document.write( "Question 204089: determine how many card holders should be sampled.\r
\n" ); document.write( "\n" ); document.write( "what is the required sample size needed to be able to estimate the mean dollars that credit card holders will spend each month. Within plus or minus $10.00 of the true mean with a 98% confidence level? Standard deviation is thought to be $500.00. how many card holders should be sampled? We then find out it will cost $5.00 per sample and the company is only planning to spend $10,000.00 on the sample. What will the trade off be when you lower the sample to $10,000.00 to meet their budget?
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Algebra.Com's Answer #154024 by stanbon(75887)\"\" \"About 
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determine how many card holders should be sampled.
\n" ); document.write( "what is the required sample size needed to be able to estimate the mean dollars that credit card holders will spend each month. Within plus or minus $10.00 of the true mean with a 98% confidence level? Standard deviation is thought to be $500.00. how many card holders should be sampled? We then find out it will cost $5.00 per sample and the company is only planning to spend $10,000.00 on the sample. What will the trade off be when you lower the sample to $10,000.00 to meet their budget
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\n" ); document.write( "n = [z*s/E]^2
\n" ); document.write( "n = [2.0537*500/10]^2 = 102.69^2 = 10544.7
\n" ); document.write( "Rounded up you get n = 10545
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\n" ); document.write( "If you lower the sample size the confidence level will decrease
\n" ); document.write( "or the margin of error will increase or both will happen.
\n" ); document.write( "That is because sample size and z* are directly related and
\n" ); document.write( "sample size and E are inversely related.
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.\r
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