document.write( "Question 265571: Can someone please answer this and help me understand how to calculate? Thank you.\r
\n" ); document.write( "\n" ); document.write( "According to a recent poll, 16.2% of credit cardholders in the United States have used their credit card to make a purchase on the internet. You will be asked to identify how large a sample should be taken if the true proportion is to be estimated to within 1.5% at the 95% reliability level? You may use the sample proportion in the calculation of the sample size. \n" ); document.write( "
Algebra.Com's Answer #195280 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
According to a recent poll, 16.2% of credit cardholders in the United States have used their credit card to make a purchase on the internet. You will be asked to identify how large a sample should be taken if the true proportion is to be estimated to within 1.5% at the 95% reliability level? You may use the sample proportion in the calculation of the sample size.
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\n" ); document.write( "Formula:
\n" ); document.write( "Since E = z*sqrt[pq/n]
\n" ); document.write( "sqrt(n) = (z/E)*sqrt(pq)
\n" ); document.write( "n = (z/E)^2(pq)
\n" ); document.write( "Your problem:
\n" ); document.write( "n = (1.96/0.015)^2*(0.162*0.83.8)
\n" ); document.write( "n = 2317.87
\n" ); document.write( "Round up to n = 2318
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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