# Error analysis

1. Dec 10, 2005

### fabsuk

i have 5 values of slit width S

0.0015m
0.0158m
0.00191m
0.0021m
0.00257m

the standard error is plus or minus 0.0005

however my teacher says she wants us to calculate ∆S error
she says its ∆S=S(subscript 2) - S(subscript 1)

she says the answer is 0.0007m

WHY IS THIS SO?

Fabian

2. Dec 11, 2005

### lightgrav

If you don't know what the "correct" value is for some quantity,
but are pretty sure you've measured what should be the same thing 5 times, you have to presume that your average value is the "correct" one.

How do you analyze your data (even after fixing #2 = 0.00158)
and obtain a "standard error" of .0005 ?
I got a mean of .001932 and a sample standard deviation of .00038 ...
but the estimate of your error in the "correct" value is the population standard deviation of .00043 .
If you have some reason to think that your variations are normally distributed, you can report the 50%-level rather than the 1-sigma ...
that would get you pretty close to .0005 , I guess.

The only way I can think of to get a .0007
would be to use the two-sigma range (mean - sigma) to (mean + sigma),
with the sample standard deviations. I'd call it unusual.

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