6
$\begingroup$

I have run two psychological experiments. The dependent variable is a rating provided by the participant, that is, an integer number from 0 to 100.

The first experiment is a mixed 2x2 design, with one between-subject factor (treatment) and one within-subject factor (question). Each of these factors has two levels. That is, participants are asked two questions in the experiment (Q1 and Q2), while a factor varies systematically between the two groups (Treatment1 and Treatment2).

The second experiment is identical, except for that the design becomes completely within-subjects. That is, it is a repeated measures 2x2 design; participants are asked four questions that encode the two manipulations in a factorial form: Q1Treatment1, Q1Treatment2, Q2Treatment1, and Q2Treatment2.

I am interested in the potential interactions between the two factors (question and treatment).

Normally, I would conduct repeated measures ANOVAs for these experiments. However, the problem is that my data deviate strikingly from normality (in fact, so do the residuals). Therefore, I am looking for nonparametric equivalents to ANOVA for these two designs.

I can use both SPSS and R.

Improve this question
$\endgroup$
7
  • $\begingroup$ Just wondering, but did you try doing the ANOVA anyway? $\endgroup$
    – Behacad
    Jan 7, 2014 at 17:33
  • 2
    $\begingroup$ These aren't exactly nonparametric, but there are nonlinear mixed effect models that may suit your problem. $\endgroup$
    – Peter Flom
    Jan 7, 2014 at 18:21
  • 3
    $\begingroup$ You may be able to come up with permutation procedures that encode your null hypotheses, in which case you can use a permutation test. In general, however, testing interactions via nonparametrics is difficult, because it intrinsically involves a deviation from additivity, which seems (??) impossible to reconcile with the kind of arbitrary rescalability that goes with nonparametric procedures. In your position I would be inclined to see if there's some parametric model that works. What are your responses? Likert? Binary? $\endgroup$
    – Ben Bolker
    Jan 8, 2014 at 1:27
  • $\begingroup$ @Behacad: I tried, but when I look at the distributions (Q-Q plot, etc.) they seem to deviate a lot from normality. $\endgroup$
    – Experimental Psychologist
    Jan 8, 2014 at 9:25
  • $\begingroup$ @Peter: It's a good suggestion, but I have never conducted that type of analysis, so I'll look for some literature. Apparently there is plenty of introductory texts. It's worth trying. </br> $\endgroup$
    – Experimental Psychologist
    Jan 8, 2014 at 9:40

1 Answer 1

Reset to default
1
$\begingroup$

In a 2 x 2 design, it is fairly easy to run a bootstrap test of the interaction. Let define the four conditions as a, b, c and d. The conditions $a$ and $b$ are the question factor for the treatment and $c$ and $d$ are the question factor for the control condition. The mean interaction contrast (MIC) is given by

$ (a + d) - (b + c).$

It quantifies the amount of non-additivity in the dataset. If MIC is zero, it means that there is a main effect of questions (there is an increase--or decrease-- from Q1 to Q2), a main effect of conditions (there is an increase --or decrease-- from control to treatment) and no interaction. If such is the case, mean in $b$ is a few points above mean in $a$, and mean in $c$ is also the same amount of points above the mean in $d$. As of treatment, the same occur (treatment measures are a few point above control measures). Defining the first increment as $d_1$ and the second as $d_2$, the means are thus

$ \left(\begin{matrix}M_a \;\;\; M_b \\ M_c\;\;\;M_d \end{matrix}\right) = \left(\begin{matrix}M_a \;\;\;M_a+d_1 \\ M_c \;\;\; M_c+d_1 \end{matrix}\right) = \left(\begin{matrix}M_a \;\;\; M_a+d_1 \\ M_a+d_2\;\;\;M_a+d_2+d1 \end{matrix}\right) $

so that

$MIC = (M_a+(M_a+d_1+d_1)) - ((M_a+d_1)+(M_a+d_2)) = 0$.

Thus, to do a boostrap estimate, sub-samples in the groups with replacement, and compute MIC. Repeat this a very large number of times (say 5,000). Finally, find the range in which 95% of the MIC found are located. If this interval includes 0, then the interaction is not significantly different from zero.

This reasoning works for a fully between group design. In a mixed design, you have to select pairs of scores randomly before computing MIC (preserving subjects' two measures).

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.