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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