1
$\begingroup$

I've chosen to compare the results of three different machine-learning models outputting the mouth animation of a talking 3D model for my thesis using a user preference study.

Users will see two videos at a time, each from one of the models, and indicate which they think look the most natural, or if they are neutral. The videos are synchronized. For each user, one comparison will be made for each sentence (out of a set of sentences) and for each pair of models to be compared.

Now, I could arrange the data as simply A vs. B, B vs. C, and A vs. C. Then, let's say A was preferred a majority of the time over B (a majority out of a total of sentences*subjects comparisons between them). What test should I use to see if this is statistically significant? Could I even just compare them all at the same time by treating the "wins" as "points" to get three sums or means representing "scores" and then use a repeated measures ANOVA followed by e.g. Tukey HSD?

A lot of methods (Tukey's HSD, Student's t-test...) have the assumption that the score in A needs to be independent to the score in B, but since this is a preference study, they aren't really independent, right? E.g. if there's a total of 20 un-neutral comparisons, if A wins 15, then B wins 5.

I would greatly appreciate any insight!

$\endgroup$
3
  • $\begingroup$ Can you give a little more detail? How many sentences are there? How many subjects? Do all subjects see all models? Do all subjects see all sentences? Is the pairing between models and sentences and comparisons exhaustive? You're probably going to want something like this: stat.wisc.edu/~larget/Stat998/Fall2015/… $\endgroup$
    – triddle
    Commented Aug 23, 2018 at 14:58
  • $\begingroup$ Also, it's worth pointing out that human preferences are not transitive, but it sounds like you want a strict (i.e. transitive) ordering of the models. So you're going to have to make an assumption somewhere that might not hold true in reality. $\endgroup$
    – triddle
    Commented Aug 23, 2018 at 15:00
  • $\begingroup$ Thanks, I'll check it out! Haven't decided yet, but there are 16 sentences left for testing. I could make more if I wanted but I think it'd be too much for the subjects. As for them, it's going to have to be as many as I can find basically, but maybe 20-30 or so (might have to use e.g. Mechanical Turk if I don't find enough willing friends/family). Yep, all subjects see all comparisons of an exhaustive pairing, so all sentences and all models, but in random order. It's basically the same thing as in this e.g. paper: disneyresearch.com/publication/deep-learning-speech-animation $\endgroup$
    – Vaering
    Commented Aug 23, 2018 at 15:06

1 Answer 1

2
$\begingroup$

Right, I'd treat this as a multilevel logistic regression. Formally:

$y_{ijk} = \beta_0 + \beta_1V_{ijk} + \alpha_i^p + \alpha_i^{p*v}V_{ijk} + \alpha_j^s + \alpha_j^{s*v}V_{ijk} + \alpha_{ij}^{p*s*v}V_{ijk} + \epsilon_{ijk}$

where you're modeling the response (y) to the $i^{th}$ sentence from the $j^{th}$ subject to the $k^{th}$ ml model.

$\beta_0$ is the overall intercept (i.e. mean response)

$\beta_1$ reflect the change expected overall moving from one model to another (this depends on how you encode the models - here I'm assuming dummy coding)

$V_{ijk}$ is the ml model $k$ for subject $j$, sentence $i$

The $\alpha_i^P$'s and $\alpha_j^s$'s represent "random intercepts", or how much each individual's ($p$) or sentence's ($i$) mean differs from the overall mean.

The $\alpha_i^{p*v}$ $\alpha_j^{s*v}$ represent "random slopes", or how much each individual or sentence shows a different pattern of preferences among the ML models.

Finally, the $\alpha_{ij}^{p*s*v}$ represents an interaction between the possible subject-sentence pairings.

To estimate this kind of model in R, you would use the following code:

library(lme4)
m <- glmer(y ∼ v + (v|subject) + (v|sentence) + (v|subject:sentence), 
           data=df, family='binomial')

Here's a some other good references on these models:

https://www.annualreviews.org/doi/abs/10.1146/annurev-psych-122414-033702 http://jakewestfall.org/publications/JWK.pdf https://www.sciencedirect.com/science/article/pii/S0749596X08000193

Edit

Based on your comments below, and points I initially overlooked, I think the model should be reparameterized slightly. Basically, I don't think you have the data to estimate the final $\alpha_{ij}^{p*s*v}$ from above. Additionally, I overlooked the point that there's no clearly defined 'baseline' in your case, since the levels of your $V_{ijk}$ should be m1_vs_m2, m1_vs_m3, and m2_vs_m3, and it isn't clear how to interpret an estimated difference between m1_vs_m2 and m2_vs_m3.

Relatedly, I was a little bit vague about the meaning of $y_{ijk}$. This should be coded as 0/1, where 1 means the right-hand model was selected. For completeness, this means that for the category m1_vs_m2, a 0 would indicate that m1 was selected, and a 1 would indicated m2 was selected; for the category m1_vs_m3, a 0 would indicate that m1 was selected, and a 1 would indicated m3 was selected; for the category m2_vs_m3, a 0 would indicate that m2 was selected, and a 1 would indicated m3 was selected.

Your data would be set up to look something like this:

  y           v subject sentence
1 1       m1vm2       1        8
2 1       m1vm2       1        1
3 1       m1vm3       1        4
4 0       m2vm3       1        6
5 1       m1vm2       1        1
6 0       m1vm2       1       12

The model using lme4 syntax would look like this:

library(lme4)
m <- glmer(y ∼ v + (v|subject) + (v|sentence), 
           data=df, family='binomial')

The (v|subject) and (v|sentence) terms are the parts of the model that take care of your concerns of non-independence. You're partitioning the variance into variance attributable to characteristic differences between subjects and sentences. Basically, you're allowing the intercept and the effect of v to vary by each of those factors.

For interpretation, you're going to want to examine the intercept of this model and the coefficients for the levels of v. Assuming m1_vs_m2 is your baseline, the intercept would reflect the log of the odds that m2 is preferred. In a null-hypothesis testing framework, if the intercept is statistically significant, then that corresponds to a preference for m2 over m1. The coefficients for v will represent deflections from this intercept for your other two comparison categories.

I'm not sure what software you were planning to use to analyze your data, but that's a bit beyond the scope of the original question. If you think this is a reasonable approach, I'd then post on stackoverflow to figure out how to program this in your language of choice.

$\endgroup$
3
  • $\begingroup$ Cool, I haven't encountered this kind of model before, and haven't even used R either. So if I understand correctly, your model assumes I have, as data, the following columns: sentence, subject-, model, wins(0, 1 or 2 since there are two other models) i.e. i, j, k, y_ijk So for each i,j,k we have the number of wins, which you defined as a function of i,j,k, so that one could later see if one can predict it given k. If so, there might be a significant enough difference between the number of wins for the different models to claim that one is better than the other? And V_ijk is just k? $\endgroup$
    – Vaering
    Commented Aug 24, 2018 at 14:22
  • $\begingroup$ Or well, looking at your R code, I'm trying to understand this lme4 syntax but the v corresponds to the B1*V_ijk term I think? Then, I can't find what (v|subject) means... is it that we get v random intercept terms per subject? $\endgroup$
    – Vaering
    Commented Aug 24, 2018 at 15:20
  • $\begingroup$ I edited the question to clarify these questions. These models are pretty complicated, but if you need to use them, you'd probably do well to read a couple of the papers I linked above, or others that hit on similar ideas. $\endgroup$
    – triddle
    Commented Aug 31, 2018 at 18:22

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.