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 Mar19 answered Difference between marginal and conditional models Mar18 comment Can somebody offer an example of a unimodal distribution which has a skewness of zero but which is not symmetrical? It's not bimodal... it's horribly multi -modal. Try plotting the density; curve(0.2*(dnorm(x, 1, .1) + dnorm(x, 3.122, .1) + dnorm(x, 5, .1) + dnorm(x, 4, .1) + dnorm(x, 1.1, .1)), 0,10) Mar18 comment How do you interpret the coefficients in PROC GLIMMIX and PROC GENMOD? Coefficients of a specific covariate represent measures of difference in mean response, in the same cluster, at observations for which the specific covariates differ by one unit and all the other covariates are identical. Depending on the link function, the "measure of difference" can be a difference, or a log-ratio, or a log odds-ratio. An exception is the intercept, which doesn't describe a difference, but instead gives the mean response in observations for which all covariates and the random effect(s) are zero. Mar18 comment How do I satisfy the nonlinearity of my regression function with this plot image? Alstat; Please add a 'homework' tag to your question. For prediction, ask your instructor about using cross-validation to pick a predictive model, then actually validating it in some independent data. Mar18 answered Combined odds ratio Mar18 comment How do I satisfy the nonlinearity of my regression function with this plot image? What are you trying to do with this data? For many purposes, you need not have linearity, so it's not clear why you want to force the relationship to be linear. Please edit your question to make the purpose of the analysis clearer. Mar16 comment What is the decision-theoretic justification for Bayesian credible interval procedures? Another way to motivate this is to re-write the loss function as a (weighted) sum of the width of the interval plus the distance, if any, by which the interval fails to cover the true $\theta$. Mar16 comment What is the decision-theoretic justification for Bayesian credible interval procedures? Very nice. But are they the only loss functions justifying these procedures? Mar15 comment Robust variance/covariance matrix in Poisson regression If you're happy with proportional hazards assumptions, you don't need to use clustering in this way; contributions to the partial likelihood at each time point only reflect people in the risk set at that time point, and (in your case) what covariates they had at that time point. Robust variance estimators (i.e. , robust in Stata) can still be useful though; they give you standard errors that appropriately summarize variability in your point estimates, regardless of whether the hazard ratio is truly log-linear in the covariates. Mar15 answered Meta-analysis in R with multiple SNPs Mar15 comment How to reference GLM? Maybe you don't need to mention GLM at all. If you did e.g. linear regression of the DV on several covariates, and based on that regression's output performed Wald tests of particular hypotheses, then just say that. (In fact there are so many ways to use GLMs that you need to say that much, to be precise about what was actually done.) Citing the piece of software you used to do the fitting would be courteous, but for use of GLM-fitting software is unlikely to be a big issue. Mar13 comment How to compare joint distribution to product of marginal distributions? The 10x10 matrix of counts in each cell are the input you would use to the Pearson $\chi^2$ test. They form a contingency table. The null hypothesis for the Pearson $\chi^2$ test can be (re)stated as saying that the distribution of entries in all columns is the same, regardless of which row you consider; this is equivalent to your definition in terms of products of marginals. (Another definition reverses "rows" and "columns" in what I wrote above). One Matlab coding of the test is in NAG toolbox (nag.co.uk/numeric/MB/manual_22_1/pdf/G11/g11aa.pdf) Mar12 comment How to compare joint distribution to product of marginal distributions? I really think you'd be better off testing correlation, and making a scatterplot. Under the null there is no correlation, so it's a valid test. However, you can use the 10x10 matrix as the input to a Pearson Chi-squared test (chisq.test() in R) of independence; the null hypothesis being tested is that the joint distribution of the cell counts in your 2-dimensional contingency table is the product of the row and column marginals. Mar12 comment How to compare joint distribution to product of marginal distributions? A test of correlation between $X_1$ and $X_2$ will give the check you want; if $X_1$ and $X_2$ are independent, their correlation is zero, and e.g. cor.test() in R will give an appropriate test; I'm sure there are Matlab commands to do the same. For the plot, one simple approach is to plot several histograms of $X_1$, each using only data where $X_2$ lies in some specified range. If those histograms look different, this suggests dependence. Alternatively, just scatterplot $X_1$ vs $X_2$ and look for a trend; increasing or decreasing ones are easiest to spot. Mar11 comment Problem with lagged covariates Sorry, should be $\geq 5$ Mar11 comment Problem with lagged covariates Do you mean that, at time $t$, $Y_t = 4-(X_{t-1} + X_{t-2} + X_{t-3} + X_{t-4})$? This would be well defined for times $>5$. (Analysis of how it is associated with heart attack incidence is perhaps best asked in another question) Mar11 comment Residualize a binary variable to remedy multicollinearity? You can run linear regression with binary outcomes if you like. (Though I'd recommend ensuring that Gender is indeed binary, and not stored as a factor.) But why do you want to do this? In your example the Gender coefficient in glm(RT~Predictability+Length+Gender, family=binomial) tells you about the (log) ratio of odds of RT, comparing Genders, in observations among which Predictability and Length have the same values. There's no need to adjust or rezidualize the Gender variable before doing logistic regression Mar10 comment How to interpret variance and correlation of random effects in a mixed-effects model? Zeda, it would be helpful to see more of the R output here, including the output's summary of the fixed effects Mar10 comment How to impute data without missing at random? If the data is completely absent at four time points, I don't think imputation is going to help. Multiple imputation helps when some observations are missing. But if there's nothing special about those four years, other than that data was not recorded in them, leaving them out of the analysis would be okay. Mar10 answered Visualizing many p-values