document.write( "Question 1162956: The manager of a pizza chain in Albuquerque, New Mexico, wants to determine the average size of their advertised 24-inch pizzas. She takes a random sample of 36 pizzas and records their mean and standard deviation as 24.60 inches and 1.90 inches, respectively. She subsequently computes the 99% confidence interval of the mean size of all pizzas as [23.78, 25.42]. However, she finds this interval to be too broad to implement quality control and decides to reestimate the mean based on a bigger sample. Using the standard deviation estimate of 1.90 from her earlier analysis, how large a sample must she take if she wants the margin of error to be under 0.5 inch? (You may find it useful to reference the z table. Round intermediate calculations to at least 4 decimal places and \"z\" value to 3 decimal places. Round up your answer to the nearest whole number.)
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Algebra.Com's Answer #786866 by Boreal(15235) $\"\"$ $\"About$

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margin of error= z(.995)* sigma/sqrt(n)=0.5
\n" ); document.write( "2.576*1.90/sqrt(n)=0.5
\n" ); document.write( "or 2.576*1.90/0.5=sqrt (n)=9.789
\n" ); document.write( "n=95.82 or 96, rounding upward
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