"Hai"

I hope Somebody help me

2.0±0.2 - 1.0±0.2

:rolleyes:

(All related to weight")

Thank you !

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- Apr 23rd 2006, 02:44 AMmiavarostandard dev
"Hai"

I hope Somebody help me

2.0±0.2 - 1.0±0.2

:rolleyes:

(All related to weight")

Thank you ! - Apr 23rd 2006, 03:18 AMCaptainBlackQuote:

Originally Posted by**miavaro**

from this but one interpretation (not the most likely, but this is the

interval arithmetic solution) will give:

$\displaystyle (2 \pm 0.2) - (1 \pm 0.2)=1 \pm 0.4$

RonL - Apr 23rd 2006, 07:16 AMtopsquarkQuote:

Originally Posted by**CaptainBlack**

1) $\displaystyle 1.0 \pm 0.2$ assuming the standard error in the measuring instrument is 0.2

2) $\displaystyle 1.0 \pm 0.3$ using the propagation of errors technique.

(Define a function f = f(a, b), where we have measurements $\displaystyle a = \bar{a}+\delta a$ and $\displaystyle b=\bar{b}+ \delta b$. Then $\displaystyle \bar{f} = f(\bar{a}, \bar{b})$ and $\displaystyle \delta f = \sqrt{ \left ( \frac{\partial f}{\partial a} \delta a \right ) ^2 + \left ( \frac{\partial f}{\partial b} \delta b \right ) ^2}$. )

Lacking any other information, my best guess is that the propagation of errors answer (2) is the answer, though there are many statistical reasons for choosing either of the other two. It depends entirely on how the original error estimates were made.

-Dan - Apr 24th 2006, 04:04 PMmiavaro
thank you