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I was looking at this page and noticed the methods for confidence intervals for lme and lmer in R. For those who don't know R, those are functions for generating mixed effects or multi-level models. If I have fixed effects in something like a repeated measures design what would a confidence interval around the predicted value (similar to mean) mean? I can understand that for an effect you can have a reasonable confidence interval but it seems to me a confidence interval around a predicted mean in such designs seems to be impossible. It could either be very large to acknowledge the fact that the random variable contributes to uncertainty in the estimate, but in that case it wouldn't be useful at all in an inferential sense comparing across values. Or, it would have to be small enough to use inferentially but useless as an estimate of the quality of the mean (predicted) value that you could find in the population.

Am I missing something here or is my analysis of the situation correct?... [and probably a justification for why it isn't implemented in lmer (but easy to get in SAS). :)]

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  • $\begingroup$ Since in essence the nesting in a lmer makes it a repeated measures design is there a way in which your question about the appropriate confidence interval around the effect size is related to the question in repeated-measures ANOVA about which measure of effect size to report? Specifically, it is unclear whether the error term should include subject variance or not (etc)? $\endgroup$
    – russellpierce
    Commented Aug 18, 2010 at 14:55
  • $\begingroup$ Nevermind - I didn't think that all the way through. $\endgroup$
    – russellpierce
    Commented Aug 18, 2010 at 15:06

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It has the same meaning as any other confidence interval: under the assumption that the model is correct, if the experiment and procedure is repeated over and over, 95% of the time the true value of the quantity of interest will lie within the interval. In this case, the quantity of interest is the expected value of the response variable.

It is probably easiest to explain this in the context of a linear model (mixed models are just an extension of this, so the same ideas apply):

The usual assumption is that:

$y_i = X_{i1} \beta_1 + X_{i2} \beta_2 + \ldots X_{ip} \beta_p + \epsilon $

where $y_i$ is the response, $X_{ij}$'s are the covariates, $\beta_j$'s are the parameters, and $\epsilon$ is the error term which has mean zero. The quantity of interest is then:

$E[y_i] = X_{i1} \beta_1 + X_{i2} \beta_2 + \ldots X_{ip} \beta_p $

which is a linear function of the (unknown) parameters, since the covariates are known (and fixed). Since we know the sampling distribution of the parameter vector, we can easily calculate the sampling distribution (and hence the confidence interval) of this quantity.

So why would you want to know it? I guess if you're doing out-of-sample prediction, it could tell you how good your forecast is expected to be (though you'd need to take into account model uncertainty).

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  • $\begingroup$ That's my second scenario, the confidence interval is too large to have any inferential value within the design of the experiment since differences between conditions are based on effects with between S variability removed. It seems it always has a compromise meaning and needs it's own special name because you can't use it like a regular CI. $\endgroup$
    – John
    Commented Aug 18, 2010 at 23:19
  • $\begingroup$ Blouin & Riopelle (2005) called them narrow and broad inference confidence intervals but given that the general scientific populace outside stats has a hard enough time with regular ones... $\endgroup$
    – John
    Commented Aug 18, 2010 at 23:25
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Maybe this makes sense in the Bayesian framework. Consider for instance the one-way random effect ANOVA model : $$(y_{ij} | \mu_i) \sim {\cal N}(\mu_i, \sigma^2_w), \quad \mu_i \sim {\cal N}(\mu, \sigma_b^2),$$ and a prior distribution on the overall mean $\mu$ and the variance components $\sigma^2_w$ and $\sigma^2_b$. Then each $\mu_i$ has a posterior distribution, and a $95\%$ dispersion interval of this distribution could play the role of a $95\%$ "confidence" interval.

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