Two-way ANOVA
Example - Capsule Dissolving Experiment (Capsule.JMP)
In this experiment researchers are interested in studying the effect of two factors on the time to begin dissolving a capsule which is recorded as the time until bubbles first appear (seconds). The factors of interest to the researchers are juice type - gastric or duodenal (Factor A) and capsule type - C or V (Factor B).
To conduct the experiment 5 capsules of each type are randomly assigned to each juice giving us 5 observations or replicates for each of four different treatment combinations (Gastric & C, Gastric & V, Duodenal & C, Duodenal & V). The data obtained from the experiment are shown below:
Capsule TypeType of Digestive Juice / C / V / Juice Type Means
Gastric / 39.5
45.7
49.8
50.2
63.8
/ 47.4
43.5
39.8
36.1
41.2
/
Duodenal / 33.5
36.7
42
38.1
31.2
/ 44
41.2
47.3
45.3
42.7
/
Capsule Type Means / / / Grand Mean
We can construct plots to visualize the effects of each factor.
Digestive Juice Capsule Type
Our preliminary conclusions would be first that fluid type has a small effect on the dissolution time with duodenal juice dissolving capsules about 5 seconds quicker on average, and secondly that capsule type has little or no effect.
These conclusions are completely WRONG!! Why?
When considering the effect of two factors on the response we cannot do so marginally, i.e. individually. It is possible, for example, that the effect of digestive juice is not the same for both capsule types. If we consider the means for each of the treatment combinations above we see that for type C capsules the duodenal juice dissolves the capsule quicker, while the exact opposite is true for type V capsules, gastric juice dissolves the capsules faster.
A better display shows the means for each treatment combination. Here we have a separate profile for each digestive juice showing how the capsule effect depends on the type of digestive juice we are using. This is what we call an interaction.
Questions of Interest:
1) Is there a significant interaction between the two factors being studied?
This question needs to answered first, because if we conclude there is a
significant interaction then both effects are important and there effects can
not be discussed individually. If we conclude there isn’t a significant
interaction between the factors being studied then we can test the effects
individually.
2) Is there a significant Factor A effect?
3) Is there a significant Factor B effect?
As always it is important to quantify any significant differences using pair-wise comparisons and CI’s for the differences in the population/treatment means.
Analysis in JMP
To fit the two-way model for these data select Fit Model from the Analyze menu and put the response Time to Bubbles in the Y box and then highlight both Fluid & Capsule and select Full Factorial from the Macros pull-down menu as shown below.
Then click Run Model to obtain the results on the next page.
The p-values for the effects suggests that the Fluid*Capsule interaction is significant (p = .0049), which implies the main effect tests for Fluid and Capsule are of little interest. It is interesting to note that the main effect of Capsule is not significant (p = .9361). This happens because the presence of the Fluid*Capsule interaction "masks" the main effect of capsule as we have seen in marginal effect plots above. The main effect of fluid is only partially masked by the Fluid*Capsule interaction and so it still tests as significant.
Because the interaction is significant we need to quantify the treatment effects by comparing the treatment combinations to one another. To do this select LSMeans Tukey HSD from the Fluid*Capsule interaction pull-down menu.
Results of the treatment mean comparisons are shown below.
Here we see that Gastric,C and Duodenal,C mean dissolution times significantly differ. In particular we estimate that the type C capsules in gastric fluid take between 3.57 and 23.429 seconds longer to dissolve on average.
If the interaction between the two factors is not significant we can use the Tukey’s procedure to compare the means across the levels of each factor individually.
Checking Two-way ANOVA Assumptions (Normality and constant variance)
Assumptions:
1. The observations between and within the treatment combinations are
independent.
2. The response is normally distributed for each treatment combination.
3. The variance of the response is the same for each treatment combination.
To check the constant variance assumption we can examine the residuals plotted vs. the fitted values andeach factor. The fitted values are simply the observed mean response at each of the four treatment combinations and the residuals are the deviations from the treatment combination means. The spread of the residuals should be uniform indicating constant response variation for the different treatment combinations.
A plot of the residuals vs. the fitted values is given each time we fit a model in JMP. The resulting plot is shown below:
There appears to be a potential outlier in this plot, otherwise this plot looks fine.
To examine the normality assumption we assess the normality of the residuals. Save the residuals to the spreadsheet as shown below and use Analyze > Distribution to examine them. With the exception of two mild outliers, normality seems satisfied.
STATISTICAL DETAILS
Two-way ANOVA Model
where,
kth observed response value when level i of factor A and level j of factor B is used.
effect due to the fact level i of Factor A was used.
effect due to the fact level j of Factor B was used.
effect due to the interaction of ith level of Factor A and the jth level of Factor B.
the random error, represents the variation in the response values when the ith level of Factor A and the jth level of Factor B are used.
We assume that , i.e. the errors are normal and their variation is constant.
See your text for formulae used to estimate these quantities and those used to test the hypotheses. The three questions of interest in a two-way ANOVA can be formulated in terms of these parameter values.
1. For testing the interaction between Factors A and B we have:
2. For testing the Factor A effect we have:
3. For testing the Factor B effect we have:
As in one-way ANOVA the test procedures decomposes total response variation into components that measure how much variation in the response is due to Factor A, Factor B, the interaction between Factors A & B, and random error.
Sum of Squares:
Degrees of Freedom:
SUM OF SQUARES FORMULAE:
= + +
+
MEAN SQUARES
The mean square for an effect is the effect sum of squares divided by the degrees of freedom.
The expected value of the mean squares are as follows:
For testing the main effects (A & B) and the interaction effect (AB) we simply compare the size of the to the . If the > we have evidence that the effect is significant. If then we have little evidence that the effect is significant. To compare the mean squares we use the ratio, which has an F-distribution.
F-distribution (numerator df = df for the effect , denominator df = df for error)
> 1 will lead to the conclusion that the effect in question significantly impacts the response. Large values lead to small p-values which support effect significance.
Example 2 – Apple Nutrient Level by Region Grown and Variety
Does the level of a certain nutrient found in apples differ significantly across regions and by variety? The regions are labeled as A,B, or C and the varieties are labeled as X, W, Y, and Z.
In JMP select Analyze > Fit Model and set up the dialog box as shown below.
The output is shown below.
Multiple Comparisons for Region and Variety Effects
We see that the mean nutrient level of apples grown in region A, regardless of variety, is significantly higher than that for apples grown in regions B and C. We estimate the mean nutrient level for apples grown in region A is between .303 and 2.66 units larger than the mean nutrient level of apples grown in region B, and is between .136 and 2.49 unit larger than the mean nutrient level of apples grown in region C.
The mean nutrient levels found in varieties Y and Z, regardless of region grown, significantly differ from those found in varieties X and W. We estimate that the mean nutrient level found in variety Z apples exceeds that for variety W by between 3.57 and 8.55 units, etc....
Additional notes
12