6.47- A company wishes to be 95% confident that the sample mean is correct to witin +/- $5 ....?

A survey is planned to determine the mean annual family medical expenses of employees of a large company. The management of the company wishes to be 95% confident that the sample mean is correct to witin +/- $50 of the true population mean annual family medical expenses. A pilot study indicates that the standard deviation can be estimated as $400.





a)How large a sample size is necessary?





b)If managment wants to be correct to within +/- $25, what sample size is necessary?








thanks so much!

6.47- A company wishes to be 95% confident that the sample mean is correct to witin +/- $5 ....?
I'm assuming the value of $400 is supposed to be assumed to be the population standard deviation, σ.





Recall the formula for determining the endpoints of a confidence interval for a mean, given the population standard deviation:


μ +- z* (σ / √n)





If we want to be correct within +- $50, then, it stands to reason that we want z* (σ / √n) to be equal to $50 (more specifically, less than or equal to $50).





z* = 1.96 for a 95% CI and σ = $400 as given.





Therefore,


1.96 ($400 / √n) %26lt;= $50


n %26gt;= 245.9





Note the inequality: sample size must be "greater than or equal" to 245.9, so we would round up to 246.





Then, for +- $25 accuracy, we notice that the value of z* doesn't change (still 95% CI), only the error does, so the inequality we need to solve is:


1.96 ($400 / √n) %26lt;= $25


n %26gt;= 983.4





I know it's tempting to use n = 983, but that would be incorrect. The inequality specifically says that n should be "greater than or equal to" 983.4 and so we must use n = 984. This concept makes sense conceptually as well; if n=983.4 creates a interval with error +- $25, we would expect that lowering the sample size increases the error (why else would we spend the time and effort to increase the sample size?).