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First, it is important that $Y_i |X$ is normally distributed. Also, a linear combination of independent normal random variables is normal. With that knowledge, all that is left is to show that $\hat{\beta_1}$ is a linear combination of $Y_i | X$. From the textbook Applied Linear Regression Models (4th ed) by Kutner, Nachtsheim, and Neter it states $$ b_1 ...


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If a one-unit increase in $X$ results in an increase in log-odds of $2.5$, then a $0.1$-unit increase in $X$ results in an increase in log-odds of $0.25$. If you drive a tenth as long, you go a tenth as far. That $2.5$ is analogous to your speed.


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Assume the following simple logistic regression model $$ \mathrm{logit}(y)=\alpha + \beta\cdot x $$ The predicted probabilities are given by $$ \hat{P}(Y=1)=\frac{1}{1 + \exp(-(\hat{\alpha} + \hat{\beta}\cdot x))} $$ The turning point - and the steepest slope - of the logistic curve (your red curve) is attained at $x=-\frac{\hat{\alpha}}{\hat{\beta}}$ ...


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You can just change what the reference level is for the training variable by using relevel() and refit the model. When you do so, the coefficient on timefollowup will be the simple effect of time at the no trained level of training. If for some reason you can't refit the model (e.g., because the dataset is huge), you can use contrasts to compute the simple ...


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Although $\hat\beta$ is a function of both $X$ and $y$, $\mathbb E[\hat\beta]$ is a constant because expectation is taken over all RVs inside. Let's call it $c$, and take the inside of the outermost expected value only. The expression becomes the following: $$\begin{align}\mathbb E[\overbrace{(\mathbb E[\hat\beta|X]-c)}^{\text{given X this is const, call it ...


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I believe this should be "saturated," rather than "satiated." This just means you use regression to calculate four means for all the possible combinations of $x_1$ and $x_2$: $(0,0),(1,0),(0,1),(1,1)$ This is a kind of a non-parametric model, which is non-linear since it allows the means to be very flexible.


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The contrast matrix approach works very well. But if you are struggling with the concept, one more basic approach is to use a plug-in technique. We can actually write out the model fit to your data as: $$ log[Odds(learning)] = 0.47 + 1.68\times trained + 2.56\times time - 1.39\times trained \times time $$ You asked about the estimated difference in log-...


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The function according to the vignetter say "The first argument passed will always be the original data. The second will be a vector of indices, frequencies or weights which define the bootstrap sample." Putting it into a function below, you define data as the data and idx and the so called index to sample with replacement from: library(boot) func = ...


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Let's go into the derivation of slope $ = \hat \beta /4 $. In the curve there are following things to note Max or min (depending on the shape of the curve) of slope happens at inflection point (where slope changes in shape) We know second order derivate at inflection point is 0. We will use this to find x at inflection point Derivative of the logit ...


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