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By Issa Bass
The Analysis
of Variance (ANOVA) is a good tool to determine if there is a
difference between several sample means, but it does not
determine where the difference comes from (if there is any
difference), It does not show what samples are so
disparate that the null hypothesis has to be rejected. To know
where the difference originates from, it is necessary to conduct
further analyses after rejecting the null hypothesis. Tukey’s,
Fisher’s and Dunnett’s are examples of comparisons that can help
situate the sources of variations between means.
A simpler way to determine if the sample means are equal and at
the same time visually determine where the difference is coming
from (if there is any) would be the Analysis Of Means (ANOM).
The Analysis Of Means is a lot simpler and easier to conduct
than ANOVA and it provides an easy to interpret visual
presentation of the results.
When conducting ANOM, what we want to achieve is to determine
the Upper and Lower Decision Limits. If all the sample means
fall within those boundaries we can say with confidence that
there is no ground to reject the null hypothesis, i.e. there no
significant difference between the samples’ means. If at least
one mean falls outside the limits, we reject the null
hypothesis.
The Upper and Lower Decision Limits depend on several factors:
·
The
samples’ means
·
The
mean of all the observed data (The mean of the samples’ means)
·
The
standard deviation
·
The
Alpha level
·
The
number of samples
·
The
sample sizes (to determine the degree of freedom)
ANOM compares
the natural variability of every sample mean with the mean of
all the sample means.
If we have
j samples and n treatment levels then the sample
means are given as:

And the mean
of all the sample means is:

Let’s call
N the number of all observed data and s is the standard deviation.
Then the
variance for the treatments would be

the overall Standard deviation

and the Upper and Lower Decision Limits would be:

Where
represents
the significance level and
is
found on the ANOM table at the intersection between the degree
of freedom and the number of samples.
In
our ANOVA
example, we wanted to know if there was a difference between the
productivity of the three shifts; after conducting the test, we
concluded that there was not a significant difference between
them. Let’s take the same example and this time use the Analysis
Of Means.
Unfortunately, Excel does not have the capabilities to conduct
ANOM, so we will use Minitab.
|
|
First Shift |
Second Shift |
Third shift |
|
Monday |
78 |
77 |
88 |
|
Tuesday |
88 |
75 |
86 |
|
Wednesday |
90 |
80 |
79 |
|
Thursday |
77 |
83 |
93 |
|
Friday |
85 |
87 |
79 |
|
Saturday |
88 |
90 |
83 |
|
Sunday |
79 |
85 |
79 |
Using Minitab, we
need first of all to stack the data

We select Column

Note that we did
not select Column C1. Click OK and a new worksheet appears with two
stacked columns.

Now that we have
the stacked data, we can conduct the ANOM. We click on Stat, we
select ANOVA and then “Analysis of means”

For “Response” we
select C2 because that’s where the data we are looking for resides
and we double-click on “Subscripts” for Factor 1 if we have selected
“Normal”. Remember that “Subscript is the default column title for
the treatments titles.

The default for
the Alpha level is 0.05. We can change it, but for the sake of this
example, we leave it as it is
When we click on OK, the following graph pops up.

Since
all the dots are within the decision boundaries, we conclude that
there is not enough evidence to reject the null hypothesis. The
difference between the three means is insignificant at Alpha level
of 0.05.
This is the same conclusion we reached when we conducted and
Analysis Of Variance with the sane
data.
About the author
Issa Bass is the managing editor of SixSigmaFirst. He can be reached at issa@sixsigmafirst.com
Tell us what you think about this article. Send a note to the Editor.
www.manorhouseassociates.com
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