document.write( "Question 1155770: An employment counselor found that in a sample of 100 unemployed workers, 65% were not
\n" );
document.write( "interested in returning to work. Find the 95% confidence interval of the true proportion of
\n" );
document.write( "workers who do not wish to return to work. \n" );
document.write( "
Algebra.Com's Answer #778476 by Boreal(15235) You can put this solution on YOUR website! The half-interval is +/- 1.96*sqrt(p*(1-p)/n. and the sample proportion is 0.65 \n" ); document.write( ".65+/-1.96 sqrt (.65*.35/100)=0.65+/-0.93 \n" ); document.write( "(0.557, 0.743) \n" ); document.write( " |