# Section 15 ```STAT 305: Chapter 15 – Two-way ANOVA
Spring 2014
Example 15.1 - 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 the four treatment
combinations (Gastric &amp; C, Gastric &amp; V, Duodenal &amp; C, Duodenal &amp; V). The
data obtained from the experiment are shown below:
Capsule Type
Type of Digestive Juice C
V
39.5
31.2
45.7
33.5
Gastric
49.8
36.7
50.2
42
63.8
38.1
Duodenal
Capsule Type Means
Juice Type Means
Y1  43.05
Y11  49.8
Y12  36.3
47.4
43.5
39.8
36.1
41.2
44
41.2
47.3
45.3
42.7
Y21  41.6
Y22  44.1
Y1  45.7
Y2  40.20
Y2  42.85
Grand Mean
Y  42.95
We can construct plots to visualize the effects of each factor.
Digestive Juice
Capsule Type
By plotting the mean time until bubbles
for both juice types we see that the mean
dissolution times for the digestive juice
types are approximately equal.
By plotting the mean time until bubbles for both
capsule types, we can see that mean dissolution time
for type V capsules is slightly smaller than that for
type C capsules (about 5.5 seconds).
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Spring 2014
Our preliminary conclusions would be first that capsule type has a small effect
on the dissolution time with type V capsules dissolving about 5.5 seconds
quicker on average, and secondly that digestive juice type has little or no effect.
These conclusions are completely WRONG!! Why? Because 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 in Two-way ANOVA:
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.
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Spring 2014
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 &amp; Capsule
and select Full Factorial from the Macros pull-down menu as shown below.
Then click Run to obtain the results on the next page.
These sections of output can be shut
off as our interest is in primarily
identifying which effects are
significant. These results are in the
Effect Tests box.
The Fluid*Capsule interaction is
significant (p=.0049), so we know
both fluid and capsule type
significantly effect the response.
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 Fluid is not significant (p = .9361). This
happens because the presence of the Fluid*Capsule interaction &quot;masks&quot; the main effect of
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Spring 2014
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.
Select LSMeans Plot from the
pull-down menu to obtain a plot
of the interaction between the
two experimental factors.
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 Capsule*Fluid interaction pull-down menu.
Results of the treatment mean comparisons are shown below.
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Spring 2014
Here we see that C,Gastric and V,Gastric 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 than type V
capsules in gastric fluid. In contrast type the capsules appear to dissolve equally
well in Duodenal digestive juice. When comparing treatment combination
means in the presence of a significant interaction we should generally only make
comparisons between means across one factor while the other is held constant.
For example, here we could compare juices for type C capsules or compare juices
for type V capsules. Also we could compare capsule types when dissolved in
Gastric juice or Duodenal juice. We generally wouldn’t compare the mean
response for type C capsules in Gastric juice to the mean response for type V
capsules in Duodenal juice.
If the interaction between the two factors is not significant we can use the
Tukey’s HSD procedure to compare the means across the levels of each factor
individually, similar to what we did for one-way ANOVA.
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 and each 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, i.e. the spread of the
observed response values about their respective treatment combination means, 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.
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To examine the normality assumption we assess the normality of the residuals. Save the
residuals to the spreadsheet using the Save Columns &gt; Residuals option as shown
below and then use Analyze &gt; Distribution to examine them. With the exception of two
mild outliers, normality seems satisfied.
Example 15.2 – Apple Nutrient Level by Region Grown and Variety
(Datafile: Fruit-nutrient.JMP)
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.
The data in JMP is entered as shown on
the left. The first column contains the
Region grown, the second column
contains the apple Variety, and the last
column contains the response the
nutrient level.
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In JMP select Analyze &gt; Fit Model and set up the dialog box as shown below.
Put the response Nutrient in the Y
box, then highlight both Region &amp;
Variety and select Full Factorial
from the Macros pull-down.
The output is shown below.
Test results:
Interaction
Region
Variety
The interaction plot shown
on the left shows some
signs of non-parallelism
and hence interaction,
however the p-value in the
ANOVA table suggests we
have very weak evidence
for its significance
(p=.0917).
The mean nutrient levels differ across region (p = .0065). In particular, the
mean nutrient level found in apples grown in region A appears to be
significantly higher than that for the other two regions.
The mean nutrient levels also differ significantly across variety (p&lt;.0001).
It appears that apple varieties Y and Z have higher mean nutrient levels
than varieties W and X.
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Multiple Comparisons for Region and Variety Effects
Because the interaction between Region and Variety is NOT significant we should look
at multiple comparisons for both Region and Variety individually. Because both main
effects have more than 2 levels we use LSMeans Tukey’s HSD to compare the mean
nutrient levels across region and variety. If a factor has only 2 levels we select LSMeans
Student’s t to compare the means for the two levels.
Discussion:
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....
Checking Required Assumptions
Constant Variance
Normality
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STATISTICAL DETAILS (FYI only)
Two-way ANOVA Model
y ijk     i   j  ( ) ij   ijk
i  1,..., a
j  1,..., b
k  1,..., n
where,
X ijk  kth observed response value when level i of factor A and level j of factor B is used.
 i  effect due to the fact level i of Factor A was used.
 j  effect due to the fact level j of Factor B was used.
( ) ij  effect due to the interaction of ith level of Factor A and the jth level of Factor B.
 ijk  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  ijk ~ N (0,  2 ) , 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:
H o : ( ) ij  0 for all treatment combinations
H a : ( ) ij  0 for all treatment combinations
2. For testing the Factor A effect we have:
H o :  i  0 for all i
H a :  i  0 for all i
3. For testing the Factor B effect we have:
H o :  j  0 for all j
H a :  j  0 for all j
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 &amp; B, and random error.
Sum of Squares:
SSTotal  SS A  SS B  SS A B  SS Error
Degrees of Freedom: N  1  (a  1)  (b  1)  (a  1)(b  1)  ab(n  1)
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SUM OF SQUARES FORMULAE:
a
b
n
a
b
i 1
j 1
 (Yijk  Y ) 2 = nb (Yi  Y ) 2 + an (Y j  Y ) 2 +
i 1 j 1 k 1
a
b
n (Yij  Yi  Y j   Y ) 2 +
i 1 j 1
a
b
n
 (Y
i 1 j 1 k 1
ijk
 Yij ) 2
MEAN SQUARES (measures of variation due to each of the possible effects)
The mean square for an effect is the effect sum of squares divided by the degrees of
freedom.
MS effect 
SS effect
df effect
When the null hypothesis of “no effect” is true the mean squares are all estimates of  2 ,
the common response variance for all treatment combinations. If there is a significant
effect then we expect the MS effect  MS Error  (within treatment combination variation).
Testing Effect Significance
For testing the main effects (A &amp; B) and the interaction effect (A  B) we simply compare
the size of the MS effect to the MS Error . If the MS effect &gt;&gt; MS Error we have evidence that
the effect is significant. If MS effect  MS Error then we have little evidence that the effect is
significant. This is analogous to the comparison of the between group variation to the
within group variation in One-way ANOVA.
To compare the mean squares we use the ratio, which has an F-distribution.
MS effect
Fo 
~ F-distribution (numerator df = df for the effect , denominator df = df for error)
MS Error
Fo &gt;&gt; 1 will lead to the conclusion that the effect in question significantly impacts the
response. Large Fo values lead to small p-values which support effect significance.
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