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# Tag Info

3

Yes, there is a mistake. The CI for $\hat{\beta}$ involves degree of freedom $n-2$, not $n-1$. Note that, the explanation has more typos. In the second bullet-point, it finds $t_{n-2}$, but uses as if it was $t_{n-1}$ in the third bullet point.

2

Here is a little function which will essentially run a regression and return a coefficient. The data on which we will perform the regression are always generated from the same data generating process (same coefficients, every time).' sim<-function(){ N = 50 x = rnorm(N) y = 2*x+1+rnorm(N) model = lm(y~x) b = coef(model)[2] return(b) } To ...

1

Sounds to me like the average Spearman correlation between the observed scores and randomly generated scores (based on observed scores and margin of errors).

1

It would be nice if next time you put the fit equation into the question. As we discussed this yesterday, I believe your question in related to the question: Why are my fitted coefficients so well-determined? So the idea is as follows. I use R, because I don't have Mathematica. I hope you can translate the code. First you need to define the (population) ...

1

Actually there is never the estimator, but an estimator. But you can speak of the estimator, if it performs "good" from several statistical points of view, with no better alternative. Than it is some kind of textbook estimator. This is often the case with Maximum Likelihood estimators, where the ML principle is one of the most succeeding principles in ...

1

I don't know how Mathematica calculates the standard errors of a the fit. However, I reckon that the point estimator and its standard deviation is calculated via \begin{align} \hat{{\beta}} &= ({X}^T{X})^{-1} {X}^T {y} \\ %%% \hat{Sd}[\hat{\beta}_j] &=\sqrt{ \hat{\textrm{Cov}}[\hat{{\beta}}]_{jj}} = \hat{\sigma} \cdot \sqrt{\big[({X}^...

1

Leave-one-out does have a standard deviation and also a standard error: each of the $n$ folds returns an error estimate, so you can calculate mean as well as standard deviation and standard error over the $n$ fold estimates. You are right though, that LOO is exhaustive: it is not possible to obtain other than those $n$ surrogate models from an LOO scheme. ...

1

In your study of subjects deciding whether or not to believe 20 videos are factual, it seems easy to get a reasonable confidence interval. Roughly speaking, suppose you have $100$ subjects who viewed the $20$ videos, and that on average 25% of the videos are believable. Then you may have data similar to the 100 in the list x below (simulated using R). ...

1

Right, at the first-year undergraduate level, this is how you typically do it. Firstly, if you were to only take one measurement for x,y and z; use the instrument uncertainty (half its smallest scale or whatever is stated by the manufacturer), and propagate its error to f. So your final answer would look like: where Now, there is a better way to ...

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