The expected mean squares for unbalanced mixed effect interactive model were derived using Brute Force Method. From the expected mean squares, there are no obvious denominators for testing for the main effects when the factors are mixed. An expression for F-test for testing for the main effects was derived which was proved to be unbiased.
The problem with unbalanced fixed effect interactive model is associated with the appropriate F-test for testing for the main effects when interactions are present. The paper by [
In [
Similarly, [
Also [
However, [
Analysis of variance is straightforward when an experimental design is balanced, but unequal cell sizes affect the computation of means, hypotheses tested and F-statistics [
Several solutions have been proposed for the analysis of unbalanced data. Solutions have focused on forcing the unbalanced data to be balanced. Suggestions include imputing cell means as additional data points into the smaller cells.
Given the model
To derive the expected mean squares for Equation (1), [
According to him, in mixed models, the expected values of the sums of squares contain functions of the fixed effects that cannot be eliminated by considering linear combinations of the sums of squares. He suggested two obvious ways of overcoming the difficulties associated with unbalanced mixed effect data. The first is to ignore the fixed effects and eliminate them from the model. What remains is a random model for which the F-test can be determined. The second possibility is to assume the fixed effects as random and therefore assume the entire model as random effect models. These suggestions are in fact unsatisfactory.
From Equation (1) above, [
They derive the expected mean squares for Equation (1) as shown in
Where
S.V | d.f | SS | MS | Expected mean squares |
---|---|---|---|---|
Factor A | p − 1 | SSA | MSA | |
Factor B | q − 1 | SSB | MSB | |
AxB | (p − 1)(q − 1) | SSλ | MSλ | |
Error | N − pq | SSe | MSe | |
Total | N − 1 | SST |
and
From
With the corresponding F-ratios as
where
And
The sums of squares for factor A, factor B, the interaction between factor A and factor and the error terms are given by
The expression for testing for the presence of interaction from
From Equation (1), if factor A is fixed and factor B is random
and
Similarly, if factor A is random and factor B is fixed
and
Using Brute Force Method, the expected mean squares of Equation (1) when factor A is fixed and factor B is random the expected mean square are shown in ANOVA
From
However if we can obtain the expression
we would have the F-test as
where
Using Welch Satterthwaite Equation
where
S.V | d.f | SS | MS | Expected mean square |
---|---|---|---|---|
p-classes | p − 1 | SSα | MSα | |
q-classes | q − 1 | SSβ | MSβ | |
pq-classes | (p − 1)(q − 1) | SSλ | MSλ | |
Error | N − pq | SSɛ | MSɛ | |
Total | N − 1 | SST | - | - |
Statement 1: Equation (37) is an unbiased estimate of Equation (31).
Proof:
We take expectation on Equation (33) to have
But
Similarly, if we are interested to test for factor B we have
there would be no obvious denominator to test for the above hypothesis. However, if we can obtain the expression
We would have the F-test as
where
where
Equation (41) is also an unbiased estimate of Equation (38).
Similarly, if factor A is random, and factor B is fixed, the expected mean square are shown in the ANOVA
Where
S.V | d.f | SS | MS | Expected mean square |
---|---|---|---|---|
p-classes | p − 1 | SSα | MSα | |
q-classes | q − 1 | SSβ | MSβ | |
pq-classes | (p − 1)(q − 1) | SSλ | MSλ | |
Error | N − pq | SSɛ | MSɛ | |
Total | N − 1 | SST | - | - |
When factor A is random, the hypothesis is given by
and there would be no obvious denominator to test for the above hypothesis. If we can obtain the expression
the F-test can be shown to be
where
where
Equation (47) is also an unbiased estimate of Equation (45).
Similarly, when factor B is fixed, the hypothesis is given by
with no obvious denominator to test for the hypothesis. However, if we can obtain the expression
the F-test can be shown to be
where
Similarly Equation (51) is also an unbiased estimate of Equation (49).
Finally, the hypothesis for testing for the presence of the interaction is given by
where
Equation (2) contains the functions of the fixed effect which is
fects and eliminate them from the model, what remains is a random model for which the F-test can be determined. The second possibility is to assume the fixed effects as random and therefore assume the entire model as random effect models. This is completely unreasonable.
From
To test for the interaction effect for the mixed effect model, we have
This does not involve obtaining any expression and the degrees of freedom for both the numerator and denominator are integer valued whereas the denominator degrees of freedom for the testing for the main effects are non integer valued.
Instead of assuming both effects to be fixed or both effects to be random to enable researchers on mixed effect unbalanced interactive model analyze their data, we highly recommend our method.
This paper is limited to only an unbalanced two-way mixed effect interactive model and cannot be applied to random or fixed effect model.
Synthetic growth hormone was administered at a clinical research center to growth hormone deficient 18 short children who had not yet reached puberty. The investigator was interested in the effects of a child’s gender (factor A) and bone development (factor B) on the rate of growth induced by hormone administration. A child’s bone development was classified into one of the three categories: severely depressed, moderately depressed and mildly depressed. Three children were randomly selected for each gender-bone development group. The response variable (Y) of interest was the difference between the growth rate during hormone treatment and the normal growth rate prior to the treatment, expressed in centimeters per month. Four of the 18 children were unable to complete the study leading to unequal treatment sample sizes shown below.
Since factor A is random and factor B is assumed to fixed, we shall make use of the information in
Gender (factor A) | Bone development (factor B) | ||||
---|---|---|---|---|---|
Severely depressed | Moderately depressed | Mildly depressed | |||
Male (A1) | 1.4 2.4 2.2 | 2.1 1.7 | 0.7 1.1 | 7 | 1.66 |
Female (A2) | 2.4 | 2.5 1.8 2.0 | 0.5 0.9 1.3 | 7 | 1.63 |
4 | 5 | 5 | 14 | ||
2.1 | 2.02 | 0.9 |
Source: Netal et al. (1996) Applied Linear Statistical Models [
Our hypothesis for factor A shall be
Using Equations (20), (22) and (23) we have
Similarly, using Equations (43) and (44)
From Equations (47) and (48)
Our conclusion is that we do not reject the null hypothesis.
Similarly, our hypothesis for factor B shall be
Using Equation (21)
From Equations (51) and (52)
Our conclusion is that we do not reject the null hypothesis.
Finally, to test for the interaction we have
Our conclusion is therefore that interaction is present.
F. C. Eze,E. U. Nwankwo, (2016) Analysis of Variance in an Unbalanced Two-Way Mixed Effect Interactive Model. Open Journal of Statistics,06,310-319. doi: 10.4236/ojs.2016.62027