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): \(\sqrt{(\frac{p_1q_1}{n_1})(\frac{N_1-n_1}{N_1-1})+ (\frac{p_2q_2}{n_2})(\frac{N_2-n_2}{N_2-1})}\)