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Since your research question involves a dependent variable (number of colonies) and an independent variable (temperature) you want regression. But you a) Should include all your plates in one regression and b) Will almost certainly need more data to get useful statistical tests, unless the relationship is very strong and clear. I am not sure what "sets of ...

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Something doesn't smell right here. I don't see how you could get those AICs with those three models if they are all using the exact same data. Adding the veggies variable shouldn't increase it that much - even if the variable is useless in predicting height - because it's only adding 1 parameter. What I suspect is happening is that there is missing data in ...

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Although this software-specific question is technically off-topic here, I do note that NA values were handled differently in the two calls: na.omit for the glo_mo model, na.fail for the dredge function call. AIC and BIC are discussed in detail on this page. There is some dispute about whether these approaches are correct for comparing the non-nested models ...

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It is relatively simple to write the log-density of the negative binomial distribution in terms of its mean, and then use this to get a log-likelihood expression for the negative binomial GLM. For all values $y=0,1,2,...$ the log-density of the negative binomial distribution is: \log f(y|r,\theta) = \log {y+r-1 \choose y} + r \log (1-\theta) + y \log (\...

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The answer is: NO, negative binomial regression is NOT CONCAVE when estimated with dispersion. I do not know how to show this mathematically (since negative binomial WITH dispersion parameter does not belong to the exponential family), but I simply found a line in my parameter space, which is not concave. Here it is: Target is the log-likelihood of the ...

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In models with nonlinear link functions there is indeed a difference in the interpretation of the regression coefficients in GEEs and mixed-effects models. In short, GEEs give you the more usual interpretation of comparing groups of subjects. E.g., for dichotomous outcomes and the logit link you get the log-odds ratio between the group of males and the ...

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The geometric distribution is a special case of the negative binomial distribution, so I would try a negative binomial glm (generalized linear model.) Some details here: When to use Poisson vs. geometric vs. negative binomial GLMs for count data?. The generalization of ANOVA in glm's is called the analysis of deviance, but in some computer systems (for ...

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The X axis is actually really a (n index) set. In your case (if we ignore the restrictions that there is no human being with a negative height and height is bounded and so forth) the index set is $\mathbb{R}^2$. For every possible combination of height and weight $(h,w) \in \mathbb{R}^2$ you have a new random variable $X_{(h,w)}$. The whole GP really is the ...

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While the intuition behind it is not bad, I would not suggest using a Poisson glm for this problem, as it would have a different domain than your problem. You could find you glm predicting a value of 6 (which for you doesn't mean anything), and this can also wrongly affect the computation of the loss. I would instead suggest using a ordinal logistic (or ...

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You have too few observations to include in your initial model that many predictors. Also, note that for binary data the effective sample size is determined by the minimum of the frequencies of the zeros and the ones. Hence, you have very little information in your data to obtain any meaningfully stable results. Finally, as noted in the comments by EdM, ...

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It is unusual, even with studentization, to see less variability for greater predicted values, and yet that's what you have here. There are three causes to consider: 1) misspecified mean-variance relationship, 2) omitted regressors or 3) undetected non-independence. They imply vastly different approaches to the problem of obtaining valid inference. 1) ...

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I would suggest to only transform your data when absolutely necessary. It's not best practice in the modern era when you're using a computer that can easily handle generalised linear models. Transformations inherently produce less useful data. Best thing to do is use the glm() function as you are, but to check the residuals from the different families of ...

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Appropriate for what purpose? That is the question. Since the residual deviance is significantly large, you can conclude that either your data is overdispersed or the model does not explain all the signal in the data. On the other hand, the model does explains a deviance of $165.8 - 29.92 = 135.9$ on 5 degrees of freedom, which is most ($135.9 / 165.8 = 82$...

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No, this isn't correct. Since marital status is coded as single = 1 and married = 0 then the effect of age is For married people - the main effect of age For single people - the main effect of age plus the interaction effect of age and marital status Since you are doing Poisson regression, both these will be exponentiated (as you have in your question)....

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Overdispersion is a phenomenon that often applies to binomial-like data ($y$ successes out of $n$ cases with $n>1$), but it cannot apply to binary data. If $y$ is binary (1/0) and $E(y)=p$ then it must be that var$(y)=p(1-p)$. There is no mathematical possibility that the variance can be any greater or less than that given $E(y)$. If your response ...

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Since AIC is not defined for quasi-models, and anyhow your two models in this case would not be nested, AIC should not be used for comparison. I would go for model visualization, see for instance Dealing with Overdispersed Negative Binomial using glmmTMB for examples of use of simulated residuals for model checking (well, simulated residuals cannot be ...

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No, the t distribution is not an exponential family. Exponential family distributions do have existing moment generating functions, and the t distribution do not. See also Why doesn't the exponential family include all distributions?

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For a GLM, do I just set up the design matrix like in linear regression and select the appropriate model? Do exactly the same thing as before? Yes, the design matrix will be the same. You will need to specify the relevant distribution (eg binomial) and possibly a link function (eg logit for logistic regression and log for poisson)

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For a video I'd recommend the MIT 18.650 Statistics for Applications course see lectures 21-24. Also the Princeton website for their courses on the subject https://data.princeton.edu/wws509

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Since you say : the effect of snpA as the idea is that the AA patients are protected, AB are at low risk and BB are at high risk. this implies that you should fit snpA as a fixed effect, not random. Besides, since it has only 3 levels and the levels you have are the total population of such levels for that genotype, it would not make much sense to ...

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Briefly, as noted in comments: You can specify a random effect for logistic regression in the glmer() function in the same way as you did for linear regression in lmer(). Residuals in a logistic regression would not be expected to be normally distributed. (For reference, in linear regression it's good to visualize the residuals as a function of predicted ...

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I think there is good news and bad news on this... The bad news is that each and every choice of algorithm implicitly makes assumptions on the underlying probability distribution of the data (more precisely, it usually makes assumptions about the structure of $E[Y|X]$ where $Y$ is the target variable and $X$ is the variable responsible for producing the ...

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NO. The quasi-Poisson **is not a distribution* at all, it is an estimation method. There is no distribution model that leads to the quasi-Poisson estimating equations, but still it is found to be useful because it has good asymptotic properties, and is a way to get around the often unreasonable property of the Poisson distribution that the variance equals ...

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You can see from your diagnostic plots that you have two separate groups, which presumably correspond to the response variables unique_locality and unique_time. This occurs because you have fit a logistic regression using a binary response outcome, rather than using a multiple logistic regression that can handle the three-category outcome you actually have. ...

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Possible reasons why you get different results from the two packages include The default of glmer() is the Laplace approximation and not the adaptive Gaussian quadrature. You could try refitting with glmer() and increase the nAGQ argument. The optimization procedure in one of the two packages was not completely successful. You could try fitting the model ...

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This publication answers your question exactly "A derivation of prediction intervals for gamma regression" https://www.tandfonline.com/doi/full/10.1080/00949655.2016.1169421

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I got the value. But it still didn’t have validation of SE. I don’t know why… x<-c(1,2,3,4) y<-c(2,3,5,4) f<-expression((exp(-a-b)*(a+b)^2/2)*(exp(-a-2*b)*(a+2*b)^3/6)*(exp(-a-3*b)*(a+3*b)^5/120)*(exp(-a-4*b)*(a+4*b)^4/24)) f1<-D(f,"a") f2<-D(f,"b") g11<-D(f1,"a") g12<-D(f1,"b") g21<-D(f2,"a") g22<-D(f2,"b") a<-1.2 b<-0.7 ...

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My experience with regression modeling, in general, involves also the exercise of judgment and in that regard, it is something of an art, especially when it comes to constructing good forecasting models. My observation is that actually parsimonious models appear to be superior, overfitting a model is not the apparent optimal best approach. My experience ...

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Partially answered in comments: It seems reasonable; however, you might want to at least contemplate the possibility that the coefficient may be other-than-one (perhaps it may be that the response is not-proportional to numViews). One way to investigate that if you considered it a serious possibility would be to also have it as a predictor (as ...

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