document.write( "Question 1197465: In our Condor Cafe, the Oxnard College Culinary Program offers the most delicious and affordable breakfast and lunch choices among the 152 Community Colleges in California. This week, a random sample of 357 orders was recorded, and 77 orders were for the $5-meals. To find the 99.5% confidence interval for the true proportion of the orders for the $5-meals, you need to use which one of the following calculators?\r
\n" ); document.write( "\n" ); document.write( "Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.
\n" ); document.write( "Confidence interval =
\n" ); document.write( "Express the same answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.
\n" ); document.write( "Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places.
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Algebra.Com's Answer #830755 by ewatrrr(24785) $\"\"$ $\"About$

You can put this solution on YOUR website!

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document.write( "Hi  \r\n" );
document.write( "n = 357   p = 17/357 = .0476  and q = 1 - .0476 = .9524 \r\n" );
document.write( "99.5% confidence interval ⇒  z = 1.96  2-sided   Invnorm(.995 + .5/2 ) = 2.807\r\n" );
document.write( "ME = = .032\r\n" );
document.write( ".0476 - .032 < p < .0476 + .032\r\n" );
document.write( ".016 < p < .080\r\n" );
document.write( " 0r\r\n" );
document.write( "p = .0476 ± .032\r\n" );
document.write( "always check the Arithmetic\r\n" );
document.write( "Wish You the Best in your Studies.\r\n" );
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