Business Research Quiz
CollegeDummy
brm.zipNeal.LarryBUS457A7.docx
Question 1
Problem:
It is not certain about the relationship between age, Y, as a function of systolic blood pressure.
Goal:
To establish the relationship between age Y, as a function of systolic blood pressure.
Finding/Conclusion:
Based on the available data, the relationship is obtained and shown below:
Regression Analysis: Age versus SBP
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
Regression 1 2933 2933.1 21.33 0.000
SBP 1 2933 2933.1 21.33 0.000
Error 28 3850 137.5
Lack-of-Fit 21 2849 135.7 0.95 0.575
Pure Error 7 1002 143.1
Total 29 6783
Model Summary
S R-sq R-sq(adj) R-sq(pred)
11.7265 43.24% 41.21% 3.85%
Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant -18.3 13.9 -1.32 0.198
SBP 0.4454 0.0964 4.62 0.000 1.00
Regression Equation
Age = -18.3 + 0.4454 SBP
It is found that there is an outlier in the dataset, which significantly affect the regression equation. As a result, the outlier is removed, and the regression analysis is run again.
Regression Analysis: Age versus SBP
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
Regression 1 4828.5 4828.47 66.81 0.000
SBP 1 4828.5 4828.47 66.81 0.000
Error 27 1951.4 72.27
Lack-of-Fit 20 949.9 47.49 0.33 0.975
Pure Error 7 1001.5 143.07
Total 28 6779.9
Model Summary
S R-sq R-sq(adj) R-sq(pred)
8.50139 71.22% 70.15% 66.89%
Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant -59.9 12.9 -4.63 0.000
SBP 0.7502 0.0918 8.17 0.000 1.00
Regression Equation
Age = -59.9 + 0.7502 SBP
The p-value for the model is 0.000, which implies that the model is significant in the prediction of Age. The R-square of the model is 70.2%, implies that 70.2% of variation in age can be explained by the model
Recommendation:
The regression model Age = -59.9 +0.7502 SBP can be used to predict the Age, such that over 70% of variation in Age can be explained by the model.
Question 2
Problem:
It is not sure that whether the factors X1 to X4 which represents four different success factors have any influences on the annual savings as a result of CRM implementation.
Goal:
To determine which of the success factors are most significant in the prediction of a successful CRM program, and develop the corresponding model for the prediction of CRM savings.
Finding/Conclusion:
Based on the available data, the relationship is obtained and shown below:
Regression Analysis: Y versus X1, X2, X3, X4
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
Regression 4 2667.90 666.975 111.48 0.000
X1 1 25.95 25.951 4.34 0.071
X2 1 2.97 2.972 0.50 0.501
X3 1 0.11 0.109 0.02 0.896
X4 1 0.25 0.247 0.04 0.844
Error 8 47.86 5.983
Total 12 2715.76
Model Summary
S R-sq R-sq(adj) R-sq(pred)
2.44601 98.24% 97.36% 95.94%
Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant 62.4 70.1 0.89 0.399
X1 1.551 0.745 2.08 0.071 38.50
X2 0.510 0.724 0.70 0.501 254.42
X3 0.102 0.755 0.14 0.896 46.87
X4 -0.144 0.709 -0.20 0.844 282.51
Regression Equation
Y = 62.4 + 1.551 X1 + 0.510 X2 + 0.102 X3 - 0.144 X4
Correlation: Y, X2, X4
Y X2
X2 0.816
0.001
X4 -0.821 -0.973
0.001 0.000
Based on the analysis of VIF and the correlations, it can be seen that there is a strong negative correlation between X2 and X4. Since X4 has a stronger correlation with Y, X2 is discarded and the regression analysis is run again.
Regression Analysis: Y versus X1, X3, X4
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
Regression 3 2664.93 888.31 157.27 0.000
X1 1 124.90 124.90 22.11 0.001
X3 1 23.93 23.93 4.24 0.070
X4 1 1176.24 1176.24 208.24 0.000
Error 9 50.84 5.65
Total 12 2715.76
Model Summary
S R-sq R-sq(adj) R-sq(pred)
2.37665 98.13% 97.50% 96.52%
Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant 111.68 4.56 24.48 0.000
X1 1.052 0.224 4.70 0.001 3.68
X3 -0.410 0.199 -2.06 0.070 3.46
X4 -0.6428 0.0445 -14.43 0.000 1.18
Regression Equation
Y = 111.68 + 1.052 X1 - 0.410 X3 - 0.6428 X4
The p-value of X3 is greater than 0.05. As a result it is also discarded. The analysis is run again.
Regression Analysis: Y versus X1, X4
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
Regression 2 2641.00 1320.50 176.63 0.000
X1 1 809.10 809.10 108.22 0.000
X4 1 1190.92 1190.92 159.30 0.000
Error 10 74.76 7.48
Total 12 2715.76
Model Summary
S R-sq R-sq(adj) R-sq(pred)
2.73427 97.25% 96.70% 95.54%
Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant 103.10 2.12 48.54 0.000
X1 1.440 0.138 10.40 0.000 1.06
X4 -0.6140 0.0486 -12.62 0.000 1.06
Regression Equation
Y = 103.10 + 1.440 X1 - 0.6140 X4
Both of the coefficients for X1 and X4 are significantly different from zero, with p-value being 0.000 for both of the coefficients. The p-value of overall model is also 0.000, and thus it is significant in predicting the CRM savings. The R-square of the model is 97.25%, which implies that 97.25% of the variation in CRM savings can be explained by the model.
The prediction model is given by: 103.10 + 1.440X1 -0.6140X4
Recommendation:
Since both X1 and X4 are both strongly correlated to the CRM savings, it is essential to ensure that both X1 and X4 are present in the implementation of CRM system.
The prediction model obtained can be used in estimating the CRM savings, given that no other success factors are being incorporated, and the data used for estimation are within the ranges of the analysis here.
Question 3
Problem:
It is not sure whether any of the case load, DRG type, case severity and patient follow-up time are significant in influencing the high readmission rates, where readmissions are very expensive and produce tremendous hardship for patients.
Goal:
To determine which of the factors of case load, DRG type, case severity and patient follow-up time are significant in the prediction of readmission rates, and develop the corresponding measure to reduce the readmission rates.
Finding/Conclusion:
Based on the available data, the relationship is obtained and shown below:
From the matrix plot, it can be seen that there is a quadratic relationship between the Readmission rate and the Time. As a result, a quadratic term of Time will be included in the regression model.
Regression Analysis: ReadmitRate versus Census, Severity, Time, DRG
Method
Categorical predictor coding (1, 0)
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
Regression 6 0.013839 0.002306 63.72 0.000
Census 1 0.000115 0.000115 3.18 0.088
Severity 1 0.000095 0.000095 2.61 0.120
Time 1 0.005527 0.005527 152.69 0.000
DRG 2 0.000032 0.000016 0.44 0.649
Time*Time 1 0.005740 0.005740 158.57 0.000
Error 22 0.000796 0.000036
Total 28 0.014635
Model Summary
S R-sq R-sq(adj) R-sq(pred)
0.0060165 94.56% 93.07% 90.34%
Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant -1.0356 0.0840 -12.33 0.000
Census 0.000039 0.000022 1.78 0.088 2.39
Severity -0.001489 0.000921 -1.62 0.120 2.81
Time 0.1661 0.0134 12.36 0.000 526.86
DRG
B 0.00274 0.00294 0.93 0.361 1.57
C 0.00158 0.00297 0.53 0.600 1.41
Time*Time -0.006244 0.000496 -12.59 0.000 522.14
Regression Equation
DRG
A ReadmitRate = -1.0356 + 0.000039 Census - 0.001489 Severity + 0.1661 Time
- 0.006244 Time*Time
B ReadmitRate = -1.0329 + 0.000039 Census - 0.001489 Severity + 0.1661 Time
- 0.006244 Time*Time
C ReadmitRate = -1.0341 + 0.000039 Census - 0.001489 Severity + 0.1661 Time
- 0.006244 Time*Time
All the coefficients except Time are not significantly different from zero, with p-values of all the coefficient greater than 0.05. As a result, all of these variables will be discarded.
Regression Analysis: ReadmitRate versus Time
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
Regression 2 0.013627 0.006814 175.76 0.000
Time 1 0.011221 0.011221 289.43 0.000
Time*Time 1 0.011825 0.011825 305.03 0.000
Error 26 0.001008 0.000039
Lack-of-Fit 22 0.000975 0.000044 5.37 0.057
Pure Error 4 0.000033 0.000008
Total 28 0.014635
Model Summary
S R-sq R-sq(adj) R-sq(pred)
0.0062264 93.11% 92.58% 90.75%
Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant -1.0088 0.0654 -15.43 0.000
Time 0.16763 0.00985 17.01 0.000 264.25
Time*Time -0.006375 0.000365 -17.47 0.000 264.25
Regression Equation
ReadmitRate = -1.0088 + 0.16763 Time - 0.006375 Time*Time
All the regression coefficients in the final model is significantly different from zero, with p-value of both coefficient of Time and Time^2 being 0.000. The p-value of the overall model is also 0.000, which implies that the model is significant in predicting the readmission rate. The residual plots do not indicate any significant deviation from the assumption of linear models. The R-square of the model is 93.11%, which implies that 93.11% of variation in the readmission rate can be explained by the model.
Prediction for ReadmitRate
Regression Equation
ReadmitRate = -1.0088 + 0.16763 Time - 0.006375 Time*Time
Variable Setting
Time 13.1
Fit SE Fit 95% CI 95% PI
0.0929927 0.0017406 (0.0894149, 0.0965704) (0.0797035, 0.106282)
The median value for Time is 13.1 days. At this value, the readmission rate is estimated to be about 9.3%
The range within which we can expect the average patient readmission rate to fall with 95% confidence is between 8.9% and 9.6%
The rate within which we can expect an individual patient’s readmission rate to fall with 95% confidence is between 7.9% and 10.6%
Recommendation:
The regression model found is Readmission rate = -1.0088 + 0.16763 Time -0.006375 Time2
This model can explain more than 93% of variation in readmission rate, and simply using patient follow-up time to predict the readmission rate. It is recommended to use the model for the range of data within those being used in the analysis.
Question 4
Problem:
It is not sure whether any of the given predictors can be used to estimate the gas mileage
Goal:
To determine the best model using the predictor variables that can estimate the gas mileage.
Finding/Conclusion:
First of all, there are two observations with missing values and they are being removed.
Based on different trial of combinations of predictors, the final best model is shown below:
Regression Analysis: Y versus X1
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
Regression 2 934.00 466.998 61.47 0.000
X1 1 209.46 209.461 27.57 0.000
X1*X1 1 67.77 67.768 8.92 0.006
Error 27 205.11 7.597
Lack-of-Fit 16 142.13 8.883 1.55 0.232
Pure Error 11 62.98 5.725
Total 29 1139.11
Model Summary
S R-sq R-sq(adj) R-sq(pred)
2.75621 81.99% 80.66% 77.59%
Coefficients
Term Coef SE Coef T-Value P-Value VIF
Constant 39.98 2.56 15.61 0.000
X1 -0.1059 0.0202 -5.25 0.000 20.89
X1*X1 0.000109 0.000036 2.99 0.006 20.89
Regression Equation
Y = 39.98 - 0.1059 X1 + 0.000109 X1*X1
It is found that the best model only used X1 as the predictor variable. Both X1 and X12 have p-value of 0.000 for the estimated coefficient. The R-square of the model is 81.99%, which implies that 81.99% of variation in the gas mileage can be explained by the model. The best model is given by Gas mileage = 39.98 – 0.1059X1 + 0.000109 X12.
Recommendation:
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Neal.LarryBUS457A6.docx
Larry Neal
Assignment 6
BUS 457
11/09/2014
Question 1
Problems:
It is not sure whether meeting the daily patient discharge target is dependent upon the number of consulting MD’s available.
Goals:
To determine whether dependency exists between discharge target and number of consulting MD’s, and make adjustment for a balanced approach to staffing for discharge purposes.
Findings / Conclusion:
Chi-square test is being used in the analysis. The result of the analysis is shown below:
Rows: Met_Discharges Columns: Consulting MDs
0 1 2 3 All
No 95 148 95 54 392
92.75 152.80 91.29 55.16
Yes 95 165 92 59 411
97.25 160.20 95.71 57.84
All 190 313 187 113 803
Cell Contents: Count
Expected count
Pearson Chi-Square = 0.744, DF = 3, P-Value = 0.863
Likelihood Ratio Chi-Square = 0.744, DF = 3, P-Value = 0.863
From the result of the chi-square test, the p-value of the chi-square test is 0.863. Since the p-value is greater than 0.05, the null hypothesis cannot be rejected and there is not sufficient evidence to conclude that dependency exists between discharge target and number of consulting MD’s
Recommendation:
Other factors instead of number of consulting MD’s should be investigated to improve meeting patient discharge targets.
Question 2a
Problems:
It is not sure whether annual sales volume is dependent upon sales proposal being won or lost over 2 years period.
Goals:
To determine whether dependency exists between annual sales volume and sales proposal being won or lost.
Findings / Conclusion:
Chi-square test is being used in the analysis. The result of the analysis is shown below:
Rows: Proposal Columns: Sales$
<1M >5M 1-2M 2-5M All
Lost 25 30 31 41 127
20.95 29.21 29.21 47.63
Won 8 16 15 34 73
12.05 16.79 16.79 27.38
All 33 46 46 75 200
Cell Contents: Count
Expected count
Pearson Chi-Square = 5.023, DF = 3, P-Value = 0.170
Likelihood Ratio Chi-Square = 5.097, DF = 3, P-Value = 0.165
From the result of the chi-square test, the p-value of the chi-square test is 0.170. Since the p-value is greater than 0.05, the null hypothesis cannot be rejected and there is not sufficient evidence to conclude that dependency exists between annual sales volume and sales proposal being won or lost.
Recommendation:
Other factors instead of sales volume or in addition to volume should be investigated to identify factors that can increase the number of sales proposals won.
Question 2b
Problems:
It is not sure whether annual sales volume conditional on seniority and company car is dependent upon sales proposal being won or lost over 2 years period.
Goals:
To determine whether dependency exists between annual sales volume and sales proposal being won or lost, conditional on seniority and company car.
Findings / Conclusion:
Chi-square test is being used in the analysis. The result of the analysis conditional on seniority is shown below:
Tabulated Statistics: Proposal, Sales$, Seniority
Results for Seniority = <5years
Rows: Proposal Columns: Sales$
<1M >5M 1-2M 2-5M All
Lost 13 11 16 22 62
8.857 13.918 13.918 25.306
1.9378 0.6119 0.3113 0.4319
Won 1 11 6 18 36
5.143 8.082 8.082 14.694
3.3373 1.0539 0.5362 0.7439
All 14 22 22 40 98
Cell Contents: Count
Expected count
Contribution to Chi-square
Pearson Chi-Square = 8.964, DF = 3, P-Value = 0.030
Likelihood Ratio Chi-Square = 10.339, DF = 3, P-Value = 0.016
Results for Seniority = 5+years
Rows: Proposal Columns: Sales$
<1M >5M 1-2M 2-5M All
Lost 12 19 15 19 65
12.11 15.29 15.29 22.30
0.00096 0.89796 0.00566 0.48942
Won 7 5 9 16 37
6.89 8.71 8.71 12.70
0.00169 1.57750 0.00994 0.85978
All 19 24 24 35 102
Cell Contents: Count
Expected count
Contribution to Chi-square
Pearson Chi-Square = 3.843, DF = 3, P-Value = 0.279
Likelihood Ratio Chi-Square = 4.027, DF = 3, P-Value = 0.259
From the result of the chi-square test conditional on the layer of seniority, it can be seen that for the layer with seniority less than or equal to 5 years, the p-value of the chi-square test is 0.030. Since the p-value is smaller than 0.05, the null hypothesis is rejected and there is sufficient evidence to conclude that dependency exists between annual sales volume and sales proposal being won or lost for those with seniority less than or equal to 5 years.
The result of the analysis conditional on seniority is shown below:
Tabulated Statistics: Proposal, Sales$, CompanyCar
Results for CompanyCar = No
Rows: Proposal Columns: Sales$
<1M >5M 1-2M 2-5M All
Lost 12 17 11 24 64
9.06 16.91 11.47 26.57
0.95660 0.00053 0.01940 0.24786
Won 3 11 8 20 42
5.94 11.09 7.53 17.43
1.45768 0.00080 0.02956 0.37769
All 15 28 19 44 106
Cell Contents: Count
Expected count
Contribution to Chi-square
Pearson Chi-Square = 3.090, DF = 3, P-Value = 0.378
Likelihood Ratio Chi-Square = 3.318, DF = 3, P-Value = 0.345
Results for CompanyCar = Yes
Rows: Proposal Columns: Sales$
<1M >5M 1-2M 2-5M All
Lost 13 13 20 17 63
12.06 12.06 18.10 20.78
0.0726 0.0726 0.2004 0.6865
Won 5 5 7 14 31
5.94 5.94 8.90 10.22
0.1476 0.1476 0.4072 1.3951
All 18 18 27 31 94
Cell Contents: Count
Expected count
Contribution to Chi-square
Pearson Chi-Square = 3.130, DF = 3, P-Value = 0.372
Likelihood Ratio Chi-Square = 3.069, DF = 3, P-Value = 0.381
From the result of the Chi-square test, it can be seen that the p-values for not having company car and having company car are 0.378 and 0.372 respectively. Therefore it can be conclude that regardless of having company car, the sales volume and proposal win or loss numbers are not significantly dependent.
Recommendations:
Lower seniority salespersons are more adept at winning the proposals with less than one million dollars than higher seniority salespersons. Therefore low seniority staff should be assigned to lower payoff clients initially. Company car will not be a valid motivation factor in sales and thus should not be used.
Question 3a
Problems:
It is not sure whether authorization errors for medical services are dependent upon department.
Goals:
To determine whether dependency exists between authorization errors for medical services and department.
Findings / Conclusion:
Chi-square test is being used in the analysis. The result of the analysis is shown below:
Rows: defect Columns: dep cd
DEP EE SP All
No 16 55 12 83
17.15 52.57 13.28
0.0775 0.1126 0.1234
Yes 15 40 12 67
13.85 42.43 10.72
0.0961 0.1395 0.1528
All 31 95 24 150
Cell Contents: Count
Expected count
Contribution to Chi-square
Pearson Chi-Square = 0.702, DF = 2, P-Value = 0.704
Likelihood Ratio Chi-Square = 0.701, DF = 2, P-Value = 0.704
From the result of the Chi-square test, it can be seen that the p-value is 0.704. Since the p-value is greater than 0.05, the null hypothesis cannot be rejected. Therefore there is not sufficient evidence to conclude that dependency exists between authorization errors for medical services and department.
Recommendations:
Department is not a significant factor in influencing authorization errors, and thus other factors should be explored.
Question 3b
Problems:
It is not sure whether authorization errors for medical services are dependent upon case entry site.
Goals:
To determine whether dependency exists between authorization errors for medical services and case entry site.
Findings / Conclusion:
Chi-square test is being used in the analysis. The result of the analysis is shown below:
Rows: defect Columns: Case Enter Site
HOU PDX PHL SD SFO All
No 11 17 26 7 22 83
10.51 18.26 26.56 6.09 21.58
0.02253 0.08694 0.01181 0.13705 0.00817
Yes 8 16 22 4 17 67
8.49 14.74 21.44 4.91 17.42
0.02791 0.10771 0.01463 0.16978 0.01013
All 19 33 48 11 39 150
Cell Contents: Count
Expected count
Contribution to Chi-square
Pearson Chi-Square = 0.597, DF = 4, P-Value = 0.963
Likelihood Ratio Chi-Square = 0.601, DF = 4, P-Value = 0.963
* NOTE * 1 cells with expected counts less than 5
From the result of the Chi-square test, it can be seen that the p-value is 0.963. Since the p-value is greater than 0.05, the null hypothesis cannot be rejected. Therefore there is not sufficient evidence to conclude that dependency exists between authorization errors for medical services and case entry site.
Recommendations:
Case entry site is not a significant factors in influencing authorization errors, and thus other factors should be explored. However, it should be noted that the expected cell count for San Diego (SD) logging a “Yes” response is less than 5, it is recommended that more data to be collected with the response for San Diego (SD) logging being “Yes” such that the count of data is greater than 5. The test is suggested to be redone at that time.
Neal.LarryBUS457A5.docx
Larry Neal
Assignment 5
BUS 457
10/31/2014
Q1.
Problem:
For a supply chain project team it is not certain whether international orders placed on weekends is longer than that of those placed on a weekday, which is 3 days. The team will pursue weekend shipment time improvements only if it is proven that the median weekend order shipments are significantly longer than 4 days.
Goal:
Determine whether the median weekend order shipments are significantly longer than 4 days.
Findings / Conclusions:
A Wilcoxon signed rank test is carried out to test for the median. The result is shown below:
Wilcoxon Signed Rank Test: ShipTime
Test of median = 4.000 versus median > 4.000
N for Wilcoxon Estimated
N Test Statistic P Median
ShipTime 37 30 351.0 0.008 6.500
From the result, it can be seen that the p-value of the test is 0.008. Therefore the null hypothesis is rejected at 5% significant level. There is sufficient statistical evidence to infer the median ship time for weekend international order is significantly longer than 4.0 days.
Recommendation:
Pursue weekend international orders improvements as a means, in part, to bring weekend shipments more in line with the week day shipment median of 3.0 days.
Q2
Problem:
Verify the theory that the patient length of stay is generally longer when they miss the target of patient discharge than when they don’t.
Goal:
Determine whether the length of stay is longer when they miss the patient discharge target than when they don’t.
Findings / Conclusion:
A Mann-Whitney test is carried out to test for the hypothesis of median length of stay whether missing the patient discharge target is longer than when not. The result of the test is shown below:
Mann-Whitney Test and CI: LOS_No, LOS_Yes
N Median
LOS_No 392 2.0000
LOS_Yes 411 2.0000
Point estimate for η1 - η2 is -0.0000
95.0 Percent CI for η1 - η2 is (-0.0001,0.0001)
W = 158801.0
Test of η1 = η2 vs η1 > η2 is significant at 0.3556
The test is significant at 0.3523 (adjusted for ties)
Based on the result of the test, the p-value is 0.3523. Therefore the null hypothesis is not rejected. There is not sufficient evidence to conclude that the median length of stay when they miss the target is longer than when they don’t.
Recommendation:
Whether meeting the patient discharge targets has no influence on the patient length of stay, and thus other factors should be considered when it is intended to adjust for length of stay.
Q3
Problem:
It is uncertain that which of the factors among hospitalist group, number of consulting MDs and payer have significant influence on the excessive patient length of stay at Basin Medical Center.
Goal:
Determine which, if any of the factors of hospitalist group, number of consulting MDs and payer have significant influence on excessive patient length of stay.
Findings / Conclusion:
Since the research data are subject to outliers and are not normally distributed, Mood’s Median nonparametric test was used to carry out the hypothesis test. The result of test between length of stay and hospitalist groups is shown below:
Mood Median Test: LOS versus Hospitalist Group
Mood median test for LOS
Chi-Square = 4.30 DF = 2 P = 0.116
Hospitalist Individual 95.0% CIs
Group N≤ N> Median Q3-Q1 ---+---------+---------+---------+---
Galen 129 150 3.00 3.00 (--------------------------------*
n/a 243 211 2.00 1.00 *--------------------------------)
Pediatrix 39 31 2.00 1.00 *--------------------------------)
---+---------+---------+---------+---
2.10 2.40 2.70 3.00
Overall median = 2.00
Based on the result of the test, it is found that the p-value of the test is 0.116. Therefore the null hypothesis is not rejected, and there are not sufficient evidence to conclude that median length of stay is different among hospitalist groups.
The result of the test between the length of stay and number of consulting MDs is shown below:
Mood Median Test: LOS versus Number of Consulting MDs
Mood median test for LOS
Chi-Square = 88.39 DF = 3 P = 0.000
Number of
Consulting Individual 95.0% CIs
MDs N≤ N> Median Q3-Q1 ---+---------+---------+---------+---
0 133 57 2.00 1.25 *
1 175 138 2.00 2.00 *
2 85 102 3.00 2.00 (-------*
3 18 95 5.00 6.00 (--------*-------)
---+---------+---------+---------+---
2.4 3.6 4.8 6.0
Overall median = 2.00
Based on the result of the test, it is found that the p-value of the test is 0.000. Therefore the null hypothesis is rejected, and there are sufficient evidence to conclude that median length of stay is different among consulting MDs with three consulting MDs produced a median length of stay of 5 days.
The result of the test between the length of stay and payer is shown below:
Mood Median Test: LOS versus Payer
Mood median test for LOS
Chi-Square = 51.31 DF = 6 P = 0.000
Payer N≤ N> Median Q3-Q1
Commercial 135 108 2.00 2.00
County 50 36 2.00 1.00
MediCal 76 61 2.00 1.00
Medicare 90 162 3.00 3.00
Other General Ins 18 0 1.00 1.00
Self Pay 33 23 2.00 3.00
Sutter Select 9 2 2.00 1.00
Individual 95.0% CIs
Payer +---------+---------+---------+------
Commercial *---------)
County *---------)
MediCal *---------)
Medicare *---------)
Other General Ins *---------)
Self Pay *---------)
Sutter Select (---------*)
+---------+---------+---------+------
1.0 2.0 3.0 4.0
Overall median = 2.00
Based on the result of the test, it is found that the p-value of the test is 0.000. Therefore the null hypothesis is rejected, and there are sufficient evidence to conclude that median length of stay is different among payer type with payer type of Medicare produced a median length of stay of 3 days.
Recommendations:
In order to solve the problem of excessive length of stay, the number of consulting MDs should be limited to two at most, and means in reducing excessive length of stay should focus on those who receive Medicre benefits.
4.
Problem:
It is interested to know whether different types of Enterprise Resource Planning software influence the purchasing decision for the customer, and a survey is carried out to ask for the customers’ opinion on it.
Goal:
Determine whether features in ERP software has influence on customer purchase decisions. If so, identify which type of ERP software is most influential.
Findings / Conclusion:
Since the data are not subject to outliers and the data are not normally distributed, a Kruskal-Wallis nonparametric test is selected for the research analysis.
The result of the test is shown below:
Kruskal-Wallis Test: Response versus ERPtype
Kruskal-Wallis Test on Response
ERPtype N Median Ave Rank Z
A 1 3.000 9.5 -0.25
B 8 2.500 8.2 -1.63
C 2 2.000 5.0 -1.44
D 9 8.000 15.7 2.98
E 1 2.000 5.0 -0.99
Overall 21 11.0
H = 9.60 DF = 4 P = 0.048
H = 10.03 DF = 4 P = 0.040 (adjusted for ties)
* NOTE * One or more small samples
Based on the result of the test, the p-value is 0.040. The null hypothesis is rejected at 5% significant level. There is sufficient evidence to conclude that there is significant differences in median response on the survey question of the opinion on whether the ERP software features influence the purchase decision. From the median, it can be seen that ERP type D has the largest median response to this question.
Recommendation:
Since the ERP type D has the largest satisfaction, the marketing strategy should be on promoting the feature of ERP type D software. More resources should be placed in further development on the features of ERP type D, and less resources should be put on other ERP types of software.
