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7) In a recent Pew Research Center survey (Oct 17, 2019) of 3500 U.S. adults, 1890 adults said congress should continue with
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7) a)\large \widehat{p}=\frac{1890}{3500}=0.54

b) z value for 95% CI is 1.96 as P(-1.96<z<1.96)=0.95

So Margin of Error is \large E=z*\sqrt{\frac{pq}{n}}=1.96*\sqrt{\frac{0.54*0.46}{3500}}=0.017

c. Hence CI is \large p \pm E=0.54 \pm 0.017=(0.523,0.557)

d. Population proportion will lie in the range of 0.523 to 0.557

e. Here E=0.03

So \large n=(\frac{z}{E})^2*p*q=(\frac{1.96}{0.03})^2*0.54*0.46=1060.28=1061

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