document.write( "Question 620500: In order to test a new production method, 36 employees were selected randomly to try the new method. The mean was 75.79 parts per hour with a standard deviation of 28.25 parts per hour.\r
\n" ); document.write( "\n" ); document.write( "What is the sample size needed to obtain a margin of error 14.01 with a 90% confidence interval?
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Algebra.Com's Answer #390181 by ewatrrr(24785)\"\" \"About 
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\n" ); document.write( "Hi,
\n" ); document.write( "In order to test a new production method, 36 employees were selected randomly to try the new method.
\n" ); document.write( "The mean was 75.79 parts per hour with a standard deviation of 28.25 parts per hour.
\n" ); document.write( "What is the sample size needed to obtain a margin of error 14.01 with a 90% confidence interval?
\n" ); document.write( "SE = 14.01 =\"+1.645%2828.25%2Fsqrt%28n%29%29\"
\n" ); document.write( " \"sqrt%28n%29+=+1.645%2A28.25%2F14.01+=+3.317\" n = 11 \n" ); document.write( "
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