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Question

# Assuming that$n _ { 1 } = n _ { 2 }$find the sample sizes needed to estimate$\left( p _ { 1 } - p _ { 2 } \right)$for each of the following situations: SE = .01 with 99% confidence. Assume that$p _ { 1 } \approx .4 \ and \ p _ { 2 } \approx .7.$

Solution

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Given:

\begin{aligned} SE&=\text{Sampling error}=0.01 \\ c&=\text{Confidence level}=99\%=0.99 \\ \hat{p}_1&=\text{Sample proportion}=0.4 \\ \hat{p}_2&=\text{Sample proportion}=0.7 \end{aligned}

Formula for the sample size (use $\hat{p}_1=\hat{p}_2=0.5$ if no estimates are known):

\begin{aligned} n=n_1=n_2&=\left( \dfrac{z_c}{E}\right)^2 (\hat{p}_1\hat{q}_1+\hat{p}_2\hat{q}_2) \\ &=\left( \dfrac{z_{\alpha/2}}{SE}\right)^2 (\hat{p}_1(1-\hat{p}_1)+\hat{p}_2(1-\hat{p}_2)) \end{aligned}

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