Suppose that we are given 2 populations.
Let different samples of size \(N_1\) drawn from 1st population and statistic \(S_1\) computed. The mean and standard deviation be \(\mu_{S_1}\) and \(\sigma_{S_1}\). Similarily from population sample 2, draw samples of size \(S_2\),compute the statistic \(S_2\), whose mean and standard deviation is \(\mu_{S_2}\) and \(\sigma_{S_2}\). Then, the sampling disribution of differences of the statistics , which distribution of the difference of \(S_1 - S_2\) has a mean and standard deviation as \[\mu_{{S_1}-{S_2}} =\mu_{S_1} -\mu_{S_2}\] and \[\sigma_{{S_1}-{S_2}} = \sqrt{\sigma{_{S_1}{^2}} + \sigma{_{S_2}{^2}}}\] provided the samples are independent.
If \(S_1\) and \(S_2\) are the samples means from the two populations which means we denote by \(\bar{X_1}\) and \(\bar{X_2}\) respectively then the sampling distribution of difference of means is given for infinite popolation with mean and standard deviation ( \(\mu_1\) , \(\sigma_1\) ) and ( \(\sigma_2\), \(\sigma_2\)) \[\mu_{{X_1}-{X_2}} =\mu_{X_1} -\mu_{X_2} = \mu_1 - \mu_2\] and \[\sigma_{{X_1}-{X_2}} = \sqrt{\sigma{_{X_1}{^2}} + \sigma{_{X_2}{^2}}} = \sqrt{\frac{\sigma{_1}{^2}}{N_1} + \frac{\sigma{_2}{^2}}{N_2}}\]
This results for the sampling with distribution of difference of proportion from two binomially distributed population with parameters(\(p_1,q_1\)) and (\(p_2,q_2\)) respectively \[\mu_{{P_1}-{P_2}} =\mu_{P_1} -\mu_{P_2} = p_1 - p_2\] and \[\sigma_{{P_1}-{P_2}} = \sqrt{\sigma{_{P_1}{^2}} + \sigma{_{P_2}{^2}}} = \sqrt{\frac{p_1q_1}{N_1} + \frac{p_2q_2}{N_2}}\]