Homework SourseThe financial department at a large hospital would like to e
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 interval be no wider than $700 (total) at a confidence level of 99%. Ten files of patients carrying outstanding balances were randomly selected from a list of all of all pediatrics patients and the outstanding balance (rounded to the nearest integer) was recorded. The data are reported below in Table 1. This can be considered a pilot study.
A) The total number of patient records that must be sampled in order for the finance dept. management to meet their constraints of confidence level and bound is:
| Table 1: Outstanding balances ($) for 10 randomly selected patients | |||||||||
| 8343 | 4677 | 7469 | 6337 | 7136 | 8947 | 4719 | 8741 | 7607 | 3913 |
Solution
Getting the mean, X,
X = Sum(x) / n
Sum(x) = 67889
As n = 10
Thus,
X = 6788.9
Setting up tables,
x x - X (x - X)^2
8343 1554.1 2415226.81
4677 -2111.9 4460121.61
7469 680.1 462536.01
6337 -451.9 204213.61
7136 347.1 120478.41
8947 2158.1 4657395.61
4719 -2069.9 4284486.01
8741 1952.1 3810694.41
7607 818.1 669287.61
3913 -2875.9 8270800.81
Thus, Sum(x - X)^2 = 29355240.9
Thus, as
s^2 = Sum(x - X)^2 / (n - 1)
As n = 10
s^2 = 3261693.433
Thus,
s = 1806.015901 [STANDARD DEVIATION]
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Note that
n = z(alpha/2)^2 s^2 / E^2
where
alpha/2 = (1 - confidence level)/2 = 0.005
Using a table/technology,
z(alpha/2) = 2.575829304
Also,
s = sample standard deviation = 1806.015901
E = margin of error = width/2 = 700/2 = 350
Thus,
n = 176.6612137
Rounding up,
n = 177 [ANSWER]
