document.write( "Question 479410: The Pizza Shop wanted to determine what proportion of its customers ordered only cheese pizza. Out of 80 customers surveyed, 15 ordered cheese pizza. What is the 99% confidence interval of the true proportion of customers who order only cheese pizza? \n" ); document.write( "
Algebra.Com's Answer #328502 by edjones(8007)![]() ![]() You can put this solution on YOUR website! p=15/80=.1875, q=.8125, \n" ); document.write( " \n" ); document.write( "(a,b)=.1875 +-(2.576*.0436) [2.576 is the area under the normal curve for the 99% confidence level.] \n" ); document.write( "=.1875 +-.1123 \n" ); document.write( "=(.2998, .0752) the 99% confidence interval of the true proportion of customers who order only cheese pizza. \n" ); document.write( " |