# Tag Info

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If you have a factor with $n$ levels, you can test $n - 1$ effects (at maximum) associated with this factor in one model. Have a look at the sum contrasts. The function contr.sum(3) returns the following matrix: [,1] [,2] 1 1 0 2 0 1 3 -1 -1 The columns denote the contrasts. For a factor with $3$ levels, you obtain $2$ contrasts. Your ...

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This is not a direct answer for lme's syntax. I would argue that while in theory a specific examiner is part of the greater examiner population and it does make sense to have it as a random effect, you have only 2 (and occasionally 3) replicates. It will most probably be more sensible to use it as fixed effect (possibly as an interaction). Moreover I would ...

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d1 <- data.frame(j, y=g00+u0j+(g10+u1j)*x1+e, x1) # You need to add error terms My intuition is that when we simulatated a mixed model data set without error terms, sometimes it may be difficult to converge. This is likely to be a zero residual problem.

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The diamond pattern in the residuals is due to a combination of the "ceiling effect" you have in the observed female data (i.e., a lot of female data points clustered at the maximum value of 18 or so) and the "floor effect" you have in the observed male data (a lot of male data points clustered at the minimum value of 0). To see this, imagine taking your ...

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You can estimate the intraclass correlation coefficient (ICC) (wikipedia link). It tells you how correlated the behavioral responses are for the same individual. It is defined as: $$ICC = \tau^2 / (\tau^2 + \sigma^2),$$ where $\tau^2$ is the "intercept" variance and $\sigma^2$ is the residual variance. Substituting the estimated values into the equation ...

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Q1: Yes - just like any regression model. Q2: Just like general linear models, your outcome variable does not need to be normally distributed as a univariate variable. However, LME models assume that the residuals of the model are normally distributed. So a transformation or adding weights to the model would be a way of taking care of this (and checking ...

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You seem quite mislead about the assumptions surrounding multi-level models. There is not an assumption of homogeneity of variance in a general sense but there is one across groups. Residuals should be roughly normally distributed. And categorical predictors are used in regression all of the time (the underlying function in R that runs an ANOVA is the linear ...

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Q1: Yes, why not? Q2: I think the requirement is that the errors are normally distributed. Q3: Can be tested with Leven's test for example.

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If time is in a continuous scale originally, make 2 models: one without random slope and one with random slope. Fit with REML. Compare them with anova(model1,model2) for quick check to see if random slope is statistically significantly better than random intercept. If it is, it means the rates of growth are different, i.e. the fitted linear lines are not ...

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Every statistical method (and every other method of making decisions as well) can give false negatives and false positives. The only way to avoid mistakes is to know in advance what is correct, but, if we already knew what was correct, we wouldn't need to do any analysis! One advantages of statistical methods is that, if you follow the rules and meet the ...

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