1
$\begingroup$

I have a conceptual problem.

I want to find out if stress during the day leads to (stronger) teeth grinding (bruxism) at night. I have a number of participants. They will fill in a self-report questionnaire rating their level of stress during the day. During the night, the durating and strength of their teeth grinding are measured electronically. Each participant will do this on a number of consecutive days. As a result I will have data in this form:

participant  stress_day_1  bruxism_night_1  stress_day_2  bruxism_night_2
1            5             8                1             3
...

What I'm usure about is wether I should take all the stress/bruxism value pairs and do my analysis on them, disregarding the fact that they come from different patients, or if I should calculate means for each patient and do my analysis on these means, or something else entirely.

My variables are stress and bruxism, but I'm confused as to what place the participants should have in my analysis. I only have three nights, so cannot do a time series analysis.

I would greatly appreciate any feedback.

Also, please edit the tags as you see fit. I wasn't sure which ones to choose.

$\endgroup$

2 Answers 2

3
$\begingroup$

I guess the standard way of doing this is by using linear mixed models. This way you can evaulate the effect of (self reported) stress on bruxism, accounting for the fact that the data come from the same participants.
Have a look at the very good (and short, and accessible) tutorials by Bodo Winter:

$\endgroup$
0
1
$\begingroup$

If I understand your question well, then you want to estimate a regression $y=\beta_1 x + \beta_2$, where (a) $x$ is stress during the day and $y$ is grinding, and (b) you think that at least one of the coefficients could be dependent on the participant, in other words your model is more like $y=\beta_1^{(p)} x + \beta_2^{(p)}$ where the superscript $(p)$ indicates that the coefficient can change from participant to participant.

Let us for simplicity assume that the constant changes from participant to participant but not the slope (but the technique decribed below also works if both are depedent on the patient), so your model is: $y=\beta_1 x + \beta_2^{(p)}$. If the number of participants is limited, then you might introcuce a dummy for the participants, but that inflates the number of parameters to estimate.

A better idea is to introduce a so-called 'random intercept' (see e.g. Fitzmaurice, Laird, Ware, "Applied Longitudinal Analysis"), in that case you assume that the intercept is different for each participant, but the distribution of the intercepts for all participants is normal with a certain mean and a certain standard deviation. If you do this then you limit the number of parameters to estimate to (a) the mean of the random intercept, (b) the standard deviation of the random intercept and (c) the slope, while if you dummy encoding for the participants you will have to esimate $N$ (the number of participant) minus one dummies and the slope.

In R you can use the package nlme for this and with the above notations ($x$ is stress, $y$ is grinding) and there is a random intercept you have the syntax

lme(y  ~ x + 1, 
    random= ~ 1|participant, 
    data=myData, 
    method="REML")

where your dataset has the structure has four columns: y (grinding), x (stress), day, participant, so you use one line for each day.

$\endgroup$
5
  • $\begingroup$ random intercept is also mixed model, i think. $\endgroup$
    – Deep North
    Aug 10, 2015 at 10:27
  • $\begingroup$ @Deep North: that's right, apparently we posted the answer at the same time ... $\endgroup$
    – user83346
    Aug 10, 2015 at 11:00
  • $\begingroup$ hehe, +1 alreday $\endgroup$
    – Deep North
    Aug 10, 2015 at 11:09
  • $\begingroup$ Thank you for the explanation, @fcoppens, this helped me a lot. $\endgroup$
    – user14650
    Aug 10, 2015 at 11:26
  • 1
    $\begingroup$ @what: Note that you can have a random intercept and a random slope, then the R-code is similar except that the parameter random for the lme function should be 'random= ~ x+1|participant', for the interpretation of the output I refer to literature on the topic like the one I mentioned. $\endgroup$
    – user83346
    Aug 10, 2015 at 12:09

Your Answer

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