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The most basic assumption to check is whether the correlation pattern your model assumes is consistent with the study design. If your response variables are serially collected over a time span than is long with respect to what's happening, then you need to take into account the forward flow of time by modeling serial correlation (e.g. Markov binary or ...


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In your last question, I provided a hierarchical Bayesian mixture model. That model can be generalized quite easily (to my surprise) to accommodate your changes here. The extension is not intended to be a legitimate solution to your problem, considering you're not familiar with Bayesian modelling. In any case, I'm posting it here for posterity. I'll first ...


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There are a few solutions from a Fequentist perspective, but I'd like to offer a Bayesian perspective. The Model It would be useful to model the data generating process as a mixture of binomials; one with 0 bias and the other with some non-zero bias left to be estimated. You can then partially pool the data from all the coins to help estimation. Here is ...


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I am not sure just what your question is asking you to do, Here are some relevant computations from R: Test $H_0:p=.5$ vs $H_a: p < .5$ at level 5% (or lower). If $p = .5,$ then $P(X \le 41) = 0.0443,$ where $X$ is the number of Heads in 100 tosses. So you can have a one-sided test at level 4.4%. Changing the critical value to 42 would give a test above ...


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There are certainly many ways you could construct a test statistic giving the evidentiary ordering for this hypothesis test. Your method is coherent, but it is unlikely to be powerful. Since there is no a priori information on the degree of bias in a biased coin, a natural thing to do here would be to compute the maximum deviation from the expected value ...


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I'm the developer of DHARMa. Below answers to your questions, but note for the future that it would be preferable to ask questions about syntax problems on the DHARMa development site here. Because in your data, data$journalID is coded as a character, if you code as.factor it works No they are not, as there are different ways to make predictions (...


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This post really helped me and I wanted to thank you. In case other users ran to the same issue I had - I am adding a slight change to the simulation above. The only thing here is that this shows that Pearson corr for two times measurements is exactly the same as $\rho$. Nothing special - only nice to see the numbers match :) Also, ever so slight correction ...


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Unfortunately, a group-level variable cannot be an outcome in lme4 models. Multilevel models require that the outcome be measured at the lowest level of the data hierarchy. The multilevel model partitions the variance in the outcome across the various levels - within-group, between-group, etc. Predictors at each level can be used to explain variance at their ...


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Let's say the Hausman test favors the fixed effect model. Why might you still be interested in using the random effects model? The random effects or multilevel model allows a degree of flexibility in modeling that is much messier and in some cases impossible to implement in the fixed effect model. Examples include: Random slopes for the association between ...


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What the the Hausman test is doing is testing whether the results (i.e. the estimated coefficients) from a fixed effects and random effects model are significantly different. (I haven't ever seen people talk about it testing whether whether the fixed and "between" effects are different, although because the random effects model is estimated as a ...


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There is no right answer here. The only rule of thumb is that an intraclass correlation coefficient (ICC) > 0 is indication that you need to somehow account for the correlations of observations from the same individual or group. As an education researcher, I see ICCs ranging from .05 up to .50, depending on the outcome and whether I am modeling ...


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If the imputed datasets are in long form (dataset 2 stacked onto dataset 1) then you can use mitml the to do the pooling of the estimates from your model to give you the correct standard errors. See the code below: library(mitml) ### Define a list that mitml will link to the multiply imputed data. implist <- as.mitml.list(split(df, df$imp)) #imp is ...


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Building off Matt Barstead's comment, looking at the fit information of model 3 vs model 4, I see no evidence that model 4 is the better model. The log likelihood is higher (with 5 additional degrees of freedom used), both AIC and BIC are higher, suggesting that model 4 is not doing a better job of prediction. You can use R's anova() function to test the two ...


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You could go with a mixed formulation, which substitutes the level 2 equation into the level 1 equation: $Y_{ij} = \alpha_0 + \beta_1*age_{ij} + \beta_2*gender_{ij} + \alpha_1*living\_area_{j} + \alpha_2*province_{j} $ $\alpha_j \sim N(0, \sigma^2_b)$ This correctly signifies that living area and province are measured at the upper level of your data ...


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