Activity 5: Two-way ANOVA
SOLUTIONS
With two factors, there is an additional question beyond whether each factor separately is
significant: Is there an interaction between the two factors? That is, does the effect of
one factor depend on the level of the other factor?
To illustrate this, let’s consider the effects of ethanolic NaOH concentration and coal type
on total acidity. Each of these factors has three levels. Open the data from the website.
1. First, let’s look at an interaction plot. Go to Stat > ANOVA > Interactions
Plot…. Enter Acidity as the response and NaOH and Coal Type as the factors (in
that order). Copy and paste the interaction plot below. For Morwell coal, what
concentration of NaOH gives the highest mean acidity? for Yallourn? for
Maddingley?
For all three types of coal, .786N NaOH gives the highest mean acidity.
Since the lines (for each concentration of NaOH) are roughly parallel, it appears that the
effect of NaOH concentration does not depend on the type of coal. This suggests that
there is NO INTERACTION between the factors. Thus, when we perform a two-way
ANOVA, the interaction term will probably not be significant. (Check but do not answer:
Do you get a different intuition if you produce an interaction plot with the factors in the
other order?)
2. So let’s look at the results. Go to Stat > ANOVA > Two-way…. Enter Acidity as
the response, NaOH as the row factor and Coal Type as the column factor. Copy and
paste the ANOVA table (and ONLY the table) below. What are the test statistic and pvalue for the interaction term? What can you conclude?
Analysis of Variance for Acidity
Source
NaOH
Coal Typ
Interaction
Error
Total
DF
2
2
4
9
17
SS
0.1243
1.0024
0.0146
0.1530
1.2942
MS
0.0622
0.5012
0.0036
0.0170
F
3.66
29.49
0.21
P
0.069
0.000
0.924
The statistic and p-value for the interaction are F=.21 (on 4, 9 df) and p=.924. We can
conclude from such a large p-value that there is no significant interaction between NaOH
and Coal Type.
If the interaction term is not significant, then we can focus on the main effects…the
effects of the individual factors.
3. Since the interaction term isn’t significant, let’s not waste those 4 degrees of freedom
on the interaction when they could be used to give a more accurate estimate of the mean
square error. Refit the model without the interaction term by clicking the “Fit additive
model” box in Stat > ANOVA > Two-way…. Paste the new ANOVA table below.
Does NaOH concentration have a significant effect on acidity? What about the type of
coal?
Analysis of Variance for Acidity
Source
DF
SS
MS
NaOH
2
0.1243
0.0622
Coal Typ
2
1.0024
0.5012
Error
13
0.1675
0.0129
Total
17
1.2942
F
4.82
38.90
P
0.027
0.000
NaOH has a significant effect on mean acidity (p=.027). Type of coal also has a
significant effect on mean acidity (p=.000).
Let’s consider a different example…the effects of brand of pen and writing surface on
writing lifetime.
4. Obtain an interaction plot (paste it below). Which brand of pen lasted longest on the
first surface? the second surface? the third surface?
Pen 1 lasted longest on the first surface; pen 2 lasted longest on the second surface; and
pen 3 lasted longest on the third surface.
Here, the effect of brand depends on the surface! This suggests that THERE IS AN
INTERACTION between the two factors. Notice that the lines cross (intersect) multiple
times. Thus, we will most likely find a significant interaction term.
5. Perform a two-way ANOVA. Be sure NOT to fit the additive model so that you can
obtain an interaction term. What are the test statistic and p-value for the interaction term?
What can you conclude?
Analysis of Variance for Lifetime
Source
DF
SS
MS
Pen
3
1388
463
Surface
2
2888
1444
Interaction
6
8100
1350
Error
12
8216
685
Total
23
20592
F
0.68
2.11
1.97
P
0.583
0.164
0.149
The test statistic and p-value for the interaction are F=1.97 (on 6, 12 df) and p=.149. We
conclude that there is not a significant interaction between pen and surface. This appears
to contradict what we saw in question 4. The explanation for this apparent contradiction
is that the interaction plot in question 4 does not take the precision (variability) of the
mean estimates into account, whereas the ANOVA table does. Even though the lines are
clearly not parallel in the interaction plot, the estimates are so variable that we cannot
distinguish those lines from lines that are truly parallel.