How can I calculate this please? If I am dividing two numbers?
Say divide
333.7±7.8 by 1.08±0.09 ?
if I simply divide 7.8 by 0.089 then that is 86.66 which is surely too big?
thanks
There are probably several ways to get an estimate. This is the one I learned.
These are standard errors, defining $\displaystyle x \pm \Delta x = 333.7 \pm 7.8$, $\displaystyle y \pm \Delta y = 1.08 \pm 0.09$ and $\displaystyle z = \frac{x}{y}$ we find that
$\displaystyle z = \frac{x}{y} = \frac{333.7}{1.08} = 308.981$
and
$\displaystyle \Delta z = \sqrt{\left ( \frac{\partial z}{\partial x} \cdot \Delta x \right ) ^2 + \left ( \frac{\partial z}{\partial y} \cdot \Delta y \right ) ^2}$
$\displaystyle \Delta z = \sqrt{\left ( \frac{1}{y} \cdot \Delta x \right ) ^2 + \left ( -\frac{x}{y^2} \cdot \Delta y \right ) ^2}$
$\displaystyle \Delta z = \sqrt{\left ( \frac{\Delta x}{y} \right ) ^2 + \left ( \frac{x \Delta}{y^2} \right ) ^2}$
$\displaystyle \Delta z = \sqrt{\left ( \frac{7.8}{1.08 } \right ) ^2 + \left ( \frac{333.7 \cdot 0.09}{1.08 ^2} \right ) ^2}$
$\displaystyle \Delta z = 8.8266$
So $\displaystyle z = 308.981 \pm 8.8266 = 309.0 \pm 8.8$
-Dan
well, thanks for the reply. you could be right there, but that would mean the source of this problem is wrong. Here are their values for instance.
1.08±0.09 / 333.7±7.8 = 309±27
0.13±0.01 / 4.1±0.1 = 32±3
0.30±0.01 / 101.8±0.8 = 339±12
??????????????????????
thanks again for your efforts. Any further comments would be appreciated.
Actually, no we are probably both right, without any further information anyway. There are several methods for finding the standard error and which one you use tends to depend on the type of experiment, though they all give close to the same result anyway.
The method I showed you was introduced to me as "Propagation of Errors" though if you look it up you might find that as something else. (I never was sure if they got the name of the process right.)
Without having your source handy I can't make any further intelligent guess as to how they want you to calculate it.
-Dan
it is table 2 in this paper I am going from. Enzyme kinetics
Characterization of phenylpyruvate decarboxylase, ...[J Bacteriol. 2007] - PubMed Result