Question

As a hospital administrator of a large hospital, you are concerned with the absenteeism among nurses’ aides. The issue has been raised by registered nurses, who feel they often have to perform work normally done by their aides. To get the facts, absenteeism data were gathered for the last 15 days. This is considered a representative period for future conditions. After taking random samples of 96 personnel files each day, the following data were produced:

Day	Aides Absent	Day	Aides Absent
1	7	9	8
2	6	10	6
3	7	11	4
4	9	12	5
5	7	13	7
6	6	14	8
7	4	15	6
8	5

Assume that there is sufficient data for preparing control charts. Because your assessment of absenteeism is likely to come under careful scrutiny, you would like to use 2σ control limits. You are not interested in the reasons for the absence but just if there is an absence or not. You want to be sure to identify any instances of unusual absences. If some are present, you will have to explore them on behalf of the registered nurses.

a) Calculate the 2σ (sigma) control limits. The centerline should be based on the data from these 15 days. All calculations should be to three decimal places (x.xxx).

b) After finding the control limits, prepare and provide the appropriate control chart.

c) Using this sample information and the control chart, do the data indicate that absenteeism is a problem or not for the hospital’s nurses’ aides? Why or why not?

d) Suppose that the administrator has obtained data from an industry association that indicates the acceptable industry standards for absenteeism has upper and lower limits of 0.150 and 0.055 respectively. Considering the performance based on the daily samples from the control chart and not just the average absentee rate or the relationship between the control limits and the industry limits, what would you conclude about the hospital’s experiences compared to the industry’s standards? Explain your answer.   

Answer 1

Day	sample size	absentees	p value
1	96	7	0.072917
2	96	6	0.0625
3	96	7	0.072917
4	96	9	0.09375
5	96	7	0.072917
6	96	6	0.0625
7	96	4	0.041667
8	96	5	0.052083
9	96	8	0.083333
10	96	6	0.0625
11	96	4	0.041667
12	96	5	0.052083
13	96	7	0.072917
14	96	8	0.083333
15	96	6	0.0625

average p value pbar = 0.066

UCLp = pbar  + 2 * [ pbar * (1-pbar)/n]^1/2

=0.066 + 2* [ 0.066*0.934/96]^1/2

= 0.1166

Corresponding absenteeism value = 0.116x96 =11.13

LCLp = pbar - 2 * [ pbar * (1-pbar)/n]^1/2

=0.066 - 2* [ 0.066*0.934/96]^1/2

= 0.0153

Corresponding absenteeism value = 0.0153x96 =1.4688

The values are within 2 sigma specification limits and therefore it does not seem to be an issue.

The absenteism values according to the industry average =

upper value =0.15x96 =14.4

lower value =0.055x96 =5.28

It can be seen that there is no value which is more than the upper value. However, there are one value of 4 and two of 5, which are actually below the lim9t. Hence t can be concluded that the hospital is performing better than the industry norms.