Question 1089638
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We will use the formula 
{{{n = p(1-p)(z/E)^2}}}
as shown on <a href="https://www.ltcconline.net/greenl/courses/201/estimation/ciprop.htm">this page</a> (scroll down to the section titled "Finding n to Estimate a Proportion")


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Initially we're told that an error of E = 0.04 will lead to a sample size of n = 800. We don't know z and we don't know p. We don't need to know p. If p is unknown, then we assume that p = 0.5. The goal here is to find z given n = 800 and E = 0.04. Let's do that


{{{n = p(1-p)(z/E)^2}}}


{{{800 = 0.5(1-0.5)(z/0.04)^2}}}


{{{800 = 0.25(z/0.04)^2}}}


{{{800/0.25 = (z/0.04)^2}}}


{{{3200 = (z/0.04)^2}}}


{{{(z/0.04)^2 = 3200}}}


{{{z/0.04 = sqrt(3200)}}}


{{{z/0.04 = 56.5685424949239}}}


{{{z = 0.04*56.5685424949239}}}


{{{z = 2.26274169979696}}}


So the z critical value is roughly z = 2.2627


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We'll use z = 2.2627 and the new error threshold of E = 0.01 to find n to be...


{{{n = p(1-p)(z/E)^2}}}


{{{n = 0.5(1-0.5)(2.2627/0.01)^2}}}


{{{n = 0.25(226.27)^2}}}


{{{n = 0.25*51198.1129}}}


{{{n = 12799.528225}}}


{{{n = 12800}}} (round up to the nearest whole number)


The new sample size needed is approximately 12800 


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