Splitplot Design

The set up

Suppose that we want to conduct an agricultural study with two treatments:

Method of tilling: This has two levels, tractor and manual.
Fertiliser: This has 3 levels, compost, phosphate and uria.

Clearly we need 2×3=6 fields, but we have only 4 large paddy fields for use in the experiment, 2 in one village and 2 in another.

Here we can carry out the experiment as follows.

First we assign methods of tilling to the fields in each village randomly.

After the fields are tilled they are each split into 3 subfields.

The three fertilisers are assigned randomly to the subfields in each field.

This design is called a splitplot design.

Let us go through the steps carefully. Instead of fields and villages we shall use the standard statistical terms. First, a split plot design is typically used when there are two treatments (1 and 2) so that treatment 1 is easily applied to more than one unit simultaneously. In our example the method of tillage is treatment 1. It is easier to till a large field at a time using a single method, rather than small fields in a peiecemill fashion.

Each field is called a whole plot, and each subfield is called a subplot. We had 2 villages with 2 fields each. Each village is called a replicate. Each replicate consists of as many whole plots as there are levels of treatment 1. We randomly assign the levels of treatment 1 to the whole plots in each replicate.

Then we split each whole plots into as many subplots as there are levels of treatment 2. These levels are assigned randomly to the subplots in each whole plot.

We get one value of the response variable (yield of paddy in our example) per subplot. Thus the subplots are our experimental units.

The model

The model used in a split plot experiment is a mixed effects model. We shall use the following subscripts:

i=1,...,I for replicates,
j=1,...,J for levels of treatment 1,
k=1,...,K for levels of treatment 2.

The model is

where

y_ijk = yield from (i,j,k)-th subplot
μ = (fixed) overall mean effect
&alpha_i = (fixed) extra effect for i-th replicate
β_j = (fixed) extra effect for j-th level of treatment 1
γ_k = (fixed) extra effect for k-th level of treatment 2
δ_jk = (fixed) interaction effect for j-th level of treatment 1 and k-th level of treatment 2
t_ij = (random) interaction effect for i-th replicate and j-th level of treatment 1
ε_ijk = random error.

We have the usual assumptions. First the assumption of the random quantities.

ε_ijk's are iid N(0,σ_e²)
t_ij's are iid N(0,σ_t²)
The ε's are independent of the t's.

Then the assumptions of the fixed effect parameters:

∑_i &alpha_i = ∑_j β_j = ∑_k γ_k = 0.
∑_i δ_ij = ∑_j δ_ij = 0.

ANOVA

The ANOVA table for the above model takes the following form. Here the bulleted subscripts have been averaged over. For example,

The ANOVA table is:

The differences marked with (*) are obtained by subtracting the Treat 1 and Treat 2 rows from the (Treat 1, treat 2) row. Also the correction factor (CF) is defined as

Testing

There are various hypotheses of interest. As usual we shall emply F-tests by taking ratios of appropriate MS terms. However, we need to be a bit careful here since the denominator is not always the Error MS! Here is alist of the different F-ratios.

To test H₀: β_j's are all 0 (i.e., Treatment 1 has no effect, or both the tilling methods are equivalent w.r.t. yield) we have to divide the Treat 1 MS by the Replicate × Treat 1 MS.
To test H₀: γ_k's are all 0 (i.e., Treatment 2 has no effect, or all the fertilisers are equivalent w.r.t. yield) we have to divide the Treat 2 MS by the Error MS.
To test H₀: δ_jk's are all 0 (i.e., Treatment 1 has no interaction with treatment 2) we have to divide the Treat 1 × Treat 2 MS by the Error MS.

Finding E(MS)

Each of the MS terms is a function of the random data, and hence are random variables. It is useful to compute their expectations. The following example illustrates the technique.

Example: Find the expectation of the Treatment 1 MS.
Solution: We shall first find the expectation of the SS and then divide by the degrees of freedom. This SS is

···(1)

We shall first find and E(CF) separately.

Now

···(2)

We had the model

y_ijk = μ + &alpha_i + β_j + γ_k + δ_jk + t_ij + ε_ijk,

from which we get

For this we have used the fact that all the fixed effects vanish when summed over any of the subscripts. So

since the t's are independent of the ε's. Now

So

Also

Putting these in (2) we get

···(3)

Next we shall find

Using the model equation as before we get

and so

So

Putting this and (3) in (1) we get the expectation of the Treatment 1 SS as

since ∑_j β_j = 0. This simplifies to

(J-1)(Kσ_t² + σ_e²) + IK∑_jβ_j².

This is the expectation of the Treatment 1 SS. We get the expectation of the MS by dividing it by the degrees of freedom:

which is the required answer.