Simple ##\chi^2## Tests for Weighted Averages and Linear Regression

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Homework Statement
Making a ##\chi^2## test on arbitrary measurements.
Relevant Equations
##\chi^2## testing.
1. Suppose one has the measurements [1.20, 1.15 ,2.0 ,1.17] with uncertainties [0.2,0.1,0.8,0.07]. Then, if ##E## is the weighted average, is it correct that ##\chi^2## is simply given by

##\sum \frac{(O-E)^2}{E} \ ?##​

2. If one has

| x | y |
| -- | -- |
| 0 | 0 ##\pm## 1 |
| 1 | 1 ##\pm## 1 |
| 2 | 4 ##\pm## 1|
| 3 | 9 ##\pm## 1 |

and one would like to test if ##y=ax+b##, then what is ##\chi^2##?
 
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1) yes What am I saying. No ! For weighted averaging

$$\chi^2 = \sum_i \;{(x_i - \bar x)^2\over \sigma_i^2}$$2) evaluate
1608511528959.png

(picture borrowed from Edinburgh University)

(perhaps an old post has some references)
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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