Two Factor ANOVA with Replication

Introduction

In Two Factor ANOVA without Replication, we consider the analysis where there is only one sample item for each combination of factor A and B levels. On this webpage, we extend this analysis to the case where there are multiple samples for each such combination. Thus, in addition to the main effects corresponding to A and B, we now study the interactions between A and B, which is the main reason for performing this type of analysis.

We will restrict ourselves to the case where all the samples are equal in size (balanced model). In Unbalanced Factorial ANOVA we show how to perform the analysis where the samples are not equal (unbalanced model) via regression. 

You should not confuse ANOVA with replication with ANOVA with repeated measures as described in ANOVA with Repeated Measures.

Example introduced

As usual, we start with an example. We then provide some background information and then complete the analysis for the example.

Example 1: Repeat the analysis from Example 1 of Two Factor ANOVA without Replication, but this time with the data shown in Figure 1 where each combination of blend and crop has a sample of size 5.

Figure 1 – Data for Example 1

Structural Model

Definition 1: We extend the structural model of Definition 1 of Two Factor ANOVA without Replication as follows.

In Definition 1 of Two Factor ANOVA without Replication the r × c table contains the entries {x_ij: 1 ≤ i ≤ r, 1 ≤ j ≤ c}. We extend these tables to contain entries {X_ij: 1 ≤ i ≤ r, 1 ≤ j ≤ c},  where X_ij is a sample for level i of factor A and level j of factor B. Here X_ij = {x_ijk: 1 ≤ k ≤ n_ij}. For now, we assume the n_ij are all equal of size m.

We use terms such as x̄_i (or x̄_i.) as an abbreviation for the mean of {x_ijk: 1 ≤ j  ≤ c, 1 ≤ k ≤ m}. We also use terms such as x̄_j (or x̄_.j) as an abbreviation for the mean of {x_ijk: 1 ≤ i ≤ r, 1 ≤ k ≤ m}.

As in Definition 1 of Two Factor ANOVA without Replication, we define the effects α_i and β_j where

Similarly, we define a_i and b_j where

We use δ_ij for the effect of level i of factor A with level j of factor B, i.e. the interaction of level i of factor A and level j of factor B. Thus, δ_ij = μ_ij – μ_i – μ_j + μ. Similarly, we have

It is easy to show that

Finally, we can represent each element in the sample as

where ε_ijk denotes the error (or unexplained) amount. As before we have the sample version

where e_ijk is the counterpart to ε_ijk in the sample. Note that

and so

Also,

Null Hypotheses

As in Definition 1 of Two Factor ANOVA without Replication, the null hypotheses for the main effects are:

H₀:  μ_1. = μ_2. = … = μ_r. (Factor A)

H₀:  μ_.2 = μ_.2 = … = μ_.c (Factor B)

These are equivalent to:

H₀: α_i = 0 for all i (Factor A)

H₀: β_j = 0 for all j (Factor B)

In addition, there is a null hypothesis for the effects due to the interaction between factors A and B.

H₀: δ_ij = 0 for all i, j

More about the structural model

Definition 2: Using the terminology of Definition 1, define

We can also define the following entities:

Since the within groups terms are used as the error terms in our model, we also use the following symbols:

Properties

Property 1:

Proof: Clearly

If we square both sides of the equation, sum over i, j, and k, and then simplify (with various terms equal to zero as in the proof of Property 2 of Basic Concepts for ANOVA), we get the first result. For the second,

Property 2: Note that the between-group terms are as for the one-way ANOVA, namely

The proof is similar to the proof of Property 1. It also follows that

Property 3: If a sample is made as described in Definitions 1 and 2, with the x_ijk independently and normally distributed and with all (or ) equal, then

Proof: The proof is similar to that of Property 1 of Basic Concepts for ANOVA.

Theorem 1: Suppose a sample is made as described in Definitions 1 and 2, with the x_ijk independently and normally distributed.

If all μ_i are equal and all  are equal then

If all μ_j are equal and all  are equal then

Also, under certain circumstances,

Proof: The result follows from Property 3 and Theorem 1 of  F Distribution.

Property 4:

Statistical Tests

We use the following tests:

Assumptions

The assumptions for Two Factor ANOVA are similar to those for One Factor ANOVA, namely

• All samples are drawn from normally distributed populations
• The samples are drawn from populations that have a common variance
• All samples are drawn independently from each other
• Within each sample, the observations are sampled randomly and independently of each other

By sample, here we mean each combination of levels from the two factors.  We also want to make sure there are no outliers that can distort the results of the test. See ANOVA Assumptions for how we check these assumptions using the Real Statistics Resource Pack.

Example continued

We now return to Example 1 and show how to conduct the required analysis using Excel’s Anova: Two-factor With Replication data analysis tool.

Example 1 (continued): The summary output from the data analysis tool is given on the right side of Figure 2, with the sample data repeated on the left side of the figure.

Figure 2 – Summary output of ANOVA data analysis for Example 1

The top part of Figure 3 contains the rest of the output from the data analysis tool. We’ll explain the bottom part momentarily.

Figure 3 – ANOVA analysis for Example 1

We now draw some conclusions from the ANOVA table in Figure 3. Since the p-value (crops) = .0649 > .05 = α, we can’t reject the Factor B null hypothesis, and so conclude (with 95% confidence) that there are no significant differences between the effectiveness of the fertilizer for the different crops.

Since the p-value (blends) = .00025 < .05 = α, we reject the Factor A null hypothesis and conclude that the blends are statistically different.

Interaction Plots

We also see that the p-value (interactions) = .0456 < .05 = α, and so conclude there are significant differences in the interaction between crop and blend. We can look more carefully at the interactions by plotting the mean interactions between the levels of the two factors (see Figure 4). Lines that are roughly parallel are indications of the lack of interaction, while lines that are not roughly parallel indicate interaction.

From the first chart we can see that Blend Y has quite a different pattern from the other brands, especially since the line for Blend Y is trending down towards Soy and up towards Rice, exactly the opposite of Blend X and Z). We also see that Blend X is trending up towards Soy much more abruptly than Blend Z.

Figure 4 – Interaction plots for Example 1

Worksheet Functions

Although the analysis in Figures 2 and 3 was produced automatically by Excel’s data analysis tool, the same result can be produced using Excel formulas, just as we were able to do for Example 1 of Two Factor ANOVA without Replication. In fact, all the entries in the ANOVA table in Figure 3 can be calculated using the tables constructed in the bottom part of Figure 3 in exactly the same way as was done in Example 1 of Two Factor ANOVA without Replication.

In fact, the only thing new is the calculation of the error term SS_W. To calculate it we must first construct the table of the square deviations for all the interactions from their mean. This table appears in cells J38:N41 of Figure 3. E.g. the entry for SS_Wheat,BrandX (in cell K39) is =DEVSQ(B5:B9). SS_W is then calculated as the sum of all the terms in the table, namely =SUM(K39:N41).

Alternatively, we can use Property 2 to calculate SS_Bet and then use the fact that SS_W = SS_T – SS_Bet. To calculate SS_Bet we first construct the table of the means of the various interactions of factors A and B (range J43:N46 of Figure 3), as described below. SS_Bet is now calculated using the formula =DEVSQ(K44:N46)*H5. For Example 1, SS_Bet = 18420.5, and so SS_W = SS_T – SS_Bet = 39640.9 – 18420.5 = 21220.4.

Example using row formatting

Example 2: Repeat the analysis for the data in Example 1 by using the presentation of the data given in the table on the left of Figure 5.

Figure 5 – Alternative presentation of data in Example 1

Excel’s ANOVA data analysis tools don’t support data in this format, and so we must proceed to create the ANOVA table (i.e. the output found in Figure 3) using the formulas. This is straightforward, although tedious, with the result presented in Figure 6. As usual, the hardest part is the calculations for the SS terms, which are shown on the right side of the worksheet in Figure 6.

Figure 6 – ANOVA output for Example 2

When the assumptions are not met

In general, when the assumptions are violated, transformations and non-parametric (rank) tests are not very useful for two-way ANOVA. We can instead abandon the omnibus test and apply the various planned and unplanned tests described in Planned Comparisons for ANOVA and Unplanned Comparisons for ANOVA by treating the two-way ANOVA as a one-way ANOVA.

In particular, when the variances are not equal we can apply Welch’s correction for contrasts. We can also use the Scheirer-Ray-Hare test or Aligned Rank Transform (ART) ANOVA

Reference

Howell, D. C. (2010) Statistical methods for psychology (7^th ed.). Wadsworth, Cengage Learning.
https://labs.la.utexas.edu/gilden/files/2016/05/Statistics-Text.pdf