# Standard error?

• Sep 9th 2007, 10:02 AM
graham_keith
Standard error?
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
• Sep 9th 2007, 11:23 AM
topsquark
Quote:

Originally Posted by graham_keith
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 $x \pm \Delta x = 333.7 \pm 7.8$, $y \pm \Delta y = 1.08 \pm 0.09$ and $z = \frac{x}{y}$ we find that
$z = \frac{x}{y} = \frac{333.7}{1.08} = 308.981$

and
$\Delta z = \sqrt{\left ( \frac{\partial z}{\partial x} \cdot \Delta x \right ) ^2 + \left ( \frac{\partial z}{\partial y} \cdot \Delta y \right ) ^2}$

$\Delta z = \sqrt{\left ( \frac{1}{y} \cdot \Delta x \right ) ^2 + \left ( -\frac{x}{y^2} \cdot \Delta y \right ) ^2}$

$\Delta z = \sqrt{\left ( \frac{\Delta x}{y} \right ) ^2 + \left ( \frac{x \Delta}{y^2} \right ) ^2}$

$\Delta z = \sqrt{\left ( \frac{7.8}{1.08 } \right ) ^2 + \left ( \frac{333.7 \cdot 0.09}{1.08 ^2} \right ) ^2}$

$\Delta z = 8.8266$

So $z = 308.981 \pm 8.8266 = 309.0 \pm 8.8$

-Dan
• Sep 9th 2007, 12:04 PM
graham_keith
hmm
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

??????????????????????

• Sep 9th 2007, 12:28 PM
topsquark
Quote:

Originally Posted by graham_keith
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

??????????????????????

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
• Sep 9th 2007, 12:55 PM
graham_keith
citation
it is table 2 in this paper I am going from. Enzyme kinetics

Characterization of phenylpyruvate decarboxylase, ...[J Bacteriol. 2007] - PubMed Result
• Sep 9th 2007, 09:12 PM
topsquark
Quote:

Originally Posted by graham_keith
it is table 2 in this paper I am going from. Enzyme kinetics

Characterization of phenylpyruvate decarboxylase, ...[J Bacteriol. 2007] - PubMed Result

Waaaaay beyond my ken. Can't help any further. Sorry!

-Dan
• Sep 10th 2007, 12:26 PM
tukeywilliams
I like the book by Taylor on Error analysis. He explains prorogation of errors well.