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The financial department at a large hospital would like to estimate the average outstanding balance owed by patients who have not paid their bills in full. In order for the interval to be useful in making budgeting decisions the department's management staff requires the the true (but unknown) average balance be within \$340 at a confidence level of 99%. Ten files of patients carrying outstanding balances were randomly selected from a list of all of all pediatrics patients and the outstanding balance (rounded to the nearest integer) was recorded. The data are reported below in Table 1. This can be considered a pilot study.

Table 1: Outstanding balances (\$) for 10 randomly selected patients $$1128 \qquad1362\qquad 2202\qquad 3738\qquad 4310\qquad 5342\qquad 6126\qquad 6436\qquad 6900\qquad 7350. $$

A) The total number of patient records that must be sampled in order for the finance dept. management to meet their constraints of confidence level and bound is:

Note: please round any necessary table values to two decimal precision before making subsequent calculations

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it looks like a homework... –  Cristian Sep 16 '12 at 23:59
Has the hospital never heard of computers? If they used a database, the actual average would be available instantly... –  Robert Israel Sep 17 '12 at 0:03
Alex, welcome to math.se! I've added the homework tag to your question. There's nothing wrong with asking homework questions on the site, as long as you mark them as such. If this is not homework, feel free to remove the tag. –  Drew Christianson Sep 17 '12 at 0:36
@Alex, do you know that there is a statistics site, called CrossValidated? –  Gerry Myerson Sep 17 '12 at 6:07
Do you have any information on the assumed distribution of outstanding balances? Without that you cannot calculate anything. –  user1965813 Jan 30 '13 at 8:25
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1 Answer

The question is to find a sample size n such that the halfwidth of a 99% confidence interval for the mean is 340 or less. The halfwidth of a 99% confidence interval is kσ/√n where k is the appropriate percentile point of a standard normal distribution,

k=2.575 for p=0.99. The random sample of 10 should have no direct baring on this. But if you do not know σ but the sample can give you an unbiased estimate s$^2$ of σ$^2$

We can use 2.575 s/√n =340 to solve for n. This is not quite right. If s were based on a sample of size n we could replace 2.575 by the constant for a t distribution with n-1 degrees of freedom. But s is based on a sample of size 10. So if we used the constant for t with 9 degrees of freedom we would have k = 3.25. The resulting n is (3.25/340) s$^2$ = 5350510.04 (3.25)/340 = 51144.58. So n=51145 is adequate. Since n>>10. This is conservative.

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So basically in your last part, you mean that the appropriate value of k has to come from a t-distribution and not normal distribution, right? –  Abhimanyu Arora Feb 16 at 11:43
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