Math 217—Two-Factor ANOVA Example
An
experiment is conducted to gauge the effect of both temperature setting and
detergent type on the cleanliness of soiled t-shirts put through a washing
cycle. Eighteen identically soiled t-shirts are randomly assigned to a
treatment (that is, to a combination of temperature setting and detergent).
Three wash-cycle temperatures are used: cold, warm, and hot. Two different
detergents are considered: Detergent A and Detergent B. The response variable
is a cleanliness rating (on a 1–10 scale). Note there are a total of 6
treatments, and since there are 18 t-shirts, we have 3 replications within each
treatment. Shown below is the output from Minitab’s general linear model
procedure (which is equivalent to a two-factor ANOVA analysis).
General Linear Model: Cleanliness
Score versus Temperature, Detergent
Factor Type
Levels Values
Temperature fixed
3 Cold, Hot, Warm
Detergent fixed
2 A, B
Analysis of Variance for
Cleanliness Score, using Adjusted SS for Tests
Source DF Seq SS Adj SS Adj MS F
P
Temperature 2 22.3333 22.3333 11.1667
14.36 0.001
Detergent 1 20.0556 20.0556 20.0556 25.79
0.000
Temperature*Detergent 2
0.7778 0.7778 0.3889
0.50 0.619
Error 12 9.3333
9.3333
0.7778
Total 17 52.5000
[Note:
You can ignore the Seq SS column. The Adj SS column is simply the usual sum of squares we’ve
discussed—it just has a different name within the general linear model
procedure).
Before analyzing the results, we
must check the normality and constant-variance conditions. Appropriate graphs
of the residuals are shown below. The normality assumption is shaky, but perhaps
plausible. The residual plot shows a fairly constant variation in the
residuals. Hence, both assumptions appear to be roughly met.
The F-test for an interaction
effect has a large P-value, which indicates there is no interaction effect.
This can be seen visually via an interaction plot (also shown above).
Both main effects are
significant. Hence, it’s appropriate to use Tukey’s
procedure to get 95% simultaneous confidence intervals for the pairwise differences in means. Via Minitab:
Tukey 95.0% Simultaneous
Confidence Intervals
Temperature = Cold subtracted
from:
Temperature
Hot 1.3093 2.6667
4.024 (-----*------)
Warm -0.5240 0.8333
2.191
(------*------)
------+---------+---------+---------+
-2.0 0.0 2.0
4.0
Temperature = Hot subtracted
from:
Temperature
Warm -3.191 -1.833
-0.4760 (------*------)
------+---------+---------+---------+
-2.0 0.0
2.0 4.0
Tukey 95.0% Simultaneous
Confidence Intervals
Detergent = A subtracted
from:
B 1.205 2.111 3.017
(-----------------*-----------------)
------+---------+---------+---------+
1.50 2.00
2.50 3.00
Detergent B’s cleanliness rating
is significantly higher, on average, than Detergent A. Furthermore, the
cleanliness rating for hot water wash is higher, on average, than for both the
cold temperature and the warm temperature. Do you think these results are
practically significant? [Note: In lab we’ll discuss an example with a
significant interaction.]