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It is difficult to give good advice without more information about your data. I will give a partial discussion of what I imagine to be the simplest case for your data and best solution for what you want to accomplish. Suppose you observed the number of species $Y_i$ at $n$ elevations $x_i$ along a particular mountain slope, using the model $$Y_i = \beta_0 ...


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Consider two independently, log-normally distributed random variables $X \sim \mathcal{LN}(\mu_X, \sigma_X^2)$ and $Y \sim \mathcal{LN}(\mu_Y, \sigma_Y^2)$. The ratio is also log-normally distributed: $$ \dfrac{X}{Y} \sim \mathcal{LN}(\mu_X - \mu_Y, \sigma_X^2 + \sigma_Y^2). $$ If $\mathrm{GSD}(X) = \mathrm{e}^{\sigma_X}$ and $\mathrm{GSD}(Y) = ...


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As $Harish has already shown, the maximum uncertainty is much smaller than what you have found. Given that you are using the general formula for error propagation (using partial derivatives, etc.) I'm guessing you are not looking for the maximum uncertainty but for the "one-sigma" uncertainty. So your answer should be a smaller than what he found. Also, ...


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Notice, 1. Taking positive sign $$(\tan\theta)_{\text{max}}=\tan (59.3+1.2)^\circ=\tan 60.5^\circ$$ Taking negative sign $$(\tan\theta)_{\text{min}}=\tan (59.3-1.2)^\circ=\tan 58.1^\circ$$ Hence, the maximum possible uncertainty in the value of $\tan \theta$ $$=(\tan60.5^\circ-\tan 58.1^\circ)\approx 0.160926831$$


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You have usual regression model $$ Y_i = bX_i + \varepsilon_i$$ but you can only measure $\tilde{Y}_i = Y_i+\delta_i$, with some measurement error $\delta_i$. Now the model becomes $$ \tilde{Y}_i = bX_i + \varepsilon_i+\delta_i$$ and if $\varepsilon_i$ and $\delta_i$ are independent, the only thing that changes is that the variance of the error term ...



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