Fixed 95->90

Value: [1e-4 to 3e-3]
This commit is contained in:
Quinn Dougherty 2022-04-27 16:51:23 -04:00
parent 98bc2ddd58
commit 6bd8aecb31
3 changed files with 6 additions and 7 deletions

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@ -22,7 +22,7 @@ argument allows you to pass an environment previously created by another `run`
call. Passing this environment will mean that all previously declared variables
in the previous environment will be made available.
The return type of `run` is a bit complicated, and comes from auto generated js
The return type of `run` is a bit complicated, and comes from auto generated `js`
code that comes from rescript. We highly recommend using typescript when using
this library to help navigate the return type.

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@ -1,8 +1,9 @@
---
title: Processing Confidence Intervals
author: Nuño Sempere
---
This page explains what we are doing when we take a 95% confidence interval, and we get a mean and a standard deviation from it.
This page explains what we are doing when we take a 90% confidence interval, and we get a mean and a standard deviation from it.
## For normals
@ -21,13 +22,10 @@ module Normal = {
We know that for a normal with mean $\mu$ and standard deviation $\sigma$,
$$
a \cdot Normal(\mu, \sigma) = Normal(a\cdot \mu, |a|\cdot \sigma)
a \cdot Normal(\mu, \sigma) = Normal(a \cdot \mu, |a| \cdot \sigma)
$$
We can now look at the inverse cdf of a $Normal(0,1)$. We find that the 95% point is reached at $1.6448536269514722$. ([source](https://stackoverflow.com/questions/20626994/how-to-calculate-the-inverse-of-the-normal-cumulative-distribution-function-in-p)) This means that the 90% confidence interval is $[-1.6448536269514722, 1.6448536269514722]$, which has a width of $2 \cdot 1.6448536269514722$.
We can now look at the inverse cdf of a $Normal(0,1)$. We find that the 90% point is reached at $1.6448536269514722$. ([source](https://stackoverflow.com/questions/20626994/how-to-calculate-the-inverse-of-the-normal-cumulative-distribution-function-in-p)) This means that the 90% confidence interval is $[-1.6448536269514722, 1.6448536269514722]$, which has a width of $2 \cdot 1.6448536269514722$.
So then, if we take a $Normal(0,1)$ and we multiply it by $\frac{(high -. low)}{(2. *. 1.6448536269514722)}$, it's 90% confidence interval will be multiplied by the same amount. Then we just have to shift it by the mean to get our target normal.

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@ -13,3 +13,4 @@ Squiggle is an _estimation language_, and a syntax for _calculating and expressi
- [Squiggle functions source of truth](https://www.squiggle-language.com/docs/Features/Functions)
- [Known bugs](https://www.squiggle-language.com/docs/Discussions/Bugs)
- [Original lesswrong sequence](https://www.lesswrong.com/s/rDe8QE5NvXcZYzgZ3)
- [Author your squiggle models as Observable notebooks](https://observablehq.com/@hazelfire/squiggle)