7.
Product and Process Comparisons
7.4.
Comparisons based on data from more than two processes
7.4.3.
Are the means equal?
7.4.3.3.
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The ANOVA table and tests of hypotheses about means
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Sums of Squares help us compute the variance estimates displayed
in ANOVA Tables
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The sums of squares SST and SSE
previously computed for the one-way ANOVA are used to form two
mean squares, one for treatments and the second for
error. These mean squares are denoted by MST and
MSE, respectively. These are typically displayed in a
tabular form, known as an ANOVA Table. The ANOVA table also
shows the statistics used to test hypotheses about the population
means.
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Ratio of MST and MSE
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When the null hypothesis of equal means is true, the two mean
squares estimate the same quantity (error variance), and should be
of approximately equal magnitude. In other words, their ratio
should be close to 1. If the null hypothesis is false, MST should
be larger than MSE.
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Divide sum of squares by degrees of freedom to obtain mean squares
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The mean squares are formed by dividing the sum of squares by the
associated degrees of freedom.
Let N =
ni. Then, the degrees of freedom for treatment,
DFT = k - 1, and the degrees of freedom for error,
DFE = N - k.
The corresponding mean squares are:
MST = SST / DFT
MSE = SSE / DFE
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The F-test
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The test statistic, used in testing the equality
of treatment means is: F = MST / MSE.
The critical value is the tabular value of the F distribution, based
on the chosen
level and the degrees of freedom DFT and DFE.
The calculations are displayed in an ANOVA table, as follows:
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ANOVA table
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| Source |
SS |
DF |
MS |
F |
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| Treatments |
SST |
k-1 |
SST / (k-1) |
MST/MSE |
| Error |
SSE |
N-k |
SSE / (N-k) |
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| Total (corrected) |
SS |
N-1 |
The word "source" stands for source of variation. Some authors
prefer to use "between" and "within" instead of "treatments" and
"error", respectively.
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ANOVA Table Example
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A numerical example
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The data below resulted from measuring the difference in resistance
resulting from subjecting identical resistors to three different
temperatures for a period of 24 hours. The sample size of each
group was 5. In the language of Design of Experiments, we have an
experiment in which each of three treatments was replicated 5 times.
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Level 1 |
Level 2 |
Level 3 |
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6.9 |
8.3 |
8.0 |
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5.4 |
6.8 |
10.5 |
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5.8 |
7.8 |
8.1 |
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4.6 |
9.2 |
6.9 |
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4.0 |
6.5 |
9.3 |
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| means |
5.34 |
7.72 |
8.56 |
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The resulting ANOVA table is
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Example ANOVA table
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| Source |
SS |
DF |
MS |
F |
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| Treatments |
27.897 |
2 |
13.949 |
9.59 |
| Error |
17.452 |
12 |
1.454 |
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| Total (corrected) |
45.349 |
14 |
| Correction Factor |
779.041 |
1 |
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Interpretation of the ANOVA table
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The test statistic is the F value of 9.59. Using an
of .05, we have that F.05; 2, 12 = 3.89 (see the
F distribution table
in Chapter 1). Since the test statistic is much larger than the
critical value, we reject the null hypothesis of equal population
means and conclude that there is a (statistically) significant
difference among the population means. The p-value for
9.59 is .00325, so the test statistic is significant at that level.
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Techniques for further analysis
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The populations here are resistor readings while operating under the
three different temperatures. What we do not know at this
point is whether the three means are all different or which of the
three means is different from the other two, and by how much.
There are several techniques we might use to further analyze the
differences. These are:
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