document.write( "Question 1039096: A random sample of 1100 car owners in a particular city found 220 car owners who recieved a speeding ticket this year. Find a 95% confidence interval for the true percent of car owners in this city who recieved a speeding ticket this year. Express your results to the nearest hundredth of a percent.
\n" );
document.write( "Answer:
\n" );
document.write( "
\n" );
document.write( "______ to ____% \n" );
document.write( "
Algebra.Com's Answer #653842 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! 1 sample proportion \n" ); document.write( "ci is 1.96 sqrt{p(1-p)/n} \n" ); document.write( "1.96*sqrt (1/5)(4/5)/1100}; the sqrt term is 0.0121 \n" ); document.write( "the interval is 0.0236 \n" ); document.write( "The CI is 0.20+/-0.0236 \n" ); document.write( "(0.1764,0.2236) \n" ); document.write( "(17.64%,22.36%) \n" ); document.write( " \n" ); document.write( " |