2
$\begingroup$

I'm planning a study on interaction techniques in virtual reality. That means I want to compare the performance of the participants on different interaction forms (e.g. selecting objects with a ray or by grabbing) in different scenarios (e.g. different distances and object sizes). My study design is rather complex and I'm not that familiar with data analysis so I hope you can help me with the following two questions:

  1. I have multiple Independent variables. For simplification let's have a look on three of them. A is the interaction technique, B is the distance and C is the object size. Not all Interaction technique work for all distances. That means for some interaction techniques there will be missing data: \begin{array} {|r|r|}\hline & B_1 & B_2 & C_1 & C_2 \\ \hline A_1 & x & & x & x \\ \hline A_2 & x & x & x & x \\ \hline A_3 & x & x & x & x \\ \hline A_4 & x & & x & x \\ \hline ... & & & & \\ \hline \end{array} $X_1$ and $X_2$ indicate the different conditions of a variable $X$. After the study I want to analyse the data for $B_1$ and $B_2$ separately. If a technique does not support $B_2$ it will simple not considered in the analysis. Is this possible? I'm afraid that the two conditions are somehow connected. I think the cleanest way would be to do two studies investigate $B_1$ and $B_2$ separately but that would be more time consuming ...

  2. I want to test 12 different interaction techniques. That's a lot and it is impossible to let all participants test all techniques. Therefore each participant will test 3 random interaction techniques. That means I have some kind of a mix of within-subject and between-subject design. Unfortunately there are no fixed groups because of the randomly assigned techniques. Therefore I cannot use a split plot ANOVA for example. Are there any other models I can use? Or is it possible to assume that each technique was used by a different person even if one person used multiple techniques? Then if would be possible to use a split plot ANOVA.

$\endgroup$
2
  • $\begingroup$ What is your response variable(s)? Measured like time, or success/failure? One or many? $\endgroup$ Dec 20, 2019 at 15:37
  • $\begingroup$ I measure the time needed for a task and the number of misses (participant did not hit the target object). So two response variables. $\endgroup$
    – Matthias
    Dec 20, 2019 at 16:25

1 Answer 1

1
$\begingroup$

You have factors (predictors) A (technique), B (distance), C (object size) and a random effect (Participant), and a response variable that I now take as measured (maybe a time?, you didn't specify.) Without knowing much context I can propose a simple linear mixed model $$ Y_{ijkl}=\mu + \alpha_j + \beta_k + \gamma_l + \pi_i + \epsilon_{ijkl} $$ where $i$ indexes participant, $j$ technique, $k$ distance and $l$ object size, and not all combinations of indexes occur (maybe also some interaction terms are necessary.) $\pi_i \sim \mathcal{N}(0,\sigma^2_\pi)$ is a participant random effect, $\epsilon_{ij} \sim \mathcal{N}(0, \sigma^2)$ is experimental error.

Modern linear mixed model software like lme4 can handle this model without all index combinations being present (those non-present simply being left out from the data file.) What you should not do is hiding the non-independence created by each participant contributing three observations! The one participant random effect here represents that dependence.

The model written above should only be taken as a point of departure, there might be aspect you didn't tell us about ...

About the experimental design: The layout should not be chosen haphazardly, but systematically, maybe to optimize some criterion like D-optimality. But I am not sure about software for D-optimality (or similar) for linear mixed models.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks a lot for your answer. A linear mixed model seems to fit very well for my study design. I worked my way though this tutorial on linear mixed models with R and found it very instructive. One last question: I expect a high variance between the different interaction techniques. Is there any reason I shouldn't use the interaction techniques as a random factor? Then I would have two partially crossed random factors (interaction technique and participant). $\endgroup$
    – Matthias
    Dec 28, 2019 at 16:12

Your Answer

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

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