Lecture 19
Lecture on the process for doing ANOVA in Minitab and how
to analyze and interpret the output.
State the assumptions for ANOVA:
1.___________________________________________
2.___________________________________________
To check the assumption of equal variances, do
_______________________________________________.
To check the assumption of normality, do a test of
residuals called the ____________________________________.
(This test must be done after ANOVA is calculated because
the residuals must be stored for use in the computation of
the test!!)
A residual in this case is simply the error calculated by
subtracting the observed value – the predicted value.
If both assumptions are met, then ANOVA is _______________.
If differences occur among the means, exactly which of the
means is different can be determined by a ________________.
A flow chart for the process in Minitab is as follows:
Interpreting the output:
1.
Examine the p-value for the Bartlett’s test
p-value <  => reject Ho => variances unequal => STOP
p-value   => DNR Ho => variances => continue
2.
Examine the p-value for the Kolmogorv-Smirnov test
p-value <  => reject Ho => not normal=> STOP
p-value   => DNR Ho => normally dist => continue
3.
Examine the p-value for the ANOVA test
p-value <  => reject Ho => significant difference
among means => continue
p-value   => DNR Ho => no significant difference
among means => STOP
4. Examine the confidence intervals for each combination
of means
interval contains 0 (signs unlike) => no significant
difference
interval doesn’t contain 0 (signs same) => significant
difference
Example from text page 434, Table 11.3
Homogeneity of Variance
Bartlett's Test (normal distribution)
Test Statistic: 0.033
P-Value
: 0.984
Normal Probability Plot
.999
.99
Probability
.95
.80
.50
.20
.05
.01
.001
-3
-2
-1
0
1
2
3
RESI1
Average: 0
StDev: 2.44949
N: 10
Kolmogorov-Smirnov Normality Test
D+: 0.208 D-: 0.193 D : 0.208
Approximate P-Value > 0.15
One-way Analysis of Variance
Analysis of Variance for time
Source
DF
SS
MS
code
2
98.40
49.20
Error
7
54.00
7.71
Total
9
152.40
Level
1
2
3
N
Mean
StDev
4
3
3
17.000
12.000
20.000
2.944
2.646
2.646
Pooled StDev =
2.777
F
6.38
P
0.026
Individual 95% CIs For Mean
Based on Pooled StDev
----+---------+---------+---------+(------*------)
(-------*-------)
(-------*-------)
----+---------+---------+---------+10.0
15.0
20.0
25.0
NOTE: To fit the above output on the page, you must use
Courier New 10 pt font!!!
Tukey's pairwise comparisons
Family error rate = 0.0500
Individual error rate = 0.0214
Critical value = 4.17
Intervals for (column level mean) - (row level mean)
1
2
-1.255
11.255
3
-9.255
3.255
2
-14.687
-1.313
EXAMPLE OF HOW AN ANOVA SHOULD BE WRITTEN UP:
CHECK OF ASSUMPTION OF EQUAL VARIANCES:
1.
2.
3.
4.
Ho: The variances are equal.
Ha: The variances are not equal
Bartlett’s test
 = .05
DNR Ho, if the p-value  
Reject Ho, if the p-value < 
5.
Homogeneity of Variance
Bartlett's Test (normal distribution)
Test Statistic: 0.033
P-Value
: 0.984
6.
7.
Since the p-value = .984 > .05, we DNR Ho.
Therefore, the assumption of equal variances is met.
CHECK THE ASSUMPTION OF NORMALITY:
Normal Probability Plot
.999
.99
Probability
.95
.80
.50
.20
.05
.01
.001
-3
-2
-1
0
1
2
3
RESI1
Average: 0
StDev: 2.44949
N: 10
6.
7.
Kolmogorov-Smirnov Normality Test
D+: 0.208 D-: 0.193 D : 0.208
Approximate P-Value > 0.15
Since the p-value > .15 > .05, we DNR Ho.
Therefore, the normality assumption is met.
CHECK FOR DIFFERENCES AMONG THE MEANS:
1.
2.
3.
4.
Ho: 1 = 2 = 3
Ha: at least one inequality exists
ANOVA
 = .05
DNR Ho, if the p-value  
Reject Ho, if the p-value < 
5.
One-way Analysis of Variance
Analysis of Variance for time
Source
DF
SS
MS
code
2
98.40
49.20
Error
7
54.00
7.71
Total
9
152.40
Level
1
2
3
N
Mean
StDev
4
3
3
17.000
12.000
20.000
2.944
2.646
2.646
Pooled StDev =
6.
7.
2.777
F
6.38
P
0.026
Individual 95% CIs For Mean
Based on Pooled StDev
----+---------+---------+---------+(------*------)
(-------*-------)
(-------*-------)
----+---------+---------+---------+10.0
15.0
20.0
25.0
Since the p-value = .026 < .05, we reject Ho.
Therefore, there are significant differences among the means.
Tukey's pairwise comparisons
Family error rate = 0.0500
Individual error rate = 0.0214
Critical value = 4.17
Intervals for (column level mean) - (row level mean)
1
2
-1.255
11.255
3
-9.255
3.255
2
-14.687
-1.313
Methods 2 and 3 are significantly different. By examining the
confidence intervals displayed with the ANOVA analysis, I would
recommend Method 2 as the method which produces tax preparers since
this group requires significantly less time to prepare returns on
average.
MINITAB INSTRUCTIONS
ANOVA
Enter response (dependent variable) in one column
Enter Treatment (independent variable or factor) in another column
Select Stat
Select ANOVA
Select Homgeneity of Variance
Select Response variable
Select Factor variable
Click OK
If the variances are not significantly different proceed with ANOVA
Select Stat
Select ANOVA
Select One-Way
Select Response variable
Select Factor variable
Click Comparisons button
Check Tukey's family error rate
Enter alpha as a number of percent(usually 5)
Click OK
Check Store residuals box
Click OK
Click OK
Click Stat
Click Basic statistics
Choose Normality Test
Select RESI1 as your variable
Select the Kolmogorov-Smirnov Test
Click OK
ASSIGNMENT:
Do by computer, #11.25, pg 444
#11.27, pg 444-5