document.write( "Question 1120679: In a random sample of 200 observations, we found the proportion of successes to be 48% use a 90% degree of confidence to estimate the population proportion of successes . repeat with n 500 \n" ); document.write( "
Algebra.Com's Answer #736972 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! The half interval for 90% CI is +/- z*SE, where z is 1.645 and SE is sqrt(.48*.52/200) \n" ); document.write( "1.645*0.0353 or 0.0581 \n" ); document.write( "So the interval is 48%+/-5.81% \n" ); document.write( "or (0.4219, 0.5381)\r \n" ); document.write( "\n" ); document.write( "For n=500 the SE is 0.0223 and the half interval 0.0368\r \n" ); document.write( "\n" ); document.write( "(0.4432, 0.5168) \n" ); document.write( " |