Number of Significant Digits
If you have checked our calculation of standard deviation in Section 2.3., you know that your calculator actually doesn't give the answer for sigma as 0.2 sec but rather
There are 9 significant digits in this number: 1,7,8,8,8,5,4,3,8. (Significant digits are the digits remaining after one ignores leading zeros.) If we kept all of them, our final answer for the oscillation period would be
T= 3.6 +- 0.178885438
The above result implies that we measured T with a precision of better than a billionth of a second with a store-bought stopwatch! Clearly, it's absurd and we should round our error to fewer significant digits to reflect the limitations of our measuring device. As a general rule you should round the error to just one significant digit for your final answer and keep two significant digits for intermediate results. So if we were interested in T only, we would write
3.6+-0.2
If, in contrast, T were to be used to calculate the acceleration g due to gravity, we would keep two significant digits for the error
3.6+-0.18
In fact, it's a good idea to report errors with two significant digits for all your measurements since you can easily round from two to one significant digit when needed.
Now, what about the number of significant digits for the central value of T itself? Suppose the calculator gives us the value T = 3.57361382 sec instead of 3.6 sec. It is ridiculous to write our final answer for T as
Since our error already affects the number 5 in the tenth place of 3.57361382, we should round our value to that decimal place. So the correct answer is
3.6+-0.2
This principle is true in general. We should round our central value to the rightmost decimal place at which our error applies. Thus if we have two decimal places in our error we should round our central value to the hundredths decimal place. 3.57+-0.18
The above matter is taken from http://phys.columbia.edu/~tutorial/
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