Which option is incorrect?
mean sum of square of total = mean sum of square of treatment + mean sum of square of error
Understanding ANOVA Concepts
The question asks us to identify the incorrect statement among the given options related to ANOVA (Analysis of Variance). ANOVA is a statistical technique used to compare means of three or more groups. It partitions the total variability observed in a dataset into different components.
ANOVA Partitioning of Variability
In a one-way ANOVA, the total variation in the data is split into two parts:
- Variation between groups (often called variation due to treatment or factor)
- Variation within groups (often called variation due to error or residual)
This partitioning applies to both the Sum of Squares (SS) and the Degrees of Freedom (df).
Sum of Squares (SS)
The total sum of squares (\(SS_{Total}\)) represents the total variation in the data. It is the sum of the sum of squares between groups (\(SS_{Between}\) or \(SS_{Treatment}\)) and the sum of squares within groups (\(SS_{Within}\) or \(SS_{Error}\)).
\(SS_{Total} = SS_{Between} + SS_{Within}\)
or
\(SS_{Total} = SS_{Treatment} + SS_{Error}\)
Degrees of Freedom (df)
Similarly, the total degrees of freedom (\(df_{Total}\)) is partitioned into the degrees of freedom between groups (\(df_{Between}\) or \(df_{Treatment}\)) and the degrees of freedom within groups (\(df_{Within}\) or \(df_{Error}\)).
For k treatments (groups) and a total of N observations:
- \(df_{Total} = N - 1\)
- \(df_{Between}\) or \(df_{Treatment} = k - 1\)
- \(df_{Within}\) or \(df_{Error} = N - k\)
Check: \((k - 1) + (N - k) = N - 1\), which holds true.
Mean Sum of Squares (MS)
The Mean Sum of Squares (MS) is calculated by dividing the Sum of Squares by its corresponding Degrees of Freedom.
- Mean Sum of Squares Between (or Treatment): \(MS_{Between} = MS_{Treatment} = \frac{SS_{Treatment}}{df_{Treatment}} = \frac{SS_{Treatment}}{k-1}\)
- Mean Sum of Squares Within (or Error): \(MS_{Within} = MS_{Error} = \frac{SS_{Error}}{df_{Error}} = \frac{SS_{Error}}{N-k}\)
- Mean Sum of Squares Total: \(MS_{Total} = \frac{SS_{Total}}{df_{Total}} = \frac{SS_{Total}}{N-1}\)
Unlike the Sum of Squares and Degrees of Freedom, the Mean Sum of Squares are not additive. That is, \(MS_{Total} \neq MS_{Treatment} + MS_{Error}\).
Analyzing the Options
Let's examine each option based on these principles.
Option 1: For k treatments and N observations, the degree of freedom of variation between group is k - 1
As discussed above, the degrees of freedom for variation between groups (treatment) is indeed \(k-1\). This statement is correct.
Option 2: mean sum of square of total = mean sum of square of treatment + mean sum of square of error
The Mean Sum of Squares are calculated by dividing the Sum of Squares by their respective Degrees of Freedom. While Sum of Squares and Degrees of Freedom are additive, Mean Sum of Squares are generally not additive in this manner. \(MS_{Total}\), \(MS_{Treatment}\), and \(MS_{Error}\) do not have this simple linear relationship. This statement is incorrect.
Option 3: sum of square of total = sum of square of treatment + sum of square of error
This statement reflects the fundamental additive property of Sum of Squares in ANOVA, where the total variability is decomposed into variability between groups and variability within groups. This statement is correct.
Option 4: For k treatments and N observations, the degree of freedom of variation within group is N - k
The degrees of freedom for variation within groups (error) is calculated as the total degrees of freedom minus the between-group degrees of freedom: \((N-1) - (k-1) = N-1-k+1 = N-k\). This statement is correct.
Based on the analysis, the incorrect option is the one stating that the mean sum of square of total equals the sum of the mean sum of square of treatment and mean sum of square of error.
Conclusion
The incorrect statement is option 2.
Revision Table: Key ANOVA Components (k treatments, N observations)
| Component | Sum of Squares (SS) | Degrees of Freedom (df) | Mean Sum of Squares (MS) | Relationship |
|---|---|---|---|---|
| Total | \(SS_{Total}\) | \(df_{Total} = N - 1\) | \(MS_{Total} = \frac{SS_{Total}}{df_{Total}}\) | |
| Between Groups (Treatment) | \(SS_{Treatment}\) | \(df_{Treatment} = k - 1\) | \(MS_{Treatment} = \frac{SS_{Treatment}}{df_{Treatment}}\) | |
| Within Groups (Error) | \(SS_{Error}\) | \(df_{Error} = N - k\) | \(MS_{Error} = \frac{SS_{Error}}{df_{Error}}\) | |
| Additive Property | \(SS_{Total} = SS_{Treatment} + SS_{Error}\) | \(df_{Total} = df_{Treatment} + df_{Error}\) | \(MS_{Total} \neq MS_{Treatment} + MS_{Error}\) | MS are not directly additive |
Additional Information: ANOVA Purpose and F-test
The main purpose of ANOVA is to test the hypothesis that the means of several groups are equal. The test is based on comparing the variability between the groups to the variability within the groups.
The test statistic used in ANOVA is the F-statistic, which is the ratio of the mean sum of squares between groups to the mean sum of squares within groups.
\(F = \frac{MS_{Between}}{MS_{Within}} = \frac{MS_{Treatment}}{MS_{Error}}\)
A large F-statistic indicates that the variability between groups is much larger than the variability within groups, suggesting that the group means are likely different.
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