Question

3.53 The time from when a patient is discharged from North Shore Hospital to the time the discharged patient’s bed is ready to be assigned to a new patient is referred to as the bed assignment turnaround time. If the bed turnaround time is excessive, it can cause problems with patient flow and delay medical procedures throughout the hospital. This can cause long waiting times for physicians and patients, thus creating customer dissatisfaction. The admissions RN has assigned a patient care associate to measure the bed turnaround time for a randomly selected bed each morning, afternoon, and evening for 30 days. Following are the bed turnaround time sample observations:

DAY	BED TURNAROUND TIMES (MIN)
1	127	135	167
2	140	155	122
3	112	128	97
4	223	135	154
5	181	155	160
6	103	158	145
7	146	135	167
8	104	122	115
9	136	158	137
10	145	163	106
11	84	146	125
12	169	152	208
13	216	124	163
14	190	178	103
15	148	205	144
16	157	151	126
17	142	102	95
18	166	178	159
19	177	211	204
20	98	91	158
21	133	160	152
22	212	131	138
23	180	165	134
24	88	126	108
25	95	156	138
26	156	202	177
27	144	157	165
28	184	171	106
29	138	142	155
30	150	99	148

a.Develop an x-chart to be used in conjunction with an R-chart using 3σ limits to monitor the bed turnaround times and indicate if the process is in control using these charts.

b.Is the hospital capable of consistently achieving bed turnaround times of 120 minutes ±15 minutes without improving the process?

Answer 1

(a)

Day	x1	x2	x3	X-bar	Range
1	127	135	167	143.0	40
2	140	155	122	139.0	33
3	112	128	97	112.3	31
4	223	135	154	170.7	88
5	181	155	160	165.3	26
6	103	158	145	135.3	55
7	146	135	167	149.3	32
8	104	122	115	113.7	18
9	136	158	137	143.7	22
10	145	163	106	138.0	57
11	84	146	125	118.3	62
12	169	152	208	176.3	56
13	216	124	163	167.7	92
14	190	178	103	157.0	87
15	148	205	144	165.7	61
16	157	151	126	144.7	31
17	142	102	95	113.0	47
18	166	178	159	167.7	19
19	177	211	204	197.3	34
20	98	91	158	115.7	67
21	133	160	152	148.3	27
22	212	131	138	160.3	81
23	180	165	134	159.7	46
24	88	126	108	107.3	38
25	95	156	138	129.7	61
26	156	202	177	178.3	46
27	144	157	165	155.3	21
28	184	171	106	153.7	78
29	138	142	155	145.0	17
30	150	99	148	132.3	51
			Avg.	146.79	47.47
				X_double-bar	R-bar

Note the following coefficients from a standard 3-sigma X-bar and R table for sample size n = 3.

A₂ = 1.023
D₃ = 0
D₄ = 2.574
d₂ = 1.693

UCL_X = X_double-bar + A₂ * R-bar = 146.79 + 1.023 * 47.47 = 195.35
LCL_X = X_double-bar - A₂ * R-bar = 146.79 - 1.023 * 47.47 = 98.23

UCL_R = D₄ * R-bar = 2.574 * 47.47 = 122.18
LCL_R = D₃ * R-bar = 0 * 47.47 = 0

The process has shown out-of-control behavior in Day-19 where the sample mean has crossed the UCL_X.

(b)

We following the above process in re-constructing the control limits by removing the Day-19 data assuming that it is due to a special cause variation that has been resolved. The result is as follows:

In the modified chart, R-bar = 47.93 and X_double-bar = 145.05

We estimate the process mean (μ) and process stdev (σ) as follows:

μ = X_double-bar = 145.05
σ = R-bar / d₂ = 47.93/1.693 = 28.31

USL = 120+15 = 135
LSL = 120 - 15 = 105

C_pu = (USL - μ) / 3.σ = (135 - 145.05) / (3*28.31) = -0.12
C_pl = (μ - LSL) / 3.σ = (145.05 - 105) / (3*28.31) = 0.47

C_pk = Min(C_pu, C_pl) = -0.12 which is less than 1.0. So, the process is incapable of producing withing the specificaion for a 3-sigma standard.