Solutions – Logistic regression
Solutions – Logistic regression
Part 1
Go to crosstabs and enter Alloc as the row and Newpu60 as the column (or vice versa, will still give same results). Under statistics click on the chi-square box.
ALLOC * NEWPU60 Crosstabulation
Count
| | |NEWPU60 | |Total |
|
| | |0 |1 | |
|
|ALLOC |1 |883 |106 |989 |
|
| |2 |881 |101 |982 |
|
|Total | |1764 |207 |1971 |
|
Chi-Square Tests
| |Value |df |Asymp. Sig. |Exact Sig. |Exact Sig. |
| | | |(2-sided) |(2-sided) |(1-sided) |
|
|Pearson |.098 |1 |.754 | | |
|Chi-Square | | | | | |
|
|Continuity |.058 |1 |.810 | | |
|Correction | | | | | |
|
|Likelihood |.098 |1 |.754 | | |
|Ratio | | | | | |
|
|Fisher's Exact| | | |.769 |.405 |
|Test | | | | | |
|
|Linear-by-Line|.098 |1 |.754 | | |
|ar Association| | | | | |
|
|N of Valid |1971 | | | | |
|Cases | | | | | |
|a Computed only for a 2x2 table
b 0 cells (.0%) have expected count less than 5. The minimum expected count is 103.13.
The null hypothesis that there is no relationship between mattress and new ulcer is not rejected at the 5% significance level. The p-value from the chi-squared test is 0.754.
Part 2
Variables in the Equation
| | |B |S.E. |Wald |df |Sig. |Exp(B) |95.0% C.I.for |
| | | | | | | | |EXP(B) |
|
|Spearman's rho|TYPEAD |Correlation |1.000 |-.440 |.228 |.215 |.104 |.053 |
| | |Coefficient | | | | | | |
|
| | |Sig. |. |.000 |.000 |.000 |.000 |.019 |
| | |(2-tailed) | | | | | | |
|
| | |N |1971 |1971 |1971 |1971 |1971 |1971 |
|
| |TYPESP |Correlation |-.440 |1.000 |-.357 |.041 |-.027 |.042 |
| | |Coefficient | | | | | | |
|
| | |Sig. |.000 |. |.000 |.072 |.230 |.060 |
| | |(2-tailed) | | | | | | |
|
| | |N |1971 |1971 |1971 |1971 |1971 |1971 |
|
| |PSGRADE |Correlation |.228 |-.357 |1.000 |-.046 |-.039 |.045 |
| | |Coefficient | | | | | | |
|
| | |Sig. |.000 |.000 |. |.040 |.080 |.045 |
| | |(2-tailed) | | | | | | |
|
| | |N |1971 |1971 |1971 |1971 |1971 |1971 |
|
| |BASCORE |Correlation |.215 |.041 |-.046 |1.000 |.563 |-.005 |
| | |Coefficient | | | | | | |
|
| | |Sig. |.000 |.072 |.040 |. |.000 |.824 |
| | |(2-tailed) | | | | | | |
|
| | |N |1971 |1971 |1971 |1971 |1971 |1971 |
|
| |BMBSCORE |Correlation |.104 |-.027 |-.039 |.563 |1.000 |.013 |
| | |Coefficient | | | | | | |
|
| | |Sig. |.000 |.230 |.080 |.000 |. |.566 |
| | |(2-tailed) | | | | | | |
|
| | |N |1971 |1971 |1971 |1971 |1971 |1971 |
|
| |DIABETES |Correlation |.053 |.042 |.045 |-.005 |.013 |1.000 |
| | |Coefficient | | | | | | |
|
| | |Sig. |.019 |.060 |.045 |.824 |.566 |. |
| | |(2-tailed) | | | | | | |
|
| | |N |1971 |1971 |1971 |1971 |1971 |1971 |
|** Correlation is significant at the .01 level (2-tailed).
* Correlation is significant at the .05 level (2-tailed).
The final model is:
Variables in the Equation
| | |B |S.E. |
|
|1 |1193.860 |.064 |.131 |
|
Omnibus Tests of Model Coefficients
| | |Chi-square |df |Sig. |
|
|Step 1 |Step |130.578 |3 |.000 |
|
| |Block |130.578 |3 |.000 |
|
| |Model |130.578 |3 |.000 |
|
Hosmer and Lemeshow Test
|Step |Chi-square |df |Sig. |
|
|1 |2.588 |8 |.958 |
|
Classification Table
| | | |Predicted | | |
|
| | | |NEWPU60 | |Percentage |
| | | | | |Correct |
|
| |Observed | |0 |1 | |
|
|Step 1 |NEWPU60 |0 |1764 |0 |100.0 |
|
| | |1 |207 |0 |.0 |
|
| |Overall | | | |89.5 |
| |Percentage | | | | |
|a The cut value is .500
Step number: 1
Observed Groups and Predicted Probabilities
800 ⎮ ⎮
⌠ ⌠
⌠ ⌠
F ⌠ ⌠
R 600 ⎮ ⎮
E ⌠ ⌠
Q ⌠ 0 ⌠
U ⌠ 0 ⌠
E 400 ⎮ 00 ⎮
N ⌠ 00 ⌠
C ⌠ 00 ⌠
Y ⌠ 00 ⌠
200 ⎮ 00 1 ⎮
⌠ 00 101 ⌠
⌠ 000 00000 ⌠
⌠ 000 0000000000 ⌠
Predicted ∫∫∫∫∫∫∫∫∫∫∫∫∫∫⎮∫∫∫∫∫∫∫∫∫∫∫∫∫∫⎮∫∫∫∫∫∫∫∫∫∫∫∫∫∫⎮∫∫∫∫∫∫∫∫∫∫∫∫∫∫∫
Prob: 0 .25 .5 .75 1
Group: 000000000000000000000000000000111111111111111111111111111111
Predicted Probability is of Membership for 1
The Cut Value is .50
Symbols: 0 - 0
1 - 1
Each Symbol Represents 50 Cases.
This graph shows that this model is not very good at predicting those with a new ulcer. By assuming that everyone with a predicted probability >50% develops a new ulcer, it has not correctly identified any of the cases. It correctly classified 89.5% of the cases, but they were all those without an ulcer, which were the vast majority of patients in this trial. For a model to be a useful prognostic tool, it should clearly separate the two groups and we should see a column of zeros and a separate column of ones at each side of the 0.5 cutpoint.
Residual plot (not very useful in the same way as they are in linear regression) Values above 3 of standardized residuals are cause for concern. This plot shows a curve in the top line with some values of 4 or above, likely to be corresponding to those patients with pressure ulcers who were misclassified by the model.
[pic]
Cooks statistic plot
[pic]
Leverage plot
[pic]
For both of these plots there are no particularly large values (>1 for Cooks distance, and nearer to 1 for the leverage). It is difficult to conclude if there are any influential observations.
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