 :  12   22   32
where 1 = Experts, 2 = Readers, 3 = Darts
H1 : Not all  2j are equal
where j = 1, 2, 3
Since the version of Minitab available to the Authors of the 8th edition Instructor’s
Solution Manual evidently did not have the Levene method available, they did the whole
drill mentioned in the text and got the results below.
Decision Rule: If p-value < 0.05, reject the null hypothesis.
ANOVA
Source of
Variation
Between
Groups
Within
Groups
SS
df
MS
F
P-value
F crit
639.2517
2 319.6258 0.723278 0.511343 4.256492
3977.215
9 441.9128
Total
4616.467
11
Since p-value = 0.5113 > 0.05, do not reject the null hypothesis. There is not enough
evidence to conclude that there is any difference in the variances.
Minitab output for the unstacked version of the problem follows.
————— 11/5/2003 9:32:18 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > Retrieve "C:\Documents and Settings\RBOVE.WCUPANET\My Documents\Drive
D\MINITAB\252CONTEST.MTW".
Retrieving worksheet from file: C:\Documents and Settings\RBOVE.WCUPANET\My
Documents\Drive D\MINITAB\252CONTEST.MTW
# Worksheet was saved on Wed Nov 05 2003
Results for: 252CONTEST.MTW
MTB > Stack c1 c2 c3 c5;
SUBC>
Subscripts c6;
SUBC>
UseNames.
12
252solnF1 11/7/05
MTB > print c5 c6
Data Display
Row
C5
C6
1
2
3
4
5
6
7
8
9
10
11
12
23.4
21.6
-25.3
-45.9
54.0
2.4
-45.9
-79.8
3.9
-21.8
-46.8
-96.2
Experts
Experts
Experts
Experts
Readers
Readers
Readers
Readers
Darts
Darts
Darts
Darts
MTB > Oneway c5 c6;
SUBC>
Tukey 5.
One-way ANOVA: C5 versus C6
Analysis of Variance for C5
Source
DF
SS
C6
2
2366
Error
9
19248
Total
11
21614
Level
Darts
Experts
Readers
N
4
4
4
Mean
-40.23
-6.55
-17.33
Pooled StDev =
MS
1183
2139
StDev
42.67
34.59
58.30
46.25
F
0.55
P
0.594
Individual 95% CIs For Mean
Based on Pooled StDev
----+---------+---------+---------+-(------------*------------)
(------------*------------)
(------------*------------)
----+---------+---------+---------+--80
-40
0
40
Tukey's pairwise comparisons
Family error rate = 0.0500
Individual error rate = 0.0209
Critical value = 3.95
Intervals for (column level mean) - (row level mean)
Darts
Experts
-125.0
57.7
Readers
-114.2
68.4
Experts
-80.6
102.1
MTB > %Vartest c5 c6;
SUBC>
Confidence 95.0.
Executing from file: W:\wminitab13\MACROS\Vartest.MAC
Macro is running ... please wait
Test for Equal Variances
Response
Factors
C5
C6
13
252solnF1 11/7/05
ConfLvl
95.0000
Bonferroni confidence intervals for standard deviations
Lower
Sigma
Upper
N
Factor Levels
21.5724
17.4863
29.4714
42.6729
34.5900
58.2980
232.089
188.128
317.071
4
4
4
Darts
Experts
Readers
Bartlett's Test (normal distribution)
Test Statistic: 0.728
P-Value
: 0.695
Levene's Test (any continuous distribution)
Test Statistic: 0.723
P-Value
: 0.511
Test for Equal Variances: C5 vs C6
Test for Equal Variances for C5
95% Confidence Intervals for Sigmas
Factor Levels
Darts
Bartlett's Test
Test Statistic: 0.728
P-Value
: 0.695
Experts
Levene's Test
Test Statistic: 0.723
P-Value
: 0.511
Readers
0
100
200
300
Notice that this time all the confidence intervals include zero, verifying the fact that none of them are
significant. Notice that the p-value for the Levine test is identical to the version that appeared in the
Instructor’s Solution Manual . Also note that the p-value for the two tests are well out of the ‘reject’ zone.
14
252solnF1 11/7/05
This problem is worth studying since it is carried through to get all types of intervals except the Tukey
interval.
Exercise 14.22( From the McClave et. al. text): a) Make a diagram with an x axis marked off with
numbers between 0 and 800. If you use different symbols for the three samples, you will notice that sample
A has a very different distribution from samples B and C. the sample means also differ, so it looks like
there is a significant difference between the population means.
b) You should be able to do the calculations below. I will use 1, 2, and 3 instead of A, B and C. Only
the three columns of numbers were given to us.
Sum
nj
1
250
150
275
100
475
600
150
800
325
230
Sample
2
100
150
75
200
55
80
110
160
132
233
3
80
125
20
186
52
92
88
141
76
200
Sum
3355 +
1295 +
1060
 5710 
10 +
10 +
10
 30  n
129.5
106.0
x j
335.5
SS
1577275 +
196963 +
141590
x 2j
112560.25 
16770.25 +
11236.00
2
ij
2
j .j
 x
 1405665 .5   x
 1915828 
2
ij
2
j
2
2
2
2
ij
5710
 190 .3333  x
30
 x  nx  1915828 30190.3333  829024.667
SSB   n x  nx  10335.5  10129.5  10106.0  30190.3333
SST 
 x
2
2
2
 101405665.0  30190.33332
=318861.667
Source
SS
DF
MS
F
Between
318861.667
2
159430.833
8.436
Within
Total
510163.000
829124.667
27
29
18894.926
F.05
F 2, 27  3.35 s
H0
Column means equal
c) We reject the null hypothesis and conclude that the different classes of firms have different audit costs.
We use the p-value of .0014 in the printout in the text and we reject the null hypothesis because it is below
the 5% significance level given in the problem.
d) We assume that we have three random samples taken from Normal populations with similar variances.
15
252solnF1 11/7/05
e) I had asked that you compute all appropriate confidence intervals, using the formulas in the outline. But
before we get to these contrasts, try an interval for a single mean. Note for all the below from the ANOVA
27
table. n  m  27 , so if   .05, tnm  t.025
 2.052 , n1  n 2  n3  10, and
2
s  MSW  18894.926  137.459 . So for a single mean  j  x. j  t n  m 
2
1  335 .5  2.052
1  106 .0  2.052
137 .459
10
137 .459
 335 .5  89 .2 ,  2  129 .5  2.052
137 .459
s
.
nj
 129 .5  89 .2 ,
10
 106 .0  89 .2 .
10
The contrasts from the outline are quoted and filled in below.
i. A single 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
. so

n1 n 2
1   2  335 .5  129 .5  2.052 137 .459 
1
1

 206  126 .1
10 10
 2   3  129 .5  106 .0  126 .1 , 1   3  335 .5  106 .0  126 .1
ii. Scheffé 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
. Note that Fm1,nm 

 n

n
1
2


is the F used in the ANOVA table F 2, 27  3.35 . So

1 1 
1   2  335 .5  129 .5  23.35 137 .459

 206  159 .1 ,

10 10 

 2   3  129 .5  106 .0  159 .1 etc. Watch the values of n j , if they are not the same you have to
recompute the error part.
iii. Bonferroni Confidence Interval
If we only need k different intervals, use 1   2  x1  x2   t n  m  s
2k
1
1
Ordinarily, you need a

n1 n 2
computer to do the values of t that this requires. k is the number of intervals you intend to do. Assume that
  .03 and that you are doing 3 intervals. Then t nm  t 27  2.771 .

1   2  335 .5  129 .5  2.771137 .459 
2k
.005
1
1

 206  170 .343 etc.
10 10
16
252solnF1 11/7/05
Exercise, 14.24†: Another problem worth looking at.
a) You should be able to do the calculations below. Only the three columns of numbers were given to us.
  .10
Sample
2
1.58
1.45
0.57
1.16
1.12
0.91
0.83
1
1.06
0.79
0.82
0.89
1.05
0.95
0.65
1.15
Sum
Sum
1.12
0.43
3
0.29
0.06
0.44
0.61
0.55
0.43
0.51
0.10
0.34
0.53
0.06
0.09
0.17
0.60
0.17
8.48 +
8.05 +
4.95
9+
8+
15
nj
 21 .48 
x j
0.94222
1.00625
0.3300
21 .48
 0.67125  x
32
SS
8.21660 +
9.22500 +
2.23692
 19 .6292 
x 2j
0.8878
1.0125
0.1089
Sum is not useful.
2
ij
2
j .j
ij
 32  n
 x  nx  19.6792  320.67125  5.26075
SSB  n x  nx  90.94222  81.00625  150.3300  320.67125
SST 
 x
 x
2
ij
2
2
2
2
2
2
2
 90.8878  81.0125  150.1089  320.671
=3.30525
Source
SS
DF
MS
F
F.10
24.51
F 2,29  2.50 s
Between
3.30525
2
1.65263
Within
Total
1.95550
5.26075
29
31
0.06743
H0
Column means equal
Since the value of F we calculated is more than the table value, we reject the null hypothesis and conclude
that the different chemicals have significantly different sorption rates. The SPSS output in the text has a pvalue of .000, so we reject the null hypothesis because it is below the 10% significance level given in the
problem.
b) We assume that we have three random samples taken from Normal populations with similar variances.
c) There are a number of ways to check this. Certainly, computation of variances for the data might help. A
fast look at standard deviations on my calculator seems to indicate that (Use the column sums of squares
and means you already have.) s1  0.1863 , s2  0.4009 and s3  0.2075 . An F-test, using the first two of
these made me a bit uncomfortable. A fast test here would be to get a table of the F-max distribution and
compare the largest F ratio from the data to the table using a value of n equal to the largest number in an
individual column. This is beyond the scope of this course. Note: Since this was written 2 methods for
comparing variances were added to the course.
17
252solnF1 11/7/05
d) I had asked that you compute all appropriate confidence intervals, using the formulas in the outline. But
before we get to these contrasts, try an interval for a single mean. Note for all the below from the ANOVA
29
table. n  m  29, so if   .10 , tn  m   t.005
 1.699 , n1  9, n2  8, n3  15, and
2
s  MSW  0.06743 . So for a single mean  j  x. j  t n  m 
2
s
.
nj
0.06743
0.06743
 0.94  0.15 ,  2  1.00625  1.699
 1.01  0.09 etc.
9
8
The contrasts from the outline are quoted and filled in below.
i. A single 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  0.94222  1.699
1   2  x1  x2   t n  m  s
2
1
1
. so

n1 n 2
1
9
1
8
1   2  0.94222  1.00625   1.699 0.06743     0.064  0.214 (Note that this shows no
1 1 
significant difference), 1  3  0.94222  .33000   1.699 0.06743     0.612  0.186
 9 15 
1 1 
(Significant) and  2  3  1.00625  .33000   1.699 0.06743     0.676  0.193 (Significant).
 8 15 
ii. Scheffé 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
. Note that Fm1,nm 

 n

n
1
2 

is the F used in the ANOVA table F 2,29  2.50 . So all we have to do to get this interval is to replace
t n  m  t 29  1.699 by m  1F m 1, n  m   22.50   2.236 in the individual intervals above. So

2
.005

1
9
1
8
1   2  0.94222  1.00625   2.236 0.06743     0.064  0.282 (Note that this shows no
1 1 
significant difference), 1  3  0.94222  .33000   2.236 0.06743     0.612  0.245
 9 15 
1 1 
(Significant) and  2  3  1.00625  .33000   2.236 0.06743     0.676  0.248 (Significant).
 8 15 
iii. Bonferroni Confidence Interval (Enough Already!)
18