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)\"\" \"About 
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( "
\n" );