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7

Let $\mu_\alpha$ be the median of a $\Gamma(\alpha)$ distribution. This means the area under the density $$f_{\Gamma(\alpha)}(x) = \frac{x^{\alpha-1}}{\Gamma(\alpha)} \,e^{-x}$$ between $x=0$ and $x=\mu_\alpha$ equals $1/2.$ A graph of $f_{\Gamma(\alpha)}$ is sketched here in black (for $\alpha=0.3$), understanding the graph extends infinitely upwards as $...


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It's not analytically available. You can use numerical approaches/software. What you want to find is actually inverse CDF of quantile function in other words. In Matlab you can use gaminv, in R, you can use qgamma, or in python you can use ppf in scipy.stats. It's not easy to find a table but here is one with unit scale, with varying shape parameters (upto ...


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This answer is a good place to start. The relevant bits (emphasis my own)... The gamma has a property shared by the lognormal; namely that when the shape parameter is held constant while the scale parameter is varied (as is usually done when using either for models), the variance is proportional to mean-squared (constant coefficient of variation). ...


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I believe in SPSS the omnibus test compares the fitted model to an intercept-only model. The pseudo r-squared will tell you the percentage of variation in the data explained by the model (explained deviance, see Dobson 2002). In addition to Dobson (2002), you might want to refer to McCullagh and Nelder (1989). If your data are ecological in nature, I highly ...


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Start with making some plots, like and it is very clear that Site 5 is different ... If some additional evidence is needed, we can compute correlations, and compare the observed correlation with its permutation distribution, maybe. One way is library(tidyverse) with(met, cor(Abundance, As)) [1] -0.4893548 library(gtools) PERMS <- gtools::...


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This is an interesting observation that I am able to reproduce in lme4 (version 1.1-21). I have also implemented the Gamma mixed model in my GLMMadaptive package (version 0.6-9 available currently on GitHub), which seems to recover the correct value for the standard deviation of the random intercepts. The following code illustrates the issue and compares the ...


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They're correct. For the first one, you could also use $E[X_i^2]=\theta$ and obtain the result w/o going into gamma distribution. For the second one, an alternate solution would be using Chi-squared distribution, by defining $Z_i=X_i/\sqrt\theta$, yielding the same result since: $$\operatorname{var}\left(\frac{1}{n}\sum X_i^2\right)=\frac{\theta^2}{n^2}\sum_{...


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