Let X have the binomial distribution Bn1 p1 and let Y have t

Let X have the binomial distribution B(n_1, p_1) and let Y have the binomial distribution B(n_2, p_2). Also assume that X and Y are independent. a. Show that E(X/n_1 - Y/n_2) = p_1 - p_2 and that Var (X/n_1 - Y/n_2) = P_1(1 - P_1)/n_1 + P_2(1 - p_2)/n_2. b. As we discussed in class, since X is binomial, the Central Limit Theorem implies that the proportion p_1 = X/n_1 will be approximately normal with mean p_1 and variance_1(1 - p_1)/n_1 as long as nx_1 is reasonably large. Similarly, p_2 = Y/n_2 will be approximately normal with mean p_2 and variance P_2 (1 -p_2)/n_2 as long as n_2 is reasonably large. Use these facts to show that p_1 - p2 is approximately normal with mean p_1 - p_2 and variance P_1(1 - p_1)/n_1 + p_2(1 - p_2)/n_2 when n_1 and n_2 are both reasonably large.

Solution

a. We know that in Binomial Distribution Mean = E (x) = np

Variance = V(X) = npq

where p = probability of success and q = =probability of failure = 1 - p

E ( aX+bY) = aE(X) +b E(Y) -----(1)

Given E(X) = n₁p₁ and E(Y) = n₂p₂

using (1) E( X/ n₁ - Y/n₂) = (1/n₁) E(X) - (1/n₂) E(Y)

= (1/n₁) n₁p₁ - (1/n₂) n₂ p₂

= p₁ - p₂

V(X) = n₁ p₁q₁ , V(Y) = n₂ p₂q₂   

WE know that V(aX+bY) = a²V(X) + b²V(Y)

V( x/n₁ - Y/n₂ ) = 1/n₁² [ n₁p₁q₁ ] - 1/n₂²[ n₂p₂q₂]

p₁(1-p₁) /n₁ - p₂(1-p₂)/n₂