document.write( "Question 1119719: Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 403 drivers and find that 284 claim to always buckle up. Construct a Invalid variable% confidence interval for the population proportion that claim to always buckle up.\r
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document.write( "Enter your answer using interval notation. Example: [2.1,5.6172) \n" );
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Algebra.Com's Answer #735412 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! proportion point estimate is 284/403=0.7047=p hat \n" ); document.write( "interval for 95% CI uses 1.96*sqrt(ph*(1-ph)/n) \n" ); document.write( "this interval is 0.0445 \n" ); document.write( "CI is 0.7047 +/- 0.0445 \n" ); document.write( "(0.6602, 0.7492) \n" ); document.write( " |