Difference Between Two Sample Proportions

Consider an infinite (or very large) population, where each observation has a probability p_X of being a success, and a probability (1-p_X) of being a failure. Let the set of independent and identically distributed random variables X₁, X₂, ..., X_m represent the observations from a sample of size m, where

X_i = 1 if the ith observation is a success
= 0 if the ith observation is a failure

Consider another similar population, with each observation having a probability p_Y of being a success, and a probability (1-p_Y) of being a failure. Let the set of independent and identically distributed random variables Y₁, Y₂, ..., Y_n represent the observations from a sample of size n, where

Y_i = 1 if the ith observation is a success
= 0 if the ith observation is a failure

Let X = (X₁ + X₂ + ... + X_m) and Y = (Y₁ + Y₂ + ... + Y_n). Then X is binomially distributed with parameters p_X and m; similarly, Y is binomially distributed with parameters p_Y and n. Using the normal approximation to the binomial distribution,

X ® N(mp_X, mp_X (1-p_X)) as m ® ¥
Y ® N(np_Y, np_Y (1-p_Y)) as n ® ¥

Therefore,

As the sample proportions (X/m) and (Y/n) are both normally distributed when m and n are large, the difference (X/m) - (Y/n) is also normally distributed. In fact,

and thus