document.write( "Question 1139079: In a survey of 2383 U.S. adults, 1073 think that there should be more government
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document.write( "regulation of oil companies.
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document.write( "(a) Construct a 95% confidence interval for the population proportion p of U.S.
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document.write( "adults who think that there should be more government regulation of oil
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document.write( "companies. (Be sure to verify that the assumptions are met for the procedure
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document.write( "you use.)
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document.write( "(b) Find the minimum sample size needed to estimate p that ensures with 99%
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document.write( "confidence that the estimate is accurate within 3% (E = 0.03) of the population
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document.write( "proportion. Use p hat = q hat = 0.5 in your calculation.\r
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document.write( "I have been working on this problem all day and I do not know what to do.\r
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Algebra.Com's Answer #756861 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! assume random sample, and numbers and probability are sufficiently large to use normal approximation. \n" ); document.write( "The point estimate is 1073/2383=0.450 \n" ); document.write( "the half-interval is z*sqrt(p*1-p)/n)=1.96*sqrt(0.45*0.55/2383) \n" ); document.write( "0.45+/-1.96(1.01), the half-interval is 0.02 \n" ); document.write( "the whole interval is (0.43, 0.47)\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The half-interval is z0.995*sqrt (p*(1-p)/n) \n" ); document.write( "z is 2.576 \n" ); document.write( "p and 1-p are 0.5 \n" ); document.write( "the half-interval equals the error of 0.03 \n" ); document.write( "square both sides and z^2*p(1-p)/n=0.0009 \n" ); document.write( "so 6.64*0.25=0.0009n \n" ); document.write( "n=1844.44 or 1845 \n" ); document.write( " \n" ); document.write( " |