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While reporting the mean time of 100 pendulum oscillation with the error.What should be the should be the error range?The one which includes all the data or the shorter one one which include most of the data?And why is it so?
The data I have:$90,91,95,92$ and the least count is 1 sec.
I think the mean should be $92\pm 3$ for it includes all the data.But it is not a short range.
I calculated the standard deviation of data.It came out to be $2.16$
Martin Ueding almost has it right and in most cases this is what you might expect to see quoted in a straightforward lab report.
The standard error in the mean is indeed given by the corrected sample standard deviation divided by the square root of the number of points.
For a large sample then you can find a reasonably unbiased estimate of the sample standard deviation by dividing the sum of the variances by $N-1$ and taking the square root. This yields the figure 2.16 in your dataset. However, for $N=4$ this is unlikely to be correct. For normally distributed uncertainties, a better estimate for small $N$ is provided by division by $N-1.5$ (resulting in a slightly larger uncertainty).
Thus, in your case: The mean is 92.0
The (corrected) standard deviation of the population is 2.36 and the standard error (in the mean) would be 2.36/2 = 1.18.
I would probably quote this as $92.0 \pm 1.2$.
I believe the decimal place is warranted because the standard deviation is more than twice the "measurement" uncertainty of $\pm 0.5$ due to the fact that you (appear to) measure to the nearest second and thus the variation you see is (a) resolved and (b) unlikely to be due to measurement uncertainty. Others may disagree on the basis that sometimes one quotes only 1 significant figure in the error. In my view, this falls down when that significant figure is 1, because there is a big difference between an uncertainty of 1.0 and 1.5.
You would use the mean $\langle T \rangle$ as your resulting value.
For the error, it is probably best to use the standard error here. For $N$ measurements, it is the standard deviation divided by $\sqrt N$. So you would use $$\Delta T = \sqrt{\frac{1}{(N-1)N} \sum_{i = 1}^N (T_i - \langle T \rangle)^2 } \,.$$ Since you already have the standard deviation $s$, you can just compute $\Delta T = s / \sqrt N = s / 2 = 1.08$.
You would quote your final result as $T = 92 ± 1.08$.
