Stat 462 Feb. 27
Interaction (Section 8.2)
In statistics, the word interaction has to do with the combined effect of two predictor
variables (x-variables) on the response variable y. An interaction is said to be between the
two predictor variables.

Specifically, two x-variables, say x1 and x2, interact if the effect of x1 on y
depends upon the specific value present for x2.
In regression, the word “effect” refers to how much E(y) changes when an x is changed
by one unit.
Example 1
An experiment is done in which participants are smokers trying to quit.
x1 = 0 if participant used placebo patch
x1 = 1 if participant used nicotine patch
x2 = 0 if participant did not live with another smoker
x2 = 1 if participant lived with another smoker
Here are success rates (percent not smoking after 8 weeks).
x2 = 0
(No other smoker in home)
x1 = 0 (Placebo patch)
20%
x1 =1 (Nicotine patch)
58%



x2=1
Another smoker in home
20%
31%
When no other smoker is in the home (x2 = 0), the difference in success rates for
nicotine patch versus placebo patch is 58%20% = 38%.
When another smoker is in the home (x2 = 1), the difference in success rates for
nicotine patch versus placebo patch is 31%20% = 11%.
So the effect of x1, which is the difference in success rates for nicotine patch
(x1=1) versus placebo patch (x1=0), depends upon the specific value present for
x2 (whether another smoker is in the home or not). Thus, there is an interaction
between x1 and x2.
Example 2
An experiment is done to determine the maximum torque (circular force) that can be
applied before a mechanism used to spin logs being cut for plywood fails. Logs of two
different diameters are used. The spinning mechanism is inserted into the logs at four
different depths.
Following is a graph of y = torque (before failure) versus Depth with the two different log
sizes marked with different symbols. Notice that the slope appears to differ for two sizes.
.
Thus, the model we should consider is
E(Torque) = 0 + 1 Depth + 2Diameter + 3 (Depth*Diameter)
Minitab output follow. Notice that the interaction term is statistically significant.
The regression equation is
Torque = 13.8 - 8.68 Depth + 0.662 Diameter + 1.97 Interact
Predictor
Constant
Depth
Diameter
Interact
S = 2.697
Coef
13.795
-8.678
0.6625
1.9667
SE Coef
5.817
2.678
0.9405
0.4330
R-Sq = 90.9%
T
2.37
-3.24
0.70
4.54
P
0.028
0.004
0.489
0.000
R-Sq(adj) = 89.6%
Questions:
1. The Diameter term is not significant (p=0.489). What does this mean in this
situation?
2. Suppose we did the regression E(Torque) = 0 + 1 Depth. What would be the
appearance of a plot of residuals versus fits?
3. Suppose we did the regression E(Torque) = 0 + 1 Depth + 2Diameter. What
would be the appearance of a plot of residuals versus fits?
Example 3
College students are asked to report their actual weight and their ideal (desired) weight.
Here’s a graph of y = ideal versus x =actual with different symbols used for males and
females and a Lowess smoother used to help identify a pattern.
There may be a curved pattern and this pattern may differ for males and females. After a
few attempts, I found reasonable success with the model estimated by the following
output. The variable actwt150 = actual150, the variable sexcode = 1 if Female and 0 if
Male, and Lin_inte = sexcode*actwt150.
The regression equation is
ideal = 156 + 0.865 actwt150 - 22.3 sexcode - 0.00334 actwt150sqrd
- 0.350 Lin_interact
Predictor
Constant
acwt150
sexcode
actwt150sqrd
Lin_inte
S = 8.822
Coef
155.567
0.86516
-22.284
-0.0033351
-0.35036
SE Coef
1.575
0.06681
1.933
0.0009648
0.09383
R-Sq = 91.4%
Here’s a graph of the fits:
T
98.75
12.95
-11.53
-3.46
-3.73
P
0.000
0.000
0.000
0.001
0.000
R-Sq(adj) = 91.2%