Question

As HIM director for a critical access hospital (CAH), you track the average length of stay for the facility. This is a critical piece of information since CMS requires that a CAH have an annual average length of stay of 26 hours or less per patient Using the data below, solve for the ALOs for the month of February 2015 and test if it meets the ALOS goal 1. 2. Calculate the standard deviation for this data Length of stay (in days) Number of patients 13 23 21 6 10 12

Answer 1

Variance and standard deviation (grouped data):

Mean: $Q8AsN3sAAAAASUVORK5CYII=$ = $WOOJIHZMAAAAASUVORK5CYII=$

Variance: = - (an equivalent formula)

Length Of Stay (In Day) -	Number Of Patients -	$3BdQGfOqiH9BRyPQAAAAASUVORK5CYII=$		$aUk5Qs+ewqBkDCgGQAAAABJRU5ErkJggg==$
1	5	5	1	5
2	13	26	4	52
3	23	69	9	207
4	21	84	16	336
5	9	45	25	225
6	3	18	36	108
7	1	7	49	49
8	1	8	64	64
10	1	10	100	100
12	1	12	144	144
Total	78	284	448	1290

Here n = 78 ( $wnxlfWulmFeEWCuAAAAAASUVORK5CYII=$ )

Solution of question 1:

So the ALOS for the month of February 2015 is = 284/78 = 3.641025641 days = (3.641025641 * 24) hours = 87.38461538 hours

Yes, it meets the ALOS goal (ALOS of 96 hours or less per patient)

Solution of question 2:

Variance = 1290/78 – (3.641025641)^2 = 16.53846 – 13.25707 = 3.281394

Standard Deviation: = $uaHh+QMxaYVIMo1DLgAAAABJRU5ErkJggg==$ = 1.811462