Post on 16-Jan-2016
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Understanding Variation in Patient Satisfaction
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Variation in Patient Satisfaction Associated With Nursing
Purpose: To use appropriate graphical and statistical techniques to identify and quantify the critical factors associated with the variation in overall patient satisfaction.
Y = f(X)Patient Satisfaction associated with Nursing
= f(??????)
Step 1 - Define the Practical Problem
Step 2 -Translate to a Statistical Problem
Step 3 - Solve the Statistical Problem
Step 4 - Translate back to the Practical Problem
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Step 1: Define the Practical Problem
The hospital has made significant improvements in patient satisfaction over the past few years, but achieving the system goal of 85% excellent ratings has been difficult to achieve and maintain.
Anecdotal evidence abounds concerning reasons for variation in scores, but will these anecdotes be supported by the data?
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Step 2: Translate to a Statistical Problem
Y – Overall Patient Satisfaction with Nursing
Is a function of:
X1 – Agency Usage
X2 – Overtime Usage
X3 – Nursing Management Changes
X4 – RN Turnover Rate
X5 – Staffing Hours Variance
X6 – Skill Mix Percentage
X7 – Associate Satisfaction RN
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Step 3: Solve the Statistical Problem
First, look at aggregate data for 2004: Patient Satisfaction with Nursing
0.65 0.69 0.73 0.77 0.81 0.85
95% Confidence Interval for Mu
0.768 0.778 0.788 0.798 0.808 0.818
95% Confidence Interval for Median
Variable: Pt Sat
A-Squared:P-Value:
MeanStDevVarianceSkewnessKurtosisN
Minimum1st QuartileMedian3rd QuartileMaximum
0.770418
0.043367
0.773366
0.5120.185
0.7866050.0525952.77E-03-7.7E-01
0.38374943
0.6450000.7590000.7870000.8270000.870000
0.802791
0.066849
0.813317
Anderson-Darling Normality Test
95% Confidence Interval for Mu
95% Confidence Interval for Sigma
95% Confidence Interval for Median
Descriptive Statistics
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Step 3: Solve the Statistical Problem (continued)
Overall Satisfaction Scores are in statistical control—however, this is not surprising since aggregated data is being used—it hides the true amount of variation.
403020100
0.95
0.90
0.85
0.80
0.75
0.70
0.65
0.60
Observation Number
Indi
vidu
al V
alue
I Chart for Pt Sat
Mean=0.7866
UCL=0.9391
LCL=0.6341
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Step 3: Solve the Statistical Problem (continued)
2004 Overall Agency Usage % by Nursing Unit
0.160.120.080.040.00
95% Confidence Interval for Mu
0.060.050.040.030.02
95% Confidence Interval for Median
Variable: Agency
0.021683
0.035100
0.031318
Maximum3rd QuartileMedian1st QuartileMinimum
NKurtosisSkewnessVarianceStDevMean
P-Value:A-Squared:
0.054634
0.054106
0.057520
0.1590000.0690000.0350000.0050000.000000
430.3841330.9716701.81E-034.26E-024.44E-02
0.0021.279
95% Confidence Interval for Median
95% Confidence Interval for Sigma
95% Confidence Interval for Mu
Anderson-Darling Normality Test
Descriptive Statistics
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Step 3: Solve the Statistical Problem (continued)
Overall Agency Usage has one unit (7B at University) which falls outside the Upper Control Limit—however, 1 out of 43 units being out of control is no cause for alarm.
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0.2
0.1
0.0
-0.1
Observation Number
Indi
vidu
al V
alue
I Chart for Agency
1
Mean=0.04442
UCL=0.1571
LCL=-0.06830
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Step 3: Solve the Statistical Problem (continued)
Correlation between Satisfaction Scores (Y) and Agency Usage (X1)
P value =0.011 therefore it can be concluded with > 95% confidence that a statistically significant correlation exists. In fact, 14.6% of the variation in Pt. Sat. is explained by variation in Agency Usage.
0.150.100.050.00
0.85
0.75
0.65
Agency
Pt S
at
S = 0.0491794 R-Sq = 14.6 % R-Sq(adj) = 12.6 %
Pt Sat = 0.807609 - 0.472877 Agency
95% CI
Regression
Regression Plot
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Step 3: Solve the Statistical Problem (continued)
The following variables did not have a statistically significant correlation to Patient Satisfaction scores using 2004 aggregate data:
X2 – Overtime p-value = 0.121
X6 – Skill mix p-value = 0.216
X7 – AFS RN p-value = 0.879
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Step 3: Solve the Statistical Problem (continued)
2004 Monthly Satisfaction Scores by Nursing Unit
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as
13 T
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2 &
4 J
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2nd
3rd 3S
4th
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7B
7 C 7W
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8A
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FB
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TC
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0.4
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1.0
C2
Ser
vice
Exc
elle
nce
Boxplots of Service by C2(means are indicated by solid circles)
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Step 3: Solve the Statistical Problem (continued)
Histogram of Monthly Variation in Satisfaction Scores
Conclusion: 72% of the time nursing units fail to meet the 85% satisfaction goal
Process Capability for Service Excellence
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One-way ANOVA: Service Excellence versus Agency Usage < 5%
Analysis of Variance for Service Source DF SS MS F PAgency U 1 0.17457 0.17457 21.88 0.000Error 481 3.83771 0.00798Total 482 4.01227 Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev ------+---------+---------+---------+n 164 0.76183 0.09767 (------*------) y 319 0.80197 0.08472 (----*----) ------+---------+---------+---------+Pooled StDev = 0.08932 0.760 0.780 0.800 0.820
Step 3: Solve the Statistical Problem (continued)
Conclusion: Nursing Units that use 5% Agency or less have a statistically significantly higher average satisfaction score than units that use greater than 5% agency.
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Step 3: Solve the Statistical Problem (continued)
Result of Correlation/Regression
using monthly data for 2004X1 – Agency Usage p value = 0.000
X2 – Overtime Usage p value = 0.000
X3 – Nursing Management Changes
X4 – RN Turnover Rate p value = 0.545
X5 – Staffing Hours Variance p value = 0.117
X6 – Skill Mix Percentage p value = 0.006
Conclusion: P values less than 0.05 are statistically significant to a > than 95% Confidence Level. However, while statistically significant these variable explain only a small fraction of the variation in Patient Satisfaction.
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The regression equation isService Excellence = 0.801 - 0.241 Agency - 0.583 OT + 0.0380 Skill Mix + 0.145 Depart Rate RN_1 - 0.00210 HPPD
Predictor Coef SE Coef T PConstant 0.80123 0.01828 43.84 0.000Agency -0.24136 0.08420 -2.87 0.004OT -0.5833 0.1638 -3.56 0.000Skill Mi 0.03800 0.02600 1.46 0.145Depart R 0.14461 0.08675 1.67 0.096HPPD -0.002103 0.001086 -1.94 0.053
S = 0.08831 R-Sq = 7.3% R-Sq(adj) = 6.3%
Best Subsets Regression: Service Excellence versus Agency, OT, ... S D H k e P A i p P g l a D e l r n t V c O M a Vars R-Sq R-Sq(adj) y T i R r
1 3.8 3.6 X 1 3.1 2.9 X 2 5.6 5.2 X X 2 4.3 3.9 X X 3 6.3 5.7 X X X 3 6.1 5.5 X X X 4 6.9 6.1 X X X X 4 6.7 6.0 X X X X 5 7.3 6.3 X X X X X
Conclusion: Putting all the Factors (X’s) into the model only explains 7.3% of the total variation in satisfaction
Step 3: Solve the Statistical Problem (continued)
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Step 4: Translate Back to the Practical Problem
What actions, if any, do you take knowing this information?