This my first posting.
The Bank wants to estimate the mean dollars that each card holder will spend each month. It would like to be within plus or minus $10 of the true mean with a 98% confidence level. The standard deviation is thought to be $500. How many card holders should be sampled? After you?ve determined how many card holders should be sampled, The Bank comes back and says that it will cost $5 per sample and they were only planning on spending $10,000 on the sample. In a memo to the Bank product development team indicate how many card holders should be sampled to meet the original requirements of the sample. Then explain the trade-offs that will occur when you lower the sample to $10,000 to meet their budget. This is what I came up with is this anywhere near right.
n= (Z a/2 x $500 ) = 2000 = (2.33 x $500) = 116.50 /10sq = 3.16=.99 or 99% $10 $10
or
So I did n=13,573 x $5 = 67,865 / 2 = 33932.5
z a/2 = 2.33 t
2.33 x 33933= 79062.73 / 10 sq= 3.16 = .99 or 99%
.99 came from the table when i matched -3.0 + .16
The Bank wants to estimate the mean dollars that each card holder will spend each month. It would like to be within plus or minus $10 of the true mean with a 98% confidence level. The standard deviation is thought to be $500. How many card holders should be sampled? After you?ve determined how many card holders should be sampled, The Bank comes back and says that it will cost $5 per sample and they were only planning on spending $10,000 on the sample. In a memo to the Bank product development team indicate how many card holders should be sampled to meet the original requirements of the sample. Then explain the trade-offs that will occur when you lower the sample to $10,000 to meet their budget. This is what I came up with is this anywhere near right.
n= (Z a/2 x $500 ) = 2000 = (2.33 x $500) = 116.50 /10sq = 3.16=.99 or 99% $10 $10
or
So I did n=13,573 x $5 = 67,865 / 2 = 33932.5
z a/2 = 2.33 t
2.33 x 33933= 79062.73 / 10 sq= 3.16 = .99 or 99%
.99 came from the table when i matched -3.0 + .16