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
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
Updated On: Oct 1, 2024
- \(\sqrt{(\frac{p_1q_1}{n_1})(\frac{N_1-n_1}{N_1-1})+ (\frac{p_2q_2}{n_2})(\frac{N_2-n_2}{N_2-1})}\)
- \(\sqrt{(\frac{p_1q_q}{n_1})+(\frac{p_2q_2}{n_@})}\)
- \(\sqrt{(\frac{p_1q_1-p_2q_2}{n_1+n2}})\)
- \(\sqrt{(\frac{p_1q_1}{n_1+n_2})(\frac{N_1-n_1}{N_1-1})+ (\frac{p_2q_2}{n_1+n_2})(\frac{N_2-n_2}{N_2-1})}\)
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The Correct Option is A
Solution and Explanation
The correct option is (A): \(\sqrt{(\frac{p_1q_1}{n_1})(\frac{N_1-n_1}{N_1-1})+ (\frac{p_2q_2}{n_2})(\frac{N_2-n_2}{N_2-1})}\)
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