Let's assume that the duration of stay in HDU has an exponential
distribution of parameter theta (say theta = one day). Let's assume that the
arrival times follow a Poisson distribution throughout the year, with an
average of 60 arrivals per year. Then you can do Monte Carlo simulations to
compute:
* proportion of time when there is 2 or fewer people simultaneously present
in HDU
* proportion of time when there 3 or more people simultaneously present in
HDU
Violations of assumptions occur if for instance there is a much higher
concentration of patients in certain time periods, or if occasionally, a
patient requires a very long stay (several months).
--
Vincent Granville, Ph.D.
Data Shaping Solutions, LLC
http://www.datashaping.com
Cathy Smith writes:
> Dear All,
>
> Data was collected on paediatric admissions for 4 years:
>
> Year: 1999 2000 2001 2002 2003 AVERAGE
>
> Admissions: 2601 2670 2954 3197 4201 3124.6
>
>
>
> Out of these admissions; there were the following HDU admmission
>
> HDU Admission: 69 52 51 57 62
>
>
>
> The HDU had two beds and in each year, and on each day; one bed was
> occupied
> while the other was empty. Obvioulsy, with an average of 3124.6
> admission;
> the two beds suffice for HDU admissions out of the total admissions.
>
>
> Is there any statistical test one would do to show that with an average of
> 3000 admissions, two beds are enough?.
>
> Cathy
>
> _________________________________________________________________
> Express yourself instantly with MSN Messenger! Download today - it's FREE!
> http://messenger.msn.click-url.com/go/onm00200471ave/direct/01/
|