document.write( "Question 977142: Determine the sample size for the following time study given a standard deviation of four minutes
\n" ); document.write( "and a 98% probability that the value of the sample mean is within 1.5 minutes? \n" ); document.write( "
Algebra.Com's Answer #598665 by rothauserc(4718)\"\" \"About 
You can put this solution on YOUR website!
A 98% confidence level corresponds to alpha = 0.02 and alpha/2 = 0.01
\n" ); document.write( "The region to the left of z(alpha/2) and to the right of z = 0, is 0.5 - .01 = 0.49.
\n" ); document.write( "**************************************************************************
\n" ); document.write( "In the table of the standard normal () distribution, an area of 0.49 corresponds to a value of 2.33.
\n" ); document.write( "The critical value is therefore z(alpha/2) = 2.33
\n" ); document.write( "***************************************************************************
\n" ); document.write( "The margin of error(E) = 1.5 and the standard deviation sigma = 4
\n" ); document.write( "let n be the sample size, then
\n" ); document.write( "n = ( (z(alpha/2) * sigma) / E )^2
\n" ); document.write( "n = ( (2.33 * 4) / 1.5 )^2
\n" ); document.write( "n = 38.60 approx 39
\n" ); document.write( "our sample size is 39\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
\n" );