Using STdev wrong or apparatus error?

[CricK0es]

Homework Statement

I have 3 (n) measurements of the radius of a capillary tube, and I'm wondering whether I should take the STDev of these values and then divide by sqrt(n-1) to obtain a standard error, and use this as my error on the mean...
or
Do I propagate the apparatus error through and use that as my error on the mean?

Left ------------ Right----------Adj. Diam----Radius
0.062700 ---- 0.061140 ---- 0.00156 ---- 0.00078
0.065000 ---- 0.062330 ---- 0.00267 ---- 0.00134
0.064610 ---- 0.063100 ---- 0.00151 ---- 0.00076

Mean-------St. dev--------St error
0.00096----0.000328-----0.000232
=24.23513% error!

When using the SE, my error propagates through to give ~94% error on the value I wish to obtain. When using the apparatus error, it's only 5.16%; and I'm sure the rest of my spreadsheet is correct. So, I'm not sure what I'm doing in this regard. Any guidance would be appreciated... Seems simple, I know; just doesn't fit. I was expecting an SE on my mean of around 5%...

Homework Equations

Using excel, so =STDEV(-,-) ; where I input the 3 values for my radii
SE = STDEV / sqrt(2)

The Attempt at a Solution

Above... Sorry, realized my little table copy/paste didn't come out well [/B]

 

[mjc123]

The STDEV procedure you have used is correct; there is a rather large standard deviation on your radius values. Unless this is simply a measurement error, it looks as if your capillary radius might actually be varying along the capillary. In that case, using the apparatus error ignores a significant factor of the actual error, and is not the right way to go.

 

[kuruman]

I assume you measured two quantities, one labeled "Left" and one labeled "Right". These are your actual measurements and you should find the std on these. Then propagate the error for the calculated (as opposed to measured) radius as
$$\delta R~= \sqrt{[\delta(Left)]^2+ [\delta(Right)]^2} $$
The error is 84%, a huge number, but not surprising considering that the middle measurement results in a radius that is almost twice as large as the other two. Also, in view of the accuracy of your measurements, you should trim the number of your reported significant figures appropriately.