Question

Suppose that two random samples of sizes n1, and n2 are selected without replacement from two binomial populations with means = $\mu_1= n_1p_1, \,\,\,\mu_2= n_2p_2,$ and variances $σ_1 ^2 = n_1p_1q_!, \,\,\,σ_2^2 = n_2p_2q_2$ , respectively. Let the difference of sample proportions $\overline{P_1}$ and $\overline{P_2}$ approximate a normal distribution with mean ($p_1 - p_2$). Then the standard deviation of the difference of sample proportions  $\overline{P_1}$ and $\overline{P_2}$, is

• A. $\sqrt{(\frac{p_1q_1}{n_1})(\frac{N_1-n_1}{N_1-1})+ (\frac{p_2q_2}{n_2})(\frac{N_2-n_2}{N_2-1})}$
• B. $\sqrt{(\frac{p_1q_q}{n_1})+(\frac{p_2q_2}{n_@})}$
• C. $\sqrt{(\frac{p_1q_1-p_2q_2}{n_1+n2}})$
• D. $\sqrt{(\frac{p_1q_1}{n_1+n_2})(\frac{N_1-n_1}{N_1-1})+ (\frac{p_2q_2}{n_1+n_2})(\frac{N_2-n_2}{N_2-1})}$

Answer 1

The Correct Option is A

The correct option is (A): $\sqrt{(\frac{p_1q_1}{n_1})(\frac{N_1-n_1}{N_1-1})+ (\frac{p_2q_2}{n_2})(\frac{N_2-n_2}{N_2-1})}$