import numpy as np
Data for pendulum swings:
The standard error (standard deviation of the mean) for the time for 10 swings is $$ \alpha = \frac{\sigma}{\sqrt{N}} = \frac{0.04}{\sqrt{12}} $$ The time for one swing is the time for 10 swings divided by 10: \begin{eqnarray*} T_{1} &=& \frac{T_{10}}{10}\\ &=& \frac{28.39 \pm \frac{0.04}{\sqrt{12}}}{10}\\ &=& 2.839 \pm \frac{0.04}{10\sqrt{12}}\\ &=& 2.839 \pm 0.001155 \end{eqnarray*} The presentation form of this result is $$ T_1 = 2.839\pm 0.001\, \mbox{s}. $$
t10 = 28.39
alpha10 = 0.04/np.sqrt(12)
t1 = t10/10
alpha1 = alpha10/10
print('T1 =',t1,'+/-',alpha1)
The standard error (standard deviation of the mean) for the time for 120 swings is $$ \alpha = \frac{\sigma}{\sqrt{N}} = \frac{0.04}{\sqrt{1}} $$ The time for one swing is the time for 120 swings divided by 120: \begin{eqnarray*} T_{1} &=& \frac{T_{120}}{120}\\ &=& \frac{340.61 \pm \frac{0.04}{1}}{120}\\ &=& 2.838417 \pm \frac{0.04}{120\sqrt{1}}\\ &=& 2.838417 \pm 0.000333 \end{eqnarray*} The presentation form of this result is $$ T_1 = 2.8384\pm 0.0003\, \mbox{s}. $$
t120 = 340.61
alpha120 = 0.04/np.sqrt(1)
t1 = t120/120
alpha1 = alpha120/120
print('T1 =',t1,'+/-',alpha1)