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)\"\" \"About 
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p=15/80=.1875, q=.8125,
\n" ); document.write( "\"sigma=sqrt%28%28.1875%2A.8125%29%2F80%29=.0436\"
\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.
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